Chapter 4

Fourier Analysis for Continuous-Time Signals and Systems

Chapter Objectives

• Learn techniques for representing continuous-time periodic signals using orthogonal sets of periodic basis functions.
• Study properties of exponential, trigonometric and compact Fourier series, and conditions for their existence.
• Learn the Fourier transform for non-periodic signals as an extension of Fourier series for periodic signals.
• Study properties of the Fourier transform. Understand energy and power spectral density concepts.
• Explore frequency-domain characteristics of CTLTI systems. Understand the system function concept.
• Learn the use of frequency-domain analysis methods for solving signal-system interaction problems with periodic and non-periodic input signals.

4.1 Introduction

In Chapters 1 and 2 we have developed techniques for analyzing continuous-time signals and systems from a time-domain perspective. A continuous-time signal can be modeled as a function of time. A CTLTI system can be represented by means of a constant-coefficient linear differential equation, or alternatively by means of an impulse response. The output signal of a CTLTI system can be determined by solving the corresponding differential equation or by using the convolution operation.

In Chapter 1 we have also discussed the idea of viewing a signal as a combination of simple signals acting as “building blocks”. Examples of this were the use of unit-step, unit-ramp, unit-pulse and unit-triangle functions for constructing signals (see Section 1.3.2) and the use of the unit-impulse function for impulse decomposition of a signal (see Section 1.3.3).

Another especially useful set of building blocks is a set in which each member function has a unique frequency. Representing a signal as a linear combination of single-frequency building blocks allows us to develop a frequency-domain view of a signal that is particularly useful in understanding signal behavior and signal-system interaction problems. If a signal can be expressed as a superposition of single-frequency components, knowing how a linear and time-invariant system responds to each individual component helps us understand overall system behavior in response to the signal. This is the essence of frequency-domain analysis.

In Section 4.2 we discuss methods of analyzing periodic continuous-time signals in terms of their frequency content. Frequency-domain analysis methods for non-periodic signals are presented in Section 4.3. In Section 4.4 representation of signal energy and power in the frequency domain is discussed. System function concept is introduced in Section 4.5. The application of frequency-domain analysis methods to the analysis of CTLTI systems is discussed in Sections 4.6 and 4.7.

4.2 Analysis of Periodic Continuous-Time Signals

Most periodic continuous-time signals encountered in engineering problems can be expressed as linear combinations of sinusoidal basis functions, the more technical name for the so-called “building blocks” we referred to in the introductory section. The idea of representing periodic functions of time in terms of trigonometric basis functions was first realized by French mathematician and physicist Jean Baptiste Joseph Fourier (1768-1830) as he worked on problems related to heat transfer and propagation. The basis functions in question can either be individual sine and cosine functions, or they can be in the form of complex exponential functions that combine sine and cosine functions together.

Later in this section we will study methods of expressing periodic continuous-time signals in three different but equivalent formats, namely the trigonometric Fourier series (TFS), the exponential Fourier series (EFS) and the compact Fourier series (CFS). Before we start a detailed study of the mathematical theory of Fourier series, however, we will find it useful to consider the problem of approximating a periodic signal using a few trigonometric functions. This will help us build some intuitive understanding of frequency-domain methods for analyzing periodic signals.

4.2.1 Approximating a periodic signal with trigonometric functions

It was established in Section 1.3.4 of Chapter 1 that a signal $\tilde{x}\left(t\right)$ which is periodic with period T0 has the property

$$\begin{array}{ccc}\tilde{x}\left(t+T_0\right)=\tilde{x}\left(t\right)& & \left(4.1\right)\end{array}$$

for all t. Furthermore, it was shown through repeated use of Eqn. (4.1) that a signal that is periodic with period T0 is also periodic with kT0 for any integer k.

In working with periodic signals in this chapter, we will adopt the convention of using the tilde (~) character over the name of the signal as a reminder of its periodicity. For the sake of discussion let us consider the square-wave signal $\tilde{x}\left(t\right)$ with a period of T0 as shown in Fig. 4.1.

Suppose that we wish to approximate this signal using just one trigonometric function. The first two questions that need to be answered are:

1. Should we use a sine or a cosine?
2. How should we adjust the parameters of the trigonometric function?

The first question is the easier one to answer. The square-wave signal $\tilde{x}\left(t\right)$ (t) shown in Fig. 4.1 has odd symmetry signified by $\tilde{x}\left(-t\right)=-\tilde{x}\left(t\right)$ (see Section 1.3.6 of Chapter 1). Among the two choices for a trigonometric function, the sine function also has odd symmetry, that is, sin (− a)= −sin (a) for any real-valued parameter a. On the other hand, the cosine function has even symmetry since cos (−a) = cos(a). Therefore it would intuitively make sense to choose the sine function over the cosine function. Our approximation would be in the form

$$\begin{array}{ccc}\tilde{x}\left(t\right)\approx b_{1}sin\left(\omega t\right)& & \left(4.2\right)\end{array}$$

Since $\tilde{x}\left(t\right)$ has a fundamental period of T0, it would make sense to pick a sine function with the same fundamental period, so that

$$\begin{array}{ccc}sin\left(\omega \left(t+T_0\right)\right)=sin\left(\omega t\right)& & \left(4.3\right)\end{array}$$

For Eqn. (4.3) to work we need ωT0 = 2π and consequently

$$\begin{array}{ccc}\omega =\frac{2\pi }{T_0}={\omega }_0=2\pi f_0& & \left(4.4\right)\end{array}$$

where we have defined the frequency f0 as the reciprocal of the period, that is, f0 = 1/T0. Let ${\tilde{x}}^{\left(1\right)}\left(t\right)$ represent an approximate version of the signal $\tilde{x}\left(t\right)$, so that

$$\begin{array}{ccc}{\tilde{x}}^{\left(1\right)}\left(t\right)=b_{1}sin\left({\omega }_{0}t\right)& & \left(4.5\right)\end{array}$$

In Eqn. (4.5) we have used the superscripted signal name ${\tilde{x}}^{\left(1\right)}$ to signify the fact that we are using only one trigonometric function in this approximation. Our next task is to determine the value of the coefficient b1. How should b1 be chosen that Eqn. (4.5) represents the best approximation of the given type possible for the actual signal $\tilde{x}\left(t\right)$? There is a bit of subjectivity in this question since we have not yet defined what the “best approximation” means for our purposes.

Let us define the approximation error as the difference between the square-wave signal and its approximation:

$$\begin{array}{ccc}{\tilde{\varepsilon }}_1\left(t\right)=\tilde{x}\left(t\right)-\tilde{x}\left(t\right)-b_{1}sin\left({\omega }_{0}t\right)& & \left(4.6\right)\end{array}$$

The subscript used on the error signal $\widetilde{{\varepsilon }_1}\left(t\right)$ signifies the fact that it is the approximation error that results when only one trigonometric function is used. $\widetilde{{\varepsilon }_1}\left(t\right)$ is also periodic with period T0. One possible method of choosing the best value for the coefficient b1 would be to choose the value that makes the normalized average power of $\widetilde{{\varepsilon }_1}\left(t\right)$ as small as possible.

The best approximation to $\tilde{x}\left(t\right)$ using only one trigonometric function is

$$\begin{array}{ccc}{\tilde{x}}^{\left(1\right)}\left(t\right)=\frac{4A}{\pi }sin\left({\omega }_{0}t\right)& & \left(4.12\right)\end{array}$$

and the approximation error is

$${\tilde{\varepsilon }}_1\left(t\right)=\tilde{x}\left(t\right)-\frac{4A}{\pi }sin\left({\omega }_{0}t\right)$$

The signal $\tilde{x}\left(t\right)$, its single-frequency approximation ${\tilde{x}}^{\left(1\right)}\left(t\right)$ and the approximation error ${\tilde{\varepsilon }}_1\left(t\right)$ are shown in Fig. 4.2.

In the next step we will try a two-frequency approximation to $\tilde{x}\left(t\right)$ and see if the approximation error can be reduced. We know that the basis function sin (2ω0t) = sin (4πf0t) is also periodic with the same period T0; it just has two full cycles in the interval (0, T0) as shown in Fig. 4.3.

Let the approximation using two frequencies be defined as

$$\begin{array}{ccc}{\tilde{x}}^{\left(2\right)}\left(t\right)=b_{1}sin\left({\omega }_{0}t\right)+b_{2}sin\left(2{\omega }_{0}t\right)& & \left(4.14\right)\end{array}$$

The corresponding approximation error is

$$\begin{array}{ccc}\begin{array}{lll}{\tilde{\varepsilon }}_2\left(t\right)\hfill & =\hfill & \tilde{x}\left(t\right)-{\tilde{x}}^{\left(2\right)}\left(t\right)\hfill \\ \hfill & =\hfill & \tilde{x}\left(t\right)-b_{1}sin\left({\omega }_{0}t\right)-b_{2}sin\left(2{\omega }_{0}t\right)\hfill \end{array}& & \left(4.15\right)\end{array}$$

Using the square-wave signal $\tilde{x}\left(t\right)$ in Eqns. (4.19) and (4.20) we arrive at the following optimum values for the two coefficients:

$$\begin{array}{ccc}{b}_1=\frac{4A}{\pi }{\text{ and }b}_2=0& & \left(4.21\right)\end{array}$$

Interestingly, the optimum contribution from the sinusoidal term at radian frequency 2ω0 turned out to be zero, resulting in

$$\begin{array}{ccc}{\tilde{x}}^{\left(2\right)}\left(t\right)=\frac{4A}{\pi }sin\left({\omega }_{0}t\right)+\left(0\right)sin\left(2{\omega }_{0}t\right)={\tilde{x}}^{\left(1\right)}\left(t\right)& & \left(4.22\right)\end{array}$$

for this particular signal $\tilde{x}\left(t\right)$.

It can be shown (see Problem 4.1 at the end of this chapter) that a three-frequency approximation

$$\begin{array}{ccc}{\tilde{x}}^{\left(3\right)}\left(t\right)=b_{1}sin\left({\omega }_{0}t\right)+b_{2}sin\left(2{\omega }_{0}t\right)+b_{3}sin\left(3{\omega }_{0}t\right)& & \left(4.23\right)\end{array}$$

has the optimum coefficient values

$$\begin{array}{ccc}{b}_1=\frac{4A}{\pi },b_2=0,{\text{ and }b}_3=\frac{4A}{3\pi }& & \left(4.24\right)\end{array}$$

The signal $\tilde{x}\left(t\right)$, its three-frequency approximation ${\tilde{x}}^{\left(3\right)}\left(t\right)$ and the approximation error ${\tilde{\in }}_3\left(t\right)$ are shown in Fig. 4.4.

Some observations are in order:

1. Based on a casual comparison of Figs. 4.2(b) and 4.4(b), the normalized average power of the error signal ${\tilde{\varepsilon }}_3\left(t\right)$ seems to be less than that of the error signal ${\tilde{\varepsilon }}_3\left(t\right)$. Consequently, ${\tilde{x}}^{\left(3\right)}\left(t\right)$ is a better approximation to the signal than ${\tilde{x}}^{\left(1\right)}\left(t\right)$.
2. On the other hand, the peak value of the approximation error seems to be ±A for both ${\tilde{\varepsilon }}_1\left(t\right)$ and ${\tilde{\varepsilon }}_3\left(t\right)$. If we were to try higher-order approximations using more trigonometric basis functions, the peak approximation error would still be ±A (see Problem 4.4 at the end of this chapter). This is due to fact that, at each discontinuity of $\tilde{x}\left(t\right)$, the value of the approximation is equal to zero independent of the number of trigonometric basis functions used. In general, the approximation at a discontinuity will yield the average of the signal amplitudes right before and right after the discontinuity. This leads to the well-known Gibbs phenomenon, and will be explored further in Section 4.2.6.

4.2.2 Trigonometric Fourier series (TFS)

We are now ready to generalize the results obtained in the foregoing discussion about approximating a signal using trigonometric functions. Consider a signal $\tilde{x}\left(t\right)$ that is periodic with fundamental period T0 and associated fundamental frequency f0 = 1/T0. We may want to represent this signal using a linear combination of sinusoidal functions in the form

$$\begin{array}{ccc}\begin{array}{cc}\tilde{x}\left(t\right)& =a_0+a_{1}cos\left({\omega }_{0}t\right)\end{array}+a_{2}cos\left(2{\omega }_{0}t\right)+\cdots +a_{k}cos\left(k{\omega }_{0}t\right)...+b_{1}sin\left({\omega }_{0}t\right)+b_{1}sin\left({\omega }_{0}t\right)+b_{2}sin\left(2{\omega }_{0}t\right)+\cdots +b_{k}sin\left(k{\omega }_{0}t\right)+\cdots & & \left(4.25\right)\end{array}$$

or, using more compact notation

$$\begin{array}{ccc}\tilde{x}\left(t\right)=a_0+\sum _{k=1}^{\infty }{a}_{k}cos\left(k{\omega }_{0}t\right)+\sum _{k=1}^{\infty }{b}_{k}sin\left(k{\omega }_{0}t\right)& & \left(4.26\right)\end{array}$$

where ω0 = 2πf0 is the fundamental frequency in rad/s. Eqn. (4.26) is referred to as the trigonometric Fourier series (TFS) representation of the periodic signal $\tilde{x}\left(t\right)$, and the sinusoidal functions with radian frequencies of ω0, 2ω0, ..., kω0 are referred to as the basis functions. Thus, the set of basis functions includes

$$\begin{array}{cccc}{\varphi }_k\left(t\right)=cos\left(k{\omega }_{0}t\right),& k=0,1,2,...,\infty & & \left(4.27\right)\end{array}$$

and

$$\begin{array}{cccc}{\psi }_k\left(t\right)=sin\left(k{\omega }_{0}t\right),& k=1,2,...,\infty & & \left(4.28\right)\end{array}$$

Using the notation established in Eqns. (4.27) and (4.28), the series representation of the signal $\tilde{x}\left(t\right)$ given by Eqn. (4.25) can be written in a more generalized fashion as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=a_0+\sum _{k=1}^{\infty }{a}_k{\varphi }_k\left(t\right)+\sum _{k=1}^{\infty }{b}_k{\psi }_k\left(t\right)& & \left(4.29\right)\end{array}$$

We will call the frequencies that are integer multiples of the fundamental frequency the harmonics. The frequencies 2ω0, 3ω0, ..., kf0 are the second, the third, and the k-th harmonics of the fundamental frequency respectively. The basis functions at harmonic frequencies are all periodic with a period of T0. Therefore, considering our “building blocks” analogy, the signal $\tilde{x}\left(t\right)$ which is periodic with period T0 is represented in terms of building blocks (basis functions) that are also periodic with the same period. Intuitively this makes sense.

Before we tackle the problem of determining the coefficients of the Fourier series representation, it is helpful to observe some properties of harmonically related sinusoids. Using trigonometric identities it can be shown that

$${\int }_{t_0}^{t_0+T_0}cos\left(m{\omega }_{0}t\right)cos\left(k{\omega }_{0}t\right)dt=\{\begin{array}{cc}\begin{array}{c}{T}_0/2,\\ 0,\end{array}& \begin{array}{cc}\begin{array}{c}m=k\\ m\ne k\end{array}& \left(4.30\right)\end{array}\end{array}$$

This is a very significant result. Two cosine basis functions at harmonic frequencies ω0 and kω0 are multiplied, and their product is integrated over one full period (t0, t0 + T0). When the integer multipliers m and k represent two different harmonics of the fundamental frequency, the result of the integral is zero. A non-zero result is obtained only when m = k, that is, when the two cosine functions in the integral are the same. A set of basis functions {cos (kω0t), k = 0, ..., ∞} that satisfies Eqn. (4.30) is said to be an orthogonal set. Similarly it is easy to show that

$${\int }_{t_0}^{t_0+T_0}sin\left(m{\omega }_{0}t\right)sin\left(k{\omega }_{0}t\right)dt=\{\begin{array}{ccc}\begin{array}{c}{T}_0/2\\ 0,\end{array}& \begin{array}{c}m=k\\ m\ne k\end{array}& \left(4.31\right)\end{array}$$

meaning that the set of basis functions {sin (kω0t), k = 1, ..., ∞} is orthogonal as well. Furthermore it can be shown that the two sets are also orthogonal to each other, that is,

$$\begin{array}{ccc}{\int }_{t{}_0}^{t_0+T_0}cos\left(m{\omega }_{0}t\right)sin\left(k{\omega }_{0}t\right)dt=0& & \left(4.32\right)\end{array}$$

for any combination of the two integers m and k (even when m = k). In Eqns. (4.30) through (4.32) the integral on the left side of the equal sign can be started at any arbitrary time instant t0 without affecting the result. The only requirement is that integration be carried out over one full period of the signal.

Detailed proofs of orthogonality conditions under a variety of circumstances are given in Appendix D.

Combining the results obtained up to this point, the TFS expansion of a signal can be summarized as follows:

Example 4.1: Trigonometric Fourier series of a periodic pulse train

A pulse-train signal $\tilde{x}\left(t\right)$ with a period of T0 = 3 seconds is shown in Fig. 4.5. Determine the coefficients of the TFS representation of this signal.

Solution: In using the integrals given by Eqns. (4.41), (4.42), and (4.43), we can start at any arbitrary time instant t0 and integrate over a span of 3 seconds. Applying Eqn. (4.43) with t0 = 0 and T0 = 3 seconds, we have

$$a_0=\frac{1}{3}\left[{\int }_0^1\left(1\right)dt+{\int }_1^3\left(0\right)dt\right]=\frac{1}{3}$$

The fundamental frequency is f0 = 1/T0 = 1/3 Hz, and the corresponding value of ω0 is

$${\omega }_0=2\pi f_0=\frac{2\pi }{3}\text{ rad}/\text{s}\text{.}$$

Using Eqn. (4.41), we have

$$\begin{array}{lll}{a}_k\hfill & =\hfill & \frac{2}{3}\left[{\int }_0^1\left(1\right)cos\left(2\pi kt/3\right)dt+{\int }_1^3\left(0\right)\text{ cos}\left(2\pi kt/3\right)dt\right]\hfill \\ \hfill & =\hfill & \begin{array}{cc}\frac{sin\left(2\pi k/3\right)}{\pi k},& \begin{array}{cc}\text{for}& k=1,2,...,\infty \end{array}\end{array}\hfill \end{array}$$

Finally, using Eqn. (4.42), we get

$$\begin{array}{lll}{b}_k\hfill & =\hfill & \frac{2}{3}\left[{\int }_0^1\left(1\right)sin\left(2\pi kt/3\right)dt+{\int }_1^3\left(0\right)sin\left(2\pi kt/3\right)dt\right]\hfill \\ \hfill & =\hfill & \begin{array}{cc}\frac{1-cos\left(2\pi k/3\right)}{\pi k},& \begin{array}{cc}\text{for}& k=1,2,...,\infty \end{array}\end{array}\hfill \end{array}$$

Using these coefficients in the synthesis equation given by Eqn. (4.40), the signal x(t) can now be expressed in terms of the basis functions as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\frac{1}{3}+\sum _{k=1}^{\infty }\left(\frac{sin\left(2\pi k/3\right)}{\pi k}\right)cos\left(2\pi kt/3\right)+\sum _{k=1}^{\infty }\left(\frac{1-cos\left(2\pi k/3\right)}{\pi k}\right)sin\left(2\pi kt/3\right)& & \left(4.44\right)\end{array}$$

Example 4.2: Approximation with a finite number of harmonics

Consider again the signal $\tilde{x}\left(t\right)$ of Example 4.1. Based on Eqns. (4.25) and (4.26), it would theoretically take an infinite number of cosine and sine terms to obtain an accurate representation of it. On the other hand, values of coefficients ak and bk are inversely proportional to k, indicating that the contributions from the higher order terms in Eqn. (4.44) will decline in significance. As a result we may be able to neglect high order terms and still obtain a reasonable approximation to the pulse train. Approximate the periodic pulse train of Example 4.1 using (a) the first 4 harmonics, and (b) the first 10 harmonics.

Solution: Recall that we obtained the following in Example 4.1:

$$\begin{array}{ccc}{a}_0=\frac{1}{3},& a_k=\frac{sin\left(K\pi /3\right)}{\pi k},& b_k=\frac{1-cos\left(2\pi /3\right)}{\pi k}\end{array}$$

These coefficients have been numerically evaluated for up to k = 10, and are shown in Table 4.1.

Let ${\tilde{x}}^{\left(m\right)}\left(t\right)$ be an approximation to the signal $\tilde{x}\left(t\right)$ utilizing the first m harmonics of the fundamental frequency:

$$\begin{array}{ccc}{\tilde{x}}^{\left(m\right)}\left(t\right)=a_0+\sum _{k=1}^m{a}_k+\sum _{k=1}^m{a}_{k}cos\left(k{\omega }_{0}t\right)+\sum _{k=1}^m{b}_{k}sin\left(k{\omega }_{0}t\right)& & \left(4.45\right)\end{array}$$

Using m = 4, we have

$$\begin{array}{lll}{\tilde{x}}^{\left(4\right)}\left(t\right)\hfill & =\hfill & 0.3333+0.2757cos\left(2\pi t/3\right)-0.1378cos\left(4\pi t/3\right)+0.0689cos\left(8\pi /3\right)\hfill \\ \hfill & \hfill & +0.4775sin\left(2\pi t/3\right)+0.2387sin\left(4\pi t/3\right)+0.1194sin\left(8\pi t/3\right)\hfill \end{array}$$

A similar but lengthier expression can be written for the case m = 10 which we will skip to save space. Fig. 4.6 shows two approximations to the original pulse train using the first 4 and 10 harmonics respectively.

Example 4.3: Periodic pulse train revisited

Determine the TFS coefficients for the periodic pulse train shown in Fig. 4.7.

Solution: This is essentially the same pulse train we have used in Example 4.1 with one minor difference: The signal is shifted in the time domain so that the main pulse is centered around the time origin t = 0. As a consequence, the resulting signal is an even function of time, that is, it has the property $\tilde{x}\left(-t\right)=\tilde{x}\left(t\right)$ for −∞ < t < ∞.

Let us take one period of the signal to extend from t0 = −1.5 to t0 + T0 = 1.5 seconds. Applying Eqn. (4.43) with t0 = −1.5 and T0 = 3 seconds, we have

$$a_0=\frac{1}{3}{\int }_{t=-0.5}^{0.5}\left(1\right)dt=\frac{1}{3}$$

Using Eqn. (4.41) yields

$$a_k=\frac{2}{3}{\int }_{-0.5}^{0.5}\left(1\right)cos\left(2\pi kt/3\right)dt=\frac{2sin\left(2\pi k/3\right)}{\pi k}$$

and using Eqn. (4.42)

$$b_k=\frac{2}{3}{\int }_{-0.5}^{0.5}\left(1\right)sin\left(2\pi kt/3\right)dt=0$$

Thus, $\tilde{x}\left(t\right)$ can be written as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\frac{1}{3}+\sum _{k=1}^{\infty }\left(\frac{2sin\left(\pi k/3\right)}{\pi k}\right)cos\left(k{\omega }_{0}t\right)& & \left(4.46\right)\end{array}$$

where the fundamental frequency is f0 = 1/3 Hz. In this case, the signal is expressed using only the cos (kω0t) terms of the TFS expansion. This result is intuitively satisfying since we have already recognized that $\tilde{x}\left(t\right)$ exhibits even symmetry, and therefore it can be represented using only the even basis functions {cos (kω0t), k = 0, 1, ..., ∞}, omitting the odd basis functions {sin (kω0t), k = 1, 2,..., ∞}.

Software resources:

ex_4_3.m

Example 4.4: Trigonometric Fourier series for a square wave

Determine the TFS coefficients for the periodic square wave shown in Fig 4.8.

Solution: This is a signal with odd symmetry, that is, $\tilde{x}\left(-t\right)=-\tilde{x}\left(t\right)$ for −∞ < t < ∞. Intuitively we can predict that the constant term a0 and the coefficients ak of the even basis functions {cos (kω0t), k = 1,..., ∞} in the TFS representation of $\tilde{x}\left(t\right)$ should all be equal to zero, and only the odd terms of the series should have significance. Applying Eqn. (4.41) with integration limits t0 = −T0/2 and t0 + T0 = T0/2 seconds, we have

$$a_k=\frac{2}{T_0}\left[{\int }_{-T_{0/2}}^0\left(-A\right)cos\left(2\pi kt/T_0\right)dt+{\int }_{t=0}^{T_0/2}\left(A\right)cos\left(2\pi kt/T_0\right)dt\right]=0$$

as we have anticipated. Next, we will use Eqn. (4.43) to determine a0:

$$a_0=\frac{1}{T_0}\left[{\int }_{-T_0/2}^0\left(-A\right)dt+{\int }_0^{T_0/2}\left(A\right)dt\right]=0$$

Finally, using Eqn. (4.42), we get the coefficients of the sine terms:

$$\begin{array}{lll}bk\hfill & =\hfill & \frac{2}{T_0}\left[{\int }_{-T_0/2}^0\left(-A\right)sin\left(2\pi kt/T_0\right)dt+{\int }_0^{T_0/2}\left(A\right)sin\left(2\pi kt/T_0\right)dt\right]\hfill \\ \hfill & =\hfill & \frac{2A}{\pi k}\left[1-cos\left(\pi k\right)\right]\hfill \end{array}$$

Using the identity

$$cos\left(\pi k\right)={\left(-1\right)}^k$$

the result found for bk can be written a

$$\begin{array}{ccc}{b}_k=\{\begin{array}{c}\begin{array}{cc}\frac{A4}{\pi k},& k\text{ odd}\end{array}\\ \begin{array}{cc}0,& k\text{ even}\end{array}\end{array}& & \left(4.47\right)\end{array}$$

Compare the result in Eqn. (4.47) to the coefficients b1 through b3 we have computed in Eqn. (4.24) in the process of finding the optimum approximation to the square-wave signal $\tilde{x}\left(t\right)$ using three sine terms. Table 4.2 lists the TFS coefficients up to the 10-th harmonic.

Finite-harmonic approximations ${\tilde{x}}^{\left(m\right)}\left(t\right)$ of the signal $\tilde{x}\left(t\right)$ for m = 3 and m = 9 are shown in Fig. 4.9(a) and (b).

Software resources:

ex_4_4a.m

ex_4_4b.m

4.2.3 Exponential Fourier series (EFS)

Fourier series representation of the periodic signal $\tilde{x}\left(t\right)$ in Eqn. (4.26) can also be written in alternative forms. Consider the use of complex exponentials $\left\{e^{jk{\omega }_{0}t},n=-\infty ,...,\infty \right\}$ as basis functions so that the signal $\tilde{x}\left(t\right)$ is expressed as a linear combination of them in the form

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_{0}t}& & \left(4.48\right)\end{array}$$

where the coefficients ck are allowed to be complex-valued even though the signal $\tilde{x}\left(t\right)$ is real. This is referred to as the exponential Fourier series (EFS) representation of the periodic signal. Before we consider this idea for an arbitrary periodic signal $\tilde{x}\left(t\right)$ we will study a special case.

