Smith Chart Problems
Problem 1A 30 long lossless transmission line with Z0 = 50 ohm operating at 2 MHz is terminated with a load
ZL=60+j40 ohm. If u=0.6 c on the line, find
a) Refection coefficient
b) The standing wave ratio
c) The input impedance
Using without Smith chart and with Smith chart
Barun Gupta
2/16/2018
Smith Chart : Answer without using Smith Chart
Problem 1A 30 long lossless transmission line with Z0 = 50
ohm operating at 2 MHz is terminated with a load
ZL=60+j40 ohm. If u=0.6 c on the line, find
a) Refection coefficient
b) The standing wave ratio
c) The input impedance
Using without Smith chart and with Smith chart
Solution using Smith Chart : Load impedance
Problem 1A 30 long lossless transmission line with Z0 = 50
ohm operating at 2 MHz is terminated with a load
ZL=60+j40 ohm. If u=0.6 c on the line, find
a) Refection coefficient
b) The standing wave ratio
c) The input impedance
Using without Smith chart and with Smith chart
The normalized load impedance is
zL = (60+j40)/50 = 1.2 +j0.8
Smith Chart :Reflection and Phase
Angle = 56 degree
Q
Reflection coefficient = OQ/OP
= 0.351
P
O
Smith Chart : SWR
To find s,
Draw a circle of radius OP with center O.
 The circle is a constant reflection Circle.
The intersection of circle with horizontal
axis give SWR value. SWR=2.1 (point S)
S
Smith Chart: Input Impedance
Convert the length of the transmission line into degree or wavelength
L= 30 m
λ = velocity/ frequency = 0.6*3e8/3e6=90m
L=30 m= λ /3
zin
In Smith Chart, 360 degree corresponds to 0.5 λ.
So one λ corresponds to 720 degree.
Thus, L=720/3=240 degree.
Point P is at 56 degree. Rotate point P by 240 degree in clockwise
direction. Zin on the s circle in the required normalized input impedance
zin is the normalized load impedance.
The value from Smith Chart is 0.47+ j0.035
Hence Zin=50*(0.47+j0.335)= 23.5+j1.75
Lumped-Element Matching
Smith Chart :Single Stub Matching Problem
Antenna with load impedance ZL = 25-j50 ohm is to be attached to a 50 ohm transmission line
with a short stub.
Determine
a) The required stub admittance
b) The distance between the stub and the antenna.
c) The sub length
d
Feedline
Y0
Yin
Yd
YL
Ys
l
Y0
Shorted stub
Smith Chart :Single Stub Matching
Admittance calculation
Load yL
0.115 λ
B
The first step is to insert load impedance onto the Smith Chart
The normalized load impedance is
zL =ZL/Z0= (25-j50)/50=0.5-j1
(Represented by point A)
gL = 1 circle
O
The next step is to find normalized admittance
(draw s circle with radius OA)
yL = 0.4 +j0.8
Load zL A
Or find it by yL=1/zL= 0.4+j0.8
In admittance domain, rL circle corresponds to gL circle
and xL circle corresponds to bL circle.
Smith Chart :Single Stub Matching
Finding the location of the TL from the load
Load yL
0.063 λ difference between B and C
0.115 λ
0.178 λ
B
The intersection of gL=1 circle with the SWR circle are C
and D
C
At C
yd = 1+j1.58 and is located at 0.178 λ.
The distance between points B and C is
(0.178-0.115) = 0.063 λ.
D
Load zL A
F -0.178 λ
Now
yin = ys + yd. The requirement is yin = 1+j0.
So
1+j0=ys+1+j1.58.
Solving this yields
ys=-j1.58
Smith Chart :Single Stub Matching
Finding the length of the TL
Load yL
0.063 λ difference between B and C
Angle= 0.115 λ
Angle= 0.178 λ
B
C
Point E represents short circuit admittance.
And the position is 0.25 λ.
At C, yd = 1+j1.58
Point F has the admittance of –j1.58 and is
located at 0.34 λ.
E
Thus the length of the TL is the difference of
length between point E and F.
0.34-0.25 = 0.09 λ
D
Load zL A
F
Angle= 0.34 λ
At F, ys = -j1.58
Smith Chart :Single Stub Matching. ??
Please attempt to find the other location and length of the transmission line corresponding to the
point D.
Answer – The other location is d=0.207 λ and the admittance ys = j1.58
The length of the transmission line corresponding to the point D is l2 = (0.25+0.16) = 0.41 λ.
Pay attention to ys and yd