Complex Numbers Review for EE-201 (Hadi Saadat) in Our Numbering System, Positive Numbers Correspond to Distance Measured Along a Horizontal Line to the Right

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Complex Numbers Review for EE-201 (Hadi saadat) In our numbering system, positive numbers correspond to distance measured along a horizontal line to the right. Negative numbers are represented to the left of origin. Numbers corresponding to distances along the line shown in Figure 1 are called real numbers and the horizontal axis is known as the real axis.

−6−5−4−3−2−1 0121212

Figure 1. The system of real numbers.

Å Consider a point c in a two-dimensional plane located a distance along a line at an angle , taken counterclockwise from the positive horizontal axis.

↑ b c(a, b)

|M|

θ →

a Ý

Figure 2. Graphical representation of a point in the Ü- plane (known as complex plane). a

The projection of c along the horizontal axis or the so called real axis is shown by . This component is known

b c

as the real component of c. The projection on the vertical axis is shown by . Thus, we may show by its two

´a; bµ coordinates as c . To differentiate between the two components, it is unfortunately common practice to call the

vertical component of c the imaginary component. In this context, the vertical axis is known as the imaginary axis.

Operator j Ý Consider a circle of radius unity with center at the origin of the Ü- plane as shown in Figure 3.

↑ 1∠90 =j

1∠180 θ 1∠0 → j2 =−1 j4=1

1∠270=j3=−j

Figure 3. Graphical representation of the operator j.

6 ¼

The radius along the positive real axis may be written as ½ . If this line of unity length is rotated counterclockwise

Æ Æ

½ 9¼ j by 9¼ , it will be located along the vertical axis, i.e., . This operation is represented by the operator ,andwe

write

j =½ 9¼

Æ 9¼ This means that if any real number is multiplied by j it is rotated by in a counterclockwise direction.

Rotating through another 9¼,wehave

Æ Æ Æ

6 6 6

´j µ´j µ=´½ 9¼ µ´½ 9¼ µ=½ ½8¼ = ½

i.e.,

j = ½

Taking the square root, we may write

j = ½

= ½ The equation j is a statement that the two operations are equivalent. However, and that is the beauty of the

concept, in all algebraic computation imaginary numbers can be handled as if j had a numerical value.

Æ Æ ¾7¼

Continuing the rotation by another 9¼ , i.e., a total of , we write

¿ ¾ Æ Æ

6 6

j =´j µ´j µ= j =½ ¾7¼ =½ 9¼

and

4 ¾ ¾ Æ Æ

6 6

j =´j µ´j µ=´ ½µ´ ½µ = ½ ¿6¼ =½ ¼

We can write the reciprocal operation ½ as

½ ´j µ j j

= = = j =½ 9¼ =

j ´j µ´j µ j ½

Rectangular and Polar Forms

c´a; bµ

With the introduction of the operator j , the point in Figure 2 may be represented as

c = a · jb

This representation indicates that the real part, a, is measured along the real axis (the abscissa) and the so called imag-

inary part, b, is reckoned along the imaginary axis (the ordinate). This representation is known as the rectangular form

of a complex number c.

= a · jb jÅ j The complex number c may also be represented as the length or magnitude of a line segment, ,atan

angle, , as indicated in Figure 2. Thus,

c = a · jb = jÅ j

= jÅ j jÅ j c The form c is called the polar form, is the magnitude and is called the angle or the argument of . The conversion from rectangular to polar form can be deduced immediately from Figure 2 in conjunction with the

Pythagorean theorem, i.e.,

¾ ¾

jÅ j = a · b

and the angle is given by

= ØaÒ a

For conversion from polar to rectangular from,

a = jÅ j cÓ×

b = jÅ j ×iÒ

Thus, c can be written as

c = jÅ j´cÓ× · j ×iÒ µ 3

The Euler’s identity – Exponential Form

Å j

Consider a complex number with the Magnitude j =1,

c =½ =cÓ× · j ×iÒ

Taking derivative of c with respect to ,resultin

= ×iÒ · j cÓ× = j ´cÓ× · j ×iÒ µ d

= jc d

Separating the variable,

dc = jd c

Integrating the above equation, we get

ÐÒ c = j · Ã

c = ¼

where Ã is the constant of integration. The magnitude of is unity regardless of the angle, therefore, at ,

j ´¼µ · Ã Ã =¼

ÐÒ´½µ = ,or ,and

ÐÒ c = j

c = e

c = cÓ× · j ×iÒ

With c given by the equation ,theexponential form also known as the Euler’s identity is

e =cÓ× · j ×iÒ

for an angle , we obtain

e =cÓ× j ×iÒ

×iÒ

Adding and subtracting the above two equations lead to the representation for cÓ× and ,

j j

e · e

cÓ× and

j j

e e

×iÒ =

¾j With the addition of the Euler’s identity, the three way of representing a complex number, rectangular, polar and

exponential forms ar all equivalent and we may write

c = a · jb = jÅ j = jÅ je

Mathematical Operations

= a · jb c

The conjugate of a complex number c denoted by and is deﬁned as

c = a jb

= jÅ j

or in polar form for c ,wehave

c = jÅ j 4

To add or subtract two complex numbers, we add (or subtract) their real parts and their imaginary parts. For two

c = a · jb c = a · jb

½ ½ ¾ ¾ ¾

complex numbers designated by ½ and ,theirsumis

c · c =´a · a µ·j ´b · b µ

½ ¾ ½ ¾ ½ ¾

c c j = ½ ¾

The multiplication of ½ and in rectangular form is obtained as follows (note )