Single-tone signals:

Let us first consider the simplest of periodic signals: a single-tone signal in the form of a cosine or a sine waveform. We know that Euler’s formula can be used for expressing such a signal in terms of two complex exponential functions:

$$\begin{array}{ccc}\begin{array}{lll}\tilde{x}\left(t\right)\hfill & =\hfill & Acos\left({\omega }_{0}t+\theta \right)\hfill \\ \hfill & =\hfill & \frac{A}{2}{e}^j\left({\omega }_{0}t\_\theta \right)+\frac{A}{2}{e}^{-j\left({\omega }_{0}t+\theta \right)}\hfill \\ \hfill & =\hfill & \frac{A}{2}{e}^{j\theta }{e}^{j{\omega }_{0}t}+\frac{A}{2}{e}^{-j\theta }{e}^{-j{\omega }_{0}t}\hfill \end{array}& & \left(4.49\right)\end{array}$$

Comparing Eqn. (4.49) with Eqn. (4.48) we conclude that the cosine waveform can be written in the EFS form of Eqn. (4.48) with coefficients

$$\begin{array}{ccc}{c}_1=\frac{A}{2}{e}^{j\theta },& c_{\mathrm{-1}}=\frac{A}{2}{e}^{-j\theta },& \begin{array}{ccc}\text{and}& c_k=0& \begin{array}{ccc}\text{for all other }k& & \left(4.50\right)\end{array}\end{array}\end{array}$$

If the signal under consideration is $\tilde{x}\left(t\right)=A$ sin (ω0t + θ), a similar representation can be obtained using Euler’s formula:

$$\begin{array}{ccc}\begin{array}{lll}\tilde{x}\left(t\right)\hfill & =\hfill & Asin\left({\omega }_{0}t+\theta \right)\hfill \\ \hfill & =\hfill & \frac{A}{2j}{e}^{j\left({\omega }_{0}t+\theta \right)}-\frac{A}{2j}{e}^{-j\left({\omega }_{0}t+\theta \right)}\hfill \end{array}& & \left(4.51\right)\end{array}$$

Using the substitutions

$$\frac{1}{j}=-j=e^{-j\pi /2}\text{ and }-\frac{1}{j}=j=e^{j\pi /2}$$

Eqn. (4.51) can be written as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\frac{A}{2}{e}^{j\left(\theta -\pi /2\right)}{e}^{j{\omega }_{0}t}+\frac{A}{2}{e}^{-j\left(\theta -\pi /2\right)}{e}^{-j{\omega }_{0}t}& & \left(4.52\right)\end{array}$$

Comparison of Eqn. (4.52) with Eqn. (4.48) leads us to the conclusion that the sine waveform in Eqn. (4.52) can be written in the EFS form of Eqn. (4.48) with coefficients

$$\begin{array}{ccc}{c}_1=\frac{A}{2}{e}^{j\left(\theta -\pi /2\right)},c_{\mathrm{-1}}=\frac{A}{2}{e}^{-j\left(\theta -\pi /2\right)},{\text{ and c}}_k=0\text{ for all other }k& & \left(4.53\right)\end{array}$$

The EFS representations of the two signals are shown graphically, in the form of a line spectrum, in Fig. 4.10(a) and (b).

The magnitudes of the coefficients are identical for the two signals, and the phases differ by π/2. This is consistent with the relationship

$$\begin{array}{ccc}Asin\left({\omega }_{0}t+\theta \right)=Acos\left({\omega }_{0}t+\theta -\pi /2\right)& & \left(4.54\right)\end{array}$$

The general case:

We are now ready to consider the EFS representation of an arbitrary periodic signal $\tilde{x}\left(t\right)$. In order for the series representation given by Eqn. (4.48) to be equivalent to that of Eqn. (4.26) we need

$$\begin{array}{ccc}{c}_0=a_0& & \left(4.55\right)\end{array}$$

and

$$\begin{array}{ccc}c{-}_k{e}^{-jk{\omega }_{0}t}+c_k{e}^{jk{\omega }_{0}t}=a_{k}cos\left(k{\omega }_{0}t\right)+b_{k}sin\left(k{\omega }_{0}t\right)& & \left(4.56\right)\end{array}$$

for each value of the integer index k. Using Euler’s formula in Eqn. (4.56) yields

$$\begin{array}{ccc}\left[c_{-k}+c_k\right]cos\left(k{\omega }_{0}t\right)+\left[-j{c}_{-k}+j{c}_k\right]sin\left(k{\omega }_{0}t\right)=a_{k}cos\left(k{\omega }_{0}t\right)+b_{k}sin\left(k{\omega }_{0}t\right)& & \left(4.57\right)\end{array}$$

Since Eqn. (4.57) must be satisfied for all t, we require that coefficients of sine and cosine terms on both sides be identical. Therefore

$$\begin{array}{ccc}{c}_k+c_{-k}=a_k& & \left(4.58\right)\end{array}$$

and

$$\begin{array}{ccc}j\left(c_k-c_{-k}\right)=b_k& & \left(4.59\right)\end{array}$$

Solving Eqns. (4.58) and (4.59) for ck and c−k we obtain

$$\begin{array}{ccc}{c}_k=\frac{1}{2}\left(a_k-j{b}_k\right)& & \left(4.60\right)\end{array}$$

and

$$\begin{array}{ccc}{c}_{-k}=\frac{1}{2}\left(a_k+j{b}_k\right)& & \left(4.61\right)\end{array}$$

for k = 1, ..., ∞. Thus, EFS coefficients can be computed from the knowledge of TFS coefficients through the use of Eqns. (4.60) and (4.61).

Comparison of Eqns. (4.60) and (4.61) reveals another interesting result: For a real-valued signal $\tilde{x}\left(t\right)$, positive and negative indexed coefficients are complex conjugates of each other, that is,

$$\begin{array}{ccc}{c}_k=c_{-k}^{*}& & \left(4.62\right)\end{array}$$

What if we would like to compute the EFS coefficients of a signal without first having to obtain the TFS coefficients? It can be shown (see Appendix D) that the exponential basis functions also form an orthogonal set:

$${\int }_{t_0}^{t_0+T_0}{e}^{jm{\omega }_{0}t}{e}^{-jk{\omega }_{0}t}dt=\{\begin{array}{ccc}\begin{array}{c}{T}_0,\\ 0,\end{array}& \begin{array}{c}m=k\\ m\ne k\end{array}& \left(4.63\right)\end{array}$$

EFS coefficients can be determined by making use of the orthogonality of the basis function set.

Example 4.5: Exponential Fourier series for periodic pulse train

Determine the EFS coefficients of the signal $\tilde{x}\left(t\right)$ of Example 4.3, shown in Fig. 4.7, through direct application of Eqn. (4.72).

Solution: Using Eqn. (4.72) with t0 = −1.5 s and T0 = 3 s, we obtain

$$c_k=\frac{1}{3}{\int }_{-0.5}^{0.5}\left(1\right)e^{-j2\pi kt/3}dt=\frac{sin\left(\pi k/3\right)}{\pi k}$$

For this particular signal $\tilde{x}\left(t\right)$, the EFS coefficients ck are real-valued. This will not always be the case. The real-valued result for coefficients {ck} obtained in this example is due to the even symmetry property of the signal $\tilde{x}\left(t\right)$. (Remember that, in working with the same signal in Example 4.3, we found bk = 0 for all k. As a result, we have ck = c−k = ak/2 for all k.)

Before the coefficients {ck} can be graphed or used for reconstructing the signal $\tilde{x}\left(t\right)$ the center coefficient c0 needs special attention. Both the numerator and the denominator of the expression we derived in Eqn. (4.73) become zero for k = 0. For this case we need to use L’Hospital’s rule which yields

$$c_0=\frac{\frac{d}{dk}\left[sin\left(\pi k/3\right)\right]}{\frac{d}{dk}\left[\pi k\right]}{|}_{k=0}=\frac{\left(\pi /3\right)cos\left(\pi k/3\right)}{\pi }{|}_{k=0}=\frac{1}{3}$$

The signal $\tilde{x}\left(t\right)$ can be expressed in terms of complex exponential basis functions as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\sum _{k=-\infty }^{\infty }\left(\frac{sin\left(\pi k/3\right)}{\pi k}\right)e^{j2\pi kt/3}& & \left(4.73\right)\end{array}$$

A line graph of the set of coefficients ck is useful for illustrating the make-up of the signal $\tilde{x}\left(t\right)$ in terms of its harmonics, and is shown in Fig. 4.11.

Note that this is not quite in the magnitude-and-phase form of the line spectrum discussed earlier. Even though the coefficients {ck} are real-valued, some of the coefficients are negative. Consequently, the graph in Fig. 4.11 does not qualify to be the magnitude of the spectrum. Realizing that ejπ = −1, any negative-valued coefficient ck < 0 can be expressed as

$$c_k=\left|c_k\right|e^{j\pi }$$

using a phase angle of π radians to account for the negative multiplier. The line spectrum is shown in Fig. 4.12 in its proper form.

Example 4.6: Periodic pulse train revisited

In Example 4.5 the EFS coefficients of the pulse train in Fig. 4.7 were found to be purely real. As discussed, this is due to the even symmetry of the signal. In this example we will remove this symmetry to see how it impacts the EFS coefficients. Consider the pulse train shown in Fig. 4.13.

Using Eqn. (4.66) with t0 = 0 and T0 = 3 seconds we obtain

$$c_k=\frac{1}{3}{\int }_0^1\left(1\right)e^{-j2\pi kt/3}dt=\frac{-1}{j2\pi k}\left[e^{-j2\pi k/3}-1\right]$$

After some simplification, it can be shown that real and imaginary parts of the coefficients can be expressed as

$$\begin{array}{ccc}Re\left\{c_k\right\}=\frac{sin\left(2\pi k/3\right)}{2\pi k}& \text{and}& Im\left\{c_k\right\}=\frac{cos\left(2\pi k/3\right)-1}{2\pi k}\end{array}$$

Contrast these results with the TFS coefficients determined in Example 4.1 for the same signal. TFS representation of the signal $\tilde{x}\left(t\right)$ was given in Eqn. (4.44). Recall that the EFS coefficients are related to TFS coefficients by Eqns. (4.60) and (4.61). Using Eqns. (4.68) and (4.69), magnitude and phase of ck can be computed as

$$\begin{array}{ccc}\left|c_k\right|=\frac{1}{\sqrt{2}\pi \left|k\right|}\sqrt{1-cos\left(2\pi k/3\right)}& & \left(4.74\right)\end{array}$$

and

$$\begin{array}{ccc}{\theta }_k={tan}^{-1}\left(\frac{cos\left(2\pi k/3\right)-1}{sin\left(2\pi k/3\right)}\right)& & \left(4.75\right)\end{array}$$

The expression for magnitude can be further simplified. Using the appropriate trigonometric identity 1 we can write

$$cos\left(2\pi k/3\right)=2{cos}^2\left(\pi k/3\right)-1=1-2{sin}^2\left(\pi k/3\right)$$

Substituting this result into Eqn. (4.74) yields

$$\left|c_k\right|=\frac{1}{3}\left|sin\text{c}\left(k/3\right)\right|$$

where we have used the sinc function defined as

$$sin\text{c}\left(\alpha \right)\triangleq \frac{sin\left(\pi \alpha \right)}{\pi \alpha }$$

The line spectrum is graphed in Fig. (4.14).

Software resources:

ex_4_6.m

Example 4.7: Effects of duty cycle on the spectrum

Consider the pulse train depicted in Fig. 4.15.

Each pulse occupies a time duration of τ within a period length of T0. The duty cycle of a pulse train is defined as the ratio of the pulse-width to the period, that is,

$$\begin{array}{ccc}d\triangleq \frac{\tau }{T_0}& & \left(4.76\right)\end{array}$$

The duty cycle plays an important role in pulse-train type waveforms used in electronics and communication systems. Having a small duty cycle means increased blank space between individual pulses, and this can be beneficial in certain circumstances. For example, we may want to take advantage of the large gap between pulses, and utilize the target system to process other signals using a strategy known as time-division multiplexing. On the other hand, a small duty cycle does not come without cost as we will see when we graph the line spectrum. Using Eqn. (4.76) with t0 = −τ/2 we obtain

$$c_k=\frac{1}{T_0}\left[{\int }_{-\tau /2}^{\tau /2}\left(1\right)e^{-j2\pi kt/T_0}dt\right]=\frac{sin\left(\pi kd\right)}{\pi k}$$

The result in Eqn. (4.77) can be written in a more compact form as

$$\begin{array}{cc}{c}_k=d\text{ sinc }\left(kd\right)& \left(4.77\right)\end{array}$$

In Eqn. (4.77) values of coefficients ck depend only on the duty cycle and not on the period T0. On the other hand, the period T0 impacts actual locations of the coefficients on the frequency axis, and consequently the frequency spacing between successive coefficients. Since the fundamental frequency is f0 = 1/T0, the coefficient ck occurs at the frequency kf0 = k/T0 in the line spectrum.

Magnitudes of coefficients ck are shown in Fig. 4.16 for duty cycle values of d = 0.1, d = 0.2, and d = 0.3 respectively.

We observe that smaller values of the duty cycle produce increased high-frequency content as the coefficients of large harmonics seem to be stronger for d = 0.1 compared to the other two cases.

Software resources:

ex_4_7.m

Example 4.8: Spectrum of periodic sawtooth waveform

Consider a periodic sawtooth signal $\tilde{x}\left(t\right)$, with a period of T0 = 1 s, defined by

$$\begin{array}{ccc}\tilde{x}\left(t\right)=at,& 0<t<1& \begin{array}{cc}\text{and}& \tilde{x}\left(t+1\right)=\tilde{x}\left(t\right)\end{array}\end{array}$$

and shown in Fig. 4.17. The parameter a is a real-valued constant. Determine the EFS coefficients for this signal.

Solution: The fundamental frequency of the signal is

$$\begin{array}{cc}{f}_0=\frac{1}{T_0}=1\text{ Hz},& {\omega }_0=\frac{2\pi }{T_0}=2\pi \text{ rad}\end{array}/\text{s}$$

Using Eqn. (4.72) we obtain

$$\begin{array}{ccc}\begin{array}{lll}{c}_k\hfill & =\hfill & {\int }_0^{1}at{e}^{-j^{2\pi kt}}dt\hfill \\ \hfill & =\hfill & a\left[\frac{t{e}^{-j2\pi kt}}{-j2\pi k}{|}_0^1+{\int }_0^1\frac{e^{-j2\pi kt}}{j2\pi k}dt\right]\hfill \\ \hfill & =\hfill & a\left[-\frac{e^{-j2\pi k}}{j2\pi k}+\frac{e^{-j2\pi k}-1}{4{\pi }^2{k}^2}\right]\hfill \end{array}& & \left(4.78\right)\end{array}$$

where we have used integration by parts. 2 Using Euler’s formula on Eqn. (4.78), real and imaginary parts of the EFS coefficients are obtained as

$$\begin{array}{ccc}Re\left\{c_k\right\}=a\left[\frac{sin\left(2\pi k\right)}{2\pi k}+\frac{cos\left(2\pi k\right)-1}{4{\pi }^2{k}^2}\right]& & \left(4.79\right)\end{array}$$

and

$$\begin{array}{ccc}Im\left\{c_k\right\}=a\left[\frac{cos\left(2\pi k\right)}{2\pi k}-\frac{sin\left(2\pi k\right)}{4{\pi }^2{k}^2}\right]& & \left(4.80\right)\end{array}$$

respectively. These two expressions can be greatly simplified by recognizing that sin (2πk) = 0 and cos(2πk) = 1 for all integers k. For k = 0, the values of Re{ck} and Im{ck} need to be resolved by using L’Hospital’s rule on Eqns. (4.79) and (4.80). Thus, the real and imaginary parts of ck are

$$Re\left\{c_k\right\}=\{\begin{array}{cc}\begin{array}{c}\begin{array}{cc}\frac{a}{2},& k=0\end{array}\\ \begin{array}{cc}0,& k\ne 0\end{array}\end{array}& \left(4.81\right)\end{array}$$

$$Im\left\{c_k\right\}=\{\begin{array}{cc}\begin{array}{cc}\begin{array}{c}0,\\ \frac{a}{2\pi k},\end{array}& \begin{array}{c}k=0\\ k\ne 0\end{array}\end{array}& \left(4.82\right)\end{array}$$

Combining Eqns. (4.81) and (4.82) the magnitudes of the EFS coefficients are obtained as

$$\left|c_k\right|=\{\begin{array}{ccc}\begin{array}{c}\frac{\left|a\right|}{2},\\ \frac{a}{2\pi k},\end{array}& \begin{array}{c}k=0\\ k\ne 0\end{array}& \end{array}$$

For the phase angle we need to pay attention to the sign of the parameter a. If a ≥ 0, we have

$${\theta }_k=\{\begin{array}{cc}\begin{array}{c}0,\\ \pi /2,\\ -\pi /2,\end{array}& \begin{array}{c}k=0\\ k>0\\ k<0\end{array}\end{array}$$

If a < 0, then θk needs to be modified as

$${\theta }_k=\{\begin{array}{cc}\begin{array}{c}\pi ,\\ -\pi /2,\\ \pi /2,\end{array}& \begin{array}{c}k=0\\ k>0\\ k<0\end{array}\end{array}$$

If a < 0, then θk needs to be modified as

$${\theta }_k=\{\begin{array}{cc}\begin{array}{c}\pi ,\\ -\pi /2,\\ \pi /2,\end{array}& \begin{array}{c}k=0\\ k>0\\ k<0\end{array}\end{array}$$

Magnitude and phase spectra for $\tilde{x}\left(t\right)$ are graphed in Fig. 4.18(a) and (b) with the assumption that a > 0.

Software resources:

ex_4_8a.m

ex_4_8b.m

Example 4.9: Spectrum of multi-tone signal

Determine the EFS coefficients and graph the line spectrum for the multi-tone signal

$$\overline{x}\left(t\right)=cos\left(2\pi \left[10f_0\right]t\right)+0.8cos\left(2\pi f_{0}t\right)cos\left(2\pi \left[10f_0\right]t\right)$$

shown in Fig. 4.19.

Solution: Using the appropriate trigonometric identity 3 the signal $\tilde{x}\left(t\right)$ can be written as

$$\overline{x}\left(t\right)=cos\left(2\pi \left[10f_0\right]t\right)+0.4cos\left(2\pi \left[11f_0\right]t\right)+0.4cos\left(2\pi \left[9f_0\right]t\right)$$

Applying Euler’s formula, we have

$$\begin{array}{ccc}\begin{array}{lll}\tilde{x}\left(t\right)\hfill & =\hfill & 0.5e^{j2\pi \left(10f_0\right)t}+0.5e^{-j2\pi \left(10f_0\right)t}\hfill \\ \hfill & \hfill & +0.2e^{j2\pi \left(11f_0\right)t}+0.2e^{-j2\pi \left(11f_0\right)t}+0.2e^{-j2\pi \left(9f_0\right)t}\hfill \end{array}& & \left(4.83\right)\end{array}$$

By inspection of Eqn. (4.83) the significant EFS coefficients for the signal $\tilde{x}\left(t\right)$ are found as

$$\begin{array}{lllll}{c}_9\hfill & =\hfill & c_{-9}\hfill & =\hfill & 0.2,\hfill \\ c_{10}\hfill & =\hfill & c_{-10}\hfill & =\hfill & 0.5,\hfill \\ c_{11}\hfill & =\hfill & c_{-11}\hfill & =\hfill & 0.2\hfill \end{array}$$

and all other coefficients are equal to zero. The resulting line spectrum is shown in Fig. 4.20.

Software resources:

ex_4_9.m

Example 4.10: Spectrum of half-wave rectified sinusoidal signal

Determine the EFS coefficients and graph the line spectrum for the half-wave periodic signal $\tilde{x}\left(t\right)$ defined by

$$\tilde{x}\left(t\right)=\{\begin{array}{ccc}\begin{array}{c}sin\left({\omega }_{0}t\right),\\ 0,\end{array}& \begin{array}{c}0\le t<T_0/2\\ T_0/2\le t<T_0\end{array},& \begin{array}{cc}\text{and}& \begin{array}{ccc}\overline{x}\left(t+T_0\right)=\tilde{x}\left(t\right)& & \left(4.84\right)\end{array}\end{array}\end{array}$$

and shown in Fig. 4.21.

Solution: Using the analysis equation given by Eqn. (4.72), the EFS coefficients are

$$c_k=\frac{1}{T_0}{\int }_0^{T_0/2}sin\left({\omega }_{0}t\right)e^{-jk{\omega }_{0}t}dt$$

In order to simplify the evaluation of the integral in Eqn. (4.85) we will write the sine function using Euler’s formula, and obtain

$$\begin{array}{c}{c}_k=\frac{1}{j2T_0}{\int }_0^{T_0/2}\left[e^{j{\omega }_{0}t}-e^{-j{\omega }_{0}t}\right]e^{-jk{\omega }_{0}t}dt\\ =\frac{1}{j2T_0}\left(\frac{e^{-j{\omega }_0\left(k-1\right)}}{-j{\omega }_0\left(k-1\right)}\right){|}_{t=0}^{T_0/2}-\frac{1}{j2T_0}\left(\frac{e^{-j{\omega }_0\left(k+1\right)}}{-j{\omega }_0\left(k+1\right)}\right){|}_{t=0}^{T_0/2}\end{array}$$

which can be simplified to yield

$$\begin{array}{ccc}{c}_k=\frac{1}{4\pi \left(k-1\right)}\left[e^{-j\pi \left(k-1\right)}-1\right]-\frac{1}{4\pi \left(k+1\right)}\left[e^{-j\pi \left(k+1\right)}-1\right]& & \left(4.85\right)\end{array}$$

We need to consider even and odd values of k separately.

Case 1: k odd and k ≠ ∓ 1

If k is odd, then both k − 1 and k + 1 are non-zero and even. We have e−jπ(k − 1) = e−jπ(k+1) = 1, and therefore ck = 0.

Case 2: k = 1

If k = 1, c1 has an indeterminate form which can be resolved through the use of L’Hospital’s rule:

$$c_1=\frac{-j\pi e^{-j\pi \left(k-1\right)}}{4\pi }{|}_{k=1}=-\frac{j}{4}$$

Case 3: k = −1

Similar to the case of k = 1 an indeterminate form is obtained for k = −1. Through the use of L’Hospital’s rule the coefficient c−1 is found to be

$$c_{-1}=\frac{j\pi e^{-j\pi \left(k+1\right)}}{4\pi }{|}_{k=-1}=\frac{j}{4}$$

Case 4: k even

In this case both k − 1 and k + 1 are odd, and we have e−jπ(k−1) = e−jπ(k+1) = −1.

$$e^{-j\pi \left(k-1\right)}-1=e^{-j\pi \left(k+1\right)}-1=-2$$

Using this result in Eqn. (4.85) we get

$$c_k=-\frac{1}{2\pi \left(k-1\right)}+\frac{1}{2\pi \left(k+1\right)}=\frac{-1}{\pi \left(k^2-1\right)}$$

Combining the results of the four cases, the EFS coefficients are

$$c_k=\{\begin{array}{cc}\begin{array}{c}0,\\ -j/4,\\ j/4,\\ \frac{-1}{\pi \left(k^2-1\right)},\end{array}& \begin{array}{c}k\text{ odd and }k\ne \mp 1\\ k=1\\ k=-1\\ k\text{ even}\end{array}\end{array}$$

The resulting line spectrum is shown in Fig. 4.22.

Software resources:

ex 4_10a.m

ex_4_10b.m

4.2.4 Compact Fourier series (CFS)

Yet another form of the Fourier series representation of a periodic signal is the compact Fourier series (CFS) expressed as

$$\begin{array}{ccc}\tilde{x}\left(t\right)=d_0+\sum _{k=1}^{\infty }{d}_{k}cos\left(k{\omega }_{0}t+{\varphi }_k\right)& & \left(4.86\right)\end{array}$$

Using the appropriate trigonometric identity 4 Eqn. (4.86) can be written as

$$\begin{array}{lll}\tilde{x}\left(t\right)\hfill & =\hfill & d_0+\sum _{k=1}^{\infty }\left[d_{k}cos\left(k{\omega }_{0}t\right)cos\left({\varphi }_k\right)-d_{k}sin\left(k{\omega }_{0}t\right)sin\left({\varphi }_k\right)\right]\hfill \\ \hfill & =\hfill & d_0+\sum _{k=1}^{\infty }{d}_{k}cos\left(k{\omega }_{0}t\right)cos\left({\varphi }_k\right)-\sum _{k=1}^{\infty }{d}_{k}sin\left(k{\omega }_{0}t\right)sin\left({\varphi }_k\right)\hfill \end{array}$$

Recognizing that this last equation for $\tilde{x}\left(t\right)$ is in a format identical to the TFS expansion of the same signal given by Eqn. (4.26), the coefficients of the corresponding terms must be equal. Therefore the following must be true:

$$\begin{array}{ccc}{a}_0=d_0& & \left(4.87\right)\end{array}$$

$$\begin{array}{ccc}{a}_k=d_{k}cos\left({\varphi }_k\right),& k=1,...,\infty & \begin{array}{cc}& \left(4.88\right)\end{array}\end{array}$$

$$\begin{array}{ccc}{b}_k=-d_{k}sin\left({\varphi }_k\right),& k=1,...,\infty & \begin{array}{cc}& \left(4.89\right)\end{array}\end{array}$$

Solving for dk and ϕk from Eqns. (4.87), (4.88) and (4.89) we obtain

$$\begin{array}{ccc}{d}_k=\sqrt{\left\{a_k^2+b_k^2\right\},}& k=1,...,\infty & \begin{array}{cc}& \left(4.90\right)\end{array}\end{array}$$

and

$$\begin{array}{ccc}{\varphi }_k=-{tan}^{-1}\left(\frac{b_k}{a_k}\right),& k=1,...,\infty & \begin{array}{cc}& \left(4.91\right)\end{array}\end{array}$$

with d0 = a0, and ϕ0 = 0. CFS coefficients can also be obtained from the EFS coefficients by using Eqns. (4.88) and (4.89) in conjunction with Eqns. (4.60) and (4.61), and remembering that c−k = c*k for real-valued signals:

$$\begin{array}{ccc}\begin{array}{cc}{d}_k=2{\left[{\left(Re\left\{c_k\right\}\right)}^2+{\left(Im\left\{c_k\right\}\right)}^2\right]}^{1/2}=2\left|c_k\right|,& k=1,...,\infty \end{array}& & \left(4.92\right)\end{array}$$

and

$$\begin{array}{ccc}{\varphi }_k={tan}^{-1}\left(\frac{Im\left\{c_k\right\}}{Re\left\{c_k\right\}}\right)={\theta }_k,& k=1,...,\infty & \begin{array}{cc}& \left(4.93\right)\end{array}\end{array}$$

In the use of Eqn. (4.93), attention must be paid to the quadrant of the complex plane in which the coefficient ck resides.

