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There are two key ideas behind the phasor representation of a signal:
- a real, time-varying signal may be represented by a complex, time-varying signal; and
- 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)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)
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.
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.
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).
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
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
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.
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 "90∘90∘ out of phase"?
(MATLAB) Modify Demo 2.1 in "The Function ex and ejθ" 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.)
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 sjθ", 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.
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.
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)."
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)dt∫Acos(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. 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)
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.
(What does "Endorsed by" mean?)
![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.](pic004.png)
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