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Phasors: Phasor Representation of Signals

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

There are two key ideas behind the phasor representation of a signal:

  1. a real, time-varying signal may be represented by a complex, time-varying signal; and
  2. a complex, time-varying signal may be represented as the product of a complex number that is independent of time and a complex signal that is dependent on time.

Let's be concrete. The signal

x(t)=Acos(ωt+φ),x(t)=Acos(ωt+φ),
(1)

illustrated in Figure 1, is a cosinusoidal signal with amplitude A, frequency ω, and phase φ. The amplitude A characterizes the peak-to-peak swing of 2A2A, the angular frequency ω characterizes the period T=2πωT=2πω between negative- to-positive zero crossings (or positive peaks or negative peaks), and the phase φ characterizes the time τ=-φωτ=-φω when the signal reaches its first peak. With τ so defined, the signal x(t)x(t) may also be written as

x(t)=Acos ω(t-τ).x(t)=Acosω(t-τ).
(2)
Figure 1: A Cosinusoidal Signal
Figure one is a cartesian graph with a sinusoidal function and a couple labeled arrows. The sinusoidal curve is labeled Acos(ωt+Φ) versus t. The amplitude of the curve, or the distance from the middle of the curve at the horizontal axis to its peak is labeled A. The horizontal distance from the beginning of an upward part of the curve to the end of a downward portion of the curve, or a complete wave, is labeled T = 2π/ω. The distance from the vertical axis to the first peak in the first quadrant of the graph is labeled 𝞃 = -Φ/ω.

When τ is positive, then τ is a “time delay” that describes the time (greater than zero) when the first peak is achieved. When τ is negative, then τ is a “time advance” that describes the time (less than zero) when the last peak was achieved. With the substitution ω=2πTω=2πT we obtain a third way of writing x(t)x(t):

x ( t ) = A cos 2 π T ( t - τ ) . x(t)=Acos 2 π T (t-τ).
(3)

In this form the signal is easy to plot. Simply draw a cosinusoidal wave with amplitude A and period T; then strike the origin (t=0)(t=0) so that the signal reaches its peak at τ. In summary, the parameters that determine a cosinusoidal signal have the following units:

A, arbitrary (e.g., volts or meters/sec, depending upon the application)

ω, in radians/sec (rad/sec)

T, in seconds (sec)

φ, in radians (rad)

τ, in seconds (sec)

Exercise 1

Show that x(t)=Acos2πT(t-τ)x(t)=Acos2πT(t-τ) is “periodic with period TT," meaning that x(t+mT)=x(t)x(t+mT)=x(t) for all integer mm.

Exercise 2

The inverse of the period T T is called the “temporal frequency” of the cosinusoidal signal and is given the symbol f f; the units of f=1Tf=1T are (seconds)-1(seconds)-1 or hertz (Hz). Write x(t)x(t) in terms of f f. How is ff related to ω ω? Explain why f f gives the number of cycles of x(t)x(t) per second.

Exercise 3

Sketch the function x(t)=110cos[2π(60)t-π8]x(t)=110cos[2π(60)t-π8] versus t t. Repeat for x(t)=5cos[2π(16×106)t+π4]x(t)=5cos[2π(16×106)t+π4] and x(t)=2cos[2π10-3(t-10-38)]x(t)=2cos[2π10-3(t-10-38)]. For each function, determine A,ω,T,f,φA,ω,T,f,φ, and τ τ. Label your sketches carefully.

The signal x(t)=Acos(ωt+φ)x(t)=Acos(ωt+φ) can be represented as the real part of a complex number:

x ( t ) = Re [ A e j ( ω t + φ ) ] = Re [ A e j φ e j ω t ] . x ( t ) = Re [ A e j ( ω t + φ ) ] = Re [ A e j φ e j ω t ] .
(4)

We call AejφejωtAejφejωt the complex representation of x(t)x(t) and write

x(t)Aejφejωt,x(t)Aejφejωt,
(5)

meaning that the signal x(t)x(t) may be reconstructed by taking the real part of AejφejωtAejφejωt. In this representation, we call AejφAejφ the phasor or complex amplitude representation of x(t)x(t) and write

x ( t ) A e j φ , x ( t ) A e j φ ,
(6)

meaning that the signal x(t)x(t) may be reconstructed from AejφAejφ by multiplying with ejωtejωt and taking the real part. In communication theory, we call AejφAejφ the baseband representation of the signal x(t)x(t).

