# 5.2: Phasor Representation of Signals

- Page ID
- 9962

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

\[x(t)=A\cos(ωt+φ)\]

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

\[x(t)=A\cosω(t−τ)\]

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 \(ω=\frac {2π} T\) we obtain a third way of writing \(x(t)\):

\[x(t)=A\cos\frac {2π} {T} (t−τ)\]

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) 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 \(\PageIndex{1}\)

Show that \(x(t)=A\cos2πT(t−τ)\) is “periodic with period T," meaning that \(x(t+mT)=x(t)\) for all integer \(m\).

Exercise \(\PageIndex{2}\)

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

Exercise \(\PageIndex{3}\)

Sketch the function \(x(t)=110\cos[2π(60)t−\frac π 8]\) versus \(t\). Repeat for \(x(t)=5\cos[2π(16×10^6)t+\frac π 4]\) and \(x(t)=2cos[\frac {2π} {10^{−3}}(t−\frac {10^{−3}} 8)]\). For each function, determine \(A,ω,T,f,φ\), and \(τ\). Label your sketches carefully.

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

\[x(t)=\mathrm{Re}[Ae^{j(ωt+φ)}]=\mathrm{Re}[Ae^{jφ}e^{jωt}]\]

We call \(Ae^{jφ}e^{jωt}\) the complex representation of \(x(t)\) and write

\[x(t)↔Ae^{jφ}e^{jωt}\]

meaning that the signal \(x(t)\) may be reconstructed by taking the real part of \(Ae^{jφ}e^{jωt}\). In this representation, we call \(Ae^{jφ}\) the phasor or complex amplitude representation of \(x(t)\) and write

\[x(t)↔Ae^{jφ}\]

meaning that the signal \(x(t)\) may be reconstructed from \(Ae^{jφ}\) by multiplying with \(e^{jωt}\) and taking the real part. In communication theory, we call \(Ae^{jφ}\) the baseband representation of the signal \(x(t)\).

Exercise \(\PageIndex{4}\)

For each of the signals in Problem 3, give the corresponding phasor representation \(Ae^{jφ}\).

## Geometric Interpretation

Let's call

\[Ae^{jφ}e^{jωt}\]

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

\[Ae^{jφ}\]

This phasor is illustrated in the Figure. In the figure, \(φ\) is approximately \(\frac {−π} {10}\) If we let \(t\) increase to time \(t_1\), then the complex representation produces the phasor

We know from our study of complex numbers that \(e^{jωt_1}\) just rotates the phasor \(Ae^{jφ}\) through an angle of \(ωt_1\)! See Figure. Therefore, as we run t from 0, indefinitely, we rotate the phasor \(Ae^{jφ}\) indefinitely, turning out the circular trajectory of the Figure. When \(t=\frac {2π} ω\) then \(e^{jωt} = e^{j2π} = 1\). Therefore, every \((2πω)\) seconds, the phasor revisits any given position on the circle of radius A. We sometimes call \(Ae^{jφ}e^{jωt}\) a rotating phasor whose rotation rate is the frequency \(ω\):

\[\frac {d} {dt} ωt=ω\]

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

In summary, \(Ae^{jφ}e^{jωt}\) is the complex, or rotating phasor, representation of the signal \(A\cos(ωt+φ)\). In this representation, \(e^{jωt}\) rotates the phasor \(Ae^{jφ}\) through angles \(ωt\) at the rate \(ω\). The real part of the complex representation is the desired signal \(A\cos(ωt+φ)\). This real part is read off the rotating phasor diagram as illustrated in the Figure. In the figure, the angle \(φ\) is about \(\frac {−2π} {10}\). As we become more facile with phasor representations, we will write \(x(t)=\mathrm{Re}[Xe^{jωt}]\) and call \(Xe^{jωt}\) the complex representation and \(X\) the phasor representation. The phasor \(X\) is, of course, just the phasor \(Ae^{jφ}\).

Exercise \(\PageIndex{5}\)

Sketch the imaginary part of \(Ae^{jφ}e^{jωt}\) to show that this is \(A\sin(ωt+φ)\). What do we mean when we say that the real and imaginary parts of \(Ae^{jφ}e^{jωt}\) are "90^{∘} out of phase"?

Exercise \(\PageIndex{6}\)

(MATLAB) Modify Demo 2.1 in "The Function \(e^x\) and \(e^{jθ}\)" so that \(θ=ωt\), with \(ω\) an input frequency variable and \(t\) a time variable that ranges from \(−2(\frax {2π} {ω})\) to+\(2(\frac {2π} ω)\) in steps of \(0.02 (\frac {2π} ω)\). In your modified program, compute and plot \(e^{jωt}\), \(\mathrm{Re{[e^{jωt}]\), and \(\mathrm{Im}[e^{jωt}]\) for \(−2(\frac {2π} ω)≤t<2(\frac {2π} ω)\) in steps of \(0.02 (\frac {2π} ω)\). Plot \(e^{jωt}\) in a two-dimensional plot to get a picture like the Figure and plot \(\mathrm{Re}[e^{jωt}]\) and \(\mathrm{Im}[e^{jωt}]\) versus \(t\) to get signals like those of the Figure. You should observe something like the Figure using the subplot features discussed in An Introduction to MATLAB. (In the figure, \(w\) represents Greek \(ω\).)

