12

CHAPTER

Phase

IN AN AC WAVE, EVERY FULL CYCLE REPLICATES EVERY OTHER FULL CYCLE; THE WAVE REPEATS INDEFInitely. In this chapter, you’ll learn about the simplest possible shape (or waveform) for an AC disturbance. We call it a sine wave or sinusoid.

Instantaneous Values

When we graph its instantaneous amplitude as a function of time, an AC sine wave has the characteristic shape shown in Fig. 12-1. This illustration shows how the graph of the function y = sin x looks on an (x,y) coordinate plane. (The abbreviation sin stands for sine in trigonometry.) Imagine that the peak voltages equal +1.0 V and −1.0 V. Further imagine that the period equals exactly one second (1.0 s) so the wave has a frequency of 1.0 Hz. Let’s say that the wave begins at time t = 0.0 s. In this scenario, each cycle begins every time the value of t “lands on” a whole number. At every such instant, the voltage is zero and positive-going.

12-1   A sine wave with a period of 1 second has a frequency of 1 Hz.

If you freeze time at, say, t = 446.00 s and then measure the instantaneous voltage, you’ll find that it equals 0.0 V. Looking at the diagram, you can see that the instantaneous voltage will also equal 0.0 V every so-many-and-a-half seconds, so it will equal 0.0 V at, say, t = 446.50 s. But instead of getting more positive at the “second-and-a-half” instants, the voltage trends negative. If you freeze time at so-many-and-a-quarter seconds, say, t = 446.25 s, the instantaneous voltage will equal +1.0 V. The wave will rest exactly at its positive peak. If you stop time at so-many-and-three-quarter seconds, say, t = 446.75 s, the instantaneous voltage will rest exactly at its negative peak, −1.0 V. At intermediate time points, such as so-many-and-three-tenths seconds, the voltage will have intermediate values.

Rate of Change

Figure 12-1 reveals the fact that the instantaneous voltage sometimes increases and sometimes decreases. Increasing, in this context, means “getting more positive,” and decreasing means “getting more negative.” In the situation shown by Fig. 12-1, the most rapid increases in voltage occur when t = 0.00 s and t = 1.00 s. The most rapid voltage decrease takes place when t = 0.50 s. When t = 0.25 s, and also when t = 0.75 s, the instantaneous voltage neither increases nor decreases. But these “unchanging voltages” exist only for vanishingly small instants in time.

Let n equal some positive whole number of seconds. No matter what whole number we choose for n, the situation at t = n.25 s appears the same as it does for t = 0.25 s. Also, for t = n.75 s, things appear the same as they are when t = 0.75 s. The single cycle shown in Fig. 12-1 represents every possible condition of an AC sine wave having a frequency of 1.0 Hz and peak values of +1.0 V and −1.0 V. The entire wave cycle repeats for as long as AC continues to flow in the circuit, assuming that we don’t change the voltage or the frequency.

Now imagine that you want to observe the instantaneous rate of change in the voltage of the wave in Fig. 12-1, as a function of time. A graph of this function turns out as a sine wave, too—but it appears displaced to the left of the original wave by ¼ of a cycle. If you plot the instantaneous rate of change of a sine wave as a function of time (Fig. 12-2), you get the derivative of the waveform. The derivative of a sine wave turns out as a cosine wave because the mathematical derivative (in calculus) of the sine function equals the cosine function. The cosine wave has the same shape as the sine wave, but the phase differs by ¼ of a cycle.

12-2   A sine wave representing the rate of change in the instantaneous voltage of the wave in Fig. 12-1.

Circles and Vectors

An AC sine wave represents the most efficient manner in which an electrical quantity can alternate. It has only one frequency component. All the wave energy concentrates into a single, smooth, swinging variation—a single frequency. When an AC wave has this characteristic, we can represent its fluctuations by comparing it to the motion of an object that follows a circular orbit at a constant speed around a fixed central point.

Circular Motion

Imagine that you revolve a ball around and around at the end of a string, at a rate of one revolution per second (1 r/s) so that the ball describes a horizontal circle in space, as shown in Fig. 12-3A. If a friend stands some distance away, with his or her eyes in the plane of the ball’s path, she sees the ball oscillating back and forth, as shown in Fig. 12-3B, with a frequency of 1 Hz. That’s one complete cycle per second because you cause the ball to “orbit” at 1 r/s.