Example 4.11: Compact Fourier series for a periodic pulse train

Determine the CFS coefficients for the periodic pulse train that was used in Example 4.6. Afterwards, using the CFS coefficients, find an approximation to $\tilde{x}\left(t\right)$ using m = 4 harmonics.

Solution: CFS coefficients can be obtained from the EFS coefficients found in Example 4.6 along with Eqns. (4.92) and (4.93):

$$\begin{array}{cccc}{d}_0=c_0& \text{and}& d_k=2\left|c_k\right|=\frac{2}{3}\left|\text{sinc}\left(k/3\right)\right|,& k=1,\cdots ,\infty \end{array}$$

and

$${\varphi }_k={tan}^{-1}\left(\frac{cos\left(2\pi k/3\right)-1}{sin\left(2\pi k/3\right)}\right),k=1,...,\infty$$

Table 4.3 lists values of some of the compact Fourier series coefficients for the pulse train x(t).

Let ${\tilde{x}}^{\left(m\right)}\left(t\right)$ be an approximation to the signal $\tilde{x}\left(t\right)$ utilizing the first m harmonics of the fundamental frequency, that is,

$$\begin{array}{ccc}{\tilde{x}}^{\left(m\right)}\left(t\right)=d_0+\sum _{k=1}^m{d}_{k}cos\left(k{\omega }_{0}t+\varphi k\right)& & \left(4.94\right)\end{array}$$

Using m = 4 and f0 = 1/3 Hz we have

$$\begin{array}{ccc}\begin{array}{lll}{x}^{\left(4\right)}\left(t\right)\hfill & =\hfill & 0.3333+0.5513cos\left(2\pi \left(\frac{1}{3}\right)t-1.0472\right)\hfill \\ \hfill & \hfill & +0.2757cos\left(2\pi \left(\frac{2}{3}\right)t-2.0944\right)+0.1378\text{ cos }\left(2\pi \left(\frac{4}{3}\right)t-1.0472\right)\hfill \end{array}& & \left(4.95\right)\end{array}$$

The individual terms in Eqn. (4.95) as well as the resulting finite-harmonic approximation are shown in Fig. 4.23.

4.2.5 Existence of Fourier series

For a specified periodic signal $\tilde{x}\left(t\right)$, one question that needs to be considered is the existence of the Fourier series: Is it always possible to determine the Fourier series coefficients? The answer is well-known in mathematics. A periodic signal can be uniquely expressed using sinusoidal basis functions at harmonic frequencies provided that the signal satisfies a set of conditions known as Dirichlet conditions named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805–1859). A thorough mathematical treatment of Dirichlet conditions is beyond the scope of this text. For the purpose of the types of signals we will encounter in this text, however, it will suffice to summarize the three conditions as follows:

Most signals we encounter in engineering applications satisfy the existence criteria listed above. Consequently, they can be represented in series expansion forms given by Eqns. (4.26), (4.48) or (4.86). At points where the signal $\tilde{x}\left(t\right)$ is continuous, its Fourier series expansion converges perfectly. At a point of discontinuity, however, the Fourier series expansion (TFS, EFS or CFS representation of the signal) yields the average value obtained by approaching the discontinuity from opposite directions. If the signal $\tilde{x}\left(t\right)$ has a discontinuity at t = t0, and if {ck} are the EFS coefficients for it, we have

$$\begin{array}{ccc}\sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_0{r}_0}=\frac{1}{2}\underset{x\to t_0^{-}}{lim}\left[\tilde{x}\left(t\right)\right]+\frac{1}{2}\underset{t\to t_0^{+}}{lim}\left[\tilde{x}\left(t\right)\right]& & \left(4.97\right)\end{array}$$

This is illustrated in Fig. 4.24.

4.2.6 Gibbs phenomenon

Let us further explore the issue of convergence of the Fourier series at a discontinuity. Consider the periodic square wave shown in Fig. 4.25.

The TFS coefficients of this signal were found in Example 4.4, and are repeated below:

$$\begin{array}{ccc}{a}_k=0,& \text{and}& b_k=\end{array}\{\begin{array}{cc}\begin{array}{c}\frac{4}{\pi k},\\ 0,\end{array}& \begin{array}{c}k\text{ odd}\\ k\text{ even}\end{array}\end{array}$$

Finite-harmonic approximation to the signal $\tilde{x}\left(t\right)$ using m harmonics is

$${\tilde{x}}^{\left(m\right)}\left(t\right)=\sum _{\begin{array}{c}k=1\\ k\text{ odd}\end{array}}^m\left(\frac{4}{\pi k}\right)sin\left(\pi t\right)$$

and the approximation error that occurs when m harmonics is used is

$${\tilde{\varepsilon }}^{\left(m\right)}\left(t\right)=\tilde{x}\left(t\right)-{\tilde{x}}^{\left(m\right)}\left(t\right)$$

Several finite-harmonic approximations to $\tilde{x}\left(t\right)$ are shown in Fig. 4.26 for m = 1, 3, 9, 25. The approximation error is also shown for each case.

We observe that the approximation error is non-uniform. It seems to be relatively large right before and right after a discontinuity, and it gets smaller as we move away from discontinuities. The Fourier series approximation overshoots the actual signal value at one side of each discontinuity, and undershoots it at the other side. The amount of overshoot or undershoot is about 9 percent of the height of the discontinuity, and cannot be reduced by increasing m. This is a form of the Gibbs phenomenon named after American scientist Josiah Willard Gibbs (1839-1903) who explained it in 1899. An enlarged view of the approximation and the error right around the discontinuity at t = 1 s are shown in Fig. 4.27 for m = 25. A further enlarged view of the approximation error corresponding to the shaded area in Fig. 4.27(b) is shown in Fig. 4.28. Notice that the positive and negative lobes around the discontinuity have amplitudes of about ±0.18 which is 9 percent of the amplitude of the discontinuity.

One way to explain the reason for the Gibbs phenomenon would be to link it to the inability of sinusoidal basis functions that are continuous at every point to approximate a discontinuity in the signal.

4.2.7 Properties of Fourier series

Some fundamental properties of the Fourier series will be briefly explored in this section using the exponential (EFS) form of the series. A more thorough discussion of these properties will be given in Section 4.3.5 for the Fourier transform.

Linearity

For any two signals $\tilde{x}\left(t\right)$ and $\tilde{y}\left(t\right)$ periodic with T0 = 2π/ω0 and with their respective series expansions

$$\begin{array}{lll}\tilde{x}\left(t\right)\hfill & =\hfill & \sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_{0}t}\hfill \\ \tilde{y}\left(t\right)\hfill & =\hfill & \sum _{k=-\infty }^{\infty }{d}_k{e}^{jk{\omega }_{0}t}\hfill \end{array}$$

and any two constants α1 and α2, it can be shown that the following relationship holds:

Symmetry of Fourier series

Conjugate symmetry and conjugate antisymmetry properties were defined for discrete-time signals in Section 1.4.6 of Chapter 1. Same definitions apply to Fourier series coefficients as well. EFS coefficients ck are said to be conjugate symmetric if they satisfy

$$\begin{array}{ccc}{c}_{-k}=c_k^{*}& & \left(4.99\right)\end{array}$$

for all k. Similarly, the coefficients form a conjugate antisymmetric set if they satisfy

$$\begin{array}{ccc}{c}_{-k}=c_k^{*}& & \left(4.100\right)\end{array}$$

for all k.

If the signal $\tilde{x}\left(t\right)$ is real-valued, it can be shown that its EFS coefficients form a conjugate symmetric set. Conversely, if the signal is purely imaginary, its EFS coefficients form a conjugate antisymmetric set.

Polar form of EFS coefficients

Recall that the EFS coefficients can be written in polar form as

$$\begin{array}{ccc}{c}_k=\left|c_k\right|e^{j{\theta }_k}& & \left(4.103\right)\end{array}$$

If the set {ck} is conjugate symmetric, the relationship in Eqn. (4.99) leads to

$$\begin{array}{ccc}\left|c_{-k}\right|e^{j{\theta }_{-k}}=\left|c_k\right|e^{-j{\theta }_k}& & \left(4.104\right)\end{array}$$

using the polar form of the coefficients. The consequences of Eqn. (4.104) are obtained by equating the magnitudes and the phases on both sides.

It was established in Eqn. (4.101) that the EFS coefficients of a real-valued $\tilde{x}\left(t\right)$ are conjugate symmetric. Based on the results in Eqns. (4.105) and (4.106) the magnitude spectrum is an even function of k, and the phase spectrum is an odd function.

Similarly, if the set {ck} is conjugate antisymmetric, the relationship in Eqn. (4.100) reflects on polar form of ck as

$$\begin{array}{ccc}\left|c_{-k}\right|e^{j{\theta }_{-k}}=-\left|c_k\right|e^{-j{\theta }_k}& & \left(4.107\right)\end{array}$$

The negative sign on the right side of Eqn. (4.107) needs to be incorporated into the phase since we could not write |c−k| = − |ck| (recall that magnitude must to be non-negative). Using e∓jπ = −1, Eqn. (4.107) becomes

$$\begin{array}{ccc}\begin{array}{lll}\left|c_{-k}\right|e^{j{\theta }_k}\hfill & =\hfill & \left|c_k\right|e^{-j{\theta }_k}{e}^{\mp j\pi }\hfill \\ \hfill & =\hfill & \left|c_k\right|e^{-j\left({\theta }_k\pm \pi \right)}\hfill \end{array}& & \left(4.108\right)\end{array}$$

The consequences of Eqn. (4.108) are summarized below.

A purely imaginary signal $\tilde{x}\left(t\right)$ leads to a set of EFS coefficients with conjugate antisymmetry. The corresponding magnitude spectrum is an even function of k as suggested by Eqn. (4.109). The phase is neither even nor odd.

Fourier series for even and odd signals

If the real-valued signal $\tilde{x}\left(t\right)$ is an even function of time, the resulting EFS spectrum ck is real-valued for all k.

Conversely it can also be proven that, if the real-valued signal $\tilde{x}\left(t\right)$ has odd-symmetry, the resulting EFS spectrum is purely imaginary.

Time shifting

The consequence of time shifting $\tilde{x}\left(t\right)$ is multiplication of its EFS coefficients by a complex exponential function of frequency.

4.3 Analysis of Non-Periodic Continuous-Time Signals

In Section 4.2 of this chapter we have discussed methods of representing periodic signals by means of harmonically related basis functions. The ability to express a periodic signal as a linear combination of standard basis functions allows us to use the superposition principle when such a signal is used as input to a linear and time-invariant system. We must also realize, however, that we often work with signals that are not necessarily periodic. We would like to have similar capability when we use non-periodic signals in conjunction with linear and time-invariant systems. In this section we will work on generalizing the results of the previous section to apply to signals that are not periodic. These efforts will lead us to the Fourier transform for continuous-time signals.

4.3.1 Fourier transform

In deriving the Fourier transform for non-periodic signals we will opt to take an intuitive approach at the expense of mathematical rigor. Our approach is to view a non-periodic signal as the limit case of a periodic one, and make use of the exponential Fourier series discussion of Section 4.2. Consider the non-periodic signal x(t) shown in Fig. 4.29.

What frequencies are contained in this signal? What kind of a specific mixture of various frequencies needs to be assembled in order to construct this signal from a standard set of basis functions?

We already know how to represent periodic signals in the frequency domain. Let us construct a periodic extension $\tilde{x}\left(t\right)$ of the signal x(t) by repeating it at intervals of T0.

This is illustrated in Fig. 4.30. The periodic extension $\tilde{x}\left(t\right)$ can be expressed as a sum of time-shifted versions of x(t) shifted by all integer multiples of T0 to yield

$$\begin{array}{ccc}\tilde{x}\left(t\right)=...+x\left(t+T_0\right)+x\left(t\right)+x\left(t-T_0\right)+x\left(t-2T_0\right)+...& & \left(4.114\right)\end{array}$$

The selected period T0 should be sufficiently large so as not to change the shape of the signal by causing overlaps. Putting Eqn. (4.114) in summation form we obtain

$$\begin{array}{ccc}\tilde{x}\left(t\right)=\sum _{k=-\infty }^{\infty }x\left(t-k{T}_0\right)& & \left(4.115\right)\end{array}$$

Since $\tilde{x}\left(t\right)$ is periodic, it can be analyzed in the frequency domain by using the methods developed in the previous section. The EFS representation of a periodic signal was given by Eqn. (4.71) which is repeated here:

$$\begin{array}{cc}\text{Eqn }\left(4.71\right):& \tilde{x}\left(t\right)=\sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_{0}t}\end{array}$$

The coefficients ck of the EFS expansion of the signal $\tilde{x}\left(t\right)$ are computed as

$$\begin{array}{cc}\text{Eqn}\text{. }\left(4.72\right):& c_k=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}\tilde{x}\left(t\right)e{-}^{jk{\omega }_{0}tdt}\end{array}$$

We have used the starting point t0 = −T0/2 for the integral in Eqn. (4.116) so that the lower and upper integration limits are symmetric. Fundamental frequency is the reciprocal of the period, that is,

$$f_0=\frac{1}{T_0}$$

and the fundamental radian frequency is

$${\omega }_0=2\pi f_0=\frac{2\pi }{T_0}$$

Realizing that $\tilde{x}\left(t\right)=x\left(t\right)$ within the span −T0/2 < t < T0/2 of the integral, let us write Eqn. (4.116) as

$$\begin{array}{ccc}{c}_k=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x\left(t\right)e^{-jk{\omega }_{0}t}dt& & \left(4.116\right)\end{array}$$

If the period T0 is allowed to become very large, the periodic signal $\tilde{x}\left(t\right)$ would start to look more and more similar to x(t). In the limit we would have

$$\begin{array}{ccc}\underset{T_0\to \infty }{lim}\left[\tilde{x}\left(t\right)\right]=x\left(t\right)& & \left(4.117\right)\end{array}$$

As the period T0 becomes very large, the fundamental frequency f0 becomes very small. In the limit, as we force T0 → ∞, the fundamental frequency becomes infinitesimal:

$$\begin{array}{ccc}\begin{array}{cccc}{T}_0\to \infty & \text{implies that}& \Delta f=\frac{1}{T_0}& \to 0\text{ and }\Delta \omega \end{array}=\frac{2\pi }{T_0}\to 0& & \left(4.118\right)\end{array}$$

In Eqn. (4.118) we have switched to the notation Δf and Δω instead of f0 and ω0 to emphasize the infinitesimal nature of the fundamental frequency. Applying this change to the result in Eqn. (4.116) leads to

$$\begin{array}{ccc}{c}_k=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}x\left(t\right)e^{-jk\Delta \omega t}dt& & \left(4.119\right)\end{array}$$

where ck is the contribution of the complex exponential at the frequency ω = k Δω. Because of the large T0 term appearing in the denominator of the right side of Eqn. (4.119), each individual coefficient ck is very small in magnitude, and in the limit we have ck → 0. In addition, successive harmonics k Δω are very close to each other due to infinitesimally small Δω. Let us multiply both sides of Eqn. (4.119) by T0 to obtain

$$\begin{array}{ccc}{c}_k{T}_0={\int }_{-T_0/2}^{T_0/2}x\left(t\right)e^{-jk\Delta \omega t}dt& & \left(4.120\right)\end{array}$$

If we now take the limit as T0 → ∞, and substitute ω = k Δω we obtain

$$\begin{array}{ccc}\begin{array}{lll}X\left(\omega \right)\hfill & =\hfill & \underset{T_0\to \infty }{lim}\left[c_k{T}_0\right]\hfill \\ \hfill & =\hfill & \underset{T_0\to \infty }{lim}\left[{\int }_{-T_0/2}^{T_0/2}x\left(t\right)e^{-jk\Delta \omega t}dt\right]\hfill \\ \hfill & =\hfill & {\int }_{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}dt\hfill \end{array}& & \left(4.121\right)\end{array}$$

The function X (ω) is the Fourier transform of the non-periodic signal x(t). It is a continuous function of frequency as opposed to the EFS line spectrum for a periodic signal that utilizes only integer multiples of a fundamental frequency. We can visualize this by imagining that each harmonic of the fundamental frequency comes closer and closer to its neighbors, finally closing the gaps and turning into a continuous function of ω.

At the start of the discussion it was assumed that the signal x(t) has finite duration. What if this is not the case? Would we be able to use the transform defined by Eqn. (4.121) with infinite-duration signals? The answer is yes, as long as the integral in Eqn. (4.121) converges.

Before discussing the conditions for the existence of the Fourier transform, we will address the following question: What does the transform X (ω) mean? In other words, how can we use the new function X (ω) for representing the signal x(t)? Recall that the Fourier series coefficients ck for a periodic signal ${\tilde{x}}_p\left(t\right)$ were useful because we could construct the signal from them using Eqn. (4.116). Let us apply the limit operator to Eqn. (4.116):

$$\begin{array}{ccc}x\left(t\right)=\underset{T_0\to \infty }{lim}\left[\tilde{x}\left(t\right)\right]=\underset{x\to \infty }{lim}\left[\sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_{0}t}\right]& & \left(4.122\right)\end{array}$$

If we multiply and divide the term inside the summation by T0 we obtain

$$\begin{array}{ccc}x\left(t\right)=\underset{T_0\to \infty }{lim}\left[\tilde{x}\left(t\right)\right]=\underset{T_0\to \infty }{lim}\left[\sum _{k=-\infty }^{\infty }{x}_{c_k{T}_0}{e}^{jk{\omega }_{0}t}\left(\frac{1}{T_0}\right)\right]& & \left(4.123\right)\end{array}$$

We know that

$$\frac{1}{T_0}=\frac{{\omega }_0}{2\pi }$$

and for large values of T0 we can write

$$\begin{array}{ccc}{\omega }_0\to \Delta \omega ,& \text{and}& \frac{1}{T_0}\to \frac{\Delta \omega }{2\pi }\end{array}$$

In the limit, Eqn. (4.123) becomes

$$\begin{array}{ccc}x\left(t\right)=\underset{T_0\to \infty }{lim}\left[\tilde{x}\left(t\right)\right]=\underset{T_0\to \infty }{lim}\left[\sum _{k=-\infty }^{\infty }{c}_k{T}_0{e}^{jk\Delta \omega t}\left(\frac{\Delta \omega }{2\pi }\right)\right]& & \left(4.124\right)\end{array}$$

Also realizing that lim

$$\underset{T_0\to \infty }{lim}\left[c_k{T}_0\right]=X\left(k\Delta \omega \right)=X\left(\omega \right)$$

and changing the summation to an integral in the limit we have

$$\begin{array}{ccc}x\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}d\omega & & \left(4.125\right)\end{array}$$

Eqn. (4.125) explains how the transform X (ω) can be used for constructing the signal x(t). We can interpret the integral in Eqn. (4.125) as a continuous sum of complex exponentials at harmonic frequencies that are infinitesimally close to each other. Thus, x(t) and X(ω) represent two different ways of looking at the same signal, one by observing the amplitude of the signal at each time instant and the other by considering the contribution of each frequency component to the signal.

In summary, we have derived a transform relationship between x(t) and X(ω) through the following equations:

Often we will use the Fourier transform operator ℱ and its inverse ℱ−1 in a shorthand notation as

$$X\left(\omega \right)=ℱ\left\{x\left(t\right)\right\}$$

for the forward transform, and

$$x\left(t\right)={ℱ}^{-1}\left\{X\left(\omega \right)\right\}$$

for the inverse transform. An even more compact notation for expressing the relationship between x (t) and X(ω) is in the form

$$\begin{array}{ccc}x\left(t\right)\stackrel{ℱ}{\leftrightarrow }X\left(\omega \right)& & \left(4.128\right)\end{array}$$

In general, the Fourier transform, as computed by Eqn. (4.127), is a complex function of ω. It can be written in Cartesian complex form as

$$X\left(\omega \right)=X_r\left(\omega \right)+j{X}_i\left(\omega \right)$$

or in polar complex form as

$$X\left(\omega \right)=\left|X\left(\omega \right)\right|e^{j\measuredangle X\left(\omega \right)}$$

Sometimes it will be more convenient to express the Fourier transform of a signal in terms of the frequency f in Hz rather than the radian frequency ω in rad/s. The conversion is straightforward by substituting ω = 2π f and d ω = 2π df in Eqns. (4.126) and (4.127) which leads to the following equations:

Note the lack of the scale factor 1/2π in front of the integral of the inverse transform when f is used. This is consistent with the relationship dω = 2π df.

4.3.2 Existence of Fourier transform

The Fourier transform integral given by Eqn. (4.127) may or may not converge for a given signal x(t). A complete theoretical study of convergence conditions is beyond the scope of this text, but we will provide a summary of the practical results that will be sufficient for our purposes. Let $\hat{x}\left(t\right)$ be defined as

$$\begin{array}{ccc}\hat{x}\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}d\omega & & \left(4.131\right)\end{array}$$

and let ε (t) be the error defined as the difference between x (t) and $\hat{x}\left(t\right)$:

$$\begin{array}{ccc}\varepsilon \left(t\right)=x\left(t\right)\hat{x}\left(t\right)=x\left(t\right)-\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}d\omega & & \left(4.132\right)\end{array}$$

For perfect convergence of the transform at all time instants, we would naturally want ε (t) = 0 for all t. However, this is not possible at time instants for which x (t) exhibits discontinuities. German mathematician Johann Peter Gustav Lejeune Dirichlet showed that the following set of conditions, referred to as Dirichlet conditions are sufficient for the convergence error ε (t) to be zero at all time instants except those that correspond to discontinuities of the signal x (t):

From a practical perspective, all signals we will encounter in our study of signals and systems will satisfy the second and third conditions regarding the number of discontinuities and the number of extrema respectively. The absolute integrability condition given by Eqn. (4.133) ensures that the result of the integral in Eqn. (4.126) is equal to the signal x (t) at all time instants except discontinuities. At points of discontinuities, Eqn. (4.126) yields the average value obtained by approaching the discontinuity from opposite directions. If the signal x (t) has a discontinuity at t = t0, we have

$$\begin{array}{ccc}\hat{x}\left(t_0\right)=\frac{1}{2}\underset{t\to t_0^{-}}{lim}\left[x\left(t\right)\right]+\frac{1}{2}\underset{t\to t_0^{+}}{lim}\left[x\left(t\right)\right]& & \left(4.134\right)\end{array}$$

An alternative approach to the question of convergence is to require

$$\begin{array}{ccc}{\int }_{-\infty }^{\infty }{\left|\varepsilon \left(t\right)\right|}^{2}dt={\int }_{-\infty }^{\infty }{\left|x\left(t\right)-\hat{x}\left(t\right)\right|}^{2}dt=0& & \left(4.135\right)\end{array}$$

which ensures that the normalized energy in the error signal ε (t) is zero even if the error signal itself is not equal to zero at all times. This condition ensures that the transform X (ω) as defined by Eqn. (4.127) is finite. It can be shown that, the condition stated by Eqn. (4.135) is satisfied provided that the signal x (t) is square integrable, that is

$$\begin{array}{ccc}{\int }_{-\infty }^{\infty }{\left|x\left(t\right)\right|}^{2}dt<\infty & & \left(4.136\right)\end{array}$$

We have seen in Chapter 1 (see Eqn. (1.81)) that a signal that satisfies Eqn. (4.136) is referred to as an energy signal. Therefore, all energy signals have Fourier transforms. Furthermore, we can find Fourier transforms for some signals that do not satisfy Eqn. (4.136), such as periodic signals, if we are willing to accept the use of the impulse function in the transform. Another example of a signal that is not an energy signal is the unit-step function. It is neither absolute integrable nor square integrable. We will show, however, that a Fourier transform can be found for the unit-step function as well if we allow the use of the impulse function in the transform.

4.3.3 Developing further insight

The interpretation of the Fourier transform for a non-periodic signal as a generalization of the EFS representation of a periodic signal is fundamental. We will take the time to apply this idea step by step to an isolated pulse in order to develop further insight into the Fourier transform. Consider the signal x (t) shown in Fig. 4.31, an isolated rectangular pulse centered at t = 0 with amplitude A and width τ.

The analytical definition of x (t) is

$$x\left(t\right)=A\pi \left(t/\tau \right)=\{\begin{array}{cc}\begin{array}{c}A,\\ 0,\end{array}& \begin{array}{c}\left|t\right|<\tau /2\\ \left|t\right|>\tau /2\end{array}\end{array}$$

Let the signal x (t) be extended into a pulse train $\tilde{x}\left(t\right)$ with a period T0 as shown in Fig. 4.32.

By adapting the result found in Example 4.7 to the problem at hand, the EFS coefficients of the signal $\tilde{x}\left(t\right)$ can be written as

$$\begin{array}{ccc}{c}_k=Ad\text{ sinc}\left(kd\right)& & \left(4.137\right)\end{array}$$

where d is the duty cycle of the pulse train, and is given by

$$\begin{array}{ccc}d=\frac{\tau }{T_0}& & \left(4.138\right)\end{array}$$

Substituting Eqn. (4.138) into Eqn. (4.137)

$$\begin{array}{ccc}{c}_k=\frac{A\tau }{T_0}\text{ sinc }\left(k\tau /T_0\right)& & \left(4.139\right)\end{array}$$

Let us multiply both sides of Eqn. (4.139) by T0 and write the scaled EFS coefficients as

$$\begin{array}{ccc}{c}_k{T}_0=A\tau \text{ sinc }\left(k{f}_0\tau \right)& & \left(4.140\right)\end{array}$$

where we have also substituted 1/T0 = f0 in the argument of the sinc function. The outline (or the envelope) of the scaled EFS coefficients ckT0 is a sinc function with a peak value of A τ at k = 0. The first zero crossing of the sinc-shaped envelope occurs at

$$\begin{array}{cccc}k{f}_0\tau =1& \Rightarrow & k=\frac{1}{f_0\tau }& \begin{array}{cc}& \left(4.141\right)\end{array}\end{array}$$

which may or may not yield an integer result. Subsequent zero crossings occur for

$$k{f}_0\tau =2,3,4...$$

or equivalently for

$$k=\frac{2}{f_0\tau },\frac{3}{f_0\tau },\frac{4}{f_0\tau },...$$

The coefficients ckT0 are graphed in Fig. 4.33(a) for A = 1, τ = 0.1 s and T0 = 0.25 s corresponding to a fundamental frequency of f0 = 4 Hz. Spectral lines for k = 1, k = 2 and k = 3 represent the strength of the frequency components at f0 = 4 Hz, 2f0 = 8 Hz and 3f0 = 12 Hz respectively. For the example values given, the first zero crossing of the sinc envelope occurs at k = 2.5, between the spectral lines for k = 2 and k = 3.

