9.3 Phasors

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9.3 Phasors

Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions.

A phasor is a complex number that represents the amplitude and phase of a sinusoid.

Phasors provide a simple means of analyzing linear circuits e xcited by sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be thoroughly f amiliar with complex numbers.

A complex number z can be written in rectangular form as

$$z = x + jy \tag{9.14a}$$

where j = √ ___ −1 ; x is the real part of z; y is the imaginary part of z. In this context, the variables x and y do not represent a location as in tw odimensional vector analysis but rather the real and imaginary parts of z in the complex plane. Nevertheless, we note that there are some resemblances between manipulating complex numbers and manipulating twodimensional vectors.

The complex number z can also be written in polar or e xponential form as

$$z = r/\phi = re^{j\phi} \tag{9.14b}$$

Charles Proteus Steinmetz (1865–1923) was a German-Austrian mathematician and electrical engineer.

Appendix B presents a short tutorial on complex numbers.

Historical

Charles Proteus Steinmetz (1865–1923), a German-Austrian mathematician and engineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He is also noted for his w ork on the theory of hysteresis.

Steinmetz was born in Breslau, Germany, and lost his mother at the age of one. As a youth, he was forced to leave Germany because of his political activities just as he w as about to complete his doctoral dis sertation in mathematics at the Uni versity of Breslau. He migrated to Switzerland and later to the United States, where he w as employed by General Electric in 1893. That same year, he published a paper in which complex numbers were used to analyze ac circuits for the first time. This led to one of his man y textbooks, Theory and Calculation of ac Phenomena, published by McGra w-Hill in 1897. In 1901, he became the president of the American Institute of Electrical Engineers, which later became the IEEE.

where r is the magnitude of z, and ϕ is the phase of z. We notice that z can be represented in three ways:

z = x + jy	Rectangular form
z = r⧸ϕ	Polar form	(9.15)
z = rejϕ	Exponential form

The relationship between the rectangular form and the polar form is shown in Fig. 9.6, where the x axis represents the real part and the y axis represents the imaginary part of a comple x number. Given x and y, we can get r and ϕ as

$$r = \sqrt{x^2 + y^2}$$

, $\phi = \tan^{-1} \frac{y}{x}$ (9.16a)

On the other hand, if we know r and ϕ, we can obtain x and y as

$$x = r \cos \phi, \qquad y = r \sin \phi \tag{9.16b}$$

Thus, z may be written as

$$z = x + jy = r/\underline{\phi} = r(\cos\phi + j\sin\phi)$$

(9.17)

Addition and subtraction of complex numbers are better performed in rectangular form; multiplication and division are better done in polar form. Given the complex numbers

$$z = x + jy = r/\underline{\phi}, \qquad z_1 = x_1 + jy_1 = r_1/\underline{\phi}_1$$ $$z_2 = x_2 + jy_2 = r_2/\underline{\phi}_2$$

the following operations are important. Addition:

$$z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)$$

\n(9.18a)

Subtraction:

$$z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2)$$

\n(9.18b)

Multiplication:

$$z_1 z_2 = r_1 r_2 / \phi_1 + \phi_2 \tag{9.18c}$$

Division:

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} / \phi_1 - \phi_2 \tag{9.18d}$$

Reciprocal:

$$\frac{1}{z} = \frac{1}{r} \angle -\phi \tag{9.18e}$$

Square Root:

$$\sqrt{z} = \sqrt{r} \sqrt{\phi/2}$$

(9.18f)

Complex Conjugate:

$$z^* = x - jy = r'_\text{p} = re^{-j\phi} \tag{9.18g}$$

Note that from Eq. (9.18e),

$$\frac{1}{j} = -j \tag{9.18h}$$

These are the basic properties of complex numbers we need. Other properties of complex numbers can be found in Appendix B.

The idea of phasor representation is based on Euler’ s identity. In general,

$$e^{\pm i\phi} = \cos\phi \pm j\sin\phi \qquad (9.19)$$

which shows that we may re gard cos ϕ and sin ϕ as the real and imagi nary parts of e ^jϕ ; we may write

$$\cos \phi = \text{Re}(e^{j\phi}) \tag{9.20a}$$ $$\sin \phi = \text{Im}(e^{j\phi})\tag{9.20b}$$

where Re and Im stand for the real part of and the imaginary part of. Given a sinusoid v(t) = Vm cos(ωt + ϕ), we use Eq. (9.20a) to e xpress v(t) as

$$v(t) = V_m \cos(\omega t + \phi) = \text{Re}(V_m e^{j(\omega t + \phi)})$$

(9.21)

or

$$v(t) = \text{Re}(V_m e^{j\phi} e^{j\omega t})$$

\n(9.22)

Thus,

$$v(t) = \text{Re}(\mathbf{V}e^{j\omega t})$$

(9.23)

where

$$\mathbf{V} = V_m e^{j\phi} = V_m / \phi \tag{9.24}$$

V is thus the phasor representation of the sinusoid v(t), as we saidearlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to develop the phasor, but the standard convention is to use Eq. (9.20a).