Exercise 4

For each of the signals in Problem 3.3, give the corresponding phasor representation AejφAejφ.

Geometric Interpretation. Let's call

A e j φ e j ω t A e j φ e j ω t
(7)

the complex representation of the real signal Acos(ωt+φ)Acos(ωt+φ). At t=0t=0, the complex representation produces the phasor

A e j φ . A e j φ .
(8)

This phasor is illustrated in Figure 2. In the figure, φ is approximately -π10-π10 If we let t t increase to time t 1 t 1 , then the complex representation produces the phasor

Figure 2: Rotating Phasor
Figure two is a dashed unit circle on a cartesian graph. It is titled with the expression Ae^jΦ e^jωt_1. Along the unit circle are various line segments extending from the origin to the edge of the circle at various points. Horizontally to the right is the first line segment, labeled A (t = -Φ/ω). There is one line segment in the first quadrant, in what looks to be approximately 45 degrees from the horizontal axis. This line is labeled Ae^(jΦ) e^(jωt_1)  (t = t_1). An arrow between this line and the aforementioned line on the horizontal axis indicates movement in the clockwise direction. There is one line segment in the second quadrant beginning at the origin and extending to a point on the unit circle, in what looks to be approximately 150 degrees from the original horizontal axis on the right side of the graph. This line is labeled Ae^(jΦ) e^(jωt_2)  (t = t_2). In between this line and the previously described line in the first quadrant is an arrow pointing both in the clockwise and counter-clockwise directions, labeled ω(t_2 - t_1).  There are two line segments in the fourth quadrant, both extending from the origin to a point on the circle. The first is approximately 20 degrees to the right from the lower vertical axis. It is labeled Ae^(jΦ) e^(jωt_3)  (t = t_3). The second is approximately 20 degrees below the positive side of the horizontal axis, and is labeled Ae^(jΦ) (t = 0). In between the two line segments in the fourth quadrant is an arrow indicating movement in the counter-clockwise direction.

We know from our study of complex numbers that ejωt1ejωt1 just rotates the phasor AejφAejφ through an angle of ωt1ωt1 ! See Figure 2. Therefore, as we run t from 0, indefinitely, we rotate the phasor AejφAejφ indefinitely, turning out the circular trajectory of Figure 2. When t=2πωt=2πω then ejωt=ej2π=1ejωt=ej2π=1. Therefore, every (2πω)(2πω) seconds, the phasor revisits any given position on the circle of radius A. We sometimes call AejφejωtAejφejωt a rotating phasor whose rotation rate is the frequency ω ω:

d d t ω t = ω . d d t ω t = ω .
(9)

This rotation rate is also the frequency of the cosinusoidal signal Acos(ωt+φ)Acos(ωt+φ).

In summary, AejφejωtAejφejωt is the complex, or rotating phasor, representation of the signal Acos(ωt+φ)Acos(ωt+φ). In this representation, ejωtejωt rotates the phasor AejφAejφ through angles ωtωt at the rate ω ω. The real part of the complex representation is the desired signal Acos(ωt+φ)cos(ωt+φ). This real part is read off the rotating phasor diagram as illustrated in Figure 3. In the figure, the angle φ is about -2π10-2π10. As we become more facile with phasor representations, we will write x(t)= Re [Xejωt]x(t)= Re [Xejωt] and call XejωtXejωt the complex representation and X X the phasor representation. The phasor X X is, of course, just the phasor AejφAejφ.