In Fig. 4.33(b) the same line spectrum is shown with actual frequencies f on the horizontal axis instead of the integer index k. The spectral line for index k now appears at the frequency f = kf0 = 4k Hz. For example, the spectral lines for k = 1, 2, 3 appear at frequencies f = 4, 8, 12 Hz respectively. The first zero crossing of the sinc envelope is at the frequency

$$f=\frac{1}{\tau }=10\text{ Hz}$$

and subsequent zero crossings appear at frequencies

$$f=\frac{2}{\tau },\frac{3}{\tau },\frac{4}{\tau },...=20,30,40,...\text{Hz}$$

What would happen if the period T0 is gradually increased while keeping the pulse amplitude A and the pulse width τ unchanged? The spectral lines in 4.33(b) are at integer multiples of the fundamental frequency f0. Since f0 = 1/T0, increasing the period T0 would cause f0 to become smaller, and the spectral lines to move inward, closer to each other. The zero crossings of the sinc envelope would not change, however, since the locations of the zero crossings depend on the pulse width τ only. Figs. 4.34(a)–(c) show the line spectra for T0 = 5, 10, 20 s respectively. In Fig. 4.34(a), the first zero crossing of the sinc envelope is still at 1/τ = 10 Hz, yet the frequency spacing of the spectral lines is now reduced to 2 Hz; therefore the first zero crossing coincides with the 5-th harmonic. In Fig. 4.34(b), the frequency spacing of the spectral lines is further reduced to 1 Hz and, consequently, the first zero crossing of the sinc envelope coincides with the 10-th harmonic, and is still at 10 Hz.

We can conclude that, as the period becomes infinitely large, the distances between adjacent spectral lines become infinitesimally small, and they eventually converge to a continuous function in the shape of the sinc envelope. It appears that the Fourier transform of the isolated pulse with amplitude A and width τ is

$$\begin{array}{ccc}X\left(f\right)=A\tau \text{ sinc }\left(f\tau \right)& & \left(4.142\right)\end{array}$$

or, using the radian frequency variable ω,

$$\begin{array}{ccc}X\left(\omega \right)=A\tau \text{ sinc}\left(\frac{\omega \tau }{2\pi }\right)& & \left(4.143\right)\end{array}$$

4.3.4 Fourier transforms of some signals

In this section we will work on examples of determining the Fourier transforms of some fundamental signals.

Example 4.12: Fourier transform of a rectangular pulse

Using the forward Fourier transform integral in Eqn. (4.127), find the Fourier transform of the isolated rectangular pulse signal

$$x\left(t\right)=A\pi \left(\frac{t}{\tau }\right)$$

shown in Fig. 4.35.

Solution: Recall that this is the same isolated pulse the Fourier transform of which was determined in Section 4.3.3 as the limit case of the EFS representation of a periodic pulse train. In this example we will take a more direct approach to obtain the same result through the use of the Fourier transform integral in Eqn. (4.127):

$$X\left(\omega \right)={\int }_{-\tau /2}^{\tau /2}\left(A\right)e^{-j\omega t}dt=A\frac{e^{-j\omega t}}{-j\omega }{|}_{-\tau /2}^{\tau /2}=\frac{2A}{\omega }sin\left(\frac{\omega \tau }{2}\right)$$

In order to use the sinc function, the result in Eqn. (4.144) can be manipulated and written in the form

$$\begin{array}{ccc}X\left(\omega \right)=A_{\tau }\frac{sin\left(\omega \tau /2\right)}{\left(\omega \tau /2\right)}=A\tau \text{ sinc}\left(\frac{\omega \tau }{2\pi }\right)& & \left(4.144\right)\end{array}$$

In this case it will be easier to graph the transform in terms of the independent variable f instead of ω. Substituting ω = 2πf into Eqn. (4.46) we obtain

$$\begin{array}{ccc}X\left(f\right)=A\tau \text{ sinc}\left(f\tau \right)& & \left(4.145\right)\end{array}$$

The spectrum X (f), shown in Fig. 4.36, is purely real owing to the fact that the signal x (t) exhibits even symmetry. The peak value of the spectrum is Aτ, and occurs at the frequency f = 0. The zero crossings of the spectrum occur at frequencies that satisfy fτ = k where k is any non-zero integer.

Software resources:

ex_4_12.m

The particular spectrum obtained in Example 4.12 is of special interest especially in digital systems. For example, in digital communications, rectangular pulses such as the one considered in Example 4.12 may be used for representing binary 0’s or 1’s. Let us observe the relationship between the pulse width and the shape of the spectrum:

Fig. 4.37 illustrates the effects of changing the pulse width on the frequency spectrum.

In all three cases illustrated in Fig. 4.37, the product of the pulse width and the pulse amplitude is fixed, i.e., Aτ = 1. Consequently, the peak of the frequency spectrum is equal to unity in each case. The placements of the zero crossings of the spectrum depend on the pulse width. In Fig. 4.37(a), the zero crossings are 1 Hz apart, consistent with the pulse width of τ = 1 s. In parts (b) and (c) of the figure, the spacing between adjacent zero crossings is 2 Hz and $\frac{1}{2}$ Hz, corresponding to pulse widths of τ = 0.5 s and τ = 2 s respectively.

Example 4.13: Fourier transform of a rectangular pulse revisited

Using the forward Fourier transform integral in Eqn. (4.127), find the Fourier transform of the isolated rectangular pulse given by

$$x\left(t\right)=A\prod \left(\frac{t-\tau /2}{\tau }\right)$$

which is shown in Fig. 4.38.

Solution: The signal x(t) is essentially a time-shifted version of the pulse signal used in Example 4.12. Using the Fourier transform integral in Eqn. (4.127) we obtain

$$\begin{array}{ccc}X\left(\omega \right)={\int }_0^{\tau }\left(A\right)e^{-j\omega t}dt=A{\frac{e^{-j\omega t}}{-j\omega }|}_0^{\tau }=\frac{A}{-j\omega }\left[e^{-j\omega \tau }-1\right]& & \left(4.146\right)\end{array}$$

At this point we will use a trick which will come in handy in similar situations in the rest of this text as well. Let us write the part of Eqn. (4.146) in square brackets as follows:

$$\begin{array}{ccc}\left[1-e^{j\omega t}\right]=\left[e^{-j\omega \tau /2}-e^{j\omega \tau /2}\right]e^{-j\omega \tau /2}& & \left(4.147\right)\end{array}$$

The use of Euler’s formula on the result of Eqn. (4.147) yields

$$\begin{array}{ccc}\left[1-e^{j\omega t}\right]=-j2sin\left(\omega \tau /2\right)e^{-j\omega \tau /2}& & \left(4.148\right)\end{array}$$

Substituting Eqn. (4.148) into Eqn. (4.146), the Fourier transform of x(t) is found as

$$\begin{array}{ccc}\begin{array}{ll}X(\omega )\hfill & =\frac{2A}{\omega }sin\left(\frac{\omega \tau }{2}\right)e^{-j\omega \tau /2}\hfill \\ \hfill & =A\tau \text{sinc}\left(\frac{\omega \tau }{2\pi }\right)e^{-j\omega \tau /2}\hfill \end{array}& & (4.149)\end{array}$$

The transform found in Eqn. (4.149) is complex-valued, and is best expressed in polar form. In preparation for that, we will find it convenient to write the transform in terms of the frequency variable f through the substitution ω = 2πf:

$$\begin{array}{ccc}X\left(f\right)=A\tau \text{ sinc}\left(f\tau \right)e^{-j\pi \pi f\tau }& & \left(4.150\right)\end{array}$$

We would ultimately like to express the transform of Eqn. (4.150) in the form

$$X\left(f\right)=\left|X\left(f\right)\right|e^{j\theta \left(f\right)}$$

Let the functions B (f) and β (f) be defined as

$$B\left(f\right)=A\tau \text{ sinc}\left(f\tau \right)$$

and

$$B\left(f\right)=-\pi f\tau$$

so that

$$X\left(f\right)=B\left(f\right)e^{-j\beta \left(f\right)}$$

The functions B (f) and β (f) are shown in Fig. 4.39.

A quick glance at Fig. 4.39 reveals the reason that keeps us from declaring the functions B (f) and β (f) as the magnitude and the phase of the transform X (f): The function B (f) could be negative for some values of f, and therefore cannot be the magnitude of the transform. However, |X (ω)| can be writteninterms of B (f) by paying attention to its sign. In order to outline the procedure to be used, we will consider two separate example intervals of the frequency variable f:

Case 1: −1/τ < f < 1/τ

This is an example of a frequency interval in which B (f) ≥ 0, and

$$B\left(f\right)=\left|B\left(f\right)\right|$$

Therefore, we can write the transform as

$$X(f)=B (f)e^{j\beta (f)}=|B (f)|e^{j\beta (f)}$$

resulting in the magnitude

$$\left|X\left(f\right)\right|=\left|B\left(f\right)\right|=B\left(f\right)$$

and the phase

$$\theta \left(f\right)=\beta \left(f\right)$$

for the transform.

Case 2: 1/τ < f < 2/τ

For this interval we have B (f)< 0. The function B (f) can be expressed as

$$B\left(f\right)=-\left|B\left(f\right)\right|=\left|B\left(f\right)\right|e^{\pm j\pi }$$

Using this form of B (f) in the transform leads to

$$X\left(f\right)=\left|B\left(f\right)\right|e^{\pm j\pi }{e}^{j\beta \left(f\right)}$$

which results in the magnitude

$$\left|X\left(f\right)\right|=\left|B\left(f\right)\right|=-B\left(f\right)$$

and the phase

$$\theta \left(f\right)=\beta \left(f\right)\pm \pi$$

for the transform X (f).

Comparison of the results in the two cases above leads us to the following generalized expressions for the magnitude and the phase of the transform X (f) as

$$\begin{array}{ccc}\left|X\left(f\right)\right|=\left|B\left(f\right)\right|& & \left(4.151\right)\end{array}$$

and

$$\begin{array}{ccc}\theta \left(f\right)=\{\begin{array}{ll}\beta \left(f\right)\hfill & \text{if }B\left(f\right)\ge 0\hfill \\ \beta \left(f\right)\pm \pi \hfill & \text{if }B\left(f\right)>0\hfill \end{array}& & \left(4.152\right)\end{array}$$

In summary, for frequencies where B (f) is negative, we add ±π radians to the phase term to account for the factor (−1). This is illustrated in Fig. 4.40.

Phase values outside the interval (−π, π) radians are indistinguishable from the corresponding values within the interval since we can freely add any integer multiple of 2π radians to the phase. Because of this, it is customary to fit the phase values inside the interval (−π, π), a practice referred to as phase wrapping. Fig. 4.41 shows the phase characteristic after the application of phase wrapping.

Software resources:

ex_4_13.m

Example 4.14: Transform of the unit-impulse function

The unit-impulse function was defined in Section 1.3.2 of Chapter 1. The Fourier transform of the unit-impulse signal can be found by direct application of the Fourier transform integral along with the sifting property of the unit-impulse function.

$$\begin{array}{ccc}ℱ\left\{\delta \left(t\right)\right\}={\int }_{-\infty }^{\infty }{\delta \left(t\right)e^{-j\omega t}dt=e^{-j\omega t}|}_{t=0}=1& & \left(4.153\right)\end{array}$$

In an effort to gain further insight, we will also take an alternative approach to determining the Fourier transform of the unit-impulse signal. Recall that in Section 1.3.2 we expressed the unit-impulse function as the limit case of a rectangular pulse with unit area. Given the pulse signal

$$q\left(t\right)=\frac{1}{a}\prod \left(\frac{t}{a}\right)$$

the unit-impulse function can be expressed as

$$\delta \left(t\right)=\underset{a\to 0}{lim}\left\{q\left(t\right)\right\}$$

If the parameter a is gradually made smaller, the pulse q (t) becomes narrower and taller while still retaining unit area under it. In the limit, the pulse q (t) becomes the unit impulse function δ (t). Using the result obtained in Example 4.12 with A = 1/a and τ = a,the Fourier transform of q (t) is

$$\begin{array}{ccc}Q\left(f\right)=ℱ\left\{q\left(t\right)\right\}=\text{sinc}\left(fa\right)& & \left(4.154\right)\end{array}$$

Using an intuitive approach, we may also conclude that the Fourier transform of the unit-impulse function is an expanded, or stretched-out, version of the transform Q (f) in Eqn. (4.154), i.e.,

$$ℱ\left\{\delta \left(t\right)\right\}=\underset{a\to 0}{lim}\left\{Q\left(f\right)\right\}=\underset{a\to 0}{lim}\left\{\text{sinc}\left(fa\right)\right\}=1$$

This conclusion is easy to justify from the general behavior of the function sinc (fa). Zero crossings on the frequency axis appear at values of f that are integer multiples of 1/a.As the value of the parameter a is reduced, the zero crossings of the sinc function move further apart, and the function stretches out or flattens around its peak at f = 0. In the limit, it approaches a constant of unity. This behavior is illustrated in Fig. 4.42.

Example 4.15: Fourier transform of a right-sided exponential signal

Determine the Fourier transform of the right-sided exponential signal

$$x\left(t\right)=e^{-at}u\left(t\right)$$

with a > 0 as shown in Fig. 4.43.

Solution: Application of the Fourier transform integral of Eqn. (4.127) to x (t) yields

$$X\left(\omega \right)={\int }_{-\infty }^{\infty }{e}^{-at}u\left(t\right)e^{-j\omega t}dt$$

Changing the lower limit of integral to t = 0 and dropping the factor u (t) results in

$$X\left(\omega \right)={\int }_0^{\infty }{e}^{-at}{e}^{-j\omega t}dt={\int }_0^{\infty }{e}^{-\left(a+j\omega \right)t}dt=\frac{1}{a+j\omega }$$

This result in Eqn. (4.155) is only valid for a > 0 since the integral could not have been evaluated otherwise. The magnitude and the phase of the transform are

$$\begin{array}{c}\left|X\left(\omega \right)\right|=\left|\frac{a}{a+j\omega }\right|=\frac{1}{\sqrt{a^2+{\omega }^2}}\\ \theta \left(\omega \right)=\measuredangle X\left(\omega \right)=-{tan}^{-1}\left(\frac{\omega }{a}\right)\end{array}$$

Magnitude and phase characteristics are graphed in Fig. 4.44.

Some observations are in order: The magnitude of the transform is an even function of ω since

$$\left|X\left(-\omega \right)\right|=\frac{1}{\sqrt{a^2+{\left(-\omega \right)}^2}}=\frac{1}{\sqrt{a^2+{\omega }^2}}=X\left(\omega \right)$$

The peak magnitude occurs at ω = 0 and its value is |X (0)| = 1/a. At the radian frequency ω = a the magnitude is equal to $\left|X\left(a\right)\right|=1/\left(\sqrt{2}a\right)$ which is $1/\sqrt{2}$ times its peak value. On a logarithmic scale this corresponds to a drop of 3 decibels (dB). In contrast with the magnitude, the phase of the transform is an odd function of ω since

$$\theta \left(-\omega \right)=-{tan}^{-1}\left(\frac{-\omega }{a}\right)={tan}^{-1}\left(\frac{\omega }{a}\right)=-\theta \left(\omega \right)$$

At frequencies ω → ±∞ the phase angle approaches

$$\underset{\omega \to \pm \infty }{lim}\left[\theta \left(\omega \right)\right]=\mp \frac{\pi }{2}$$

Furthermore, at frequencies ω = ±a the phase angle is

$$\theta \left(\pm a\right)=\mp \frac{\pi }{4}$$

We will also compute real and imaginary parts of the transform to write its Cartesian form representation. Multiplying both the numerator and the denominator of the result in Eqn. (4.155) by (a − jω) we obtain

$$X\left(\omega \right)=\frac{1}{\left(a+j\omega \right)}\frac{\left(a-j\omega \right)}{\left(a-j\omega \right)}=\frac{a-j\omega }{a^2+{\omega }^2}$$

from which real and imaginary parts of the transform can be extracted as

$$\begin{array}{ccc}{X}_r\left(\omega \right)=\frac{a}{a^2+{\omega }^2}& & \left(4.155\right)\end{array}$$

and

$$\begin{array}{ccc}{X}_i\left(\omega \right)=\frac{-\omega }{a^2+{\omega }^2}& & \left(4.156\right)\end{array}$$

respectively. Xr (ω) and Xi (ω) are shown in Fig. 4.45(a) and (b).

In this case the real part is an even function of ω, and the imaginary part is an odd function.

Software resources:

ex_4_15a.m

ex_4_15b.m

Example 4.16: Fourier transform of a two-sided exponential signal

Determine the Fourier transform of the two-sided exponential signal given by

$$x\left(t\right)=e^{-a\left|t\right|}$$

where a is any non-negative real-valued constant. The signal x (t) is shown in Fig. 4.46.

Solution: Applying the Fourier transform integral of Eqn. (4.127) to our signal we get

$$X\left(\omega \right)={\int }_{-\infty }^{\infty }{e}^{-a\left|t\right|}{e}^{-j\omega t}dt$$

Splitting the integral into two halves yields

$$X\left(\omega \right)={\int }_{-\infty }^0{e}^{-a\left|t\right|}{e}^{-j\omega t}dt+{\int }_0^{\infty }{e}^{-a\left|t\right|}{e}^{-j\omega t}dt$$

Recognizing that

$$\begin{array}{lll}t\le 0\hfill & \Rightarrow \hfill & e^{-a\left|t\right|}=e^{at}\hfill \\ t\ge 0\hfill & \Rightarrow \hfill & e^{-a\left|t\right|}=e^{-at}\hfill \end{array}$$

the transform is

$$\begin{array}{ccc}\begin{array}{ll}X(\omega )\hfill & ={\int }_{-\infty }^0{e}^{at}{e}^{-j\omega t}dt+{\int }_0^{\infty }{e}^{-at}{e}^{-j\omega t}dt\hfill \\ \hfill & =\frac{1}{a-j\omega }+\frac{1}{a+j\omega }\hfill \\ \hfill & =\frac{2a}{a^2+{\omega }^2}\hfill \end{array}& & (4.157)\end{array}$$

X (ω) is shown in Fig. 4.47.

The transform X (ω) found in Eqn. (4.157) is purely real for all values of ω. This is a consequence of the signal x (t) having even symmetry, and will be explored further as we look at the symmetry properties of the Fourier transform in the next section. In addition to being purely real, the transform also happens to be non-negative in this case, resulting in a phase characteristic that is zero for all frequencies.

Software resources:

ex_4_16.m

Example 4.17: Fourier transform of a triangular pulse

Find the Fourier transform of the triangular pulse signal given by

$$x\left(t\right)=A\Lambda \left(\frac{t}{\tau }\right)=\{\begin{array}{cc}A+At/\tau ,& -\tau <t<0\\ A-At/\tau ,& 0\le t<\tau \\ 0,& \left|t\right|\ge \tau \end{array}$$

where Λ (t) is the unit-triangle function defined in Section 1.3.2 of Chapter 1. The signal x (t) is shown in Fig. 4.48.

Solution: Using the Fourier transform integral on the signal x (t), we obtain

$$\begin{array}{ccc}\begin{array}{ll}X(\omega )\hfill & ={\int }_{-\tau }^0\left(A+\frac{At}{\tau }\right)e^{-j\omega t}dt+{\int }_0^{\tau }\left(A-\frac{At}{\tau }\right)e^{-j\omega t}dt\mathrm{ }\hfill \\ \hfill & =A{\int }_{-\tau }^0{e}^{-j\omega t}dt+\frac{A}{\tau }{\int }_{-\tau }^{0}t e^{-j\omega t}dt+A{\int }_0^{\tau }{e}^{-j\omega t}dt-\frac{A}{\tau }{\int }_0^{\tau }t e^{-j\omega t}dt\hfill \\ \hfill & =A{\int }_{-\tau }^{\tau }{e}^{-j\omega t}dt+\frac{A}{\tau }{\int }_{-\tau }^{0}t e^{-j\omega t}dt-\frac{A}{\tau }{\int }_0^{\tau }t e^{-j\omega t}dt\hfill \end{array}& & (4.158)\end{array}$$

The last two integrals in Eqn. (4.158) are similar in form to entry (B.16) of the table of indefinite integrals in Appendix B, repeated here for convenience:

$$\begin{array}{cc}\text{Eqn}.\left(\text{B}.16\right):& \int t{e}^{at}dt=\frac{1}{a^2}{e}^{at}\left[at-1\right]\end{array}$$

Setting a = −jω results in

$$\int t{e}^{-j\omega t}dt=\frac{1}{{\omega }^2}{e}^{-j\omega t}\left[j\omega t+1\right]$$

and the Fourier transform is

$$\begin{array}{ll}X(\omega )\hfill & =A{\left(\frac{e^{-j\omega t}}{-j\omega }\right)|}_{-\tau }^{\tau }+{\frac{A}{{\omega }^2\tau }\left(e^{-j\omega t}[j\omega t+1]\right)|}_{-\tau }^0-{\frac{A}{{\omega }^2\tau }\left(e^{-j\omega t}[j\omega t+1]\right)|}_0^{\tau }\hfill \\ \hfill & =\frac{2Asin(\omega t)}{\omega }+\frac{2A}{{\omega }^2\tau }[1-\omega \tau sin(\omega \tau )-cos(\omega \tau )]\hfill \\ \hfill & =\frac{A}{{\omega }^2\tau }[1-cos(\omega \tau )]\hfill \end{array}$$

Using the appropriate trigonometric identity5 it can be shown that

$$\begin{array}{ccc}X\left(\omega \right)=\frac{4A}{{\omega }^2\tau }{sin}^2\left(\frac{\omega \tau }{2}\right)& & \left(4.159\right)\end{array}$$

It would be convenient to use the sinc function with the result in Eqn. (4.159). Recognizing that

$$\text{sinc }\left(\frac{\omega \tau }{2}\right)=\frac{sin\left(\omega \tau /2\right)}{\left(\omega \tau /2\right)}=\frac{2}{\omega \tau }sin\left(\frac{\omega \tau }{2}\right)$$

the transform in Eqn. (4.159) can be written as

$$\text{X}\left(\omega \right)=A\tau {\text{ sinc}}^2\left(\frac{\omega \tau }{2\pi }\right)$$

If the transform is written in terms of f instead of ω, we obtain

$$\text{X}\left(f\right)=A\tau {\text{ sinc}}^2\left(f\tau \right)$$

Thus, the Fourier transform of a triangular pulse is proportional the square of the sinc function. Contrast this with the Fourier transform of a rectangular pulse which is proportional to the sinc function as we have seen in Example 4.12. The transform of the triangular pulse, shown in Fig. 4.49, seems to be concentrated more heavily around low frequencies compared to the transform of the rectangular pulse.

We will have occasion to look at the transform of a triangular pulse in more depth when we study the properties of the Fourier transform in the next section. Specifically, a much simpler method of obtaining X (ω) for a triangular pulse will be presented after studying the convolution property of the Fourier transform.

Software resources:

ex_4_17.m

Example 4.18: Fourier transform of the signum function

Determine the Fourier transform of the signum function defined as

$$\begin{array}{ccc}x\left(t\right)=sgn\left(t\right)=\{\begin{array}{cc}-1,& t<0\\ 1,& t>0\end{array}& & \left(4.160\right)\end{array}$$

and shown graphically in Fig. 4.50.

Solution: If we were to apply the Fourier transform integral directly to this signal, we would obtain

$$X\left(\omega \right)={\int }_{-\infty }^0\left(-1\right)e^{-j\omega t}dt+{\int }_0^{\infty }\left(1\right)e^{-j\omega t}dt$$

in which the two integrals cannot be evaluated. Instead, we will define an intermediate signal p (t) as

$$p\left(t\right)\{\begin{array}{cc}-e^{at},& t<0\\ e^{-at},& t>0\end{array}$$

where a ≥ 0. We will first determine the Fourier transform of this intermediate signal, and then obtain the Fourier transform of the signum function from this intermediate result. The Fourier transform of p (t) is found as

$$\begin{array}{cc}P(\omega )\hfill & ={\int }_{-\infty }^0(-e^{at})e^{-j\omega t}dt+{\int }_0^{\infty }(-e^{at})e^{-j\omega t}dt\hfill \\ \hfill & ={\left(\frac{-e^{at}{e}^{-j\omega t}}{a-j\omega }\right)|}_{-\infty }^0+\left(\frac{e^{-at}{e}^{-j\omega t}}{-a-j\omega }\right){|}_0^{\infty }\hfill \\ \hfill & =\frac{-j 2\omega }{a^2+{\omega }^2}\hfill \end{array}$$

The result is purely imaginary for all ω due to the odd symmetry of p (t). As the value of a is reduced, the intermediate signal p (t) starts to look more and more similar to the signum function, and becomes the signum function in the limit as a → 0. The transition of p (t) to sgn (t) is illustrated in Fig. 4.51 for a = 0.4, 0.15, and 0.04.

Setting a = 0 in the transform P (ω), we obtain the Fourier transform of the original signal x (t):

$$X\left(\omega \right)=ℱ\left\{sgn\left(t\right)\right\}=\underset{a\to 0}{lim}\left[\frac{-j\text{ 2}w}{a^2+{\omega }^2}\right]=\frac{2}{j\omega }$$

The magnitude of the transform is

$$\left|X\left(\omega \right)\right|=\frac{2}{\left|\omega \right|}$$

and the phase is

$$\theta \left(\omega \right)=\measuredangle X\left(\omega \right)=\{\begin{array}{cc}\pi /2,& \omega <0\\ -\pi /2,& \omega >0\end{array}$$

Magnitude and phase characteristics are shown in Fig. 4.52.

Software resources:

ex_4_18.m

4.3.5 Properties of the Fourier transform

In this section we will explore some of the fundamental properties of the Fourier transform that provide further insight into its use for understanding and analyzing characteristics of signals. We will also see that some of these properties greatly simplify the computation of the Fourier transform for certain types of signals.

