ECE 3025 Problem Set #1 Solutions

1. A sinusoidal voltage wave at radian frequency ! propagates in the negative z direction on a lossytransmission line. The wave has real amplitude, V0. The line has attenuation coe�cient, ↵, andphase constant, �.

a. Write down the voltage function for this wave in phasor form.

Vs(z) = V0 e+↵ze+j�z

b. Express the wave in real instantaneous form.

V(z, t) = Re�Vs(z) ej!t

= V0 e+↵z cos(!t + �z)

c. The parameters of the line, operating at ! = 6.0 ⇥ 108 rad/s are L = 0.50µH/m, C = 50.0pF/m, G = 45.0µS/m and R = 8.0 ⌦/m. Find ↵, �, �, and Z0 (in magnitude and phase).

We use

� =p

ZY =p

(R + j!L)(G + j!C)

=p

[8.0 + j(6⇥ 108)(0.50⇥ 10�6)][4.5⇥ 10�5 + j(6⇥ 108)(50⇥ 10�12)]

= 0.042 + j3.0 m�1 = ↵ + j�

Therefore, ↵ = 0.042 Np/m, � = 3.0 rad/m, and � = 2⇡/� = 2.09 m. Finally,

Z0 =r

Z

Y=

sR + j!L

G + j!C=

s8.0 + j3.00⇥ 102

(4.5⇥ 10�5) + j3.00⇥ 10�2= 100.0� j1.26 ⌦ = 100.0 e�j0.013

d. Write down the phasor and real instantaneous expressions for the current in the line, using yourresults of part c.

As the voltage wave travels in the backward z direction, the phasor current will be

Is(z) = �Vs(z)Z0

= � V0

100.0e+j0.013 e+0.042z e+j3.0z

The real instantaneous current will be

I(z, t) = Re�Is(z)ej!t

= � V0

100.0e+0.042z cos(6⇥ 108t + 3.00z + 0.013)

e. Calculate the phase velocity of the wave, and find an expression for the time-average power inthe line, as a function of distance, z.

The phase velocity is vp = !/� = 6⇥ 108/3.00 = 2.00⇥ 108 m/s. The power is

< P >=12Re {VsI

⇤s } =

V 20

200.0e+0.084z W

where the power decrease with decreasing z, as the wave propagates in the backward z direction.

f. Find the distance from z = 0 over which the wave power is reduced to one-half its initial value:We set e+0.084z = 0.5 ) z = (1/0.084) ln(0.5) = �8.25m.

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2. A sinusoidal wave on a transmission line is specified by voltage and current in real instantaneousform:

V(z, t) = V0 e�↵z cos(!t� �z) and I(z, t) = I0 e�↵z cos(!t� �z + �)

a) In which direction does this wave propagate and why? Propagation is in the forward z direction,because of the minus sign in the argument of cos(!t� �z).

b) It is found that ↵ = 0, Z0 = 50⌦, and the wave velocity is vp = 2.5 ⇥ 108 m/s, with f = 100MHz. Evaluate R, G, L, C, �, and �: First, the fact that ↵ = 0 means that the line is lossless,from which we immediately conclude that R = G = 0. As this is true it follows that Z0 =

pL/C

and vp = 1/p

LC, from which

C =1

Z0vp=

150(2.5⇥ 108)

= 8.0⇥ 10�11 F = 80 pF

ThenL = CZ2

0 = (8.0⇥ 10�11)(50)2 = 2.0⇥ 10�7 = 0.20 µHNow,

� =vp

f=

2.5⇥ 108

1.00⇥ 108= 2.5 m

Finally, the current phase is found through

I0ej� =

V0

Z0

Since V0, I0, and Z0 are all real, it follows that � = 0.

3. Two voltage waves of equal amplitude, V0, and frequency, !, propagate in opposite directions in alossless transmission line of characteristic impedance, Z0 One wave is phase-shifted from the otherby � radians.

a) Find an expression for the net voltage wave formed by the superposition of the given voltages.To do this, express both given waves in phasor form. Combine the two algebraically, and convertthe result back to real instantaneous form, to produce a single wave function. The total phasorvoltage will be:

VsT (z) = V0e�j�z + V0e

+j�zej�

= V0ej�/2

he�j�ze�j�/2 + ej�zej�/2

i= 2V0e

j�/2 cos(�z + �/2)In real instantaneous form, this becomes:

V(z, t) = Re�VsT (z)ej!t

= Re

n2V0e

j�/2 cos(�z + �/2)ej!to

= 2V0 cos(�z + �/2) cos(!t + �/2) V

b) Find an expression for the net current in the line: Using the phasor voltage as found in part a,and noting that Z0 is real (lossless line):

Is(z) =V0

Z0ej�/2

he�j�ze�j�/2 � ej�zej�/2

i=�j2V0

Z0sin(�z + �/2)ej�/2

ThenI(z, t) = Re

�Is ej!t

=

2V0

Z0sin(�z + �/2) sin(!t + �/2) A

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