Figure 3: Reading a Real Signal from a Complex, Rotating Phasor
Figure three is labeled Ae^(jΦ) e^(jωt). At the top of a figure is a circle, with a horizontal and vertical line cutting the circle into four. At what looks to be the 45 degree, 135, and 315-degree marks (measured from the right horizontal axis like a unit circle) are arrows extending from the origin to points on the circle with arrows pointing away from the origin. Tangent to the rightmost and leftmost portions of the circle are vertical lines that point downward. To the right of the tangent lines is the expression Ae^(jΦ). Additionally, at the end of the arrow pointing in the 315-degree direction of the circle is another line segment pointing vertically downward. Halfway down the page is a horizontal line labeled on the left as -A and on the right as A. From this point is a sinusoidal wave-like function rotated 90 degrees so that its waves move left and right rather than up and down. This wave is bound by the tangent lines from the circle above. Two points on the vertical axis in the middle of this wave are marked. The first is near the second horizontal line down the page, is labeled -Φ/ω, and is located approximately at a peak of the wave-like curve to the right. The second is furthest down the page, is labeled 2π/ω, and is located just before another peak to the right of the wave-like curve. Below the long vertical line that is drawn from top to bottom on this graph, splitting the circle and located in the middle of the wave-like curve, is a label, t.

Exercise 5

Sketch the imaginary part of AejφejωtAejφejωt to show that this is Asin(ωt+φ)Asin(ωt+φ). What do we mean when we say that the real and imaginary parts of AejφejωtAejφejωt are "9090 out of phase"?

Exercise 6

(MATLAB) Modify Demo 2.1 in "The Function ex and e" so that θ=ωtθ=ωt, with ωω an input frequency variable and t t a time variable that ranges from -2(2πω)-2(2πω)to+2(2πω)to+2(2πω) in steps of 0.02 (2πω)(2πω). In your modified program, compute and plot ejωt, Re [ejωt]ejωt, Re [ejωt], and Im [ejωt] Im [ejωt] for -2(2πω)t<2(2πω)-2(2πω)t<2(2πω) in steps of 0.02 (2πω)(2πω). Plot ejωtejωt in a two-dimensional plot to get a picture like Figure 2 and plot Re [ejωt] Re [ejωt] and Im [ejωt] Im [ejωt] versus t to get signals like those of Figure 1. You should observe something like Figure 4 using the subplot features discussed in An Introduction to MATLAB. (In the figure, w w represents Greek ω.ω.)

Figure 4: The Functions ejωtejωt, Re[ejωt]Re[ejωt], and Im[ejωt]Im[ejωt]
Figure four is comprised of two graphs. The first plots the horizontal axis as real, and the vertical axis as imaginary. Above the graph is the expression exp(jwt). The horizontal and vertical values both range from -1 to 1 in increments of 1. Inside the graph is a horizontal dotted line across the vertical value 0, a vertical dotted line across the horizontal value 0, and a circle centered at (0,0) with radius 1. The second graph is much wider. Its horizontal axis is labeled, time, and above the graph is the title Re[exp(jwt)] and Im[exp(jwt)] vs t  w = 1000*2*pi. The horizontal axis ranges in value from -2 to 2 in increments of 0.5. The vertical axis ranges in value from -1 to 1 in increments of 1. Inside the graph is one horizontal line along the vertical value of 0, one sinusoidal wave function marked by a solid line, and one sinusoidal wave function marked by a dotted line. The solid sinusoidal line begins at (-2, 1) downward and completes four troughs and three and a half peaks reaching a minimum vertical value of -1 and extending completely across the graph. The dotted curve begins at (-2, 0) and moves sinusoidally with the same wavelength and amplitude as the solid curve, completing four troughs and peaks until it also terminates at the far-right end of the graph. Below this second graph is an extra label written as x10^-3.

Positive and Negative Frequencies. There is an alternative phasor representation for the signal x(t)=Acos(ωt+φ)x(t)=Acos(ωt+φ). We obtain it by using the Euler formula of "The Function ex s", namely, cos θ=12(ejθ+e-jθ)θ=12(ejθ+e-jθ). When this formula is applied to x(t)x(t), we obtain the result

x ( t ) = A 2 [ e j ( ω t + φ ) + e - j ( ω t + φ ) ] = A 2 e j φ e j ω t + A 2 e - j φ e - j ω t . x ( t ) = A 2 [ e j ( ω t + φ ) + e - j ( ω t + φ ) ] = A 2 e j φ e j ω t + A 2 e - j φ e - j ω t .
(10)