c c =á · jb µá · jb µ=a a b b · j á b · b a µ

½ ¾ ½ ½ ¾ ¾ ½ ¾ ½ ¾ ½ ¾ ½ ¾

6 6

c = jÅ j c = jÅ j c c

½ ½ ¾ ¾ ¾ ½ ¾

or in polar form for ½ ,and , The multiplication of and is

6 6

c c = ´jÅ j µ´jÅ j µ

½ ¾ ½ ½ ¾ ¾

j j

½ ¾

= ´jÅ je µ´jÅ je µ

½ ¾

j ´ · µ

½ ¾

= jÅ jjÅ je

½ ¾

= jÅ jjÅ j ·

½ ¾ ½ ¾

= a · jb c = a jb

if a complex number c is multiplied by its conjugate , the result is

£ ¾ ¾ ¾

cc =´a · jbµ´a jbµ=a · b = jÅ j

or in polar form

£ ¾

6 6

cc = jÅ j jÅ j = jÅ j

c c ¾ For the division of ½ by in rectangular form, we multiply numerator and denominator by the conjugate of the

denominator, this results in

c ´a · jb µ

½ ½ ½

c ´a · jb µ

¾ ¾ ¾

´a · jb µ´a jb µ

½ ½ ¾ ¾

´a · jb µ´a jb µ

¾ ¾ ¾ ¾

a a · b b b a a b

½ ¾ ½ ¾ ½ ¾ ½ ¾

= · j

¾ ¾

a · b a · b

¾ ¾ ¾ ¾

6 6

c = jÅ j c = jÅ j c c

½ ½ ¾ ¾ ¾ ½ ¾

or in polar form for ½ ,and , the division of by is

c jÅ j

½ ½ ½

c jÅ j

¾ ¾ ¾

jÅ je

jÅ j

j ´ µ

½ ¾

= e

jÅ j

½ ¾

jÅ j ¾

Example 1

For the complex numbers

Æ Æ

6 6 6 6

c =¾¼ ¿6:87 ; c =4¼ 5¿:½¿ ; c =4¼·j 8¼; c =½¾ ; c =5 ; c =¿¼·j 4¼

¾ ¿ 4 5 6

½ and

6 6

c =´c · c · c µ=´c £ c c µ

¾ ¿ 4 5 6

Find ½

c c ¾

Substituting for the values and converting ½ and to rectangular form, we have

Æ Æ

6 6

¾¼ ¿6:87 ·4¼ 5¿:½¿ ·4¼·j 8¼

c =

6 6

´½¾ µ´5 µ ´¿¼ · j 4¼µ

6 6

´½6 · j ½¾µ · ´¾4 j ¿¾µ · ´4¼ · j 8¼µ 8¼ · j 6¼

= =

´6¼ · j ¼µ ´¿¼ · j 4¼µ ¿¼ j 4¼

½¼¼ ¿6:87

= =¾ 9¼ = j ¾

5¼ 5¿:½¿ 5

Use MATLAB To evaluate the complex number c described in Example 1. In MATLAB, we use the following commands: c1 = 20*exp(j*36.87*pi/180); % In MATLAB angles must be in radian c2 = 40*exp(-j*53.13*pi/180); c3 = 40 + j*80; c4 = 12*exp(j*pi/6); c5 = 5*exp(-j*pi/6); c6 = 30 + j*40; c = (c1+c2+c3)/(c4*c5-c6) M=abs(c) % Magnitude theta = angle(c)*180/pi % Angle in degree

Save in a ﬁle with extension m, and run to get the result c= 0.0000 + 2.0000i M= 2.0000 theta = 90.0000

Example 2

A complex function is described by

¾5¼¼´j!µ

g =

´¾5 · j!µ´½¼¼ · j!µ ! Write an script m-ﬁle to evaluate the magnitude and phase angle of g for from 0 to 200 in step of 1, and obtain the

magnitude and phase angle plots versus ! . We use the following statements: w=0:1:200; g= (2500*j*w)./((25+j*w).*(100+j*w)); %Array Multiplication & division use .* & ./ M=abs(g); % Magnitude theta = angle(g)*180/pi; % Angle in degree subplot(2,1,1), plot(w, M), grid ylabel(’M’), xlabel(’\omega’) subplot(2,1,2), plot(w, theta), grid ylabel(’\theta, degree’), xlabel(’\omega’)

The result is 6

M 10

0 0 50 100 150 200 ω 100

0 , degree θ −50

−100 0 50 100 150 200 ω Figure 4. Magnitude and phase angle plots for complex function in Example 2.

Homework Problems

! = ½½ ! = 5¼ ! = ½½¾ 1. Using your calculator evaluate g for the function in Example 2 for (a) ,(b) ,and(c) . Express your answers both in rectangular and polar forms.

2. A complex function is described by

½¼¼¼¼

g =

´½¼ · j!µ´½¼¼ ! ·½¼´j!µµ ! Write an script m-ﬁle to evaluate the magnitude and phase angle of g for from 0 to 50 in step of 1, and obtain the

magnitude and phase angle plots versus ! .

! =½¼ ! =½4:½4¾ ! =¾½:5 3. Using your calculator evaluate g for the function in Problem 2 for (a) ,(b) ,and(c) . Express your answers both in rectangular and polar forms.

Prepared for EE-201 by H. Saadat, October 17, 99