Complex Numbers and Phasor Technique

RaoApp-Av3.qxd 12/18/03 5:44 PM Page 779

APPENDIX A Complex Numbers and Phasor Technique

In this appendix, we discuss a mathematical technique known as the phasor technique, pertinent to operations involving sinusoidally time-varying quanti- ties.The technique simplifies the solution of a differential equation in which the steady-state response for a sinusoidally time-varying excitation is to be determined, by reducing the differential equation to an algebraic equation involving phasors. A phasor is a complex number or a complex variable. We first review complex numbers and associated operations. A complex number has two parts: a real part and an imaginary part. Imag- inary numbers are square roots of negative real numbers. To introduce the concept of an imaginary number, we define

2-1 = j (A.1a)

;j 2 =-1 (A.1b) 1 2

Thus, j5 is the positive square root of -25, -j10 is the negative square root of -100, and so on.A complex number is written in the form a + jb, where a is the real part and b is the imaginary part. Examples are 3 + j4 -4 + j1 -2 - j2 2 - j3 A complex number is represented graphically in a complex plane by using Rectangular two orthogonal axes, corresponding to the real and imaginary parts, as shown in form Fig.A.1, in which are plotted the numbers just listed. Since the set of orthogonal axes resembles the rectangular coordinate axes, the representation a + jb is 1 2 known as the rectangular form.

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780 Appendix A Complex Numbers and Phasor Technique

Imaginary (3 ϩ j4) 4

(Ϫ4 ϩ j1) 1 Ϫ2 2

Ϫ 4 0 3 Real

Ϫ2 (Ϫ2 Ϫj2) Ϫ3 Ϫ FIGURE A.1 (2 j3) Graphical representation of complex numbers in rectangular form.

Exponential An alternative form of representation of a complex number is the expo- and polar nential form Aejf, where A is the magnitude and f is the phase angle. To con- forms vert from one form to another, we first recall that

x2 x3 ex = 1 + x + + + Á (A.2) 2! 3!

Substituting x = jf, we have jf 2 jf 3 ejf = 1 + jf + 1 2 + 1 2 + Á 2! 3! 2 3 f f Á = 1 + jf - - j + 2! 3! (A.3) f2 f3 = 1 - + Á + j f - + Á a 2! b a 3! b = cos f + j sin f

This is the so-called Euler’s identity. Thus,

Aejf = A cos f + j sin f 1 2 (A.4) = A cos f + jA sin f

Now, equating the two forms of the complex numbers, we have

a + jb = A cos f + jA sin f

a = A cos f (A.5a) b = A sin f (A.5b) RaoApp-Av3.qxd 12/18/03 5:44 PM Page 781

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These expressions enable us to convert from exponential form to rectangular form. To convert from rectangular form to exponential form, we note that a2 + b2 = A2 a b b cos f = sin f = tan f = A A a Thus, A = 2a2 + b2 (A.6a)

-1 b f = tan (A.6b) a

Note that in the determination of f, the signs of cos f and sin f should be considered to see if it is necessary to add p to the angle obtained by taking the in- verse tangent of b/a. In terms of graphical representation, A is simply the distance from the origin of the complex plane to the point under consideration, and f is the angle measured counterclockwise from the positive real axis f = 0 to the line 1 2 drawn from the origin to the complex number, as shown in Fig. A.2. Since this representation is akin to the polar coordinate representation of points in two- dimensional space, the complex number is also written as Alf, the polar form. Turning now to Euler’s identity, we see that for f =;p/2, Ae;jp/2 = A cos p/2 ; jA sin p/2 =;jA. Thus, purely imaginary numbers correspond to f =;p/2. This justifies why the vertical axis, which is orthogonal to the real (horizontal) axis, is the imaginary axis. The complex numbers in rectangular form plotted in Fig. A.1 may now be Conversion converted to exponential form (or polar form): from rectangular to 2 2 j tan-1 4/3 j0.295p 3 + j4 = 23 + 4 e 1 2 = 5e = 5l53.13° exponential 2 2 j[tan-1 -1/4 +p] j0.922p or polar form -4 + j1 = 24 + 1 e 1 2 = 4.12e = 4.12l165.96° 2 2 j[tan-1 1 +p] j1.25p -2 - j2 = 22 + 2 e 1 2 = 2.83e = 2.83l225° 2 2 j tan-1 -3/2 -j0.313p 2 - j3 = 22 + 3 e 1 2 = 3.61e = 3.61l -56.31°.