In this formula, the term A2ejφejωtA2ejφejωt is a rotating phasor that begins at the phasor value AejφAejφ (for t=0t=0) and rotates counterclockwise with frequency ω. The term A2e-jφe-jωtA2e-jφe-jωt is a rotating phasor that begins at the (complex conjugate) phasor value A2e-jφA2e-jφ (for t=0t=0) and rotates clockwise with (negative) ffequency ω. The physically meaningful frequency for a cosine is ω, a positive number like 2π(60)2π(60) for 60 Hz power. There is no such thing as a negative frequency. The so-called negative frequency of the term A2e-jφe-jωtA2e-jφe-jωt just indicates that the direction of rotation for the rotating phasor is clock-wise and not counterclockwise. The notion of a negative frequency is just an artifact of the two-phasor representation of Acos(ωt+φ)cos(ωt+φ). In the one-phasor representation, when we take the “real part,” the artifact does not arise. In your study of circuits, systems theory, electromagnetics, solid-state devices, signal processing, control, and communications, you will encounter both the one- and two-phasor representations. Become facile with them.

Exercise 7

Sketch the two-phasor representation of Acos(ωt+φ)Acos(ωt+φ). Show clearly how this representation works by discussing the counterclockwise rotation of the positive frequency part and the clockwise rotation of the negative frequency part.

Adding Phasors. The sum of two signals with common frequencies but different amplitudes and phases is

A 1 cos ( ω t + φ 1 ) + A 2 cos ( ω t + φ 2 ) . A 1 cos ( ω t + φ 1 ) + A 2 cos ( ω t + φ 2 ) .
(11)

The rotating phasor representation for this sum is

( A 1 e j φ 1 + A 2 e j φ 2 ) e j ω t . ( A 1 e j φ 1 + A 2 e j φ 2 ) e j ω t .
(12)

The new phasor is A1ejφ1+A2ejφ2A1ejφ1+A2ejφ2, and the corresponding real signal is x(t)= Re [(A1ejφ1+A2ejφ2)ejωt]x(t)= Re [(A1ejφ1+A2ejφ2)ejωt]. The new phasor is illustrated in Figure 5.

Figure 5: Adding Phasors
Figure five is a cartesian graph with two concentric circles made of dashed lines centered about the origin and three arrows pointing from the origin to various points in the first quadrant. The first arrow points with the shallowest positive slope, reaches the edge of the outer circle, and is labeled A_2 e^(jΦ_2). The second arrow with a stronger positive slope extends beyond both circles and is labeled A_3 e^(jΦ_3). The third arrow points with the strongest positive slope extends only to the edge of the inside circle, and is labeled A_1 e^(jΦ_1). All arrows are pointing away from the origin.

Exercise 8

Write the phasor A1ejφ1+A2ejφ2A1ejφ1+A2ejφ2 as A3ejφ3A3ejφ3 ; determine A 3 A 3 and φ 3 φ 3 in terms of A1,A2,φ1A1,A2,φ1, and φ 2 φ 2 . What is the corresponding real signal?

Differentiating and Integrating Phasors. The derivative of the signal Acos(ωt+φ)Acos(ωt+φ) is the signal

d d t A cos ( ω t + φ ) = - ω A sin ( ω t + φ ) = - Im [ ω A e j φ e j ω t ] = Re [ j ω A e j φ e j ω t ] = Re [ ω e j π / 2 A e j φ e j ω t ] . d d t A cos ( ω t + φ ) = - ω A sin ( ω t + φ ) = - Im [ ω A e j φ e j ω t ] = Re [ j ω A e j φ e j ω t ] = Re [ ω e j π / 2 A e j φ e j ω t ] .
(13)

This finding is very important. It says that the derivative of Acos(ωt+φ)Acos(ωt+φ) has the phasor representation

d d t A cos ( ω t + φ ) j ω A e j φ ω e j π / 2 A e j φ . d d t A cos ( ω t + φ ) j ω A e j φ ω e j π / 2 A e j φ .
(14)

These two phasor representations are entirely equivalent. The first says that the phasor AejφAejφ is complex scaled by jωjω to produce the phasor for ddtAcos(ωt+ddtAcos(ωt+φ)φ), and the second says that it is scaled by ω ω and phased by +π/2+π/2. The phasor representations of Acos(ωt+φ)Acos(ωt+φ) and ddtAcos(ωt+φ)ddtAcos(ωt+φ) are illustrated in Figure 3.6. Note that the derivative “leads by π/2π/2 radians (90)."