## Positive and Negative Frequencies

There is an alternative phasor representation for the signal \(x(t)=A\cos(ωt+φ)\). We obtain it by using the Euler formula of "The Function e^{x} e^{jθ}", namely, \(\cos θ=\frac 1 2 (e^{jθ}+e^{−jθ})\). When this formula is applied to \(x(t)\), we obtain the result

\[x(t)=\frac A 2 [e^{j(ωt+φ)}+e^{−j(ωt+φ)}]=\frac A 2 e^{jφ}e^{jωt} + \frac A 2 e^{−jφ}e^{−jωt}\]

In this formula, the term \(\frac A 2 e^{jφ}e^{jωt}\) is a rotating phasor that begins at the phasor value \(\frac A 2 e^{jφ}\) (for t=0) and rotates counterclockwise with frequency \(ω\). The term \(\frac A 2 e^{−jφ}e^{−jωt}\) is a rotating phasor that begins at the (complex conjugate) phasor value \(\frac A 2 e^{−jφ}\) (for t=0) and rotates clockwise with (negative) fequency \(ω\). The physically meaningful frequency for a cosine is \(ω\), a positive number like \(2π(60)\) for 60 Hz power. There is no such thing as a negative frequency. The so-called negative frequency of the term \(\frac A 2 e^{−jφ}e^{−jωt}\) 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 \(Acos(ωt+φ)\). 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 \(\PageIndex{7}\)

Sketch the two-phasor representation of \(A\cos(ωt+φ)\). 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

\[A_1\cos(ωt+φ_1)+A_2\cos(ωt+φ_2)\]

The rotating phasor representation for this sum is

\[(A_1e^{jφ_1}+A_2e^{jφ_2})e^{jωt}\]

The new phasor is \(A_1e^{jφ_1}+A_2e^{jφ_2}\), and the corresponding real signal is \(x(t)=Re[(A_1e^{jφ_1}+A_2e^{jφ_2})e^{jωt}]\). The new phasor is illustrated in the Figure.

Exercise \(\PageIndex{8}\)

Write the phasor \(A_1e^{jφ_1}+A_2e^{jφ_2}\) as \(A_3e^{jφ_3}\) ; determine \(A_3\) and \(φ_3\) in terms of \(A_1,A_2,φ_1,\) and \(φ_2\). What is the corresponding real signal?

## Differentiating and Integrating Phasors

The derivative of the signal \(A\cos(ωt+φ)\) is the signal

\[\frac d {dt} A\cos(ωt+φ) = −ωA\sin(ωt+φ)\]

\[=−\mathrm{Im}[ωAe^{jφ}e^{jωt}]=\mathrm{Re}[jωAe^{jφ}e^{jωt}]\]

\[=\mathrm{Re}[ωe^{jπ/2}Ae^{jφ}e^{jωt}]\]

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

\[\frac d {dt} A\cos(ωt+φ) ↔ jωAe^{jφ}↔ωe^{jπ/2}Ae^{jφ}\]

These two phasor representations are entirely equivalent. The first says that the phasor \(Ae^{jφ}\) is complex scaled by \(jω\) to produce the phasor for \(\frac d {dt} A\cos(ωt+φ)\), and the second says that it is scaled by \(ω\) and phased by \(+π/2\). The phasor representations of \(A\cos(ωt+φ)\) and \(\frac d {dt} A\cos(ωt+φ)\) are illustrated in Figure 3.6. Note that the derivative “leads by \(π/2\) radians (90)."

The integral of \(A\cos(ωt+φ)\) is

\[∫A\cos(ωt+φ)dt=\frac A ω \sin(ωt+φ)=\mathrm{Im}[\frac A ω e^{jφ}e^{jωt}]\]

\[=\mathrm{Re}[−j\frac A ω e^{jφ}e^{jωt}] = \mathrm{Re}[\frac A {jω} e^{jφ}e^{jωt}] = \mathrm{Re}[\frac 1 ω e^{−jπ/2}Ae^{jφ}e^{jωt}]\]

This finding shows that the integral of \(A\cos(ωt+φ)\) has the phasor representation

\[∫A\cos(ωt+φ)dt↔\frac 1 {jω} Ae^{jφ}↔\frac 1 ω e^{−jπ/2} Ae^{jφ}\]

The phasor \(Ae^{jφ}\) is complex scaled by \(\frac 1 {jω}\) or scaled by \(\frac 1 ω\) and phased by \(e^{−jπ/2}\) to produce the phasor for \(∫A\cos(ωt+φ)dt\). This is illustrated in the Figure. Note that the integral “lags by \(π/2\) radians (90". Keep these geometrical pictures of leading and lagging by \(π/2\) in your mind at all times as you continue your more advanced study of engineering.

## An Aside: The Harmonic Oscillator

The signal \(A\cos(ωt+φ)\) 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

\[\frac {d^2x(t)} {dt^2} + ω^2x(t)=0\]

has the solution

\[x(t)=A\cos(ωt+φ)\]

Try it:

\[\frac {d^2} {dt^2} x(t)=\frac d {dt}[−Aω\sin(ωt+φ)]=−ω2A\cos(ωt+φ)\]

The constants \(A\) and \(φ\) are determined from the initial conditions

\[x(0)=A\cosφ ⇔ x^2(0)+x^2(\frac π {2ω})=A^2\]

\[x(\frac π {2ω})=−A\sinφ ⇔ −\frac {x(π/2ω)} {x(0)} = \tanφ\]

Exercise \(\PageIndex{9}\)

Show how to compute \(A\) and \(φ\) in the equation \(x(t)=A\cos(ωt+φ)\) from the initial conditions \(x(O)\) and \(\frac d {dt} x(t)|_{t=0}\).

## 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!