Linearity

Fourier transform is a linear operator. For any two signals x1 (t) and x2 (t) with their respective transforms

$$\begin{array}{l}{x}_1\left(t\right)\stackrel{ℱ}{\leftrightarrow }{X}_1\left(\omega \right)\\ x_2\left(t\right)\stackrel{ℱ}{\leftrightarrow }{X}_2\left(\omega \right)\end{array}$$

and any two constants α1 and α2, it can be shown that the following relationship holds:

Duality

Consider the transform pair

$$x\left(t\right)\stackrel{ℱ}{\leftrightarrow }X\left(\omega \right)$$

The transform relationship between x (t) and X (ω) is defined by the inverse and forward Fourier transform integrals given by Eqns. (4.126) and (4.127) and repeated here for convenience:

$$\begin{array}{cc}\text{Eqn}\text{. }\left(4.126\right):& x\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X\left(\omega \right)e^{j\omega t}d\omega \\ \text{Eqn}\text{. }\left(4.122\right):& X\left(\omega \right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}dt\end{array}$$

Suppose that we swap the roles of the independent variables to construct a new time-domain signal X (t) and a new frequency-domain function x (ω). What would be the relationship between X (t) and x (ω)?

The time-domain function X (t) and the frequency-domain function 2π x (−ω) form a Fourier transform pair. Thus, if X (ω) is the Fourier transform of x (t), then 2π x (−ω) is the Fourier transform of X (t).

We may find it more convenient to express the duality property using the frequency f instead of the radian frequency ω since this approach eliminates the 2π factor from the equations:

Duality property will be useful in deriving some of the other properties of the Fourier transform.

Example 4.19: Fourier transform of the sinc function

Find the Fourier transform of the signal

$$x\left(t\right)=\text{sinc}\left(t\right)$$

Solution: The Fourier transform of a rectangular pulse was found in Example 4.12 as a sinc function. Specifically we obtained

$$\begin{array}{ccc}ℱ\left\{A\prod \left(\frac{t}{\tau }\right)\right\}=A\tau \text{sinc}\left(\frac{\omega \tau }{2\pi }\right)& & \left(4.170\right)\end{array}$$

This relationship can be used as a starting point in finding the Fourier transform of x (t) = sinc (t). Let τ = 2π so that the argument of the sinc function in Eqn. (4.170) becomes ω:

$$\begin{array}{ccc}ℱ\left\{A\prod \left(\frac{t}{\tau }\right)\right\}=2\pi \text{sinc}\left(\omega \right)& & \left(4.171\right)\end{array}$$

In the next step, let A = 1/2π to obtain

$$\begin{array}{ccc}ℱ\left\{\frac{1}{2\pi }\prod \left(\frac{t}{2\pi }\right)\right\}= \text{sinc}\left(\omega \right)& & \left(4.172\right)\end{array}$$

Applying the duality property given by Eqn. (4.169) to the transform pair in Eqn. (4.172) leads to the result

$$\begin{array}{ccc}ℱ \left\{\text{sinc}\left(t\right)\right\}=\prod \left(\frac{-\omega }{2\pi }\right)=\prod \left(\frac{\omega }{2\pi }\right)& & \left(4.173\right)\end{array}$$

Using f instead of ω in Eqn. (4.173) yields a relationship that is easier to remember:

$$\begin{array}{ccc}ℱ \left\{\text{sinc}\left(t\right)\right\}=\prod \left(f\right)& & \left(4.174\right)\end{array}$$

The Fourier transform of x (t)= sinc(t) is a unit pulse in the frequency domain. The signal x (t) and its transform are graphed in Fig. 4.53(a) and (b).

Example 4.20: Transform of a constant-amplitude signal

Find the Fourier transform of the constant-amplitude signal

$$\begin{array}{cc}x\left(t\right)=1,& \text{all }t\end{array}$$

Solution: Finding the Fourier transform of x (t) by direct application of Eqn. (4.127) is not possible since the resulting integral

$$X\left(\omega \right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}dt={\int }_{-\infty }^{\infty }\left(1\right)e^{-j\omega t}dt$$

could not be evaluated. On the other hand, in Example 4.14 we have concluded that the Fourier transform of the unit-impulse signal is a constant for all frequencies, that is

$$\begin{array}{cccc}ℱ\left\{\delta \left(t\right)\right\}=1,& \text{all }\omega & & \left(4.175\right)\end{array}$$

Applying the duality property to the transform pair in Eqn. (4.175) we obtain

$$ℱ\left\{1\right\}=2\pi \delta \left(-\omega \right)=2\pi \delta \left(\omega \right)$$

If f is used instead of ω in the transform, then we have

$$ℱ\left\{1\right\}=\delta \left(f\right)$$

This example also presents an opportunity for two important observations:

1. The signal x (t) = 1 does not satisfy the existence conditions discussed in Section 4.3.2; it is neither absolute integrable nor square integrable. Therefore, in a strict sense, its Fourier transform does not converge. By allowing the impulse function to be used in the transform, we obtain a function X (ω) that has the characteristics of a Fourier transform, and that can be used in solving problems in the frequency domain.

2. Sometimes we express the Fourier transform in terms of the radian frequency ω; at other times we use the frequency f. In general, the conversion from one format to another is just a matter of using the relationship ω = 2πf, so that

   $$\begin{array}{cc}{X\left(\omega \right)=X\left(f\right)|}_{f=\omega /2\pi },& {X\left(f\right)=X\left(\omega \right)|}_{\omega =2\pi f}\end{array}$$

One exception to this rule is seen when the transform contains a singularity function. Compare Eqns. (4.176) and (4.176) in the example above.X (ω) has a 2π factor that X (f) does not have. This behavior is consistent with scaling of the impulse function (see Problem 1.6 of Chapter 1).

Example 4.21: Another example of using the duality property

Using the duality property, find the Fourier transform of the signal

$$x\left(t\right)=\frac{1}{3+2t^2}$$

which is graphed in Fig. 4.54.

Solution: While it is possible to solve this problem by a direct application of the Fourier transform integral given by Eqn. (4.127), it would be difficult to evaluate the integral unless we resort to the use of integration tables. Instead, we will simplify the solution by using another transform pair that we already know, and adapting that transform pair to our needs. In Example 4.16 wehavefound that

$$e^{-a\left|t\right|}\stackrel{ℱ}{\leftrightarrow }\frac{2a}{a^2+{\omega }^2}$$

where the form of the transform bears a similarity to the time-domain expression for the signal x (t). In both functions the numerator is constant, and the denominator has a second-order term next to a constant term. We will make use of this similarity through the use of the duality property of the Fourier transform. The duality property allows us to write

$$\begin{array}{ccc}\frac{2a}{a^2+t^2}\stackrel{ℱ}{\leftrightarrow }2\pi e^{-a\left|\omega \right|}& & \left(4.176\right)\end{array}$$

as another valid transform pair. By taking steps that will preserve the validity of this transform pair, we will work toward making the time-domain component of the relationship in Eqn. (4.176) match the signal x (t) given in the problem statement. The signal x (t) has a 2t2 term in its denominator. Let us multiply both the numerator and the denominator of the time-domain component of Eqn. (4.176) by 2:

$$\frac{4a}{2a^2+2t^2}\stackrel{ℱ}{\leftrightarrow }2\pi e^{-a\left|\omega \right|}$$

To match the constant term in the denominator of x (t) we will choose

$$\begin{array}{ccc}2a^2=3& \Rightarrow & a=\sqrt{\frac{3}{2}}\end{array}$$

and obtain

$$\begin{array}{ccc}\frac{2\sqrt{6}}{3+2t^2}\stackrel{ℱ}{\leftrightarrow }2\pi e^{-\sqrt{\frac{3}{2}}\left|\omega \right|}& & \left(4.177\right)\end{array}$$

Afterwards, by scaling both sides of Eqn. (4.177) with $2\sqrt{6}$ we get

$$\frac{1}{3+2t^2}\stackrel{ℱ}{\leftrightarrow }\frac{\pi }{\sqrt{6}}{e}^{-\sqrt{\frac{3}{2}}\left|\omega \right|}$$

The transform found in Eqn. (4.178) is graphed in Fig. 4.55 using f as the independent variable instead of ω.

Software resources:

ex_4_21.m

Example 4.22: Signum function revisited

In Example 4.18 the Fourier transform of the signal x (t) = sgn(t) was found to be

$$X\left(f\right)=ℱ\left\{sgn\left(t\right)\right\}=\frac{2}{j\omega }$$

Determine the time-domain signal, the Fourier transform of which is

$$X\left(\omega \right)=-jsgn\left(\omega \right)$$

Solution: This is another problem that could be simplified by the use of the duality property. Starting with the already known transform pair

$$sgn\left(t\right)\stackrel{ℱ}{\leftrightarrow }\frac{2}{j\omega }$$

and applying the duality property as expressed by Eqn. (4.168) we conclude that the following must also be a valid transform pair:

$$\frac{2}{jt}\stackrel{ℱ}{\leftrightarrow }2\pi sgn\left(-\omega \right)$$

Recognizing that the sgn function has odd symmetry, that is, sgn (−ω) = −sgn(ω),

$$\begin{array}{ccc}\frac{2}{jt}\stackrel{ℱ}{\leftrightarrow }-2\pi sgn\left(\omega \right)& & \left(4.178\right)\end{array}$$

Multiplying both sides of Eqn. (4.178) by j/2π, we have

$$\begin{array}{ccc}\frac{1}{\pi t}\stackrel{ℱ}{\leftrightarrow }-jsgn\left(\omega \right)& & \left(4.179\right)\end{array}$$

The signal we are seeking is

$$x\left(t\right)=\frac{1}{\pi t}$$

which is graphed in Fig. 4.56.

It is interesting to look at the magnitude and the phase of the transform X (ω) in Eqn. (4.178). The transform can be written as

$$X\left(\omega \right)=-jsgn\left(\omega \right)=\{\begin{array}{rr}\hfill -j,& \hfill \omega >0\\ \hfill j,& \hfill \omega <0\end{array}$$

Realizing that j = ejπ/2 we can write

$$X\left(\omega \right)=\{\begin{array}{rr}\hfill e^{-j\pi /2},& \hfill \omega >0\\ \hfill e^{j\pi /2},& \hfill \omega <0\end{array}$$

The magnitude and the phase of X (ω) are shown in Fig. 4.57.

The magnitude of the transform is equal to unity for all frequencies. The phase of the transform is −π/2 radians for positive frequencies and +π/2 radians for negative frequencies.

Example 4.23: Fourier transform of the unit-step function

Determine the Fourier transform of the unit-step function x (t) = u (t).

Solution: The unit-step function does not satisfy the conditions for the existence of the Fourier transform. An attempt to find the transform by direct application of Eqn. (4.127) would result in

$$ℱ\left\{u\left(t\right)\right\}={\int }_{-\infty }^{\infty }u\left(t\right)e^{-j\omega t}dt={\int }_0^{\infty }{e}^{-j\omega t}dt$$

which does not converge. Instead, we will try to make use of some of the earlier results found along with the linearity property of the Fourier transform. The unit-step function can be expressed in terms of the signum function and a constant offset as

$$u\left(t\right)=\frac{1}{2}+\frac{1}{2}sgn\left(t\right)$$

This is illustrated in Fig. 4.58.

Using the linearity of the Fourier transform we obtain

$$\begin{array}{ccc}\begin{array}{ll}ℱ\{u(t)\}\hfill & =ℱ\left\{\frac{1}{2}+\frac{1}{2}sgn(t)\right\}\hfill \\ \hfill & =\frac{1}{2}ℱ\{1\}+\frac{1}{2}ℱ\{sgn(t)\}\hfill \end{array}& & (4.180)\end{array}$$

The transform of a constant signal was found in Example 4.20:

$$ℱ\left\{1\right\}=2\pi \delta \left(\omega \right)$$

Furthermore, in Example 4.18 we found

$$ℱ\left\{sgn\left(t\right)\right\}=\frac{2}{j\omega }$$

Using these two results in Eqn. (4.180), Fourier transform of the unit-step function is

$$\begin{array}{ccc}ℱ\left\{u\left(t\right)\right\}=\pi \delta \left(\omega \right)+\frac{1}{j\omega }& & \left(4.181\right)\end{array}$$

Fig. 4.59 shows the unit-step signal and the magnitude of its transform.

If the transform of the unit-step function is needed in terms of f instead of ω, the result in Eqn. (4.181) needs to be modified as

$$\begin{array}{ccc}ℱ\left\{u\left(t\right)\right\}=\frac{1}{2}\delta \left(f\right)+\frac{1}{j2\pi f}& & \left(4.182\right)\end{array}$$

Symmetry of the Fourier Transform

Conjugate symmetry and conjugate antisymmetry properties were defined for time-domain signals in Section 1.3.6 of Chapter 1. Same definitions apply to a frequency-domain transform as well: A transform X (ω) is said to be conjugate symmetric if it satisfies

$$\begin{array}{ccc}{X}^{*}\left(\omega \right)=X\left(-\omega \right)& & \left(4.183\right)\end{array}$$

for all ω. Similarly, a transform X (ω) is said to be conjugate antisymmetric if it satisfies

$$\begin{array}{ccc}{X}^{*}\left(\omega \right)=-X\left(-\omega \right)& & \left(4.184\right)\end{array}$$

for all ω. A transform does not have to be conjugate symmetric or conjugate antisymmetric. If it has any symmetry properties, however, they can be explored for simplifying the analysis in the frequency domain.

If the signal x (t) is real-valued,itcan be shown thatits Fourier transform X (ω) is conjugate symmetric. Conversely, if the signal x (t) is purely imaginary, its transform is conjugate antisymmetric.

Cartesian and polar forms of the transform

A complex transform X (ω) can be written in polar form as

$$\begin{array}{ccc}X\left(\omega \right)=\left|X\left(\omega \right)\right|e^{j\Theta \left(\omega \right)}& & \left(4.191\right)\end{array}$$

and in Cartesian form as

$$\begin{array}{ccc}X\left(\omega \right)=X_r\left(\omega \right)+j{X}_i\left(\omega \right)& & \left(4.192\right)\end{array}$$

If X (ω) is conjugate symmetric, the relationship in Eqn. (4.183) can be written as

$$\begin{array}{ccc}\left|X\left(-\omega \right)\right|e^{j\Theta \left(-\omega \right)}=\left|X\left(\omega \right)\right|e^{-j\Theta \left(\omega \right)}& & \left(4.193\right)\end{array}$$

using the polar form of the transform, and

$$\begin{array}{ccc}{X}_r\left(-\omega \right)+j{X}_i\left(-\omega \right)=X_r\left(\omega \right)-j{X}_i\left(\omega \right)& & \left(4.194\right)\end{array}$$

using its Cartesian form. The consequences of Eqns. (4.193) (4.194) can be obtained by equating the magnitudes and the phases on both sides of Eqn. (4.193) and by equating real and imaginary parts on both sides of Eqn. (4.194). The results can be summarized as follows:

We have already established the fact that the transform of a real-valued x (t) is conjugate symmetric. The results in Eqns. (4.195) and (4.196) suggest that, for such a transform, the magnitude is an even function of ω, and the phase is an odd function. Furthermore, based on Eqns. (4.197) and (4.198), the real part of the transform exhibits even symmetry while its imaginary part exhibits odd symmetry.

Similarly, if X (ω) is conjugate antisymmetric, the relationship in Eqn. (4.183) reflects on polar form of X (ω) as

$$\begin{array}{ccc}\left|X\left(-\omega \right)\right|e^{j\Theta \left(-\omega \right)}=-\left|X\left(-\omega \right)\right|e^{-j\Theta \left(\omega \right)}& & \left(4.199\right)\end{array}$$

The negative sign on the right side of Eqn. (4.199) needs to be incorporated into the phase sincewecould notwrite |X (−ω)| = −|X (ω)| (recall that magnitude needs to be a nonnegative function for all ω). Using e∓jπ = − 1, Eqn. (4.199) can be written as

$$\begin{array}{ccc}\begin{array}{ll}\left|X\left(-\omega \right)\right|e^{j\Theta \left(-\omega \right)}\hfill & =\left|X\left(\omega \right)\right|e^{-j\Theta \left(\omega \right)}{e}^{\mp j\pi }\hfill \\ \hfill & =\left|X\left(\omega \right)\right|e^{-j\left[\Theta \left(\omega \right)\pm \pi \right]}\hfill \end{array}& & \left(4.200\right)\end{array}$$

Conjugate antisymmetry property of the transform can also be expressed in Cartesian form as

$$\begin{array}{ccc}{X}_r\left(-\omega \right)+j{X}_i\left(-\omega \right)=-X_r\left(\omega \right)+j{X}_i\left(\omega \right)& & \left(4.201\right)\end{array}$$

The consequences of Eqns. (4.199) and (4.201) are given below.

A purely imaginary signal leads to a Fourier transform with conjugate antisymmetry. For such a transform the magnitude is still an even function of ω as suggested by Eqn. (4.202). The phase is neither even nor odd. The real part is an odd function of ω, and the imaginary part is an even function.

Example 4.24: Symmetry properties for the transform of right-sided exponential signal

The Fourier transform of the right-sided exponential signal x (t) = e−at u (t) was found in Example 4.15 to be

$$X\left(f\right)=ℱ\left\{e^{-at}u\left(t\right)\right\}=\frac{1}{a+j\omega }$$

Elaborate on the symmetry properties of the transform.

Solution: Since the signal x (t) is real-valued, its transform must be conjugate-symmetric. The complex conjugate of the Fourier transform is

$$X^{*}\left(\omega \right)={\left(\frac{1}{a+j\omega }\right)}^{*}=\frac{1}{a-j\omega }$$

and the folded version of the transform, obtained by replacing ω with −ω in Eqn. (4.206), is

$$X\left(-\omega \right)={\left(\frac{1}{a+j\omega }\right)|}_{\omega \to -\omega }=\frac{1}{a-j\omega }$$

Thus we verify that X* (ω) = X (−ω), and the transform is conjugate symmetric. The magnitude and the phase of the transform are

$$\left|X\left(\omega \right)\right|=\left|\frac{1}{a+j\omega }\right|=\frac{1}{\sqrt{a^2+{\omega }^2}}$$

and

$$\Theta \left(\omega \right)=\measuredangle X\left(\omega \right)=-{tan}^{-1}\left(\frac{\omega }{a}\right)$$

respectively. We observe that the magnitude is an even function of frequency, that is,

$$\left|X\left(-\omega \right)\right|=\frac{1}{\sqrt{a^2+{\left(-\omega \right)}^2}}=\frac{1}{\sqrt{a^2+w^2}}=\left|X\left(\omega \right)\right|$$

consistent with the conclusion reached in Eqn. (4.195). On the other hand, the phase has odd symmetry since

$$\Theta \left(-\omega \right)=-{tan}^{-1}\left(\frac{-\omega }{a}\right)={tan}^{-1}\left(\frac{\omega }{a}\right)=-\Theta \left(\omega \right)$$

consistent with Eqn. (4.196). By multiplying both the numerator and the denominator of the transform in Eqn. (4.206) by (a − jω) we have

$$\begin{array}{ccc}X\left(\omega \right)=\frac{\left(1\right)}{\left(a+j\omega \right)}\frac{\left(a-j\omega \right)}{\left(a-j\omega \right)}=\frac{a}{a^2+w^2}-j\frac{\omega }{a^2+w^2}& & \left(4.206\right)\end{array}$$

Real and imaginary parts of the transform can be found as

$$\begin{array}{ccc}{X}_r\left(\omega \right)=Re\left\{X\left(\omega \right)\right\}=\frac{a}{a^2+w^2}& & \left(4.207\right)\end{array}$$

and

$$\begin{array}{ccc}{X}_i\left(\omega \right)=Im\left\{X\left(\omega \right)\right\}=\frac{-\omega }{a^2+w^2}& & \left(4.208\right)\end{array}$$

It is atrivial matter to show that Xr (ω) is an even function of ω, and Xi (ω) is an odd function.

Software resources:

ex_4_24a.m

ex_4_24b.m

Transforms of even and odd signals

If the real-valued signal x (t) is an even function of time, the resulting transform X (ω) is real-valued for all ω.

Conversely it can also be proven that, if the real-valued signal x (t) has odd-symmetry, the resulting Fourier transform is purely imaginary. This can be mathematically stated as follows:

Conjugating a purely imaginary transform is equivalent to negating it, so an alternative method of expressing the relationship in Eqn. (4.215) is

$$\begin{array}{ccc}{X}^{*}\left(\omega \right)=-X\left(\omega \right)& & \left(4.216\right)\end{array}$$

The procedure for proving Eqn. (4.216) is very similar to the one given above for the transform of an even symmetric signal. (See Problem 4.25 at the end of this chapter.)

Example 4.25: Transform of a two-sided exponential signal

The two-sided exponential signal x (t) = e−a|t| was shown in Fig. 4.46, and has even symmetry. Its Fourier transform was found in Example 4.16 as

$$X\left(\omega \right)=\frac{2a}{a^2+{\omega }^2}$$

and was graphed in Fig. 4.47. As expected, the transform is real-valued for all ω.

Example 4.26: Transform of a pulse with odd symmetry

Consider the signal

$$x\left(t\right)=\{\begin{array}{cc}\begin{array}{c}-1,\\ 1,\\ 0,\end{array}& \begin{array}{l}-1<t<0\hfill \\ 0<t<1\hfill \\ t<-1\text{ or }t>1\hfill \end{array}\end{array}$$

shown in Fig. 4.60. Determine the Fourier transform X (ω) and show that it is purely imaginary.

Solution: The transform can be found by direct use of the forward transform integral.

$$\begin{array}{ccc}\begin{array}{ll}X\left(\omega \right)\hfill & ={\int }_{-1}^0{e}^{-j\omega t}dt+{\int }_0^1\left(1\right)e^{-j\omega t}dt\hfill \\ \hfill & =\frac{j2}{\omega }\left[cos\left(\omega \right)-1\right]\hfill \end{array}& & \left(4.217\right)\end{array}$$

As expected, the transform is purely imaginary. Its magnitude, phase and imaginary part are graphed in Fig. 4.61.

Software resources:

ex_4_26.m

Time shifting

The consequence of shifting, or delaying, a signal in time is multiplication of its Fourier transform by a complex exponential function of frequency.

Example 4.27: Time shifting a rectangular pulse

The transform of a rectangular pulse with amplitude A, width τ and center at t = 0 was found in Example 4.12 as

$$A\prod \left(\frac{t}{\tau }\right)\stackrel{ℱ}{\leftrightarrow }A\tau \text{ sinc }\left(\frac{\omega \tau }{2\pi }\right)$$

In Example 4.13 the transform of a time shifted version of this pulse was determined by direct application of the Fourier transform integral as

$$A\prod \left(\frac{t-\tau /2}{\tau }\right)\stackrel{ℱ}{\leftrightarrow }A\tau \text{ sinc }\left(\frac{\omega \tau }{2\pi }\right)e^{-j\omega t/2}$$

The result in Eqn. (4.149) could easily have been obtained from the result in Eqn. (4.144) through the use of the time shifting property.

Example 4.28: Time shifting a two-sided exponential signal

Consider again the two-sided exponential signal used in earlier examples. A time shifted version of it is

$$x\left(t\right)=e^{-a\left|t-\tau \right|}$$

The signal x (t) is shown in Fig 4.62 with the assumption that a > 0.

In Example 4.16 its Fourier transform was determined to be

$$ℱ\left\{e^{-a\left|t\right|}\right\}=\frac{2a}{a^2+{\omega }^2}$$

Using the time shifting property of the Fourier transform we obtain

$$ℱ\left(\omega \right)=ℱ\left\{e^{-a\left|t-\tau \right|}\right\}=\frac{2a{e}^{-j\omega \tau }}{a^2+{\omega }^2}$$

The transform X (ω) is shown in Fig. 4.63.

Compare this with the transform of the signal of Example 4.16. Note that time shifting the signal affects only the phase of the frequency spectrum, and not its magnitude.

Software resources:

ex_4_28.m

Frequency shifting

Modulation property

Modulation property is an interesting consequence of the frequency shifting property combined with the linearity of the Fourier transform. As its name implies, it finds significant use in modulation techniques utilized in analog and digital communications. We will have the occasion to use it when we study amplitude modulation in Chapter 11.

Multiplication of a signal by a cosine waveform causes its spectrum to be shifted in both directions by the frequency of the cosine waveform, and to be scaled by ½. Multiplication of the signal by a sine waveform causes a similar effect with an added phase shift of −π/2 radians for positive frequencies and π/2 radians for negative frequencies.

Example 4.29: Modulated pulse

Find the Fourier transform of the modulated pulse given by

$$x\left(t\right)=\{\begin{array}{ll}cos\left(2\pi f_{0}t\right),\hfill & \left|t\right|<\tau \hfill \\ 0,\hfill & \left|t\right|>\tau \hfill \end{array}$$

Solution: Let p (t) be defined as a rectangular pulse centered around the time origin with a pulse width of 2τ and a unit amplitude, i.e.,

$$p\left(t\right)=\prod \left(\frac{t}{2\tau }\right)$$

Using p (t), the signal x (t) can be expressed in a more compact form as

$$x\left(t\right)=p\left(t\right)cos\left(2\pi f_{0}t\right)$$

This is illustrated in Fig. 4.64.

The transform of the pulse p (t) is

$$\begin{array}{ccc}P\left(f\right)=ℱ\left\{p\left(t\right)\right\}=2\tau \text{ sinc }\left(2\tau f\right)& & \left(4.237\right)\end{array}$$

We will apply the modulation property to this result to obtain the transform we seek:

$$\begin{array}{ccc}\begin{array}{ll}X\left(\omega \right)\hfill & =\frac{1}{2}P\left(f-f_0\right)+\frac{1}{2}P\left(f-f_0\right)\hfill \\ \hfill & =\tau \text{ sinc }\left(2\tau \left(f+f_0\right)\right)+\tau \text{ sinc }\left(2\tau \left(f-f_0\right)\right)\hfill \end{array}& & \left(4.238\right)\end{array}$$

Construction of this spectrum is illustrated in Fig. 4.65.

Software resources:

ex_4_29.m

Time and frequency scaling

This property is significant in signal processing. The implication of Eqn. (4.239) is that, if the time variable is scaled by a factor a, the frequency variable is scaled by the same factor, but in the opposite direction.

Example 4.30: Working with scaled and shifted pulses

Determine the Fourier transform of the signal

$$x\left(t\right)=\{\begin{array}{cc}\begin{array}{c}-1,\\ 0.75,\end{array}& \begin{array}{l}-0.5<t<1.5\hfill \\ 1.5<t<5.5\hfill \end{array}\end{array}$$

which is shown in Fig 4.66.

