# 1.7 Review of Phasors

Phasor analysis is a useful mathematical tool for solving problems involving linear systems in which the excitation is a periodic time function. Many engineering problems are cast in the form of linear integro-differential equations. If the excitation, more commonly known as the forcing function, varies sinusoidally with time, the use of phasor notation to represent time-dependent variables allows us to convert a linear integro-differential equation into a linear equation with no sinusoidal functions, thereby simplifying the method of solution. After solving for the desired variable, such as the voltage or current in a circuit, conversion from the phasor domain back to the time domain provides the desired result.

The phasor technique also can be used to analyze linear systems when the forcing function is a (nonsinusoidal) periodic time function, such as a square wave or a sequence of pulses. By expanding the forcing function into a Fourier series of sinusoidal components, we can solve for the desired variable using phasor analysis for each Fourier component of the forcing function separately. According to the principle of superposition, the sum of the solutions due to all of the Fourier components gives the same result as one would obtain had the problem been solved entirely in the time domain without the aid of Fourier representation. The obvious advantage of the phasor–Fourier approach is simplicity. Moreover, in the case of nonperiodic source functions, such as a single pulse, the functions can be expressed as Fourier integrals, and a similar application of the principle of superposition can be used as well.

The simple RC circuit shown in Fig. 1-20 contains a sinusoidally time-varying voltage source given by

υ s(t)=V 0 sin(ω t+ϕ 0),(1.55)

where V 0 is the amplitude, ω is the angular frequency, and ϕ 0 is a reference phase. Application of Kirchhoff’s voltage law gives the following loop equation:

Our objective is to obtain an expression for the current i(t). We can do this by solving Eq. (1.56) in the time domain, which is somewhat cumbersome because the forcing function υ s(t) is a sinusoid. Alternatively, we can take advantage of the phasor-domain solution technique as follows.

# Figure 1-20 RC circuit connected to a voltage source υ s(t).

# 1.7.1 Solution Procedure

1. Step 1: Adopt a cosine reference

   To establish a phase reference for all time-varying currents and voltages in the circuit, the forcing function is expressed as a cosine (if not already in that form). In the present example,

   υ s(t)=V 0 sin(ω t+ϕ 0)=V 0 cos π 2−ω t−ϕ 0=V 0 cos ω t+ϕ 0−π 2,(1.57)

   where we used the properties sin x=cos(π/2−x) and cos(−x)=cos x.

2. Step 2: Express time-dependent variables as phasors

   Any cosinusoidally time-varying function z(t) can be expressed as

   z(t)=R e Z˜e j ω t,(1.58)

   where Z˜ is a time-independent function called the phasor of the instantaneous function z(t). To distinguish instantaneous quantities from their phasor counterparts, a tilde (∼) is added over the letter representing a phasor. For the voltage υ s(t) given by Eq. (1.57), its phasor equivalent V˜s is given by

   V˜s=V 0 e j(ϕ 0−π/2).(1.59)

   Thus, V˜s has the same amplitude as υ s(t), and its exponential function is j(ϕ 0−π/2), where (ϕ 0−π/2) is the phase part of the cosine function in Eq. (1.57) without the ω t term. To verify that the expression for V˜s given by Eq. (1.59) is indeed the phasor equivalent of υ s(t), we apply the right-hand side of Eq. (1.58):

   R e[V˜s e j ω t]=R e[V 0 e j(ϕ 0−π/2)⋅e j ω t]=V 0 cos(ω t+ϕ 0−π/2),

   which is, indeed, the expression for υ s(t) given by Eq. (1.57).

   The time-domain and phasor-domain correspondence can be expressed in the general form

   for any phase angle ϕ. Hence, ϕ may be a constant, such as ϕ 0−π/2, or it may be a function of one or more spatial variables, such as ϕ=β x.

   The phasor V˜s, corresponding to the time function υ s(t), contains amplitude and phase information but it is independent of the time variable t. Next we define the unknown variable i(t) in terms of a phasor equivalent I˜ as

   i(t)=R e(I˜e j ω t).(1.61)

   If the equation we are trying to solve contains derivatives or integrals, we use the following two properties:

   d i d t=d d t R e(I˜e j ω t)=R e d d t(I˜e j ω t)=R e[j ω I˜e j ω t],(1.62)

   and

   ∫i d t=∫R e(I˜e j ω t)d t=R e∫I˜e j ω t d t=R e I˜j ω e j ω t. (1.63)

3. Step 3: Recast the differential/integral equation in phasor form

   Upon using the form of Eq. (1.58) to represent i(t) and υ s(t) in Eq. (1.56) in terms of their phasor equivalents, we have

   R R e(I˜e j ω t)+1 C R e I˜j ω e j ω t=R e(V˜s e j ω t).(1.64)

   Combining all three terms under the same real-part (R e) operator leads to

   R e R+1 j ω C I˜−V˜s e j ω t=0. (1.65a)

   Had we adopted a sine reference—instead of a cosine reference—to define sinusoidal functions, the preceding treatment would have led to the result

   I m R+1 j ω C I˜−V˜s e j ω t=0. (1.65b)

   Since both the real and imaginary parts of the expression inside the curly brackets are zero, the expression itself must be zero. Moreover, since e j ω t≠0, it follows that

   The time factor e j ω t has disappeared because it was contained in all three terms. Equation (1.66) is the phasor-domain equivalent of Eq. (1.56).

