Sanzhar Askaruly Nazarbayev University, School of Engineering

Homework 2

Exercise 2.6

Results from Formula:

Solutions from CD Module:

Figure 1. Impedance vs. Radius b

Results from Formula: R = 0.788 Ohms/m; L = 139nH/m; G = 9.1 ms/m; C = 362 pF/m

Solutions from CD Module: R = 0.777 Ohms/m; L = 139nH/m; G = 9.1 ms/m; C = 361 pF/m

Results obtained are almost the same, with maximum of 1.42% relative error (R)

Sanzhar Askaruly Nazarbayev University, School of Engineering

Exercise 2.8

Results from Formula:

Solutions from CD Module:

Figure 2. Impedance vs. Radius a

Results from Formula: α = 0.109 Np/m; β = 44.5 rad/m; up = 1.41 108 m/s; Z0 = (19.6 + j0.03) Ohms

Solutions from CD Module: α = 0.109 Np/m; β = 44.4 rad/m; up = λ

f = 1.41 108 m/s; Z0 = (19.6 +

j0.03) Ohms

Results obtained are almost the same, with maximum of 0.23% relative error (β)

Sanzhar Askaruly Nazarbayev University, School of Engineering

Exercise 2.10

Figure 3. 100 Ohms Transmission Line has width of 1.542 mm

Exercise 2.12

MATLAB code for a plot of Z0 as a function of strip width w:

x = [0.05, 0.25, 0.55, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5];

y = [111.0622, 73.3852, 56.5466, 40.3344, 31.7293, 26.2072, 22.3446, 19.4841, 17.2779, 15.5235,

14.0948, 12.9088];

figure

hold on

grid on

title('Impedance as a function of strip width w for a microstrip line'); % title

ylabel('Zo, in Ohms'); % label for y axis

xlabel('Width, in mm'); % label for x axis

plot(x,y);

Z0 as a function of strip width w

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

20

40

60

80

100

120Impedance as a function of strip width w for a microstrip line

Zo,

in O

hm

s

Width, in mm

Sanzhar Askaruly Nazarbayev University, School of Engineering

Exercise 2.33

Exercise 2.43

Sanzhar Askaruly Nazarbayev University, School of Engineering

Exercise 2.47

a) ZL = 3Z0;

Analytic Solution:

CD Module Smith Chart:

Sanzhar Askaruly Nazarbayev University, School of Engineering

b) ZL = (2 – j2)Z0;

Analytic Solution:

CD Module Smith Chart:

c) ZL = j2Z0;

Analytic Solution:

Sanzhar Askaruly Nazarbayev University, School of Engineering

CD Module Smith Chart:

d) ZL = 0 (short circuit)

CD Module Smith Chart:

Sanzhar Askaruly Nazarbayev University, School of Engineering

Exercise 2.72

ZL = (50 + j50)Z0; zL = 1 + j1; d(max) = 0.0881 λ

ZL = (50 - j50)Z0; zL = 1 - j1; d(max) = 0.4119 λ

Exercise 2.50

Exercise 2.55