These are shown plotted in Fig.A.3. It can be noted that in converting from rectangular to exponential (or polar) form, the angle f can be correctly determined if the number is first plotted in the complex plane to see in which quadrant it lies. Also note that angles traversed in the clockwise sense from the

Imaginary

b (a ϩ jb) ϭ Ae jf A FIGURE A.2 f Real Graphical representation of a complex number a in exponential form or polar form. RaoApp-Av3.qxd 12/18/03 5:44 PM Page 782

782 Appendix A Complex Numbers and Phasor Technique

Imaginary (3 ϩ j4)

5 (Ϫ4 ϩ j1) 165.96 4.12 53.13

Real 225 56.31

3.61 2.83 (Ϫ2Ϫ j2) FIGURE A.3 Polar form representation of the (2 Ϫ j3) complex numbers of Fig. A.1.

positive real axis are negative angles. Furthermore, adding or subtracting an integer multiple of 2p to the angle does not change the complex number. Arithmetic of Complex numbers are added (or subtracted) by simply adding (or sub- complex tracting) their real and imaginary parts separately as follows: numbers 3 + j4 + 2 - j3 = 5 + j1 1 2 1 2 2 - j3 - -4 + j1 = 6 - j4 1 2 1 2 Graphically, this procedure is identical to the parallelogram law of addition (or subtraction) of two vectors. Two complex numbers are multiplied by multiplying each part of one complex number by each part of the second complex number and adding the four products according to the rule of addition as follows:

2 3 + j4 2 - j3 = 6 - j9 + j8 - j 12 1 21 2 1 2 = 6 - j9 + j8 + 12 = 18 - j1

Two complex numbers whose real parts are equal but whose imaginary parts are the negative of each other are known as complex conjugates. Thus, a - jb is the complex conjugate of a + jb , and vice versa. The product of 1 2 1 2 two complex conjugates is a real number:

2 2 2 2 a + jb a - jb = a - jab + jba + b = a + b (A.7) 1 21 2 This property is used in division of one complex number by another by multiplying both the numerator and the denominator by the complex conjugate of the denominator and then performing the division by real number. For example,

3 + j4 3 + j4 2 + j3 -6 + j17 = 1 21 2 = =-0.46 + j1.31 2 - j3 2 - j3 2 + j3 13 1 21 2 RaoApp-Av3.qxd 12/18/03 5:44 PM Page 783

Appendix A 783

The exponential form is particularly useful for multiplication, division, and other operations, such as raising to the power, since the rules associated with exponential functions are applicable. Thus,

+ jf1 jf2 = j f1 f2 A e A e A A e 1 2 (A.8a) 1 1 21 2 2 1 2 jf1 A e A - 1 1 j f1 f2 = e 1 2 (A.8b) jf2 A2 e A2 Aejf n = Anejnf (A.8c) 1 2 Let us consider some numerical examples:

(a) 5ej0.295p 3.61e-j0.313p = 18.05e-j0.018p 1 21 2 5ej0.295p (b) = 1.39ej0.608p 3.61e-j0.313p (c) 2.83ej1.25p 4 = 64.14ej5p = 64.14ejp 1 2 j0.922p j 0.922p+2kp 1/2 (d) 24.12e = [4.12e 1 2] , k = 0, 1, 2, Á j 0.461p+kp = 24.12 e 1 2, k = 0, 1 = 2.03ej0.461p, or 2.03ej1.461p