One way of looking at Eqs. (9.23) and (9.24) is to consider the plot of the sinor Ve jωt = Vme j(ωt+ϕ) on the comple x plane. As time increases, the sinor rotates on a circle of radius Vm at an angular velocity ω in the counterclockwise direction, as shown in Fig. 9.7(a). We may regard v(t) as the projection of the sinor Ve jωt on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time t = 0 is the phasor V of the sinusoid v(t). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term e jωt is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency ω of the phasor; otherwise we can make serious mistakes.

A phasor may be regarded as a mathematical equivalent of a sinusoid with the time dependence dropped.

If we use sine for the phasor instead of cosine, then ^v (t) = ^{V m} sin(ωt + ϕ) = Im(Vmej(ωt+ϕ) ) and the corresponding phasor is the same as that in Eq. (9.24).

Equation (9.23) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time f actor ejω^t and tak e the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Because a phasor has magnitude and phase (“direction”), it behaves as a vector and is printed in boldface. For example, phasors V = Vm⧸ϕ and I = Im⧸−θ are graphically represented in Fig. 9.8. Such a graphical representation of phasors is known as a phasor diagram.

Equations (9.21) through (9.23) re veal that to get the phasor cor responding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we tak e out the time f actor e^jω^t , and whate ver is left is the pha sor corresponding to the sinusoid. By suppressing the time f actor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:

$$v(t) = V_m \cos(\omega t + \phi) \qquad \Leftrightarrow \qquad \mathbf{V} = V_m / \underline{\phi}$$

(9.25)
\n(Time-domain representation) (Phasor-domain representation)

We use lightface italic letters such as ^z to represent complex numbers but boldface letters such as V to represent phasors, because phasors are vectorlike quantities.

A phasor diagram showing V = Vm⧸ϕ and I = Im⧸−θ.

Given a sinusoid v(t) = Vm cos(ωt + ϕ), we obtain the corresponding phasor as V = Vm ^⧸ϕ. Equation (9.25) is also demonstrated in Table 9.1, where the sine function is considered in addition to the cosine function. From Eq. (9.25), we see that to get the phasor representation of a sinu soid, we e xpress it in cosine form and tak e the magnitude and phase. Given a phasor, we obtain the time domain representation as the cosine function with the same magnitude as the phasor and the ar gument as ωt plus the phase of the phasor. The idea of expressing information in alternate domains is fundamental to all areas of engineering.

TABLE 9.1

Sinusoid-phasor transformation.

Phasor domain representation
Vm⧸ϕ
Vm⧸ϕ − 90°
Im⧸θ
Im⧸θ − 90°

Note that in Eq. (9.25) the frequenc y (or time) f actor ejω^t is sup pressed, and the frequency is not explicitly shown in the phasor domain representation because ω is constant. However, the response depends on ω. For this reason, the phasor domain is also known as the frequency domain.

From Eqs. (9.23) and (9.24), v(t) = Re(Ve jωt ) = Vm cos(ωt + ϕ), so that

$$\frac{dv}{dt} = -\omega V_m \sin(\omega t + \phi) = \omega V_m \cos(\omega t + \phi + 90^\circ)$$

= Re( $\omega V_m e^{j\omega t} e^{j\phi} e^{j90^\circ}$ ) = Re( $j\omega V e^{j\omega t}$ ) (9.26)

This shows that the deri vative v(t) is transformed to the phasor domain as jωV

$$\frac{dv}{dt} \qquad \Leftrightarrow \qquad j\omega V \qquad (9.27)$$

\n(Time domain)

\n(Phasor domain)

Similarly, the inte gral of v(t) is transformed to the phasor domain as V∕jω

$$\int v \, dt \qquad \Leftrightarrow \qquad \frac{V}{j\omega} \qquad (9.28)$$

\n(Time domain)

\n(Phasor domain)

Equation (9.27) allows the replacement of a derivative with respect to time with multiplication of jω in the phasor domain, whereas Eq. (9.28) allows the replacement of an inte gral with respect to time with di vision by jω in the phasor domain. Equations (9.27) and (9.28) are useful in finding the steady-state solution, which does not require knowing the initial values of the variable involved. This is one of the important applications of phasors.