12-3   A revolving ball and string as seen from above (A) and from the side (B).

If you graph the position of the ball, as seen by your friend, with respect to time, you’ll get a sine wave, as shown in Fig. 12-4. This wave has the same fundamental shape as all sine waves. One sine wave might exhibit a greater distance between the peaks (peak-to-peak amplitude) than another, and one sine wave might appear “stretched out” lengthwise (wavelength) more than another. But every sine wave has the same general nature as every other. If we multiply or divide the peak-to-peak amplitude and/or the wavelength of any sine wave by the right numbers, we can make that sine wave fit exactly along the curve of any other sine wave. A standard sine wave has the following function in the (x,y) coordinate plane:

y = sin x

12-4   Position of ball (horizontal axis) as seen from the side, graphed as a function of time (vertical axis).

In the situation of Fig. 12-3A, you might make the string longer or shorter. You might whirl the ball around faster or slower. These changes would alter the peak-to-peak amplitude and/or the frequency of the sine wave graphed in Fig. 12-4. But you can always portray a sine wave as the equivalent of constant circular revolution. Mathematicians and engineers call this trick the circular-motion model of a sine wave.

The Rotating Vector

In Chap. 9, you learned about degrees of phase. If you wondered why we discussed phase in terms of angles going around a circle, you should have a better grasp of the idea now! A circle contains 360 angular degrees (360°), as you know from your courses in basic geometry. Points along a sine wave intuitively correspond to angles, or positions, around a circle.

Figure 12-5 shows how we can use a rotating vector to represent a sine wave in a system of polar coordinates. In the polar coordinate system, we plot a point according to its distance (called the radius) from the origin (center of the graph) and its angle expressed counterclockwise from “due east” (called the direction). Compare this system with the more traditional system of rectangular coordinates, where we plot a point according to its horizontal and vertical displacement from the origin.

12-5   Rotating-vector representation of a sine wave. Vector A portrays the start of the cycle (0°); vector B portrays the wave ¼ of the way through the cycle (90°); vector C portrays the wave halfway through the cycle (180°); vector D portrays the wave ¾ of the way through the cycle (270°). The vector length never changes.

A vector constitutes a mathematical quantity with two independent properties, called magnitude (also called length or amplitude) and direction (or angle). Vectors lend themselves perfectly to polar coordinates. In the circular-motion model of Fig. 12-5, we can note the following specific situations:

•   Vector A has a direction angle of 0°; it portrays the instant at which the wave amplitude equals zero and increases positively.

•   Vector B points “north,” representing the 90° phase angle, at which the wave has attained its maximum positive amplitude.

•   Vector C points “west,” representing a phase angle of 180°, the instant when the wave has gone back to zero amplitude while growing more negative.

•   Vector D points “south,” representing 270° of phase, the instant at which the wave has attained its maximum negative amplitude.

•   When the vector has rotated counterclockwise through a full circle (360°), it once again becomes vector A, and the wave begins its next cycle.

If you’re astute, you’ll notice that while the vector’s direction constantly changes, its length always remains the same.

Vector “Snapshots”

Figure 12-6 shows the four points, on a sine wave, representing the instantaneous vectors A, B, C, and D from Fig. 12-5. Think of these four points as “snapshots” of the wave vector as it rotates counterclockwise at a constant angular speed that corresponds to one revolution per cycle of the wave. If the wave has a frequency of 1 Hz, the vector revolves at a rate of 1 r/s. We can increase or decrease the frequency and still use this model. If the wave has a frequency of 100 Hz, the speed of the vector will equal 100 r/s, or a revolution every 0.01 s. If the wave has a frequency of 1 MHz, then the speed of the vector will equal 1,000,000 r/s (10⁶ r/s), and it will go once around the circle every 0.000001 s (10⁻⁶ s).

12-6   The four points for the vector model of Figure 12-5, shown in the standard amplitude-versus-time graphical manner for a sine wave.

