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Appendix D

Orthogonality of Basis Functions

Orthogonality properties are used extensively in the development of Fourier series representations for the analysis of continuous-time and discrete-time signals. The use of an orthogonal set of basis functions ensures that the relationship between a periodic signal and its Fourier series representation is one-to-one, that is, each periodic signal has a unique set of Fourier series coefficients, and each set of coefficients corresponds to a unique signal. In this appendix we will summarize several forms of orthogonality properties and carry out their proofs.

D.1 Orthogonality for Trigonometric Fourier Series

Consider the two real-valued basis function sets defined as

\begin{array}{l} φ_{k} (t) = cos (k ω_{0} t); & k = 1, ..., \infty & (D .1) \end{array}

$\begin{array}{l} φ_{k} (t) = cos (k ω_{0} t); & k = 1, ..., \infty & (D .1) \end{array}$

and

\begin{array}{l} ψ_{k} (t) = sin (k ω_{0} t); & k = 1, ..., \infty & (D .2) \end{array}

$\begin{array}{l} ψ_{k} (t) = sin (k ω_{0} t); & k = 1, ..., \infty & (D .2) \end{array}$

where ω0 is the fundamental frequency in rad/s. The period that corresponds to the fundamental frequency ω0 is

T_{0} = \frac{2 π}{ω_{0}}

$T_{0} = \frac{2 π}{ω_{0}}$

It can be shown that each of the two sets in Eqns. (D.1) and (D.2) is orthogonal within itself, that is,

\begin{matrix} \int_{0}^{T_{0}} ϕ_{k} (t) ϕ_{m} (t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2 \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .3) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} ϕ_{k} (t) ϕ_{m} (t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2 \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .3) \end{matrix}$

and

\begin{matrix} \int_{0}^{T_{0}} ψ_{k} (t) ψ_{m} (t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2 \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .4) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} ψ_{k} (t) ψ_{m} (t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2 \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .4) \end{matrix}$

Functions Furthermore, the two sets are orthogonal to each other:

\begin{matrix} \int_{0}^{T_{0}} ϕ_{k} (t) ψ_{m} (t) d t = 0, & for any k, m & (D .5) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} ϕ_{k} (t) ψ_{m} (t) d t = 0, & for any k, m & (D .5) \end{matrix}$

Writing Eqns. (D.3), (D.4) and (D.5) using the definitions of the basis function sets in Eqns. (D.1) and (D.2), we arrive at the orthogonality properties

\begin{matrix} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2, & k = m \end{matrix} \\ \begin{matrix} 0, & k \neq m \end{matrix} \end{matrix} & (D6) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2, & k = m \end{matrix} \\ \begin{matrix} 0, & k \neq m \end{matrix} \end{matrix} & (D6) \end{matrix}$

\begin{matrix} \int_{0}^{T_{0}} sin (k ω_{0} t) sin (m ω_{0} t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2, \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .7) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} sin (k ω_{0} t) sin (m ω_{0} t) d t = {\begin{matrix} \begin{matrix} T_{0} / 2, \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .7) \end{matrix}$

\begin{matrix} \int_{0}^{T_{0}} cos (k ω_{0} t) sin (m ω_{0} t) d t = 0, & for any k, m & (D .8) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} cos (k ω_{0} t) sin (m ω_{0} t) d t = 0, & for any k, m & (D .8) \end{matrix}$

Proof:

Using the trigonometric identity in Eqn. (B.4), Eqn. (D.6) becomes

\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t & = & \frac{1}{2} \int_{0}^{T_{0}} cos ((k + m) ω_{0} t) d t \\ + \frac{1}{2} \int_{0}^{T_{0}} cos ((k - m) ω_{0} t) d t \end{array} & (D .9) \end{matrix}

$\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t & = & \frac{1}{2} \int_{0}^{T_{0}} cos ((k + m) ω_{0} t) d t \\ + \frac{1}{2} \int_{0}^{T_{0}} cos ((k - m) ω_{0} t) d t \end{array} & (D .9) \end{matrix}$

Let us first assume k ≠ m . The integrand of the first integral on the right side is periodic with period

\frac{2 π}{| (k + m) | ω_{0}} = \frac{T_{0}}{| k + m |}

$\frac{2 π}{| (k + m) | ω_{0}} = \frac{T_{0}}{| k + m |}$

The integrand cos ((k +m) ω0t)has (k + m) full periods in the interval from 0 to T0. Therefore the result of the first integral is zero. Similarly the integrand of the second integral on the right side of Eqn. (D.9) is periodic with period

\frac{2 π}{| k - m | ω_{0}} = \frac{T_{0}}{| k - m |}

$\frac{2 π}{| k - m | ω_{0}} = \frac{T_{0}}{| k - m |}$

The integrand cos ((k − m) ω0t) has |k − m| full periods in the interval from 0 to T0. Therefore the result of the second integral is also zero.