Figure 6: Differentiating and Integrating Phasors
Figure six is a cartesian graph with one circle made with a dashed line centered about the origin and three line segments ending at labeled points in various directions on the graph. The first begins from the origin and extends with a shallow positive slope to a point on the circle in the first quadrant, and is labeled Ae^(jΦ). The second begins from the origin and extends with a strong negative slope into the second quadrant beyond the circle, and is labeled ωe^(jπ/2)Ae^(jΦ). The third segment begins from the origin with the same negative slope as the second segment, this time into the fourth quadrant but does not reach the edge of the circle. It is labeled 1/ω e^(-jπ/2) Ae^(jΦ).

The integral of Acos(ωt+φ)Acos(ωt+φ) is

A cos ( ω t + φ ) d t = A ω sin ( ω t + φ ) = Im [ A ω e j φ e j ω t ] = Re [ - j Aωejφejωt] = Re [ A j ω e j φ e j ω t ] = Re [ 1 ω e - j π / 2 A e j φ e j ω t ] . A cos ( ω t + φ ) d t = A ω sin ( ω t + φ ) = Im [ A ω e j φ e j ω t ] = Re [ - j Aωejφejωt] = Re [ A j ω e j φ e j ω t ] = Re [ 1 ω e - j π / 2 A e j φ e j ω t ] .
(15)

This finding shows that the integral of Acos(ωt+φ)Acos(ωt+φ) has the phasor representation

A cos ( ω t + φ ) d t 1 j ω A e j φ 1 ω e - j π / 2 A e j φ . A cos ( ω t + φ ) d t 1 j ω A e j φ 1 ω e - j π / 2 A e j φ .
(16)

The phasor AejφAejφ is complex scaled by 1jω1jω or scaled by 1ω1ω and phased by e-jπ/2e-jπ/2 to produce the phasor for Acos(ωt+φ)dtAcos(ωt+φ)dt. This is illustrated in Figure 6. Note that the integral “lags by π/2π/2 radians (90). Keep these geometrical pictures of leading and lagging by π/2π/2 in your mind at all times as you continue your more advanced study of engineering.

An Aside: The Harmonic Oscillator. The signal Acos(ωt+φ)Acos(ωt+φ) stands on its own as an interesting signal. But the fact that it reproduces itself (with scaling and phasing) under differentiation means that it obeys the second-order differential equation of the simple harmonic oscillator.1 That is, the differential equation

d2x(t)dt2+ω2x(t)=0d2x(t)dt2+ω2x(t)=0
(17)

has the solution

x ( t ) = A cos ( ω t + φ ) . x ( t ) = A cos ( ω t + φ ) .
(18)

Try it:

d 2 d t 2 x ( t ) = d d t [ - A ω sin ( ω t + φ ) ] = - ω 2 A cos ( ω t + φ ) . d 2 d t 2 x ( t ) = d d t [ - A ω sin ( ω t + φ ) ] = - ω 2 A cos ( ω t + φ ) .
(19)

The constants A A and φ φ are determined from the initial conditions

x ( 0 ) = A cos φ x 2 ( 0 ) + x 2 ( π 2 ω ) = A 2 x ( π 2 ω ) = - A sin φ - x ( π / 2 ω ) x ( 0 ) = tan φ . x ( 0 ) = A cos φ x 2 ( 0 ) + x 2 ( π 2 ω ) = A 2 x ( π 2 ω ) = - A sin φ - x ( π / 2 ω ) x ( 0 ) = tan φ .
(20)

Exercise 9

Show how to compute A A and φ φ in the equation x(t)=Acos(ωt+φ)x(t)=Acos(ωt+φ) from the initial conditions x(O)x(O) and ddtx(t)|t=0ddtx(t)|t=0.

Footnotes

  1. This means, also, that we have an easy way to synthesize cosines with circuits that obey the equation of a simple harmonic oscillator!

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