Solution: It certainly is possible to use the Fourier transform integral of Eqn. (4.127) directly on the signal x (t). It would be more convenient, however, to make use of the properties of the Fourier transform. The signal x (t) can be written as the sum of two rectangular pulses:

1. One with amplitude of A1 = −1, width τ1 = 2 s, center at td1 = 0.5s
2. One with amplitude of A2 = 0.75, width τ2 = 4 s, center at td2 = 3.5

Using the unit-pulse function Π (t) we have

$$\begin{array}{ll}x\left(t\right)\hfill & =A_1\Pi \left(\frac{t-t_{d1}}{{\tau }_1}\right)+A_2\Pi \left(\frac{t-t_{d2}}{{\tau }_2}\right)\hfill \\ \hfill & =-\Pi \left(\frac{t-0.5}{2}\right)+0.75\Pi \left(\frac{t-3.5}{4}\right)\hfill \end{array}$$

The Fourier transform of the unit-pulse function is

$$ℱ\left\{\prod \left(t\right)\right\}=\text{sinc }\left(\frac{\omega }{2\pi }\right)$$

Using the scaling property of the Fourier transform, the following two relationships can be written:

$$\begin{array}{c}\prod \left\{\frac{t}{2}\right\}\stackrel{ℱ}{\leftrightarrow }\text{2 sinc }\left(\frac{\omega }{\pi }\right)\\ \prod \left\{\frac{t}{4}\right\}\stackrel{ℱ}{\leftrightarrow }\text{4 sinc }\left(\frac{2\omega }{\pi }\right)\end{array}$$

In addition, the use of the time-shifting property leads to

$$\begin{array}{ccc}\prod \left(\frac{t-0.5}{2}\right)\stackrel{ℱ}{\leftrightarrow }\text{2 sinc }\left(\frac{\omega }{\pi }\right)e^{-j0.5\omega }& & \left(4.246\right)\end{array}$$

$$\begin{array}{ccc}\prod \left(\frac{t-3.5}{4}\right)\stackrel{ℱ}{\leftrightarrow }\text{4 sinc }\left(\frac{2\omega }{\pi }\right)e^{-j3.5\omega }& & \left(4.247\right)\end{array}$$

Using Eqns. (4.246) and (4.247) the Fourier transform of x (t) is

$$X\left(\omega \right)=-2\text{ sinc }\left(\frac{\omega }{\pi }\right)e^{-j0.5\omega }+3\text{ sinc }\left(\frac{2\omega }{\pi }\right)e^{-j3.5\omega }$$

which is graphed in Fig. 4.67.

Software resources:

ex_4_30.m

Differentiation in the time domain

It is obvious that each differentiation of the signal x (t) will bring up another (jω) factor in the corresponding transform. Differentiating the signal x (t) for n times corresponds to multiplying the transform by (jω)n.

Example 4.31: Triangular pulse revisited

Recall that the Fourier transform of the triangular pulse signal

$$x\left(t\right)=A\Lambda \left(\frac{t}{\tau }\right)$$

was found in Example 4.17 through direct application of the Fourier transform integral of Eqn. (4.127). Rework the problem using the differentiation-in-time property.

Solution: Instead of working with the signal x (t) directly, we will find it convenient to work with its derivative

$$w\left(t\right)=\frac{dx\left(t\right)}{dt}$$

shown in Fig. 4.68.

Using the unit-pulse function along with time-scaling and time-shifting operations, w (t) can be expressed as

$$\begin{array}{ccc}w\left(t\right)=\frac{A}{\tau }\left[\prod \left(\frac{t+\tau /2}{\tau }\right)-\prod \left(\frac{t-\tau /2}{\tau }\right)\right]& & \left(4.255\right)\end{array}$$

Using the variable f instead of ω for convenience, the transform W (f) can be computed from Eqn. (4.255) as

$$\begin{array}{ccc}\begin{array}{ll}W\left(f\right)\hfill & =A\text{ sinc }\left(f\tau \right)e^{j2\pi f\left(\tau /2\right)}-A\text{ sinc }\left(f\tau \right)e^{-j2\pi f\left(\tau /2\right)}\hfill \\ \hfill & =A\text{ sinc }\left(f\tau \right)\left[e^{j2\pi f\left(\tau /2\right)}-e^{-j2\pi f\left(\tau /2\right)}\right]\hfill \\ \hfill & =2j\text{ sinc }\left(f\tau \right)sin\left(\pi f\tau \right)\hfill \end{array}& & \left(4.256\right)\end{array}$$

The real part of W (f) is equal to zero; its imaginary part is shown in Fig. 4.69. Frequency-domain equivalent of the relationship in Eqn. (4.256) is

$$\begin{array}{ccc}W\left(f\right)=\left(j2\pi f\right)X\left(f\right)& & \left(4.257\right)\end{array}$$

Solving Eqn. (4.257) for X (f) yields

$$X\left(f\right)=\frac{W\left(f\right)}{j2\pi f}=A\tau {\text{ sinc}}^2\left(f\tau \right)$$

which is in agreement with the answer found in Example 4.17.

Example 4.32: Fourier transform of a trapezoidal pulse

Determine the Fourier transform of the trapezoidal pulse x (t) shown in Fig. 4.70

Solution: Analytical description of this signal could be given in the form

$$x\left(t\right)=\{\begin{array}{cc}\begin{array}{c}0,\\ \left(t+b\right)/\left(b-a\right),\\ 1,\\ \left(-t+b\right)/\left(b-a\right),\\ 0,\end{array}& \begin{array}{l}t<-b\hfill \\ -b\le t<-a\hfill \\ -a\le t<a\hfill \\ a\le t<b\hfill \\ t>b\hfill \end{array}\end{array}$$

Once again we will use the intermediate signal w (t) obtained by differentiating x (t) with respect to t:

$$w\left(t\right)=\frac{dx\left(t\right)}{dt}=\{\begin{array}{cc}\begin{array}{c}1/\left(b-a\right),\\ -1/\left(b-a\right),\\ 0,\end{array}& \begin{array}{l}-b\le t<-a\hfill \\ a\le t<b\hfill \\ \text{otherwise}\hfill \end{array}\end{array}$$

The signal w (t) is shown in Fig. 4.71.

Finding the Fourier transform of w (t) is much easier compared to a direct evaluation of the Fourier transform integral for x (t). For notational convenience, let two new parameters be defined as

$$\begin{array}{lllll}\tau =b-a\hfill & \text{and}\hfill & \lambda =\frac{a+b}{2}\hfill & \hfill & \left(4.258\right)\hfill \end{array}$$

The parameter τ is the width of each of the two pulses that make up w (t), and these pulses are centered at ±λ. The derivative w (t) can be written using the unit-pulse function as

$$w\left(t\right)=\frac{1}{\tau }\prod \left(\frac{\left(t+\lambda \right)}{\tau }\right)-\frac{1}{\tau }\prod \left(\frac{\left(t-\lambda \right)}{\tau }\right)$$

and the corresponding transform W (f) is

$$\begin{array}{ccc}\begin{array}{ll}W\left(f\right)\hfill & =\text{sinc }\left(f\tau \right)e^{j2\pi f\lambda }-\text{sinc }\left(f\tau \right)e^{-j2\pi f\lambda }\hfill \\ \hfill & =\text{sinc }\left(f\tau \right)\left[e^{j2\pi f\lambda }-e^{-j2\pi f\lambda }\right]\hfill \\ \hfill & =2j\text{ sinc }\left(f\tau \right)\text{ sin }\left(2\pi f\lambda \right)\hfill \end{array}& & \left(4.259\right)\end{array}$$

The transform of x (t) can now be found by scaling this result:

$$X\left(f\right)=\frac{W\left(f\right)}{j2\pi f}=2\lambda \text{ sinc }\left(f\tau \right)\text{ sinc }\left(2f\lambda \right)$$

The transform X (f) is graphed in Fig. 4.72 for various values of the parameters τ and λ.

Having found the solution for X (f) we will speculate on the differentiation in time property a bit further. What if we had differentiated w (t) one more time to obtain a new signal v (t)? Let

$$v\left(t\right)=\frac{dw\left(t\right)}{dt}=\frac{d^{2}x\left(t\right)}{d{t}^2}$$

The signal v (t) consists of some shifted and scaled impulses, and is shown in Fig. 4.73.

The transform of v (t) would be related to the transform of x (t) by

$$\begin{array}{ccc}X\left(f\right)=\frac{V\left(t\right)}{{\left(j2\pi f\right)}^2}=\frac{V\left(f\right)}{4{\pi }^2{f}^2}& & \left(4.260\right)\end{array}$$

Analytically, v (t) can be expressed as

$$v\left(t\right)=\frac{1}{b-a}\delta \left(t+b\right)-\frac{1}{b-a}\delta \left(t+a\right)-\frac{1}{b-a}\delta \left(t-a\right)+\frac{1}{b-a}\delta \left(t-b\right)$$

Knowing that ℱ{δ(t)} = 1, and utilizing the time-shifting property of the Fourier transform, V (f) is found to be

$$\begin{array}{ccc}\begin{array}{ll}V\left(f\right)\hfill & =\frac{1}{b-a}{e}^{j2\pi fb}-\frac{1}{b-a}{e}^{j2\pi fa}-\frac{1}{b-a}{e}^{-j2\pi fa}+\frac{1}{b-a}{e}^{-j2\pi fb}\hfill \\ \hfill & =\frac{2}{b-a}\left[cos\left(2\pi fb\right)-cos\left(2\pi fa\right)\right]\hfill \end{array}& & \left(4.261\right)\end{array}$$

Based on Eqn. (4.258), we have

$$\begin{array}{ccc}a=\lambda -\frac{\tau }{2}& \text{and}& b=\lambda +\frac{\tau }{2}\end{array}$$

Substituting a and b into Eqn. (4.261) we obtain

$$V\left(f\right)=\frac{2}{\tau }\begin{array}{ccc}\left[cos\left(2\pi f\left(\lambda +\frac{\tau }{2}\right)\right)-cos\left(2\pi f\left(\lambda -\frac{\tau }{2}\right)\right)\right]& & \left(4.262\right)\end{array}$$

Using the appropriate trigonometric identity,6 Eqn. (4.262) becomes

$$V\left(f\right)=-\frac{4}{\tau }sin\left(\pi f\tau \right)sin\left(2\pi f\lambda \right)$$

and the transform X (f) is obtained by using Eqn. (4.260) as

$$\begin{array}{ll}X\left(f\right)\hfill & =\frac{1}{{\pi }^2{f}^2\tau }sin\left(\pi f\tau \right)sin\left(2\pi f\lambda \right)\hfill \\ \hfill & =2\lambda \text{ sinc }\left(f\tau \right)\text{ sinc }\left(2f\lambda \right)\hfill \end{array}$$

which matches the result found earlier.

Software resources:

ex_4_32.m

Differentiation in the frequency domain

Example 4.33: Pulse with trapezoidal spectrum

Find the signal x (t) the Fourier transform of which is the trapezoidal function given by

$$X\left(f\right)=\{\begin{array}{cc}1,& \left|f\right|\le f_0\left(1-r\right)\\ \frac{1}{2r}-\left[-\frac{\left|f\right|}{f_0}+1+r\right],& f_0\left(1-r\right)<\left|f\right|\le f_0\left(1+r\right)\\ 0,& \left|f\right|>f_0\left(1+r\right)\end{array}$$

and shown in Fig. 4.74. Real-valued parameter r is adjustable in the range 0 < r < 1.

Solution: The answer can easily be found by applying the duality property to the result obtained in Eqn. (4.259) of Example 4.32 with the adjustments a = f0 (1 − r), b = f0 (1 + r), λ = f0 and τ =2f0r. (See Problem 4.32 at the end of this chapter.) Instead of using that approach, however, we will use this example as an opportunity to apply the differentiation in frequency property. Let

$$W\left(f\right)=\frac{dX\left(f\right)}{df}$$

By carrying out the differentiation we obtain W (f) as

$$W\left(f\right)=\frac{1}{2f_{0}r}\prod \left(\frac{f+f_0}{2f_{0}r}\right)-\frac{1}{2f_{0}r}\prod \left(\frac{f-f_0}{2f_{0}r}\right)$$

which is shown in Fig. 4.75.

The Fourier transform of the sinc function is (see Example 4.19)

$$ℱ\left\{\text{sinc}\left(t\right)\right\}=\prod \left(f\right)$$

Using the time and frequency scaling property of the Fourier transform given by Eqn. (4.239) we obtain

$$\begin{array}{ccc}ℱ\left\{\text{sinc}\left(2f_{0}rt\right)\right\}=\frac{1}{2f_{0}r}\prod \left(\frac{f}{2f_{0}r}\right)& & \left(4.275\right)\end{array}$$

The two frequency-domain pulses that make up the transform W (f) are simply frequency shifted and frequency scaled versions of the right side of Eqn. (4.275). Therefore, the inverse

transform of W (f) is

$$\begin{array}{ll}w\left(t\right)\hfill & =\text{sinc}\left(2f_{0}rt\right)e^{-j2\pi f_{0}t}-\text{sinc}\left(2f_{0}rt\right)e^{j2\pi f_{0}t}\hfill \\ \hfill & =\text{sinc}\left(2f_{0}rt\right)\left[e^{-j2\pi f_{0}t}-e^{j2\pi f_{0}t}\right]\hfill \\ \hfill & =-2j\text{ sin }\left(2\pi f_{0}t\right)\text{ sinc}\left(2f_{0}rt\right)\hfill \end{array}$$

Since w (t) = −j2πt x (t), we can write x (t) as

$$\begin{array}{ccc}x\left(t\right)=2f_0\text{ sinc}\left(2f_{0}t\right)\text{ sinc }\left(2f_{0}rt\right)& & \left(4.276\right)\end{array}$$

Fig. 4.76 shows the signal x (t) for various values of the parameter r.

The signal x (t) with a trapezoidal spectrum exhibits some interesting properties:

1. Zero crossings of the signal x (t) occur at uniform time intervals of 1/(2f0) regardless of the value of r.
2. Larger values of r result in a reduction in the oscillatory behavior of the signal.

These properties are very significant in communication systems in the context of pulse shaping. A signal with the properties listed above can be used for representing binary or M-ary symbols while eliminating the interference of those symbols with each other, a phenomenon known as intersymbol interference (ISI). Harry Nyquist has shown that any signal the spectrum of which has certain symmetry properties will exhibit uniformly spaced zero crossings. The trapezoidal spectrum shown in Fig. 4.74 is one example.

Software resources:

ex_4_33.m

Convolution property

Multiplication of two signals

Example 4.34: Transform of a truncated sinusoidal signal

A sinusoidal signal that is time-limited in the interval −τ< t < τ is given by

$$x\left(t\right)=\{\begin{array}{cc}cos\left(2\pi f_{0}t\right),& -\tau <t<\tau \\ 0,& \text{otherwise}\end{array}$$

Determine the Fourier transform of this signal using the multiplication property.

Solution: The signal x (t) is the same as the modulated pulse signal used in Example 4.29 and shown in Fig. 4.64. In Example 4.29 the transform was determined using the modulation property. Let x1 (t) and x2 (t) be defined as

$$\begin{array}{ccc}{x}_1\left(t\right)=cos\left(2\pi f_{0}t\right)& \text{and}& x_2\left(t\right)=\prod \left(\frac{t}{2\tau }\right)\end{array}$$

The signal x (t) can be written as the product of these two signals:

$$x\left(t\right)=x_1\left(t\right)x_2\left(t\right)=cos\left(2\pi f_{0}t\right)\prod \left(\frac{t}{2\tau }\right)$$

It will be more convenient to use f rather than ω in this case. Fourier transforms of the signals x1 (t) and x2 (t) are

$$X_1\left(f\right)=\frac{1}{2}\delta \left(f+f_0\right)+\frac{1}{2}\delta \left(f-f_0\right)$$

and

$$X_2\left(f\right)=2\tau \text{ sinc }\left(2\tau f\right)$$

respectively. Using Eqn. (4.285), the Fourier transform of the product x (t) = x1 (t) x2 (t) is

$$\begin{array}{ll}X\left(f\right)\hfill & =X_1\left(f\right)*X_2\left(f\right)\hfill \\ \hfill & =\tau \text{ sinc}\left(2\tau \left(f+f_0\right)\right)+\tau \text{ sinc}\left(2\tau \left(f-f_0\right)\right)\hfill \end{array}$$

which matches the result found in Example 4.29.

Software resources:

ex_4_34.m

Integration

Table 4.4 contains a summary of key properties of the Fourier transform. Table 4.5 lists some of the fundamental Fourier transform pairs.

4.3.6 Applying Fourier transform to periodic signals

In developing frequency-domain analysis methods in the previous sections of this chapter we have distinguished between periodic and non-periodic continuous-time signals. Periodic signals were analyzed using various forms of the Fourier series such as TFS, EFS or CFS. In contrast, Fourier transform was used for analyzing non-periodic signals. While this distinction will be appropriate when we work with one type of signal or the other, there may be times when we need to mix periodic and non-periodic signals within the same system. An example of this occurs in amplitude modulation where a non-periodic message signal may be multiplied with a periodic carrier signal. Another example is the use of a periodic signal as input to a system the impulse response of which is non-periodic. In such situations it would be convenient to use the Fourier transform for periodic signals as well. In general a periodic signal is a power signal that does not satisfy the existence conditions for the Fourier transform. It is neither absolute integrable nor square integrable. On the other hand, we have seen in Example 4.20 that a Fourier transform can be found for a constant-amplitude signal that does not satisfy the existence conditions, as long as we are willing to accept singularity functions in the transform. The next two examples will expand on this idea. Afterwards we will develop a technique for converting the EFS representation of a periodic continuous-time signal to a Fourier transform that is suitable for use in solving problems.

Example 4.35: Fourier transform of complex exponential signal

Determine the transform of the complex exponential signal x (t) = e jωot.

Solution: The transform of the constant unit-amplitude signal was found in Example 4.20 to be

$$ℱ\left\{1\right\}=2\pi \delta \left(\omega \right)$$

Using this result along with the frequency shifting property of the Fourier transform given by Eqn. (4.222), the transform of the complex exponential signal is

$$\begin{array}{cc}ℱ\left\{e^{j\omega ot}\right\}=2\pi \delta \left(\omega -{\omega }_0\right)& \left(4.299\right)\end{array}$$

This is illustrated in Fig. 4.77.

Example 4.36: Fourier transform of sinusoidal signal

Determine the transform of the sinusoidal signal x(t) = cos (ω0t).

Solution: The transform of the constant unit-amplitude signal was found in Example 4.20 to be

$$ℱ\left\{1\right\}=2\pi \delta \left(\omega \right)$$

Using this result along with the modulation property of the Fourier transform given by Eqn. (4.230), the transform of the sinusoidal signal is

$$\begin{array}{cc}ℱ\left\{\cos \left(\omega ot\right)\right\}=\pi \delta \left(\omega -{\omega }_0\right)+\pi \delta \left(\omega +{\omega }_0\right)& \left(4.300\right)\end{array}$$

This transform is shown in Fig. 4.78.

In Examples 4.35 and 4.36 we were able to obtain the Fourier transforms of two periodic signals, namely a complex exponential signal and a sinusoidal signal. Both transforms were possible through the use of impulse functions in the transform. The idea can be generalized to apply to any periodic continuous-time signal that has an EFS representation.

Using Eqn. (4.304) in Eqn. (4.303), the transform of the periodic signal $\tilde{x}\left(t\right)$ with EFS coefficients {ck} is obtained as

$$\begin{array}{cc}X\left(\omega \right)={\displaystyle \sum _{k=-\infty }^{\infty }2\pi c_k\delta \left(\omega -k{\omega }_0\right)}& \left(4.305\right)\end{array}$$

The Fourier transform for a periodic continuous-time signal is obtained by converting each EFS coefficient ck to an impulse with area equal to 2πck and placing it at the radian frequency ω = kω0. This process is illustrated in Fig. 4.79.

Example 4.37: Fourier transform of periodic pulse train

Determine the Fourier transform of the periodic pulse train with duty cycle d = τ/T0 as shown in Fig. 4.80.

Solution: The EFS coefficients for $\tilde{x}\left(t\right)$ were determined in Example 4.7 as

$$c_k=d\sin c\left(kd\right)$$

Using Eqn. 4.305 the Fourier transform is

$$\begin{array}{cc}X\left(\omega \right)={\displaystyle \sum _{k=-\infty }^{\infty }2\pi d\sin \text{c}\left(kd\right)\delta \left(\omega -k{\omega }_0\right)}& \left(4.306\right)\end{array}$$

where ω0 = 1/T0 is the fundamental radian frequency.

4.4 Energy and Power in the Frequency Domain

In this section we will discuss a very important theorem of Fourier series and transform known as Parseval’s theorem which can be used as the basis of computing energy or power of a signal from its frequency domain representation. Afterwards energy and power spectral density concepts will be introduced.

4.4.1 Parseval’s theorem

The left side of Eqn. (4.307) represents the normalized average power in a periodic signal as we have derived in Eqn. (1.92). The left side of Eqn. (4.308) represents the normalized signal energy as derived in Eqn. (1.81). The relationships given by Eqns. (4.307) and (4.308) relate signal energy or signal power to the frequency-domain representation of the signal. They are two forms of Parseval’s theorem.

4.4.2 Energy and power spectral density

Examining the two statements of Parseval’s theorem expressed by Eqns. (4.307) and (4.308), we reach the following conclusions:

1. In Eqn. (4.307) the left side corresponds to the normalized average power of the signal $\tilde{x}\left(t\right)$, and therefore the summation on the right side must also represent power. The term |ck|2 corresponds to the power of the frequency component at f = kf0, thatis, the power in the k-th harmonic only. Suppose we construct a new function Sx (f) by starting with the line spectrum ck and replacing each spectral line with an impulse function scaled by |ck|2 and time shifted to the frequency kf0, i.e.,

   $$\begin{array}{cc}{S}_x\left(f\right)={\displaystyle \sum _{k=-\infty }^{\infty }{\left|c_k\right|}^2\delta \left(f-k{f}_0\right)}& \left(4.315\right)\end{array}$$

   In the next step we will integrate Sx (f) to obtain

   $$\begin{array}{lll}{\displaystyle {\int }_{-\infty }^{\infty }{S}_x\left(f\right)df}\hfill & ={\displaystyle {\int }_{-\infty }^{\infty }\left[{\displaystyle \sum _{k=-\infty }^{\infty }{\left|c_k\right|}^2\delta \left(f-k{f}_0\right)}\right]df}\hfill & \hfill \\ \hfill & ={\displaystyle \sum _{k=-\infty }^{\infty }\left[{\displaystyle {\int }_{-\infty }^{\infty }{\left|c_k\right|}^2\delta \left(f-k{f}_0\right)df}\right]}\hfill & \left(4.316\right)\hfill \end{array}$$

   Using the sifting property of the unit-impulse function Eqn. (4.316) becomes

   $${\displaystyle {\int }_{-\infty }^{\infty }{S}_x\left(f\right)df=}\begin{array}{cc}{\displaystyle \sum _{k=-\infty }^{\infty }{\left|c_k\right|}^2}& \left(4.317\right)\end{array}$$

   Finally, substituting Eqn. (4.317) into Eqn. (4.307) we have

   $$\begin{array}{cc}\frac{1}{T_0}{\displaystyle {\int }_{t_0}^{t_0+T_0}{\left|\tilde{x}\left(t\right)\right|}^{2}dt=}{\displaystyle {\int }_{-\infty }^{\infty }{S}_x\left(f\right)df}& \left(4.318\right)\end{array}$$

   In Eqn. (4.318) the normalized average power of the signal x (t) is computed by integrating the function Sx (f) over all frequencies. Consequently, the function Sx (f) is the power spectral density of the signal x (t).

   If we use the radian frequency variable ω instead of f, Eqn. (4.318) becomes

   $$\begin{array}{cc}\frac{1}{T_0}{\displaystyle {\int }_{t_0}^{t_0+T_0}{\left|\tilde{x}\left(t\right)\right|}^{2}dt=}\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{\infty }{S}_x\left(\omega \right)d\omega }& \left(4.319\right)\end{array}$$

   As an interesting by-product of Eqn. (4.318), the power of x (t) thatis withina specific frequency range can be determined by integrating Sx (f) over that frequency range. For example, the power contained at frequencies in the range (−f0, f0) can be computed as

   $$\begin{array}{cc}{P}_x\text{ in}\left(-f_0,f_0\right)={\displaystyle {\int }_{-f_0}^{f_0}{S}_x\left(f\right)df}& \left(4.320\right)\end{array}$$

2. In Eqn. (4.308) the left side is the normalized signal energy for the signal x (t), and the right side must be the same. The integrand |X (f)|2 is therefore the energy spectral density of the signal x (t). Let the function Gx (f) be defined as

   $$\begin{array}{cc}{G}_x\left(f\right)={\left|X\left(f\right)\right|}^2& \left(4.321\right)\end{array}$$

   Substituting Eqn. (4.321) into Eqn. (4.308), the normalized energy in the signal x (t) can be expressed as

   $${\displaystyle {\int }_{-\infty }^{\infty }{\left|x\left(t\right)\right|}^{2}dt={\displaystyle {\int }_{-\infty }^{\infty }{G}_x\left(f\right)df}}$$

   If ω is used in place of f, Eqn. (4.322) becomes

   $${\displaystyle {\int }_{-\infty }^{\infty }{\left|x\left(t\right)\right|}^{2}dt=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{\infty }{G}_x\left(\omega \right)d\omega }}$$

3. In Eqn. (4.308) we have expressed Parseval’s theorem for an energy signal, and used it to lead to the derivation of the energy spectral density in Eqn. (4.321). We know that some non-periodic signals are power signals, therefore their energy cannot be computed. The example of one such signal is the unit-step function. The power of a non-periodic signal was defined in Eqn. (1.93) which will be repeated here:

   $$\begin{array}{cc}\text{Eqn}\text{.}\left(1.93\right):& P_x=\underset{T\to \infty }{\lim }\left[\frac{1}{T}{\displaystyle {\int }_{-T/2}^{T/2}{\left|x\left(t\right)\right|}^{2}dt}\right]\end{array}$$

   In order to write the counterpart of Eqn. (4.308) for a power signal, we will first define a truncated version of x (t) as

   $$\begin{array}{cc}{x}_T\left(t\right)=\{\begin{array}{ll}x\left(t\right),\hfill & -T/2<t<T/2\hfill \\ 0,\hfill & \text{otherwise}\hfill \end{array}& \left(4.322\right)\end{array}$$

   Let XT(f) be the Fourier transform of the truncated signal XT(t):

   $$\begin{array}{cc}{x}_T\left(f\right)=ℱ\left\{x_T\left(t\right)\right\}={\displaystyle {\int }_{-T/2}^{T/2}{x}_T\left(t\right)e^{-j2\pi ft}dt}& \left(4.323\right)\end{array}$$

   Now Eqn. (4.308) can be written in terms of the truncated signal and its transform as

   $$\begin{array}{cc}{\displaystyle {\int }_{-T/2}^{T/2}{\left|x_T\left(t\right)\right|}^{2}dt={\displaystyle {\int }_{-\infty }^{\infty }{\left|x_T\left(f\right)\right|}^{2}df}}& \left(4.324\right)\end{array}$$

   Scaling both sides of Eqn. (4.324) by (1/T) and taking the limit as T becomes infinitely large, we obtain

   $$\underset{T\to \infty }{\lim }\left[\frac{1}{T}{\displaystyle {\int }_{-T/2}^{T/2}{\left|x_T\left(t\right)\right|}^{2}dt}\right]\begin{array}{cc}=\underset{T\to \infty }{\lim }\left[\frac{1}{T}{\displaystyle {\int }_{-\infty }^{\infty }{\left|x_T\left(f\right)\right|}^{2}df}\right]& \left(4.325\right)\end{array}$$

   The left side of Eqn. (4.325) is the average normalized power in a non-periodic signal as we have established in Eqn. (1.93). Therefore, the power spectral density of x(t) is

   $$S_x\left(f\right)=\underset{T\to \infty }{\lim }\begin{array}{cc}\left[\frac{1}{T}{\left|X_T\left(f\right)\right|}^2\right]& \left(4.326\right)\end{array}$$

Example 4.38: Power spectral density of a periodic pulse train

The EFS coefficients for the periodic pulse train $\tilde{x}\left(t\right)$ shown in Fig. 4.81 were found in Example 4.5 to be

$$c_k=\frac{1}{3}\sin \text{c}\left(k/3\right)$$

Determine the power spectral density for $\tilde{x}\left(t\right)$. Also find the total power, the dc power, the power in the first three harmonics, and the power above 1 Hz.

