ECE 451 Homework 1 Solutions

1. Given the lossless transmission line shown above,

(a) Find the reflection coefficient R and the standing wave ratio (SWR) for this

loaded transmission line.

(b) Sketch the standing wave patterns for the magnitude of the voltage along this transmission line in terms of VR.

(c) Determine the impedance at the input of this loaded transmission line.

(d) If a sinusoidal generator 10 o V, which has a source impedance of 100 is

connected to this loaded transmission line, what is the time average power

delivered to the 75- load.

(a) R oR

R o

Z Z 75 500.2

Z Z 75 50

1 0.2SWR 1.5

1 0.2

(b)

(c) 22

oin

R

Z 50 2500Z 33.33

Z 75 75

VR

+

-

IR

l = 3/4

Zo = 50 ZR = 75

(d)

2

s in*in in 2 2

s in

V Z1 100 33.33P Re V I

2 2 Z Z 2 100 33.33

P 0.0937 watts

2. A transmission line of length 0.625 and characteristic impedance 50 is connected to an

unknown load impedance ZR. The input admittance is measured to be Yin= 0.004 - j0.008 -1.

Using the Smith chart find:

(a) the normalized and the actual load admittance

(b) the normalized and the actual load impedance

(c) the load reflection coefficient

(d) the voltage standing wave ratio (VSWR)

(e) the distance, in wavelengths, of the first voltage maximum and minimum, with

reference to the load.

Yin = 0.004 –j 0.008 mhos

Yo = 0.02 mhos

yin = 0.2 –j 0.4

(a)

yR =1 – j2

YR=0.02 – j0.04 mhos

(b) zR = 0.2 + j0.4

ZR=10 +j20

(c) R 0.7 135

(d) VSWR = 5.66

(e) first maximum is 0.188 from load

first minimum is 0.438from load

3. Consider the lossless transmission line network shown below. The operating frequency is

2000 MHz and the propagation velocity on the transmission line is 0.3 m/ns.

(a) Using the Smith chart, determine the SWR on the section of line of length l.

9

0.315

2 10o

o

vcm

f

15 4015 40 0.3 0.8

50R

jj z j

From Smith chart, VSWR=5.7

SWR = 5.7

(b) Using the Smith chart find two values for the length l such that Z(l) is equal to Zo ± jX.

l1=0.186-0.112=0.074=0.07415=1.11 cm

l2=0.313-0.112=0.201=0.20115=3.015 cm

Z(l1) = Zo + jX, l1 =1.11cm

Z(l2) = Zo - jX, l2 =3.015cm

(c) Determine the value of series inductance L and the proper length of the transmission line section (l1 or l2) that insures Zin = Zo

Impedance is capacitive at l2 with z=1-j1.9 or Z=50+j95 . Can be compensated with inductor

such that

9

957.55

2 2 10L nH

L = 7.55_ (nanoHenries)

4. Answer the following questions using the attached Smith chart. Clearly identify significant

features on the chart. A transmission line with characteristic impedance 50 is terminated by a

load of impedance Zr = 40 + j50 .

(a) What is the SWR? (5 pts)Solution: The normalized zr is 0.8 + j 1.0. Find this

point on the Smith Chart, draw the constant SWR circle, read off the SWR=3.0

(b) What is the phase of r? (5 pts)72 degrees, from the Smith Chart.

(c) What is the normalized admittance at the load? (5 pts) 0.5-j0.6, from the Smith

Chart.

(d) What is the normalized admittance at d = 12.2 toward the generator from the

load? (5 pts) 0.5+j0.6, from the Smith Chart.

(e) What is the phase of at d = 12.2 toward the generator from the load? (10 pts) -

72 degrees, from the Smith Chart. Be sure to read the phase of Gamma from the

impedance side of the constant SWR circle.

(f) What is the shortest distance from the load at which a short-circuited stub could

be attached to achieve an impedance match? (10 pts) 0.265 wavelengths, from the

Smith Chart.

(g) What would the normalized input admittance of the stub be? (10 pts) -j1.1

5. For the above figure, let V s = 1 volt, Zs = 30 , and ZR = 40 .

(a) What is the impedance at the input of the line (z = -d) ? (10 pts)

2 25062 5

40o

inR

Z ( )Z .

Z

(b) What is the phasor current through Zs? (10 pts)

1 10 01081

30 62 5inS in

I . AZ Z .

(c) What is the phasor current through ZR ? (20 pts)

Ts=5/8, s=-2/8, R= -1/9, ,

V+ = -j0.608 V, 0 6081 0 111

50

j ..

IR=-j13.51 mA

(d) What is the time-average power delivered to the load ? (10 pts)

62 5 0 01081 0 01081*in in

1 1P= Re V I Re . . .

2 2

P=3.65 mW

/4

Zo = 50

Zs

Vs

z = 0z = -d

ZR