The peak amplitude (either positive or negative without the sign) of a pure AC sine wave corresponds to the length of its vector in the circular-motion model. Therefore, the peak-to-peak amplitude corresponds to twice the length of the vector. As the amplitude increases, the vector gets longer. In Fig. 12-5, we portray time during an individual cycle as an angle going counterclockwise from “due east.”

In Fig. 12-5 and all other circular-model sine-wave vector diagrams, the vector length never changes, although it constantly rotates counterclockwise at a steady angular speed. The frequency of the wave corresponds to the speed at which the vector rotates. As the wave frequency increases, so does the vector’s rotational speed. Just as the sine-wave’s vector length remains independent of its rate of rotation, the amplitude of a sine wave is independent of its frequency.

Expressions of Phase Difference

The phase difference, also called the phase angle, between two sine waves can have meaning only when those two waves have the same frequency. If the frequencies differ, even by the slightest amount, the relative phase constantly changes, and we can’t specify a value for it. In the following discussions of phase angle, let’s assume that the two waves always have identical frequencies.

Phase Coincidence

The term phase coincidence means that two waves begin at exactly the same moment. They “line up perfectly.” Figure 12-7 illustrates a situation of this sort for two sine waves having different amplitudes. The phase difference in this case equals 0°. Alternatively, we could say that the phase difference equals some whole-number multiple of 360°, but engineers and technicians rarely speak of any phase angle of less than 0° or more than 360°.

12-7   Two sine waves in phase coincidence.

If two sine waves exist in phase coincidence, and if neither wave has any DC superimposed on it, then the resultant wave constitutes a sine wave with positive peak (pk+) or negative peak (pk−) amplitudes equal to the sum of the positive and negative peak amplitudes of the composite waves. The phase of the resultant wave coincides with the phases of the two composite waves.

Waves ½ Cycle out of Phase

When two sine waves begin exactly ½ cycle (180°) apart, we get a situation like the one shown in Fig. 12-8. In this case, engineers sometimes say that the waves are out of phase, although this expression constitutes an imprecise statement because someone might take it to mean a phase difference other than 180°.

12-8   Two sine waves that differ in phase by 180°.

If two sine waves have the same amplitudes and exist 180° out of phase, and if neither wave has DC superimposed, they cancel each other out because the instantaneous amplitudes of the two waves are equal and opposite at every moment in time.

If two sine waves have different amplitudes and exist 180° out of phase, and if neither wave has DC superimposed, then the resultant wave is a sine wave with positive or negative peak amplitudes equal to the difference between the positive and negative peak amplitudes of the composite waves. The phase of the resultant wave coincides with the phase of the stronger composite.

Phase Opposition

Any perfect sine wave without superimposed DC has the property that, if we shift its phase by precisely 180°, we get a result identical to what we get by “flipping the original wave upside-down” (inverting it), a condition called phase opposition. Not all waveforms have this property. Perfect square waves do, but most rectangular and sawtooth waves don’t, and irregular waveforms almost never do.

In most nonsinusoidal waves (waves that do not follow the sine or cosine function’s graph), a phase shift of 180° does not yield the same result as “flipping the wave upside down.” Never forget the conceptual difference between a 180° phase shift and a state of phase opposition!

Intermediate Phase Differences

Two perfect sine waves having the same frequency can differ in phase by any amount from 0° (phase coincidence), through 90° (phase quadrature, meaning a difference of a quarter of a cycle), through 180°, through 270° (phase quadrature again), and finally back to 360° (phase coincidence again).

Leading Phase

Imagine two sine waves, called wave X and wave Y, with identical frequency. If wave X begins a fraction of a cycle earlier than wave Y, then we say that wave X leads wave Y in phase. For this situation to hold true, wave X must begin its cycle less than 180° before wave Y. Figure 12-9 shows wave X leading wave Y by 90°.

12-9   Wave X leads wave Y by 90° of phase (¼ of a cycle).

When a particular wave X (the dashed line in Fig. 12-9) leads another wave Y (the solid line), then wave X lies to the left of wave Y on the time axis by some distance less than ½ wavelength. In a time-domain graph or display, displacement to the left represents earlier moments in time, and displacement to the right represents later moments in time; time “flows” from left to right.