If k = m then Eqn. (D.9) becomes

\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t & = & \int_{0}^{T_{0}} {cos}^{2} (k ω_{0} t) d t \\ = & \frac{1}{2} \int_{0}^{T_{0}} cos (2 k ω_{0} t) d t + \frac{1}{2} \int_{0}^{T_{0}} d t \\ = & \frac{T_{0}}{2} \end{array} & (D .10) \end{matrix}

$\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} cos (k ω_{0} t) cos (m ω_{0} t) d t & = & \int_{0}^{T_{0}} {cos}^{2} (k ω_{0} t) d t \\ = & \frac{1}{2} \int_{0}^{T_{0}} cos (2 k ω_{0} t) d t + \frac{1}{2} \int_{0}^{T_{0}} d t \\ = & \frac{T_{0}}{2} \end{array} & (D .10) \end{matrix}$

completing the proof of Eqn. (D.6). Eqns. (D.7) and (D.8) can be proven similarly, using the trigonometric identities given by (B.5) and (B.6) and recognizing integrals that span an integer number of periods of sine and cosine functions.

D.2 Orthogonality for Exponential Fourier Series

Consider the set of complex periodic basis functions

\begin{array}{l} ϕ_{k} (t) = e^{j k ω_{0} t}; & k = - \infty, ..., \infty & (D .11) \end{array}

$\begin{array}{l} ϕ_{k} (t) = e^{j k ω_{0} t}; & k = - \infty, ..., \infty & (D .11) \end{array}$

where the parameter ω0 is the fundamental frequency in rad/s as in the case of the trigonometric set of basis functions of the previous section, and T0 = 2π/ω0 is the corresponding period. It can be shown that this basis function set in Eqn. (D.11) is orthogonal in the sense

\begin{matrix} \int_{0}^{T_{0}} ϕ_{0} (t) ϕ_{m}^{*} (t), d t = {\begin{matrix} \begin{matrix} T_{0}, \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .12) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} ϕ_{0} (t) ϕ_{m}^{*} (t), d t = {\begin{matrix} \begin{matrix} T_{0}, \\ 0, \end{matrix} & \begin{matrix} k = m \\ k \neq m \end{matrix} \end{matrix} & (D .12) \end{matrix}$

The second term in the integrand is conjugated due to the fact that we are working with a complex set of basis functions. Writing Eqn. (D.12) in open form using the definition of the basis function set in Eqns. (D.11) we arrive at the orthogonality property

\begin{matrix} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t = {\begin{matrix} \begin{matrix} T_{0}, & k = m \end{matrix} \\ \begin{matrix} 0, & k \neq m \end{matrix} \end{matrix} & (D .13) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t = {\begin{matrix} \begin{matrix} T_{0}, & k = m \end{matrix} \\ \begin{matrix} 0, & k \neq m \end{matrix} \end{matrix} & (D .13) \end{matrix}$

Proof:

Let us combine the exponential terms in the integral of Eqn. (D.13) and then apply Euler’s formula to write it as

\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t & = & \int_{0}^{T_{0}} e^{j (k - m) ω_{0} t} d t \\ = & \int_{0}^{T_{0}} cos ((k - m) ω_{0} t) d t + j \int_{0}^{T_{0}} sin ((k - m) ω_{0} t) d t \end{array} & (D .14) \end{matrix}

$\begin{matrix} \begin{array}{l} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t & = & \int_{0}^{T_{0}} e^{j (k - m) ω_{0} t} d t \\ = & \int_{0}^{T_{0}} cos ((k - m) ω_{0} t) d t + j \int_{0}^{T_{0}} sin ((k - m) ω_{0} t) d t \end{array} & (D .14) \end{matrix}$

If k ≠ m , both integrands on the right side of Eqn. (D.14) are periodic with period

\frac{2 π}{| k - m | ω_{0}} = \frac{T_{0}}{| k - m |}

$\frac{2 π}{| k - m | ω_{0}} = \frac{T_{0}}{| k - m |}$

Both integrands cos ((k − m) ω0t) and sin ((k − m) ω0t) have exactly |k − m| full periods in the interval from 0 to T0. Therefore the result of each integral is zero.