Note that in evaluating the square roots, although k can assume an infinite number of integer values, only the first two need to be considered since the numbers repeat themselves for higher values of integers. Similar considerations apply for cube roots, and so on. Having reviewed complex numbers, we are now ready to discuss the pha- Phasor sor technique. The basis behind the phasor technique lies in the fact that since defined

Aejx = A cos x + jA sin x (A.9)

we can write

A cos x = Re[Aejx] (A.10)

where Re stands for “real part of.” In particular, if x = vt + u, then we have

j vt+u A cos vt + u = Re[Ae 1 2] 1 2 = Re[Aejuejvt] (A.11) = Re[Aejvt]

where A = Aeju is known as the phasor (the overbar denotes that A is complex) corresponding to A cos vt + u . Thus, the phasor corresponding to a co- 1 2 sinusoidally time-varying function is a complex number having magnitude same as the amplitude of the cosine function and phase angle equal to the phase of the cosine function for t = 0. To find the phasor corresponding to a sine function, RaoApp-Av3.qxd 12/18/03 5:44 PM Page 784

784 Appendix A Complex Numbers and Phasor Technique

we first convert it into a cosine function and proceed as in (A.11). Thus,

B sin vt + f = B cos vt + f - p/2 1 2 1 2 j vt+f-p/2 = Re[Be 1 2] (A.12) j f-p/2 jvt = Re[Be 1 2e ]

j f-p/2 jf -jp/2 Hence, the phasor corresponding to B sin vt + f is Be 1 2, or Be e , f 1 2 or -jBej . Addition of Let us now consider the addition of two sinusoidally time-varying func- two sine tions (of the same frequency), for example,5 cos vt and 10 sin vt - 30° , by 1 2 functions using the phasor technique. To do this, we proceed as follows:

5 cos vt + 10 sin vt - 30° = 5 cos vt + 10 cos vt - 120° 1 2 1 2 jvt j vt-2p/3 = Re[5e ] + Re[10e 1 2] = Re[5ej0ejvt] + Re[10e-j2p/3ejvt] = Re[5ej0ejvt + 10e-j2p/3ejvt] = Re[ 5ej0 + 10e-j2p/3 ejvt] 1 2 (A.13) = Re [ 5 + j0 + -5 - j8.66 ]ejvt 5 1 2 1 2 6 = Re[ 0 - j8.66 ejvt] 1 2 = Re[8.66e-jp/2ejvt] j vt-p/2 = Re[8.66e 1 2] = 8.66 cos vt - 90° 1 2 In practice, we need not write all the steps just shown. First, we express all functions in their cosine forms and then recognize the phasor corresponding to each function. For the foregoing example, the complex numbers 5ej0 and 10e-j2p/3 are the phasors corresponding to 5 cos vt and 10 sin vt - 30° , re- 1 2 spectively.Then we add the phasors and from the sum phasor write the required cosine function as one having the amplitude the same as the magnitude of the sum phasor and the argument equal to vt plus the phase angle of the sum phasor. Thus, the steps involved are as shown in the block diagram of Fig. A.4. Solution of We shall now discuss the solution of a differential equation for sinusoidal differential steady-state response by using the phasor technique. To do this, let us consider equation the problem of finding the steady-state solution for the current I(t) in the simple RL series circuit driven by the voltage source V t = V cos vt + f , as 1 2 m 1 2 shown in Fig. A.5. From Kirchhoff’s voltage law, we then have

dI t RI t + L 1 2 = V cos vt + f (A.14) 1 2 dt m 1 2

We know that the steady-state solution for the current must also be a cosine function of time having the same frequency as that of the voltage source, RaoApp-Av3.qxd 12/18/03 5:44 PM Page 785

Appendix A 785

5 cos vt ϩ 10 sin (vt Ϫ 30 )

5 cos vt ϩ 10 cos (vt Ϫ 120 )

5e j0 ϩ 10e Ϫj2p/3 (Phasors)

ϭ 8.666e Ϫjp/2 (Sum Phasor) FIGURE A.4 Block diagram of steps involved in the application of the phasor technique to 8.66 cos (vt Ϫ 90 ) the addition of two sinusoidally time-varying functions.

