# 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 \(Ae^{jφ}\).