Besides time differentiation and integration, another important use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6 provides one.

The differences between v(t) and V should be emphasized:

• 1. v(t) is the instantaneous or time domain representation, while V is the frequency or phasor domain representation.
• 2. v(t) is time dependent, while V is not. (This f act is often for gotten by students.)
• 3. v(t) is al ways real with no comple x term, while V is generally complex.

Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.

Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by jω.

Integrating a sinusoid is equivalent to dividing its corresponding phasor by jω.

Adding sinusoids of the same frequency is equivalent to adding their corresponding phasors.

Evaluate these complex numbers: Example 9.3

(a)

$$(40/50^{\circ} + 20/-30^{\circ})^{1/2}$$

\n(b)

$$\frac{10/-30^{\circ} + (3-j4)}{(2+j4)(3-j5)^{*}}$$

Solution:

(a) Using polar to rectangular transformation,

$$40/50^{\circ} = 40(\cos 50^{\circ} + j \sin 50^{\circ}) = 25.71 + j30.64$$ $$20\angle -30^{\circ} = 20[\cos(-30^{\circ}) + j\sin(-30^{\circ})] = 17.32 - j10$$

Adding them up gives

$$40/50^{\circ} + 20/-30^{\circ} = 43.03 + j20.64 = 47.72/25.63^{\circ}$$

Taking the square root of this,

$$(40/50^{\circ} + 20/-30^{\circ})^{1/2} = 6.91/12.81^{\circ}$$

(b) Using polar-rectangular transformation, addition, multiplication, and division,

(b) Using polar-rectangular transformation, a
division,

$$\frac{10(-30^\circ + (3 - j4))}{(2 + j4)(3 - j5)^*} = \frac{8.66 - j5 + (3 - j4)}{(2 + j4)(3 + j5)}$$ $$= \frac{11.66 - j9}{-14 + j22} = \frac{14.73(-37.66^\circ)}{26.08(122.47^\circ)}$$ $$= 0.565(-160.13^\circ)$$

Practice Problem 9.3	Evaluate the following complex numbers:
	(a) [(5 + j2)(−1 + j4) − 5⧸ 60°]*
	10 + j5 + 3⧸ 40° ______________ (b) + 10⧸ 30° + j5 −3 + j4
	Answer: (a) −15.5 − j13.67, (b) 8.293 + j7.2.

Example 9.4 Transform these sinusoids to phasors:

(a) i = 6 cos(50t − 40°) A (b) v = −4 sin(30t + 50°) V

Solution:

(a) i = 6 cos(50t − 40°) has the phasor

$$I = 6/–40^{\circ} A$$

(b) Since

$$-\sin A = \cos(A + 90^\circ)$$

,
$v = -4 \sin(30t + 50^\circ) = 4 \cos(30t + 50^\circ + 90^\circ)$

$$= 4\cos(30t + 140^\circ) \text{ V}$$

The phasor form of v is

$$V = 4/140^{\circ} V$$

Practice Problem 9.4 Express these sinusoids as phasors:

(a) v = −14 sin(5t − 22°) V (b) i = −8 cos(16t + 15°) A

Answer: (a) V = 14⧸ 68° V, (b) I = 8⧸−165° A.

Find the sinusoids represented by these phasors: Example 9.5

(a)

$$\mathbf{I} = -3 + j4 \text{ A}$$

\n(b) $\mathbf{V} = j8e^{-j20^{\circ}} \text{ V}$

Solution:

(a) I = −3 + j4 = 5⧸ 126.87°. Transforming this to the time domain gives

$$i(t) = 5 \cos(\omega t + 126.87^{\circ})$$

A

(b) Because j = 1⧸ 90°,

$$\mathbf{V} = j8 \underline{/ -20^{\circ}} = (1 \underline{/ 90^{\circ}})(8 \underline{/ -20^{\circ}})$$ $$= 8 \underline{/ 90^{\circ} - 20^{\circ}} = 8 \underline{/ 70^{\circ}} \text{ V}$$

Converting this to the time domain gives

$$v(t) = 8\cos(\omega t + 70^\circ)\text{V}$$

Find the sinusoids corresponding to these phasors: Practice Problem 9.5

(a) V = −25⧸ 40° V (b) I = j(12 − j5) A Answer: (a) v(t) = 25 cos(ωt − 140°) V or 25 cos(ωt + 220°) V, (b) i(t) = 13 cos(ωt + 67.38°) A.

Given i1(t) = 4 cos(ωt + 30°) A and i2(t) = 5 sin(ω^{t −} 20°) A, find Example 9.6 their sum.