Solution: The period of the signal is T0 = 3 s, and therefore the fundamental frequency is $f_0=\frac{1}{3}\text{ Hz}$. The power spectral density Sx (f) can be written using Eqn. (4.315) as

$$S_x\left(f\right)={\displaystyle \sum _{k=-\infty }^{\infty }{\left|\frac{1}{3}\sin \text{c}\left(k/3\right)\right|}^2\delta \left(f-k/3\right)}$$

which is graphed in Fig. 4.82.

The total power in the signal x (t) can be easily determined using the left side of Eqn. (4.307), i.e.,

$$P_x=\frac{1}{T_0}{\displaystyle {\int }_{t_0}^{t_0+T_0}{\left|\tilde{x}\left(t\right)\right|}^{2}dt=\frac{1}{3}{\displaystyle {\int }_{-0.5}^{0.5}{\left(1\right)}^{2}dt=0.3333}}$$

The de power in the signal x (t) is

$$P_{dc}={\left|c_0\right|}^2={\left(0.3333\right)}^2=0.1111$$

The power in the fundamental frequeney is

$$P_1={\left|c-1\right|}^2+{\left|c_1\right|}^2=2{\left|c_1\right|}^2=2\left(0.0760\right)=0.1520$$

Power values for the seeond and third harmonies ean be found similarly as

$$P_2={\left|c-2\right|}^2+{\left|c_2\right|}^2=2{\left|c_2\right|}^2=2\left(0.0190\right)=0.0380$$

and

$$P_3={\left|c-3\right|}^2+{\left|c_3\right|}^2=2{\left|c_3\right|}^2=2{\left(0\right)}^2=0$$

The third harmonie is at frequeney f = 1 Hz. Thus, the power above 1 Hz ean be determined by subtraeting the power values for the de and the first three harmonies from the total, i.e.,

$$\begin{array}{ll}{P}_{hf}\hfill & =P_x-P_{dc}-P_1-P_2-p_3\hfill \\ \hfill & =0.3333-0.1111-0.1520-0.0380-0\hfill \\ \hfill & =0.0322\hfill \end{array}$$

Example 4.39: Power spectral density of a sinusoidal signal

Find the power speetral density of the signal $\tilde{x}\left(t\right)=5\cos \left(200\pi t\right)$.

Solution: Using Euler’s formula, the signal in question ean be written as

$$x\left(t\right)=\frac{5}{2}{e}^{-j200\pi t}+\frac{5}{2}{e}^{j200\pi t}$$

from an inspeetion of whieh we eonelude that the only signifieant eoeffieients in the EFS representation of the signal are

$$c_{-1}=c_1=\frac{5}{2}$$

with all other eoeffieients equal to zero. The fundamental frequeney is f0 = 100 Hz. Using Eqn. (4.315), the power speetral density is

$$\begin{array}{lll}{S}_x\left(f\right)\hfill & =\hfill & {\displaystyle \sum _{n=-\infty }^{\infty }{\left|c_n\right|}^2\delta \left(f-100n\right)}\hfill \\ \hfill & =\hfill & {\left|c-1\right|}^2\delta \left(f+100\right)+{\left|c_1\right|}^2\delta \left(f-100\right)\hfill \\ \hfill & =\hfill & \frac{25}{4}\delta \left(f+100\right)+\frac{25}{4}\delta \left(f-100\right)\hfill \end{array}$$

whieh is shown in Fig. 4.83

Example 4.40: Energy spectral density of a rectangular pulse

Determine the energy spectral density of the rectangular pulse

$$x\left(t\right)=20\prod \left(100t\right)$$

Solution: It was determined in Example 4.12 that

$$A\prod \left(t/\tau \right)\underleftrightarrow{ℱ}A\tau \text{ sinc}\left(f\tau \right)$$

Thus the Fourier transform of the pulse given above is

$$X\left(f\right)=0.2\text{ sinc}\left(0.01f\right)$$

and the energy spectral density is

$$G_x\left(f\right)={\left|X\left(f\right)\right|}^2=0.04{\text{ sinc}}^2\left(0.01f\right)$$

This result is shown in Fig. 4.84.

Example 4.41: Energy spectral density of the sinc function

Determine the energy spectral density of the signal

$$x\left(t\right)=\text{sinc}\left(10t\right)$$

Afterwards, compute (a) the total energy in the sinc pulse, and (b) the energy in the sinc pulse at frequencies up to 3 Hz.

Solution: The spectrum of a sinc pulse is a rectangle. Through the use of the duality property of the Fourier transform, X (f) is found to be

$$X\left(f\right)=\frac{1}{10}\prod \left(\frac{f}{10}\right)$$

The energy spectral density is

$$G_x\left(f\right)=\frac{1}{100}\prod \left(\frac{f}{10}\right)$$

which is shown in Fig. 4.85. The total energy in the signal x (t) is

$$E_x={\displaystyle {\int }_{-\infty }^{\infty }{G}_x\left(f\right)df={\displaystyle {\int }_{-5}^5\left(\frac{1}{100}\right)df=0.1}}$$

The energy at frequencies up to 3 Hz is found by integrating Gx (f) in the interval −3 Hz < f < 3 Hz:

$$E_x\text{ in}\left(-3.3\text{ Hz}\right)={\displaystyle {\int }_{-3}^3{G}_x\left(f\right)df={\displaystyle {\int }_{-3}^3\left(\frac{1}{100}\right)df=0.06}}$$

Fig. 4.86 illustrates the use of Gx (f) for computing the signal energy.

4.4.3 Autocorrelation

The energy spectral density Gx (f) or the power spectral Sx (f) density for a signal can be computed through direct application of the corresponding equations derived in Section 4.4.2; namely Eqn. (4.321) for an energy signal, and either Eqn. (4.315) or Eqn. (4.326) for a power signal, depending on its type. In some circumstances it is also possible to compute both from the knowledge of the autocorrelation function which will be defined in this section. Let x (t) be a real-valued signal.

Even though we refer to rxx(τ) as a function, we will often treat it as if it is a continuous-time signal, albeit one that uses a different independent variable, τ, than the signal x (t) for which it is computed. The variable τ simply corresponds to the time shift between the two copies of x (t) used in the definitions of Eqns. (4.327) through (4.329).

The definition of the autocorrelation function for a periodic signal $\tilde{x}\left(t\right)$can be used in a similar manner in determining the power spectral density of the signal.

A relationship similar to the one expressed by Eqn. (4.341) applies to random processes that are wide-sense stationary, and is known as the Wiener-Khinchin theorem. It is one of the fundamental theorems of random signal processing.

Example 4.42: Power spectral density of a sinusoidal signal revisited

Find the autocorrelation function for the signal

$$\tilde{x}\left(t\right)=5\cos \left(200\pi t\right)$$

Afterwards determine the power spectral density from the autocorrelation function.

Solution: Using Eqn. (4.328) with T0 = 0.01 s we have

$${\tilde{r}}_{xx}\left(\tau \right)=\frac{1}{0.01}{\displaystyle {\int }_{-0.005}^{0.005}25\text{ cos}\left(200\pi t\right)}\text{ cos}\left(200\pi \left[t+\tau \right]\right)dt$$

Using the appropriate trigonometric identity 7 the integral is evaluated as

$${\tilde{r}}_{xx}\left(\tau \right)=\frac{25}{2}\cos \left(200\pi \tau \right)$$

Power spectral density is the Fourier transform of the autocorrelation function:

$$S_x\left(f\right)=ℱ\left\{{\tilde{r}}_{xx}\left(\tau \right)\right\}=\frac{25}{4}\delta \left(f+100\right)+\frac{25}{4}\delta \left(f-100\right)$$

which is in agreement with the earlier result found in Example 4.39.

Properties of the autocorrelation function

The autocorrelation function as defined by Eqns. (4.327), (4.328) and (4.329) has a number of important properties that will be listed here:

1. rxx (0) ≥ |rxx(τ)| for all τ.

   To see why this is the case, we will consider the non-negative function [x (t) ∓ x (t + τ)]2. The time average of this function must also be non-negative, so we can write

   $$\langle {\left[x\left(t\right)\mp x\left(t+\tau \right)\right]}^2\rangle \ge 0$$

   or equivalently

   $$\langle x^2\left(t\right)\rangle \mp \langle 2x\left(t\right)x\left(t+\tau \right)\rangle +\langle x^2\left(t+\tau \right)\rangle \ge 0$$

   which implies that

   $$r_{xx}\left(0\right)\mp 2r_{xx}\left(\tau \right)+r_{xx}\left(0\right)\ge 0$$

   which is the same as property 1.

2. rxx (-τ) = rxx (τ) for all τ, that is, the autocorrelation function has even symmetry. Recall that the autocorrelation function is the inverse Fourier transform of the power spectral density, i.e., $r_{xx}\left(\tau \right)={ℱ}^{-1}\left\{S_x\left(f\right)\right\}$. We know from the symmetry properties of the Fourier transform that, since Sx (f) is purely real, rxx (τ) mustbean even function of τ.

3. If the signal x (t) is periodic with period T, then its autocorrelation function ${\tilde{r}}_{xx}\left(\tau \right)$ is also periodic with the same period, that is, if $x\left(t\right)=x\left(t+kT\right)$ for all integers k, then ${\tilde{r}}_{xx}\left(\tau \right)={\tilde{r}}_{xx}\left(\tau +kT\right)$ for all integers k.

   This property easily follows from the time-average based definition of the autocorrelation function given by Eqn. (4.328). The time average of a periodic signal is also periodic with the same period.

4.5 System Function Concept

In time-domain analysis of systems in Chapter 2 we have relied on two distinct description forms for CTLTI systems:

1. A linear constant-coefficient differential equation that describes the relationship between the input and the output signals
2. An impulse response which can be used with the convolution operation for determining the response of the system to an arbitrary input signal

In this section, the concept of system function will be introduced as the third method for describing the characteristics of a system.

Recall that the impulse response is only meaningful for a system that is both linear and time-invariant, since the convolution operator could not be used otherwise. Consequently, the system function concept is valid for linear and time-invariant systems only.

In general, H (ω) is a complex function of ω, and can be written in polar form as

$$\begin{array}{cc}H\left(\omega \right)=\left|H\left(\omega \right)\right|e^{j\Theta \left(\omega \right)}& \left(4.350\right)\end{array}$$

Example 4.43: System function for the simple RC circuit

The impulse response of the RC circuit shown in Fig. 4.87 was found in Example 2.18 to be

$$h\left(t\right)=\frac{1}{RC}{e}^{-t/RC}u\left(t\right)$$

Determine the system function.

Solution: Taking the Fourier transform of the impulse response we obtain

$$H\left(\omega \right)={\displaystyle {\int }_0^{\infty }\frac{1}{RC}{e}^{-t/RC}{e}^{-j\omega t}dt=\frac{1}{1+j\omega RC}}$$

To simplify notation we will define ωc = 1/RC and write the system function as

$$\begin{array}{cc}H\left(\omega \right)=\frac{1}{1+j\left(\omega /{\omega }_c\right)}& \left(4.351\right)\end{array}$$

The magnitude and the phase of the system function are

$$\left|H\left(\omega \right)\right|=\frac{1}{\sqrt{1+{\left(\omega /{\omega }_c\right)}^2}}$$

and

$$\Theta \left(\omega \right)=\measuredangle H\left(\omega \right)=-{\tan }^{-1}\left(\omega /{\omega }_c\right)$$

respectively. The results found in Eqns. (4.352) and (4.352) are shown in Fig. 4.88.

The value of the system function at the frequency ωC is H (ωc) = 1/(1 + j), and the corresponding magnitude is $\left|H\left({\omega }_c\right)\right|=1/\sqrt{2}$. Thus, ωC represents the frequency at which the magnitude of the system function is 3 decibels below its peak value at ω = 0, that is,

$$20{\text{ log}}_{10}\left[\frac{\left|H\left({\omega }_c\right)\right|}{\left|H\left(0\right)\right|}\right]=20{\log }_{10}\left[\frac{1}{\sqrt{2}}\right]\approx -3\text{ dB}$$

The frequency ωc is often referred to as the 3-dB cutoff frequency of the system.

Software resources:

ex_4_43.m

Obtaining the system function from the differential equation

Sometimes we need to find the system function for a CTLTI system described by means of a differential equation. Two properties of the Fourier transform will be useful for this purpose; the convolution property and the time differentiation property.

1. Since the output signal is computed as the convolution of the impulse response and the input signal, that is, $y\left(t\right)=h\left(t\right)*x\left(t\right)$, the corresponding relationship in the frequency domain is $Y\left(\omega \right)=H\left(\omega \right)X\left(\omega \right)$. Consequently, the system function is equal to the ratio of the output transform to the input transform:

   $$\begin{array}{cc}H\left(\omega \right)=\frac{Y\left(\omega \right)}{X\left(\omega \right)}& \left(4.352\right)\end{array}$$

2. Using the time differentiation property the transforms of the individual terms in the differential equation are

   $$\begin{array}{ccc}\frac{d^{k}y\left(t\right)}{d{t}^k}\underleftrightarrow{ℱ}{\left(j\omega \right)}^{k}Y\left(\omega \right)& k=0,1,\cdots & \left(4.353\right)\end{array}$$

   $$\begin{array}{ccc}\frac{d^{k}x\left(t\right)}{d{t}^k}\underleftrightarrow{ℱ}{\left(j\omega \right)}^{k}X\left(\omega \right)& k=0,1,\cdots & \left(4.354\right)\end{array}$$

   The system function is obtained from the differential equation by first transforming both sides of the differential equation through the use of Eqns. (4.353) and (4.354), and then using Eqn. (4.352).

Example 4.44: Finding the system function from the differential equation

Determine the system function for a CTLTI system described by the differential equation

$$\frac{d^{2}y\left(t\right)}{d{t}^2}+2\frac{dy\left(t\right)}{dt}+26y\left(t\right)=x\left(t\right)$$

Solution: Taking the Fourier transform of both sides of the differential equation we obtain

$${\left(j\omega \right)}^{2}Y\left(\omega \right)+2\left(j\omega \right)Y\left(\omega \right)+26Y\left(\omega \right)=X\left(\omega \right)$$

or

$$\left[\left(26-{\omega }^2\right)+j2\omega \right]Y\left(\omega \right)=X\left(\omega \right)$$

The system function is

$$H\left(\omega \right)=\frac{Y\left(\omega \right)}{X\left(\omega \right)}=\frac{1}{\left(26-{\omega }^2\right)+j2\omega }$$

4.6 CTLTI Systems with Periodic Input Signals

In earlier sections of this chapter we have developed methods for representing periodic signals as linear combinations of sinusoidal or complex exponential basis functions. Using the TFS representation we have

$$\begin{array}{cc}\text{Eqn}\text{.}\left(4.26\right):& \tilde{x}\left(t\right)\end{array}=a_0+{\displaystyle \sum _{k=1}^{\infty }{a}_k\cos \left(k{\omega }_{0}t\right)+}{\displaystyle \sum _{k=1}^{\infty }{b}_k\sin \left(k{\omega }_{0}t\right)}$$

Alternatively, using the EFS representation

$$\begin{array}{cc}\text{Eqn}\text{.}\left(4.48\right):& \tilde{x}\left(t\right)\end{array}={\displaystyle \sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{\omega }_{0}t}}$$

If a periodic signal is used as input to a CTLTI system, the superposition property can be utilized for finding the output signal. The response of the CTLTI system to the periodic signal can be determined as a linear combination of its responses to the individual basis functions in the TFS or EFS representation. We will analyze this concept in two steps; first for a complex exponential input signal, and then for a sinusoidal signal.

4.6.1 Response of a CTLTI system to complex exponential signal

Consider a CTLTI system with impulse response h (t), driven by an input signal in the form of a complex exponential function of time

$$\begin{array}{cc}\tilde{x}\left(t\right)=e^{j{\omega }_{0}t}& \left(4.355\right)\end{array}$$

The response y (t) of the system can be found through the use of the convolution relationship as we have derived in Section 2.7.2 of Chapter 2.

$$\begin{array}{lll}y\left(t\right)\hfill & =h\left(t\right)*\tilde{x}\left(t\right)\hfill & \hfill \\ \hfill & ={\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)\tilde{x}\left(t-\lambda \right)d\lambda }\hfill & \left(4.356\right)\hfill \end{array}$$

For a CTLTI system driven by a complex exponential input signal, we have the following important relationship:

1. The response of a CTLTI system to a complex exponential input signal is a complex exponential output signal with the same frequency ω0.
2. The effect of the system on the complex exponential input signal is to
   1. (a) Scale its amplitude by an amount equal to the magnitude of the system function at ω = ω0
   2. (b) Shift its phase by an amount equal to the phase of the system function at ω = ω0

4.6.2 Response of a CTLTI system to sinusoidal signal

Let the input signal to the CTLTI system under consideration be a sinusoidal signal in the form

$$\begin{array}{cc}\tilde{x}\left(t\right)=\cos \left({\omega }_{0}t\right)& \left(4.361\right)\end{array}$$

The response of the system in this case will be determined by making use of the results of the previous section.

For a CTLTI system driven by a cosine input signal, we have the following important relationship:

1. When a CTLTI system is driven by single-tone input signal at frequency ω0, its output signal is also a single-tone signal at the same frequency.
2. The effect of the system on the input signal is to
   1. (a) Scale its amplitude by an amount equal to the magnitude of the system function at ω = ω0
   2. (b) Shift its phase by an amount equal to the phase of the system function at ω = ω0

Example 4.45: Steady-state response of RC circuit for single-tone input

Consider the RC circuit of Example 4.43. Let the component values be chosen to yield a 3-dB cutoff frequency of ω c = 160 π rad/s, or equivalently fc = 80 Hz. Let the input signal be in the form

$$\tilde{x}\left(t\right)=5\cos \left(2\pi ft\right)$$

Compute the steady-state output signal for the cases f1 = 20 Hz, f2 = 100 Hz, f3 = 200 Hz, and f4 = 500 Hz.

Solution: For the system under consideration the system function is

$$\begin{array}{cc}H\left(f\right)=\frac{1}{1+j\left(f/80\right)}& \left(4.367\right)\end{array}$$

The magnitude and the phase of the system function were determined in Example 4.43 Eqns. (4.352) and (4.352). Using f instead of ω we have

$$\left|H\left(f\right)\right|\begin{array}{cc}=\frac{1}{\sqrt{1+{\left(f/80\right)}^2}}& \begin{array}{cc}\text{and}& \Theta \left(f\right)=-{\tan }^{-1}\left(f/80\right)\end{array}\end{array}$$

as shown in Fig. 4.89. For the input frequency of f1 = 20 Hz, the magnitude and the phase of the system function are

$$\left|H\left(20\right)\right|\begin{array}{cc}=\frac{1}{\sqrt{1+{\left(20/80\right)}^2}}=0.9701& \left(4.368\right)\end{array}$$

and

$$\begin{array}{cc}\Theta \left(20\right)=-{\tan }^{-1}\left(20/80\right)=-0.245\text{ radians}& \left(4.369\right)\end{array}$$

The impact of the system on the 20 Hz input signal is amplitude scaling by a factor of 0.9701 and phase shift by −0.245 radians. The corresponding steady-state output signal is

$$\begin{array}{lll}{y}_1\left(t\right)\hfill & =\left(5\right)\left(0.9701\right)\cos \left(40\pi t-0.245\right)\hfill & \hfill \\ \hfill & =4.8507\cos \left(40\pi \left(t-0.0019\right)\right)\hfill & \left(4.370\right)\hfill \end{array}$$

The phase shift of −0.245 radians translates to a time-delay of about td = 1.9 ms. The steady-state output signal y1 (t) is shown in Fig. 4.90 along with the input signal for comparison.

Calculations in Eqns. (4.368) through 4.370) can easily be repeated for the other three frequency choices, but will be skipped here to save space. The results are summarized in Table 4.6 for all four frequencies. The frequency-discriminating nature of the system is evident.

Input-output signal pairs for f2 = 100 Hz, f3 = 200 Hz, and f4 = 500 Hz are shown in Fig. 4.91.

Software resources:

ex_4_45a.m

ex_4_45b.m

ex_4_45c.m

4.6.3 Response of a CTLTI system to periodic input signal

Using the development in the previous sections, we are now ready to consider the use of a general periodic signal $\tilde{x}\left(t\right)$ as input to a CTLTI system. Let $\tilde{x}\left(t\right)$ be a signal that satisfies the existence conditions for the Fourier series. Using the EFS representation of the signal, the response of the system is

$$\begin{array}{cc}{S}_{\text{ys}}\left\{\tilde{x}\left(t\right)\right\}=S_{\text{ys}}\left\{{\displaystyle \sum _{k=-\infty }^{\infty }{c}_k{e}^{jk{w}_{0}t}}\right\}& \left(4.371\right)\end{array}$$

Let us use the linearity of the system to write the response as

$$\begin{array}{cc}{S}_{\text{ys}}\left\{\tilde{x}\left(t\right)\right\}={\displaystyle \sum _{k=-\infty }^{\infty }{S}_{\text{ys}}\left\{c_k{e}^{jk{w}_{0}t}\right\}=}{\displaystyle \sum _{k=-\infty }^{\infty }{S}_{\text{ys}}\left\{e^{jk{w}_{0}t}\right\}}& \left(4.372\right)\end{array}$$

The response of the system to an exponential basis function at frequency ω = kωo is given by

$$\begin{array}{cc}{S}_{\text{ys}}\left\{e^{jk{w}_{0}t}\right\}=e^{jk{w}_{0}t}H\left(k{\omega }_0\right)& \left(4.373\right)\end{array}$$

Using Eqn. (4.373) in Eqn. (4.372), the response of the system to $\tilde{x}\left(t\right)$ is found as

$$\begin{array}{cc}{S}_{\text{ys}}\left\{\tilde{x}\left(t\right)\right\}={\displaystyle \sum _{k=-\infty }^{\infty }{c}_k}H\left(k{\omega }_0\right)e^{jk{w}_{0}t}& \left(4.374\right)\end{array}$$

Two important observations should be made based on Eqn. (4.374):

1. For a CTLTI system driven by a periodic input signal, the output signal is also periodic with the same period.
2. If the EFS coefficients of the input signal are {ck; k= 1, ..., ∞} then the EFS coefficients of the output signal are {ck H (kω0); k = 1, ..., ∞}.

Example 4.46: RC circuit with pulse-train input

Consider again the RC circuit used in Example 4.45 with system function given by Eqn. (4.368). Let the input signal be a pulse train with period T0 = 50 ms and duty cycle d = 0.2 as shown in Fig. 4.92. Determine the output signal in steady state.

Solution: The EFS coefficients were determined in Example 4.7 for the type of periodic pulse train shown in Fig. 4.92. Using the specified duty cycle the coefficients are

$$\begin{array}{cc}{c}_k=0.2\text{sinc}\left(0.2k\right)& \left(4.375\right)\end{array}$$

The fundamental frequency is f0 = 1/T0 = 20 Hz, and the EFS representation of the signal $\tilde{x}\left(t\right)$ is

$$\begin{array}{cc}\tilde{x}\left(t\right)={\displaystyle \sum _{k=-\infty }^{\infty }0.2\text{ sinc}\left(0.2k\right)e^{j40\pi kt}}& \left(4.376\right)\end{array}$$

Let {dk ; k = −∞,..., ∞] be the EFS coefficients of the output signal $\tilde{y}\left(t\right)$. Using Eqn. (4.374) we have

$$\begin{array}{ccc}{d}_k=c_{k}H\left(k{f}_0\right)=\frac{c_k}{1+j\left(20k/80\right)},& k=-\infty ,\cdots ,\infty & \left(4.377\right)\end{array}$$

The EFS representation of the output signal is

$$\begin{array}{cc}\tilde{y}\left(t\right)={\displaystyle \sum _{k=-\infty }^{\infty }\left(\frac{c_k}{1+j\left(20k/80\right)}\right)}{e}^{j40\pi kt}& \left(4.378\right)\end{array}$$

Using the expressions for the magnitude and the phase of the system function, Eqn. (4.377) can be written as

$$\begin{array}{cc}\left|d_k\right|=\left|c_k\right|\left|H\left(k{f}_0\right)\right|=\frac{\left|c_k\right|}{\sqrt{1+{\left(20k/80\right)}^2}}& \left(4.379\right)\end{array}$$

and

$$\begin{array}{cc}\measuredangle d_k=\measuredangle c_k+\Theta \left(k{f}_0\right)=\measuredangle c_k-{\tan }^{-1}\left(20k/80\right)& \left(4.380\right)\end{array}$$

The relationship between the system function H (f) and the line spectra ck and dk is illustrated in Fig. 4.93. An approximation to the output signal $\tilde{y}\left(t\right)$ using terms up to the 75-th harmonic is shown in Fig. 4.94.