If k = m we have

\begin{matrix} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t = \int_{0}^{T_{0}} e^{j (0) ω_{0} t} d t = \int_{0}^{T_{0}} d t = T_{0} & (D .15) \end{matrix}

$\begin{matrix} \int_{0}^{T_{0}} e^{j k ω_{0} t} e^{- j m ω_{0} t} d t = \int_{0}^{T_{0}} e^{j (0) ω_{0} t} d t = \int_{0}^{T_{0}} d t = T_{0} & (D .15) \end{matrix}$

which completes the proof for Eqn. (D.13).

D.3 Orthogonality for Discrete-Time Fourier Series

Consider the set of discrete-time complex periodic basis functions

\begin{matrix} w_{N}^{K} = e^{- j \frac{2 π}{N} k}; & k = 0, ..., N - 1 & (D .16) \end{matrix}

$\begin{matrix} w_{N}^{K} = e^{- j \frac{2 π}{N} k}; & k = 0, ..., N - 1 & (D .16) \end{matrix}$

Parameters N and k are both integers. It can be shown that this basis function set in Eqn. (D.16) is orthogonal in the sense

\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} = {\begin{matrix} \begin{matrix} 0, & k \neq m \end{matrix} \\ \begin{matrix} N, & k = m \end{matrix} \end{matrix} & (D .17) \end{matrix}

$\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} = {\begin{matrix} \begin{matrix} 0, & k \neq m \end{matrix} \\ \begin{matrix} N, & k = m \end{matrix} \end{matrix} & (D .17) \end{matrix}$

Proof:

The summation in Eqn. (D.17) can be put into a closed form using the finite-length geometric series formula in Eqn. (C.10) to obtain

\begin{array}{l} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} & = & \sum_{n = 0}^{N - 1} e^{j (2 π / N) (k - m)} n \\ = & \frac{1 - e^{j (2 π / N) (k - m) N}}{1 - e^{j (2 π / N) (k - m)}} \end{array}

$\begin{array}{l} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} & = & \sum_{n = 0}^{N - 1} e^{j (2 π / N) (k - m)} n \\ = & \frac{1 - e^{j (2 π / N) (k - m) N}}{1 - e^{j (2 π / N) (k - m)}} \end{array}$

which can be simplified as

\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) k n} e^{- j (2 π / N) m n} = \frac{1 - e^{j 2 π (k - m)}}{1 - e^{(j 2 π / N) (k - m)}} & (D18) \end{matrix}

$\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) k n} e^{- j (2 π / N) m n} = \frac{1 - e^{j 2 π (k - m)}}{1 - e^{(j 2 π / N) (k - m)}} & (D18) \end{matrix}$

Let us begin with the case k ≠ m . The numerator of the fraction on the right side of Eqn. (D.18) is equal to zero. Since both integers k and m are in the interval k, m = 0,...,N − 1 their absolute difference |k − m| is in the interval n = 1,...,N − 1. Therefore the denominator of the fraction is non-zero, and the result is zero, proving the first part of Eqn. (D.17).

If k = m then the denominator of the fraction in Eqn. (D.18) also becomes zero, and L’Hospital’s rule must be used leading to

\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} = N, & for k = m & \begin{matrix} (D .19) \end{matrix} \end{matrix}

$\begin{matrix} \sum_{n = 0}^{N - 1} e^{j (2 π / N) k n} e^{- j (2 π / N) m n} = N, & for k = m & \begin{matrix} (D .19) \end{matrix} \end{matrix}$

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Table of Contents for Appendix D Orthogonality of Basis Functions

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Orthogonality of Basis Functions

D.1 Orthogonality for Trigonometric Fourier Series

D.2 Orthogonality for Exponential Fourier Series

D.3 Orthogonality for Discrete-Time Fourier Series

Table of Contents for
Appendix D Orthogonality of Basis Functions