Solution:

Here is an important use of phasors—for summing sinusoids of the same frequency. Current i1(t) is in the standard form. Its phasor is

$$\mathbf{I}_1 = 4/30^\circ$$

We need to express i2(t) in cosine form. The rule for converting sine to cosine is to subtract 90°. Hence,

$$i_2 = 5\cos(\omega t - 20^\circ - 90^\circ) = 5\cos(\omega t - 110^\circ)$$

and its phasor is

I2 = 5⧸−110°

If we let i = i1 + i2, then

$$\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 = 4/30^\circ + 5/-110^\circ$$

= 3.464 + j2 - 1.71 - j4.698 = 1.754 - j2.698
= 3.218/-56.97° A

Transforming this to the time domain, we get

$$i(t) = 3.218 \cos(\omega t - 56.97^{\circ}) \text{ A}$$

Of course, we can find i1 + i2 using Eq. (9.9), but that is the hard way.

Practice Problem 9.6 If v1 = −10 sin(ωt − 30°) V and v2 = 20 cos(ωt + 45°) V, find v⁼ v1 + v2.

Answer: v(t) = 29.77 cos(ωt + 49.98°) V.

Example 9.7 Using the phasor approach, determine the current i(t) in a circuit described by the integrodifferential equation

$$4i + 8\int i\,dt - 3\frac{di}{dt} = 50\cos(2t + 75^\circ)$$

Solution:

We transform each term in the equation from time domain to phasor domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor form of the given equation as

$$4\mathbf{I} + \frac{8\mathbf{I}}{j\omega} - 3j\omega\mathbf{I} = 50/75^{\circ}$$

But ω = 2, so

$$I(4 - j4 - j6) = 50 / 75^{\circ}$$ $$I = \frac{50/75^{\circ}}{4 - j10} = \frac{50/75^{\circ}}{10.77/-68.2^{\circ}} = 4.642/143.2^{\circ}$$

A

Converting this to the time domain,

$$i(t) = 4.642 \cos(2t + 143.2^{\circ}) \text{ A}$$

Keep in mind that this is only the steady-state solution, and it does not require knowing the initial values.

Practice Problem 9.7 Find the v oltage v(t) in a circuit described by the inte grodifferential equation

$$2\frac{dv}{dt} + 5v + 10 \int v \, dt = 50 \cos(5t - 30^{\circ})$$

using the phasor approach.

Answer:

$$v(t) = 5.3 \cos(5t - 88^\circ)$$

V.

9.4 Phasor Relationships for Circuit Elements

Now that we know how to represent a voltage or current in the phasor or frequency domain, one may le gitimately ask ho w we apply this to cir cuits involving the passive elements R, L, and C. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passi ve sign convention.

We be gin with the resistor . If the current through a resistor R is i = Im cos(ωt + ϕ), the voltage across it is given by Ohm’s law as

$$v = iR = RIm \cos(\omega t + \phi)$$

(9.29)

The phasor form of this voltage is

$$\mathbf{V} = R I_m / \underline{\phi} \tag{9.30}$$

But the phasor representation of the current is I = Im^⧸ϕ. Hence,

$$V = RI \tag{9.31}$$

showing that the v oltage-current relation for the resistor in the phasor domain continues to be Ohm’s law, as in the time domain. Figure 9.9 illustrates the voltage-current relations of a resistor. We should note from Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor diagram in Fig. 9.10.

For the inductor L, assume the current through it is i = Im cos(ωt + ϕ). The voltage across the inductor is

$$v = L\frac{di}{dt} = -\omega L I_m \sin(\omega t + \phi)$$

\n(9.32)

Recall from Eq. (9.10) that −sin A = cos(A + 90°). We can write the voltage as

$$v = \omega L I_m \cos(\omega t + \phi + 90^\circ) \tag{9.33}$$

which transforms to the phasor

$$\mathbf{V} = \omega L I_m e^{j(\phi + 90^\circ)} = \omega L I_m e^{j\phi} e^{j90^\circ} = \omega L I_m / \phi + 90^\circ \tag{9.34}$$

But Im⧸ϕ = I, and from Eq. (9.19), ej90° = j. Thus,

$$V = j\omega L I \tag{9.35}$$

showing that the v oltage has a magnitude of ωLIm and a phase of ϕ + 90°. The voltage and current are 90° out of phase. Specifically, the current lags the v oltage by 90°. Figure 9.11 sho ws the v oltage-current relations for the inductor. Figure 9.12 shows the phasor diagram.

For the capacitor C, assume the voltage across it is v = Vm cos(ωt + ϕ). The current through the capacitor is

$$i = C \frac{dv}{dt}$$

(9.36)