Software resources:

ex_4_46a.m

ex_4_46b.m

4.7 CTLTI Systems with Non-Periodic Input Signals

Let us consider the case of using a non-periodic signal x (t) as input to a CTLTI system. We have established in Section 2.7.2 of Chapter 2 that the output of a CTLTI system is equal to the convolution of the input signal with the impulse response, that is

$$\begin{array}{lll}y\left(t\right)\hfill & =h\left(t\right)*x\left(t\right)\hfill & \hfill \\ \hfill & ={\displaystyle {\int }_{-\infty }^{\infty }h\left(\lambda \right)x\left(t-\lambda \right)d\lambda }\hfill & \left(4.381\right)\hfill \end{array}$$

Let us assume that

1. The system is stable ensuring that H (ω) converges
2. The input signal has a Fourier transform

We have seen in Section 4.3.5 of this chapter that the Fourier transform of the convolution of two signals is equal to the product of individual transforms:

Writing each transform involved in Eqn. (4.382) in polar form using its magnitude and phase we obtain the corresponding relationships:

Example 4.47: Pulse response of RC circuit revisited

Consider again the RC circuit used Example 4.43. Let fc = 1/RC = 80 Hz. Determine the Fourier transform of the response of the system to the unit-pulse input signal x (t) = ∏ (t).

Solution: In Example 2.21 of Chapter 2 the output signal y (t) was found through the use of convolution operation. In this example we will approach the same problem from a system function perspective. The system function of the RC circuit under consideration was found in Example 4.43 to be

$$\begin{array}{cc}\text{Eqn}\text{.}\left(4.351\right):& H\left(f\right)=\frac{1}{1+j\left(f/f_c\right)}\end{array}$$

The transform of the output signal is

$$X\left(f\right)=\text{sinc}\left(f\right)$$

Using fc = 80 Hz, the transform of the output signal is

$$\begin{array}{lll}Y\left(f\right)\hfill & =H\left(f\right)X\left(f\right)\hfill & \hfill \\ \hfill & =\frac{1}{1+j\left(\frac{f}{80}\right)}\text{sinc}\left(f\right)\hfill & \left(4.385\right)\hfill \end{array}$$

Magnitude and phase of the transform Y (ω) can be found using Eqns. (4.383) and (4.384). The magnitude is

$$\begin{array}{cc}\left|Y\left(f\right)\right|=\frac{1}{\sqrt{1+{\left(\frac{f}{80}\right)}^2}}\left|\text{sinc}\left(f\right)\right|& \left(4.386\right)\end{array}$$

and the phase is

$$\begin{array}{cc}\measuredangle Y\left(f\right)=-{\tan }^{-1}\left(\frac{f}{80}\right)+\measuredangle \left[\text{sinc}\left(f\right)\right]& \left(4.387\right)\end{array}$$

as illustrated in Fig. 4.95. With a bit of work it can be shown that Y (f) found above is indeed the Fourier transform of the signal y (t) found in Example 2.21 using the convolution integral.

Software resources:

ex_4_47.m

4.8 Further Reading

[1] R.J. Beerends. Fourier and Laplace Transforms. Cambridge University Press, 2003.

[2] R.N. Bracewell. The Fourier Transform and Its Applications. Electrical Engineering Series. McGraw-Hill, 2000.

[3] E.A. Gonzalez-Velasco. Fourier Analysis and Boundary Value Problems. Elsevier Science, 1996.

[4] R. Haberman. Applied Partial Differential Equations: With Fourier Series and Boundary Value Problems. Pearson Education, 2012.

[5] K.B. Howell. Principles of Fourier Analysis. Studies in Advanced Mathematics. Taylor & Francis, 2010.

MATLAB Exercises

21  title('Approximation to x(t) using m=1,...,5'),

22  xlabel ('Time (sec)'),

23  ylabel ('Amplitude'),

24  legend ('x^{(1)}(t) ','x^{(2)}(t)','x^{(4)}(t)','x^{(5)}(t)'),

On line 9 the matrix “xmat” is initialized as a row vector holding the constant (zero-order) approximation to $\tilde{x}\left(t\right)$. The loop construct between lines 11 and 16 is a slightly modified version of the corresponding construct in MATLAB Exercise 4.1. At each pass through the loop a new row is added to the matrix “xmat” as the next approximation to the periodic signal. When the script is finished, the rows of the matrix “xmat” correspond to finite-harmonic approximations as follows:

$$\begin{array}{c}\begin{array}{lll}\text{row 1}\hfill & \Rightarrow \hfill & {\tilde{x}}^{\left(0\right)}\left(t\right)\hfill \\ \text{row }2\hfill & \Rightarrow \hfill & {\tilde{x}}^{\left(1\right)}\left(t\right)\hfill \\ \text{row }3\hfill & \Rightarrow \hfill & {\tilde{x}}^{\left(2\right)}\left(t\right)\hfill \end{array}\\ ⋮\end{array}$$

The graph produced by this script is shown in Fig. 4.97.

Software resources:

matex 4 2.m

Problems

1. 4.1. Consider the periodic square-wave signal $\tilde{x}\left(t\right)$ shown in Fig. 4.1 on page 267.

   1. Set up the cost function for approximating this signal using three terms in the form

      $${\tilde{x}}^{\left(3\right)}\left(t\right)=b_1\sin \left({\omega }_{0}t\right)+b_2\sin \left(2{\omega }_{0}t\right)+b_3\sin \left(3{\omega }_{0}t\right)$$

   2. Show that the optimum values of the coefficients are

      $$\begin{array}{cccc}{b}_1=\frac{4A}{\pi },& b_2=0,& \text{and}& b_3=\frac{4A}{3\pi }\end{array}$$

2. 4.2. Consider the pulse train shown in Fig. P.4.2.

   Figure P. 4.2

   

   1. Determine the fundamental period T0 and the fundamental frequency ω0 for the signal.

   2. Using the technique described in Section 4.2 find an approximation to $\tilde{x}\left(t\right)$ in the form

      $${\tilde{x}}^{\left(1\right)}\left(t\right)\approx a_0+a_1\cos \left({\omega }_{0}t\right)+b_1\sin \left({\omega }_{0}t\right)$$

      Determine the optimum coefficients a0, a1 and b1.

3. 4.3. Consider again the pulse train shown in Fig. P.4.2. Using the technique described in Section 4.2 find an approximation to $\tilde{x}\left(t\right)$ in the form

   $${\tilde{x}}^{\left(2\right)}\left(t\right)\approx a_0+a_1\cos \left({\omega }_{0}t\right)+b_1\sin \left({\omega }_{0}t\right)+a_2\cos \left(2{\omega }_{0}t\right)+b_2\sin \left(2{\omega }_{0}t\right)$$

4. 4.4. Consider again the periodic square-wave signal $\tilde{x}\left(t\right)$ shown in Fig. 4.1. The approximation using M trigonometric functions would be in the form

   $${\tilde{x}}^{\left(M\right)}\left(t\right)=b_1\sin \left({\omega }_{0}t\right)+b_2\sin \left({\omega }_{0}t\right)+...+b_M\sin \left(M{\omega }_{0}t\right)={\displaystyle \sum _{k=1}^{M}bk}\sin \left(k{\omega }_{0}t\right)$$

   Show that the approximation error is equal to ±A at discontinuities of the signal $\tilde{x}\left(t\right)$ independent of the number of terms M, that is,

   $$\begin{array}{cc}{\tilde{\epsilon }}_M\left(n\frac{T_0}{2}\right)=\pm A,& n:\text{Integer}\end{array}$$

   Hint: It should not be necessary to determine the optimum coefficient values b1, b2,..., bM for this problem.

5. 4.5. Consider the pulse train $\tilde{x}\left(t\right)$ shown in Fig. P.4.5.

   Fig. P.4.5.

   

   1. Determine the fundamental period T0 and the fundamental frequency ω0 for the signal.

   2. Using the approach followed in Section 4.2 determine the coefficients of the approximation

      $${\tilde{x}}^{\left(2\right)}\left(t\right)=a_0+a_1\cos \left({\omega }_{0}t\right)+a_2\cos \left(2{\omega }_{0}t\right)$$

      to the signal $\tilde{x}\left(t\right)$ that results in the minimum mean-squared error.

6. 4.6. Refer to the TFS representation of a periodic signal $\tilde{x}\left(t\right)$ given by Eqn. (4.26). Using the orthogonality properties of the basis functions show that the coefficients of the sine terms are computed as

   $$\begin{array}{ccc}{b}_k=\frac{2}{T_0}{\displaystyle {\int }_{t_0}^{t_0+T_0}\tilde{x}\left(t\right)\sin \left(k{\omega }_{0}t\right)dt,}& \text{for}& k=1,...,\infty \end{array}$$

   where T0 is the fundamental period, ω0 = 2π/T0 is the fundamental frequency in rad/s and t0 is an arbitrary time instant.

7. 4.7. Determine the TFS coefficients for the periodic signal shown in Fig. P.4.7.

   Fig. P.4.7.

   

8. 4.8. Determine the TFS coefficients for the periodic signal shown in Fig. P. 4.8. One period of the signal is $\tilde{x}\left(t\right)=e^{-2t}$ for 0 ≤ t ≤ 1 s.

   Fig. P. 4.8.

   

9. 4.9. Consider the periodic signal $\tilde{x}\left(t\right)$ shown in Fig. P.4.7.

   1. Determine the EFS coefficients from the TFS coefficients obtained in Problem 4.7.
   2. Determine the EFS coefficients by direct application of the analysis equation given by Eqn. (4.72).
   3. Sketch the line spectrum (magnitude and phase).

10. 4.10. Consider the periodic signal $\tilde{x}\left(t\right)$ shown in Fig. P.4.8.

    1. Determine the EFS coefficients from the TFS coefficients obtained in Problem 4.8.
    2. Determine the EFS coefficients by direct application of the analysis equation given by Eqn. (4.72).
    3. Sketch the line spectrum (magnitude and phase).

11. 4.11. Consider the periodic sawtooth signal $\tilde{g}\left(t\right)$ shown in Fig. P.4.11.

    Figure P. 4.11

    

    1. Determine EFS coefficients of this signal using the analysis equation in Eqn. (4.72).
    2. Sketch the line spectrum in terms of magnitude and phase.
    3. Determine the TFS coefficients of $\tilde{g}\left(t\right)$ from the EFS coefficients using the conversion formulas given by Eqns. (4.58) and (4.59).

12. 4.12. Let the periodic signal $\tilde{x}\left(t\right)$ have the EFS coefficients ck. A new periodic signal $\tilde{g}\left(t\right)$ is defined as a time reversed version of $\tilde{x}\left(t\right)$, that is,

    $$\tilde{g}\left(t\right)=\tilde{x}\left(-t\right)$$

    1. If $\tilde{g}\left(t\right)$ has the EFS coefficients dk, find the relationship between the two sets of coefficients ck and dk.
    2. Consider the signal $\tilde{x}\left(t\right)$ shown in Fig. 4.17 on page 290. Its EFS coefficients of were determined in Example 4.8 and given by Eqns. (4.81) and (4.82). Obtain the EFS coefficients of $\tilde{g}\left(t\right)$ shown in Fig. P.4.11 from the EFS coefficients of $\tilde{x}\left(t\right)$ using the results of part (a).

13. 4.13. The signal $\tilde{x}\left(t\right)$ shown in Fig. 4.17 and the signal $\tilde{g}\left(t\right)$ shown in Fig. P4.11 add up to a constant:

    $$\tilde{x}\left(t\right)+\tilde{g}\left(t\right)=a$$

    Explain how this property can be used for finding the EFS coefficients of $\tilde{g}\left(t\right)$ from the EFS coefficients of $\tilde{x}\left(t\right)$ that were determined in Example 4.8 and given by Eqns. (4.81) and (4.82).

14. 4.14. Determine the EFS coefficients of the full-wave rectified signal shown in Fig. P.4.14.

    Figure. P.4.14

    

15. 4.15. Prove the time shifting property of the exponential Fourier series given by Eqn. (4.113). Specifically show that, if the periodic signal $\tilde{x}\left(t\right)$ has the EFS coefficients ck, then the signal $\tilde{x}\left(t-\tau \right)$ has the coefficients

    $$d_k=c_k{e}^{-jk{w}_0\tau }$$

16. 4.16. Refer to the half-wave rectified sinusoid shown in Fig. 4.21 on page 293. Its EFS coefficients were determined in Example 4.10. Explain how the EFS coefficients of the full-wave rectified sinusoid in Fig. P.4.14 could be obtained from the results of Example 4.10 through the use of the time shifting property.

17. 4.17. Refer to the half-wave rectified sinusoid shown in Fig. 4.21. Its EFS coefficients were determined in Example 4.10. Using the conversion formulas given by Eqns. (4.92) and (4.93) find the compact Fourier series (CFS) representation of the signal.

18. 4.18. Find the Fourier transform of each of the pulse signals given below:

    1. x (t) = 3 ∏ (t)
    2. x (t) = 3 ∏ (t − 0.5)
    3. $x\left(t\right)=2\prod \left(\frac{t}{4}\right)$
    4. $x\left(t\right)=2\prod \left(\frac{t-3}{2}\right)$

19. 4.19. Starting with the Fourier transform integral given by Eqn. (4.129) find the Fourier transform of the signal

    $$\prod \left(t-0.5\right)-\prod \left(t-1.5\right)$$

    shown in Fig. P.4.19.

    Figure P. 4.19

    

20. 4.20. Starting with the Fourier transform integral given by Eqn. (4.129) find the Fourier transform of the signal

    $$e^{-at}u\left(t\right)-e^{-at}u\left(-t\right)$$

    shown in Fig. P.4.20. Assume a ≥ 0.

    Figure. P.4.20

    

21. 4.21. Refer to the signal shown in Fig. P.4.19. Find its Fourier transform by starting with the transform of the unit pulse and using linearity and time shifting properties.

22. 4.22. Refer to the signal shown in Fig. P.4.20. Find its Fourier transform by starting with the transform of the right-sided exponential signal and using linearity and time scaling properties.

23. 4.23. The Fourier transform of the triangular pulse with peak amplitude A and two corners at ±τ was found in Example 4.17 as

    $$A\Lambda \left(\frac{t}{\tau }\right)\underleftrightarrow{ℱ}A\tau {\text{ sinc}}^2\left(\frac{\omega \tau }{2\pi }\right)$$

    Using this result along with linearity and time shifting properties of the Fourier transform, find the transform of the signal shown in Fig. P.4.23.

    Figure. P.4.23

    

24. 4.24. The transform pair

    $$e^{-a\left|t\right|}\underleftrightarrow{ℱ}\frac{2a}{a^2+{\omega }^2}$$

    was obtained in Example 4.16. Using this pair along with the duality property, find the Fourier transform of the signal

    $$x\left(t\right)=\frac{2}{1+4t^2}$$

25. 4.25. Prove that, for a real-valued signal with odd symmetry, the Fourier transform is purely imaginary.

    Hint: Start with the conditions x* (t) = x (t) and x (−t) = −x (t). Show that Re {X (ω)} = 0.

26. 4.26.

    1. Find the Fourier transform X (ω) of the pulse signal

       $$x\left(t\right)=\prod \left(\frac{t-1}{2}\right)$$

    2. Express x (t) as the sum of its even and odd components, that is,

       $$x\left(t\right)=x_e\left(t\right)+x_0\left(t\right)$$

       Determine and sketch each component.

    3. Find the Fourier transforms Xe (ω) and Xo (ω) of the even and odd components respectively. Show that

       $$\begin{array}{ccc}{X}_e\left(\omega \right)=\mathrm{Re}\left\{X\left(\omega \right)\right\}& \text{and}& X_0\left(\omega \right)=j\mathrm{Im}\left\{X\left(\omega \right)\right\}\end{array}$$

27. 4.27. Compute and sketch the Fourier transforms of the modulated pulse signals given below:

    1. x (t) = cos (10πt) ∏ (t)
    2. $x\left(t\right)=\cos \left(10\pi t\right)\prod \left(\frac{t}{2}\right)$
    3. x (t) = cos (10πt) ∏ (2t)
    4. x (t) = cos (10πt) ∏ (4t)

28. 4.28.

    1. Starting with the inverse Fourier transform integral given by Eqn. (4.129) prove the second form of the differentiation-in-time property stated in Eqn. (4.249) for n = 1.
    2. Show that, if Eqn. (4.249) holds true for n = k where k is any positive integer, then it must also hold true for n = k + 1.

29. 4.29.

    1. Starting with the Fourier transform integral given by Eqn. (4.130) prove the second form of the differentiation-in-frequency property stated in Eqn. (4.265) for n = 1.
    2. Show that, if Eqn. (4.265) holds true for n = k where k is any positive integer, then it must also hold true for n = k + 1.

30. 4.30. Taking Eqn. (4.249) as a starting point and employing the duality property of the Fourier transform, prove the second form of the differentiation-in-frequency property stated in Eqn. (4.265) for n = 1.

31. 4.31. Using the differentiation-in-time property of the Fourier transform, determine the transform of the signal shown in Fig. P.4.31.

    Figure. P.4.31

    

32. 4.32. The Fourier transform of the trapezoidal signal x (t) shown in Fig. 4.70 on page 355 was found in Example 4.32 to be

    $$X\left(f\right)=2\lambda \text{ sinc}\left(f\tau \right)\text{sinc}\left(2f\lambda \right)$$

    Find the inverse transform of the spectrum shown in Fig. 4.74 on page 360 by using the result above in conjunction with the duality property.

33. 4.33. Determine the Fourier transform of the signal

    $$\begin{array}{cc}X\left(t\right)=\text{sin}\left(\pi t\right)\prod \left(t-\frac{1}{2}\right)=& \{\begin{array}{cc}\sin \left(\pi t\right),& 0\le t\le 1\\ 0,& \text{otherwise}\end{array}\end{array}$$

    1. Using the modulation property of the Fourier transform
    2. Using the multiplication property of the Fourier transform

34. 4.34. Consider the periodic pulse train shown in Fig. P.4.34.

    Figure. P.4.34

    

    Recall that the EFS coefficients of this signal were found in Example 4.6. Express the Fourier transform of $\tilde{x}\left(t\right)$ using impulse functions.

35. 4.35. Consider the pulse train with duty cycle d shown in Fig. 4.15 on page 289. Its EFS coefficients were determined in Example 4.7 to be

    $$c_k=d\text{ sinc}\left(kd\right)$$

    1. Working in the time domain, compute the power of the pulse train as a function of the duty cycle d.
    2. Sketch the power spectral density based on the EFS coefficients.
    3. Let d = 0.5. Suppose this signal is processed through a lowpass system that only retains the first m harmonics and eliminates the others. How many harmonics should be retained if we want to preserve at least 90 percent of the signal power?
    4. How many harmonics should be retained to preserve at least 95 percent of the signal power?
    5. How many harmonics should be retained to preserve at least 99 percent of the signal power?

36. 4.36. Repeat parts (c)-(e) of Problem 4.35 using d = 0.2. Does it take fewerormore harmonics to preserve the same percentage of power when the duty cycle is reduced?

37. 4.37. Use Parseval's theorem to prove that

    $${\displaystyle {\int }_{-\infty }^{\infty }{\left|\text{sinc}\left(f\right)\right|}^{2}df=1}$$

38. 4.38. Determine and sketch the power spectral density of the following signals:

    1. x (t) = 3 cos (20πt)
    2. x (t) = 2 cos (20πt) + 3 cos (30πt)
    3. x (t) = 5 cos (200πt) + 5 cos (200πt) cos (30πt)

MATLAB Problems

1. 4.39. Refer to Problem 4.1 in which a periodic square-wave signal was approximated using three terms. Let the parameters of the square wave be A = 1 and T0 = 1. Using MATLAB, compute and graph the three-term approximation superimposed with the original signal, that is, graph $\tilde{x}\left(t\right)$ and ${\tilde{x}}^{\left(3\right)}\left(t\right)$ on the same coordinate system. Display the signals in the time interval −1.5 < t < 2.5 s.

2. 4.40. Refer to Problem 4.2 in which a periodic pulse train was approximated using using both cosine and sine functions. Using MATLAB, compute and graph the approximation ${\tilde{x}}^{\left(1\right)}\left(t\right)$ superimposed with the original signal $\tilde{x}\left(t\right)$ and on the same coordinate system. Show the time interval −1.5 < t < 2.5 s.

3. 4.41. Compute and graph finite-harmonic approximations to the signal $\tilde{x}\left(t\right)$ shown in Fig. P.4.7 using 3, 4, and 5 harmonics. Also graph the approximation error for each case.

4. 4.42. Compute and graph finite-harmonic approximations to the signal $\tilde{x}\left(t\right)$ shown in Fig. P.4.8 using 3, 4, and 5 harmonics. Also graph the approximation error for each case.

5. 4.43. Consider the periodic pulse train shown in Fig. 4.15 on page 289. Its EFS coefficients were determined in Example 4.7 and given in Eqn. (4.77). Write a script to compute and graph the EFS line spectrum for duty cycle values d = 0.4, 0.6, 0.8 and 1. Comment on the changes in the spectrum as d is increased.

6. 4.44. Refer to Problem 4.17 in which the CFS representation of a half-wave rectified sinusoidal signal was determined. Write a script to compute and graph finite-harmonic approximations to the signal using m = 2, 4 and 6 harmonics.

7. 4.45. Refer to Problem 4.27. Write a script to compute and graph the transform of the modulated pulse

   $$x\left(t\right)=\cos \left(2\pi f_{0}t\right)\prod \left(\frac{t}{\tau }\right)$$

   In your script, define an anonymous function that takes two arguments, namely “f” and “tau”, and returns the spectrum of a pulse with width τ centered around the time origin. Use the result with modulation property of the Fourier transform to compute the transform sought. Use the script to verify the results obtained in Problem 4.27.

8. 4.46. Refer to Problem 4.35.

   1. Write a script to compute and graph the finite-harmonic approximation to the signal $\tilde{x}\left(t\right)$ using thefirst m harmonics. Leave m as a parameter that can be adjusted.
   2. Using the values of m found in parts (c), (d) and (e) of Problem 4.35 compute and graph the finite-harmonic approximations to the signal. Comment on the correlation between the percentage of power preserved and the quality of the finite-harmonic approximation.

MATLAB Projects

1. 4.47. Consider the RC circuit shown in Fig. 4.87 on page 382. Its system function was determined in Example 4.43 to be

   $$H\left(\omega \right)=\frac{1}{1+j\left(\omega /{\omega }_c\right)}$$

   Assume that the element values of the circuit are adjusted to obtain a critical frequency of fc = 80 Hz or, equivalently, ωc = 160π rad/s. Let the input signal to this circuit be a periodic pulse train with a period of T0 = 20 milliseconds as shown in Fig. P.4.47.

   Figure P. 4.47

   

   1. Determine the EFS coefficients of the signal $\tilde{x}\left(t\right)$ as a function of the duty cycle d. Write a script to graph the EFS spectrum of the signal for duty cycle values d = 0.5, 0.3, 0.1.
   2. Let the pulse width be τ = 10 ms corresponding to a duty cycle of d = 0.5. The signal $\tilde{x}\left(t\right)$ is used as input to the RC circuit. Write a script to compute the EFS coefficients of the output signal $\tilde{y}\left(t\right)$. Also graph the EFS spectrum for the output signal.
   3. Write a script to construct and graph an approximation to the output signal $\tilde{y}\left(t\right)$ using the first 30 harmonics in the EFS spectrum.

   4. An alternative to approximating the output of the RC circuit in response to a periodic signal is to determine its response to one isolated period, and then extend the result periodically. Let one period of the input signal be f(t) so that $\tilde{x}\left(t\right)$ can be expressed as

      $$\tilde{x}\left(t\right)={\displaystyle \sum _{r=-\infty }^{\infty }f\left(t-r{T}_0\right)=}{\displaystyle \sum _{r=-\infty }^{\infty }f\left(t-0.02r\right)}$$

      For the input signal we are considering, f (t) would be

      $$f\left(t\right)=\prod \left(\frac{t}{\tau }\right)$$

      If the response of the RC circuit to one period f (t) is

      $$g\left(t\right)={\text{S}}_{ys}\left\{f\left(t\right)\right\}$$

      then the response to the periodic signal $\tilde{x}\left(t\right)$ is

      $$\tilde{y}\left(t\right)={\displaystyle \sum _{r=-\infty }^{\infty }g\left(t-r{T}_0\right)=}{\displaystyle \sum _{r=-\infty }^{\infty }g\left(t-0.02r\right)}$$

      For d = 0.5 manually determine g(t), the response to a single pulse with τ = 10 ms. Write a script to compute the response using this approach and graph it. In your script use an anonymous function to compute the values of g(t). Approximate the summation for $\tilde{y}\left(t\right)$ using the terms for r = −20,..., 20. Compare the result obtained through this approach with the one obtained in part (c). What are the differences, and how would you explain them?

   5. Repeat parts (b) and (d) for duty cycle values d = 0. 3 and 0.1.

2. 4.48. Problem 4.47 should illustrate that the signal at the output of the RC circuit is a distorted version of the input signal. It should also illustrate that the severity of the distortion is dependent on the pulse-width τ and consequently the duty cycle d.

   The amount of distortion is also dependent on the shape of the input signal. In digital communication systems that rely on pulses for transmitting information, the fidelity of the output signal is important; the received pulses should be as similar to the transmitted pulses as possible. If alternative pulse shapes can provide better fidelity, they may be preferred over rectangular pulses.

   In this problem we will explore the use of an alternative pulse train. Consider the signal shown in Fig. P.4.48 to be used as input in place of the signal of Problem 4.47 with d = 0.5.

   Figure P. 4.48

   

   A general expression for the EFS coefficients for the signal was obtained in Example 4.10. Repeat parts (a) through (c) of Problem 4.47 using this alternative signal. Comment on the results, especially in comparison to those obtained in Problem 4.47.

1cos (2a) = 2 cos2 (a) − 1 = 1 − 2 sin2 (a).

2 ${\int }_a^{b}ud\upsilon =u\upsilon {|}_a^b-{\int }_a^b\upsilon du.$

3 $cos\left(a\right)cos\left(b\right)=\frac{1}{2}cos\left(a+b\right)+\frac{1}{2}cos\left(a-b\right).$

4 cos(a + b) = cos(a) cos(b) −sin(a) sin(b).

5 ${sin}^2\left(a\right)=\frac{1}{2}\left[1-cos\left(a\right)\right]$.

6 cos (a + b) = cos (a) cos (b) − sin (a) sin (b).

7 $\cos \left(a\right)\cos \left(b\right)=\frac{1}{2}\cos \left(a+b\right)+\frac{1}{2}\cos \left(a-b\right)$.