Lagging Phase

Now imagine that some sine wave X begins its cycle more than 180° (½ cycle) but less than 360°(a full cycle) before wave Y starts. In this situation, we can imagine that wave X starts its cycle later than wave Y by some value between 0° and 180°. Then we say that wave X lags wave Y. Figure 12-10 shows wave X lagging wave Y by 90°. When a particular wave X (the dashed line in Fig. 12-10) lags another wave Y (the solid line), then wave X lies to the right of wave Y on the time axis by some distance less than ½ wavelength.

12-10   Wave X lags wave Y by 90° of phase (¼ of a cycle).

Vector Diagrams of Relative Phase

Suppose that a sine wave X leads a sine wave Y by, say, q degrees (where q represents a positive angle less than 180°). In this situation, we can draw the two waves as vectors, with vector X oriented q degrees counterclockwise from vector Y. In the opposite sense, if a sine wave X lags a sine wave Y by q degrees, then vector X appears to point in a direction going clockwise from vector Y by q degrees. If two waves X and Y coincide in phase, then their vectors X and Y point in the same direction. If two waves X and Y occur 180° out of phase, then their vectors X and Y point in opposite directions.

Figure 12-11 shows four phase relationships between two sine waves X and Y that have the same frequency but different amplitudes, as follows:

12-11   Vector representations of phase difference. At A, waves X and Y are in phase. At B, wave X leads wave Y by 90°. At C, wave X and wave Y are 180° out of phase. At D, wave X lags wave Y by 90°. We represent time as counterclockwise rotation of both vectors X and Y at a constant angular speed.

1.  At A, wave X exists in phase with wave Y, so vectors X and Y line up.

2.  At B, wave X leads wave Y by 90°, so vector X points in a direction 90° counterclockwise from vector Y.

3.  At C, waves X and Y exist 180° apart in phase, so vectors X and Y point in opposite directions.

4.  At D, wave X lags wave Y by 90°, so vector X points in a direction 90° clockwise from vector Y.

In all of these examples, we should imagine that the vectors both rotate counterclockwise at a continuous, steady rate as time passes, always maintaining the same angle with respect to each other, and always staying at the same lengths. If the frequency in hertz equals f, then the pair of vectors rotates together, counterclockwise, at an angular speed of f, expressed in complete, full-circle rotations per second (r/s).

Quiz

Refer to the text in this chapter if necessary. A good score is 18 correct. Answers are in the back of the book.

1.  A sine wave has a frequency of 50 kHz. Therefore, a complete cycle takes

(a)  0.20 microseconds (μs). (Note: 1 μs = 0.000001 second.)

(b)  2.0 μs.

(c)  20 μs.

(d)  200 μs.

2.  Which of the following statements is not characteristic of a pure sine wave with a well-defined and constant period?

(a)  The electrical energy is distributed over a wide range (band) of frequencies.

(b)  The wave can be represented as a vector that rotates at a constant angular speed.

(c)  The wave has a well-defined, constant wavelength as long as the medium that carries it doesn’t change.

(d)  The wave has a well-defined, constant frequency.

3.  If someone says that two sine waves differ in phase by an amount that constantly changes, then we know that the waves have

(a)  different periods.

(b)  different wavelengths.

(c)  different frequencies.

(d)  All of the above

4.  If wave X leads wave Y by ⅓ of a cycle, then

(a)  Y lags 120° behind X.

(b)  Y lags 90° behind X.

(c)  Y lags 60° behind X.

(d)  Y lags 30° behind X.

5.  Fill in the blank to make the following statement true: “Suppose that two pure sine waves having identical frequencies and no DC components, coincide in phase. If you change the phase of one wave by ________, you’ll get two waves in phase opposition.”

(a)  90°

(b)  180°

(c)  270°

(d)  360°

6.  We can change the phase of a pure sine wave having a constant frequency and no DC component by one of the following four phase angles, and end up with, in effect, the same wave. Which angle?

(a)  45°

(b)  90°

(c)  180°

(d)  360°

7.  A phase difference of 22.5° in the circular-motion model of a sine wave represents

(a)   of a revolution.