R I(t)

ϩ V(t) Ϫ L

FIGURE A.5 RL series circuit.

but not necessarily in phase with it. Hence, let us assume

I t = I cos vt + u (A.15) 1 2 m 1 2

The problem now consists of finding Im and u. Using the phasor concept, we write

+ = j vt+f V cos vt f Re[V e 1 2] m 1 2 m = jf jvt Re[Vm e e ] (A.16a) = Re[Vejvt]

+ = j vt+u I cos vt u Re[I e 1 2] m 1 2 m = ju jvt Re[Im e e ] (A.16b) = Re[Iejvt]

= jf – = ju = where V V e and I I e are the phasors corresponding to V t m m 1 2 V cos vt + f and I t = I cos vt + u , respectively. Substituting these m 1 2 1 2 m 1 2 into the differential equation, we have

– jvt d – jvt jvt R Re[Ie ] + L Re[Ie ] = Re[Ve ] (A.17) 5 6 dt5 6 RaoApp-Av3.qxd 12/18/03 5:44 PM Page 786

786 Appendix A Complex Numbers and Phasor Technique

Since R and L are constants, and since d/dt and Re can be interchanged, we can simplify this equation in accordance with the following steps:

– jvt d – jvt jvt Re[RIe ] + Re L Ie = Re[Ve ] c dt 1 2d – – Re[RIejvt] + Re[jvLIejvt] = Re[Vejvt] (A.18) – – Re[ RI + jvLI ejvt] = Re[Vejvt] 1 2 Let us now consider two values of vt, say,vt = 0 and vt = p/2. For vt = 0, we obtain – – Re RI + jvLI = Re V (A.19) 1 2 1 2 For vt = p/2, we obtain – – Re[j RI + jvLI ] = Re[jV] 1 2 or – – Im RI + jvLI = Im V (A.20) 1 2 1 2 where Im stands for “imaginary part of.” Now, since the real parts as well as the – – imaginary parts of RI + jvLI and V are equal, it follows that the two com- 1 2 plex numbers are equal. Thus,

– – RI + jvLI = V (A.21)

– By solving this equation, we obtain I and hence Im and u. Note that by using the phasor technique, we have reduced the problem of solving the differential equation (A.14) to one of solving the phasor (algebraic) equation (A.21). In fact, the phasor equation can be written directly from the differential equation without the necessity of the intermediate steps, by recognizing that the time functions – I(t) and V(t) are replaced by their phasors I and V, respectively, and d/dt is replaced by jv. We have here included the intermediate steps merely to illustrate the basis behind the phasor technique. We shall now consider an example.

Example A.1 Solution of differential equation using phasor technique For the circuit of Fig. A.5, let us assume that R = 1 Æ, L = 10-3 H, and V t = 1 2 10 cos 1000t + 30° V and obtain the steady-state solution for I(t). 1 2 The differential equation for I(t) is given by

-3 dI I + 10 = 10 cos 1000t + 30° dt 1 2 – Replacing the current and voltage by their phasors I and 10ejp/6, respectively, and d/dt by jv = j1000, we obtain the phasor equation

– - – I + 10 3 j1000I = 10ejp/6 1 2 RaoApp-Av3.qxd 12/18/03 5:44 PM Page 787

Appendix A 787

or – I 1 + j1 = 10ejp/6 1 2 jp/6 jp/6 – = 10e = 10e I + 1 j1 22 ejp/4 = 7.07e-jp/12 – Having determined the value of I, we now find the required solution to be – I t = Re[Iejvt] 1 2 = Re[7.07e-jp/12ej1000t] = 7.07 cos 1000t - 15° A 1 2