Notes 3 - Smith chart examples

Report
ECE 5317-6351
Microwave Engineering
Fall 2011
Prof. David R. Jackson
Dept. of ECE
Notes 3
Smith Chart
Examples
1
Example 1
I(-d)
Z0  50 
+
Z0 , 
Z L  100  j50 
-
ZL
V(-d)
d
Find Z(-d)
d  3 / 8g
at d / g  1/ 4, 3/ 8, 1/ 2
Z L ,n 
a
d
ZL
 2  j1
Z0
0
o
/
r1
2 g
Impedance chart
b
Z Ln
d  g / 4
Z n  0.4  j 0.2
a
 Z  g / 4   20  j10 
d  g / 4
2
Example 1 (cont.)
b
I(-d)
d  3 / 8g
Z n  0.5  j 0.5
Z0 , 
 Z  3 / 8g   25  j 25 
+
-
ZL
V(-d)
d
d  3 / 8g
c
d  g / 2
d
0.087g
0.212g
Z n  2  j1
 Z  g / 2   100  j50 
b
0
 g
Z Ln
c
0.5g
a
d  g / 4
0.462g
3
/2
Example 2
Z 0  50  Y0  20 mS
I(-d)
YL  8mS  j 4 mS
+
Z0 , 
-
d
Find Y(-d)
at d / g  1/ 4, 3/ 8, 1/ 2
YL,n 
ZL
V(-d)
d  3 / 8g
0
d  1 / 2 g
r
YL
 0.4  j 0.2
Y0
o
Admittance chart
b
a
YLn
c
d  g / 4
a
Yn  2  j1
 Y  g / 4   40 mS  j 20 mS
d  g / 4
4
Example 2 (cont.)
I(-d)
b
d  3 / 8g
+
Z0 , 
Yn  1  j1
 Y  3 / 8g   20 mS  j 20 mS
-
ZL
V(-d)
d
d  3 / 8g
c
d
d  g / 2
0
/
1
2 g
Admittance chart
Yn  0.4  j 0.2
 Y  g / 2   8mS  j 4 mS
b
YLn
c
a
d  g / 4
5
Which Chart to Use?
Simple answer:
* When adding elements in series use Z-chart
ZA
ZB

ZC
Z A  Z B  ZC
* When adding elements in parallel use Y-chart
YA
YB

YC
YA  YB  YC
6
Using Reactive Loads
 Use reactively loaded section of
transmission line.
most common to use open or short load
 X L  
Impedance chart
 X L  0
Z  d 
SC
OC
X Ln
d
X n  d 
Z0 , 
ZL
d
Z L  jX L
d=0
7
Using Reactive Loads (cont.)
Y  d 
Z0 , 
SC
YL
d
OC
YL  j  L
Y Ln
d
Yn  d 
Admittance chart
d=0
Y  d   Yn  d  Y0
 Yn  d  / Z 0
8
Example 3
Use a short-circuited section of air-filled TEM, 50  transmission line
( = k0, g =0) to create an impedance of Zin = -j25  at f = 10 GHz.
Z in ,n   j
Impedance chart
50
25
  j 1/ 2 
50
Zin  j25 
50 Ω , k0
SC
0g
L
-1/2
0 .4
SC
L  0.426g  0g  0.426g
L = 1.28 cm
g
26
0 
c 2
2


f k0  0 0
0  3.0 cm
9
Example 4
Use an open circuited section of 75  (Y0 = 1/75 S) air-filled
transmission line at f = 10 GHz to create an admittance of
Yin   j
1
S  j13.3mS
75
Admittance chart
Yin,n  j 1
1/75 S
Yin   j 1/ 75 
75 Ω , k0
OC
L
OC
Y0 
1

75
j1
L  0.125 0
0  3.0 cm
L
 L  0.375cm
10
Matching Circuit
Z in , Yin
Y0 
1
Z0
d
Z0
ZL
YS
Admittance chart
YA
Y0
Want to pick d and Ys such that Yin = Y0.
Gn 1
 Zin  Z0 
YL ,n
Z L,n 
1
jb
n
YA,n
ZL
Z0
 YL,n 
YL
1

Y0 Z L,n
Note: At d we have
d
Yin,n  YA,n  YS ,n
11
Matching Circuit (cont.)
Z in , Yin
d
We want
Yin,n  1  1  jbn  YS ,n
Z0
ZL
YS
YA
Choose
YS ,n   jbn
Y0
Gn 1
 Yin ,n  1  jbn   jbn  1
YL ,n
 Yin  Y0
1
 Z0
Yin
YA,n
jb
n
d
1
 Z in 
Admittance chart
12
Example 5
Z0  50[]
ZL  100  j100 []
d
Z0
In this example we will
use the “usual” Smith
chart, but as an
admittance calculator.
L 
Z L  Z0 Z Ln  1

Z L  Z0 Z Ln  1
ZL
Z L, n  2  j 2
Z0s
ls
YL ,n 
1
 0.25  j .25 
2  j2
L  0.62 e j /6  0.62 30o
13
Example 5 (cont.)
0.041  0.178  0.219g
0.041  0.322  0.363g
Solution :
0.178g
Add - j1.57 at d  0.219g
or
 j1.57 at d  0.363g
Admittance calculator
1  j1.57
(We’ll use the first choice.)
 plane
0.219g
SC
OC
0.363g
0.041g
1  j1.57
YL, n  0.25  j 0.25
Smith chart scale:
wavelengths toward load
wavelengths toward generator
0.322g
14
Example 5 (cont.)
From the Smith chart:
ls  0.09g
Admittance calculator
Analytically:
Bs , n   cot   l 
1.57   cot  l
cot  l  1.57; tan  l 
l 
2

S/ C
O/ C
1
1.57
l  0.567 [radians]
0.09
0  j1.57
ls  0.0903g
15
Quarter-Wave Transformer
g / 4
 Z  jZ 0T tan  
Z in  Z 0T  L

Z

jZ
tan

L
 0T

At f0 :
2 g 
    g / 4  

g 4 2
Z0T
Zin, in
ZL
ZL is real
For in = 0
Z 02T
 Zin 
ZL
Z 0T   Zin Z L 
Z0
Z in  Z 0
1
2
 Z 0T   Z 0 Z L 
1
2
16
Example 6
Match 100  load to 50  transmission Line at f0.
g / 4
, 50 Ω
Z0T
Z 0T  100  50
 70.7
g
f0

2

f0
0
2
2



k k0  r
r
100 Ω
 r  1
Z0T  70.7 
17

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