(b)  ⅛ of a revolution.

(c)  ¼ of a revolution.

(d)  ½ of a revolution.

8.  Two perfect sine waves exist in phase opposition. One wave has voltage peaks of +7 V pk+ and −7 V pk−, and the other wave has voltage peaks of +3 V pk+ and −3 V pk−. What’s the peak-to-peak voltage of the composite wave?

(a)  4 V pk-pk

(b)  6 V pk-pk

(c)  8 V pk-pk

(d)  12 V pk-pk

9.  A sine wave has a frequency of 60 Hz. How long does it take for 90° of phase to occur? (Note: 1 ms = 0.001 second.)

(a)  2.1 ms

(b)  4.2 ms

(c)  8.3 ms

(d)  We need more information to calculate it.

10.  In effect, a cosine wave is a sine wave shifted by

(a)  60°.

(b)  90°.

(c)  120°.

(d)  180°.

11.  Two sine waves have the same frequency but differ in phase by 60°. Neither wave has a DC component. The two waves are offset by

(a)   of a cycle.

(b)  ¼ of a cycle.

(c)  ⅓ of a cycle.

(d)  ½ of a cycle.

12.  Technically, the term phase opposition refers to two waves (whether sine or not) having the same frequency and

(a)  inverted with respect to each other.

(b)  displaced in phase by 90°.

(c)  displaced in phase by 180°.

(d)  displaced in phase by 270°.

13.  In a polar-coordinate vector diagram in which the radius represents voltage, the length of the rotating vector for a pure sine wave containing no DC component represents

(a)  half the peak voltage (positive or negative).

(b)  the peak voltage (positive or negative).

(c)  twice the peak voltage (positive or negative).

(d)  the peak-to-peak voltage.

14.  A sine wave X lags another sine wave Y by 45° of phase, so Y is

(a)   of a cycle ahead of X.

(b)   of a cycle ahead of X.

(c)  ⅛ of a cycle ahead of X.

(d)   of a cycle ahead of X.

15.  Figure 12-12 shows two sine waves X and Y that have the same frequency as a pair of polar vectors X and Y. Neither wave has a DC component. Which of the following statements holds true on the basis of this graph?

12-12   Illustration for Quiz Questions 15 and 16.

(a)  wave X lags wave Y by of a cycle.

(b)  wave X lags wave Y by ⅓ of a cycle.

(c)  wave X leads wave Y by of a cycle.

(d)  wave X leads wave Y by ⅓ of a cycle.

16.  Which, if any, of the following conclusions can we make about the relative peak-to-peak amplitudes of the two waves shown in Fig. 12-12?

(a)  The two waves have the same peak-to-peak amplitude.

(b)  The peak-to-peak amplitude of wave X exceeds the peak-to-peak amplitude of wave Y.

(c)  The peak-to-peak amplitude of wave Y exceeds the peak-to-peak amplitude of wave X.

(d)  We can’t say anything definitive about the relative peak-to-peak amplitudes of the two waves.

17.  Figure 12-13 illustrates two sine waves, neither of which has a DC component, that are in phase

12-13   Illustration for Quiz Questions 17 and 18.

(a)  coincidence.

(b)  opposition.

(c)  quadrature.

(d)  reinforcement.

18.  If we invert one of the waves in Fig. 12-13, then the two waves will be in phase

(a)  coincidence.

(b)  opposition.

(c)  quadrature.

(d)  reinforcement.

19.  Two AC waves (not necessarily sine waves) having the same frequency and both with 10-V peak-to-peak amplitude are in phase opposition. Neither wave has a DC component. What’s the peak-to-peak voltage of the composite wave?

(a)  7.071 V pk-pk

(b)  14.14 V pk-pk

(c)  0 V pk-pk

(d)  We need more information to know.

20.  Imagine that the two waves from the previous question differ in phase by 180°. What’s the peak-to-peak voltage of the composite wave? Remember that the waves don’t have to be sinusoids; they can have any imaginable form.

(a)  7.071 V pk-pk

(b)  14.14 V pk-pk

(c)  0 V pk-pk

(d)  We need more information to know.