4. Step 4: Solve the phasor-domain equation

   From Eq. (1.66), the phasor current I˜ is given by

   I˜=V˜s R+1/(j ω C).(1.67)

   Before we apply the next step, we need to convert the right-hand side of Eq. (1.67) into the form I 0 e j θ with I 0 being a real quantity. Thus,

   I˜=V 0 e j(ϕ 0−π/2)j ω C 1+j ω R C=V 0 e j(ϕ 0−π/2)ω C e j π/2 1+ω 2 R 2 C 2 e j ϕ 1=V 0 ω C 1+ω 2 R 2 C 2 e j(ϕ 0−ϕ 1), (1.68)

   where we have used the identity j=e j π/2. The phase angle associated with (1+j ω R C) is ϕ 1=tan−1(ω R C) and lies in the first quadrant of the complex plane.

5. Step 5: Convert back to the time domain

   To find i(t), we simply apply Eq. (1.61). That is, we multiply the phasor I˜ given by Eq. (1.68) by e j ω t and then take the real part:

   i(t)=R e[I˜e j ω t]=R e V 0 ω C 1+ω 2 R 2 C 2 e j(ϕ 0−ϕ 1)e j ω t=V 0 ω C 1+ω 2 R 2 C 2 cos(ω t+ϕ 0−ϕ 1). (1.69)

   In summary, we converted all time-varying quantities into the phasor domain, solved for the phasor I˜ of the desired instantaneous current i(t), and then converted back to the time domain to obtain an expression for i(t). Table 1-5 provides a summary of some time-domain functions and their phasor-domain equivalents.

   # Table 1-5 Time-domain sinusoidal functions z(t) and their cosine-reference phasor-domain counterparts Z˜, where z(t)=R e[Z˜e j ω t].

   Table 1 5 Full Alternative Text

# Example 1-4: RL Circuit

The voltage source of the circuit shown in Fig. 1-21 is given by

υ s(t)=5 sin(4×10 4 t−30°)(V). (1.70)

# Figure 1-21 RL circuit (Example 1-4).

Figure 1 21 Full Alternative Text

Obtain an expression for the voltage across the inductor.

Solution: The voltage loop equation of the RL circuit is

R i+L d i d t=υ s(t).(1.71)

Before converting Eq. (1.71) into the phasor domain, we express Eq. (1.70) in terms of a cosine reference:

υ s(t)=5 sin(4×10 4 t−30°)=5 cos(4×10 4 t−120°)(V).(1.72)

The coefficient of t specifies the angular frequency as ω=4×10 4 (rad/s). Per the second entry in Table 1-5, the voltage phasor corresponding to υ s(t) is

V˜s=5 e−j 120°(V),

and the phasor equation corresponding to Eq. (1.71) is

R I˜+j ω L I˜=V˜s.(1.73)

Solving for the current phasor I˜, we have

I˜=V˜s R+j ω L=5 e−j 120°6+j 4×10 4×2×10−4=5 e−j 120°6+j 8=5 e−j 120°10 e j 53.1°=0.5 e−j 173.1°(A).

The voltage phasor across the inductor is related to I˜ by

V˜L=j ω L I˜=j 4×10 4×2×10−4×0.5 e−j 173.1°=4 e j(90°−173.1°)=4 e−j 83.1°(V),

and the corresponding instantaneous voltage υ L(t) is therefore

υ L(t)=R e V˜L e j ω t=R e 4 e−j 83.1°e j 4×10 4 t=4 cos(4×10 4 t−83.1°)(V).

Concept Question 1-12: Why is the phasor technique useful? When is it used? Describe the process.

Concept Question 1-13: How is the phasor technique used when the forcing function is a nonsinusoidal periodic waveform, such as a train of pulses?

# 1.7.2 Traveling Waves in the Phasor Domain

According to Table 1-5, if we set ϕ 0=0, its third entry becomes

A cos(ω t+β x)⇔A e j β x.(1.74)

From the discussion associated with Eq. (1.31), we concluded that A cos(ω t+β x) describes a wave traveling in the negative x direction.