#### Transcript notes 13 3317

```ECE 3317
Prof. Ji Chen
Spring 2014
Notes 13
Transmission Lines
(Impedance Matching)
d
Z0
ZL
Z0s
ls
1
Smith Chart
Impedance matching is very important to avoid reflected power, which causes a
loss of efficiency and interference.
Z L  Z0
L 
Z L  Z0
Zg
S
Sinusoidal source
L
Z0
ZL
z=0
We will discuss two methods:
z
 Quarter-wave transformer
 Single-stub matching
2
Quarter-Wave Transformer
Quarter-Wave Transformer: First consider a real load.
l /4
Z0
Z0T
ZL = RL
Zin
 Z L  jZ 0T tan   l  
Z in  Z 0T 

Z

jZ
tan

l


L
 0T

 2
tan   l   tan 
 
2
Z 0T
Zin 
ZL

 
l   tan    

2
2
Z 0T
Hence Zin 
 real
RL
3
Quarter-Wave Transformer (cont.)
l /4
Z0
Z0T
ZL = RL
Zin
Set
Example:
Zin  Z 0
Hence
This gives us
2
Z 0T
 Z0
RL
Z 0T  Z 0 RL
Z 0  50   
Z L  100   
Z 0T 
 50 100   70.71  
4
Quarter-Wave Transformer (cont.)
Next, consider a general (complex) load impedance ZL.
l /4
Z0
Z0T
YL = 1 / ZL
YL  GL  jBL
Shunt (parallel) susceptance
Choose Ys  jBs   jBL
Bs = -BL
l /4
New model:
Z0
Z0T
YLTOT = GL
ZLTOT = 1 / GL (real)
5
Quarter-Wave Transformer (cont.)
Summary of quarter-wave transformer matching method
l /4
Z0
Z0T
YL = GL + j BL
Ys = jBs
Z 0T  Z 0 / GL
Bs   BL
6
Quarter-Wave Transformer (cont.)
Realization using a shorted stub
(An open-circuited stub could also be used.)
l /4
Z0
Z0T
YL = G L + j B L
Bs = -BL
Z0s
ls
X s  Z0s tan  sls 
Bs  Y0 s cot  sls 
7
Quarter-Wave Transformer with Extension
Z0
l /4
d
Z0T
Z0
ZL
Zin(-d)
 We choose the length d to make the input impedance Zin (-d) real.
 We then use a quarter-wave transform to change the impedance to Z0.
8
Quarter-Wave Transformer with Extension (cont.)
Example
Z0  50[]
Z L  50  j 75 []
Z0
l /4
d
Z0T
Z0
ZL
Z LN  1  j (1.5)
9
Quarter-Wave Transformer with Extension (cont.)
0.176
Z LN
Wavelengths
towards generator
0
d  0.250  0.176
ZinN  d 
0.250
ZinN  d   SWR  4.3
d  0.074
10
Quarter-Wave Transformer with Extension (cont.)
Z0  50[]
l /4
d
Z0T
Z0
Z L  50  j 75 []
d  0.074
ZinN  d   4.3

Z0T 
Zin  d   50  4.3  215[]
50 215
Z0T  103.7[]
11
Single-Stub Matching
d
ZL
Y0 = 1 / Z0
Ys  jBs
Yin
1) We choose the distance d so that at this distance from the load
Yin  Y0  jBin
(i.e., Gin = Y0)
2) We then choose the shunt susceptance so that
Bs   Bin
12
Single-Stub Matching (cont.)
d
Ys  jBs
Y0
Yin  Y0
ZL
Yin  Y0  jBin
Bs   Bin
The feeding transmission line on the left sees a perfect match.
13
Single-Stub Matching (cont.)
Realization using a shorted stub
(An open-circuited stub could also be used.)
d
Z0
ZL
Z0s
ls
14
Single-Stub Matching (cont.)
We use the Smith chart as an admittance calculator to determine the
distance d.
d
Z0
ZL
Z0s
ls
2) Determine the distance d to make the normalized input conductance equal to 1.0.
3) Determine the required value of Bs to cancel Bin.
4) If desired, we can also use the Smith chart to find the stub length ls.
15
Single-Stub Matching (cont.)
Example
Z0  50[]
Z L  100  j100 []
d
Z0
ZL
Z LN  2  j 2
Z0s
ls
YLN 
Z L  Z 0 Z LN  1
L 
 N
Z L  Z0 Z L  1
1
 0.25  j  0.25 
2  j2
 L  0.62 e j /6  0.62 30o
16
Single-Stub Matching (cont.)
0.041  0.322  0.363
Solution :
Im   z  or Im    z  
Add YsN  - j1.57 at d  0.219
or
0.041  0.178  0.219
0.178
YsN   j1.57 at d  0.363
Use this one
1  j1.57
Z LN  2  j 2
0.219
Re   z  or Re    z  
0.363
0.041
GinN  1
Gin  1
1  j1.57
YLN  0.25  j 0.25
Smith chart scale:
0.322
Wavelengths toward generator
17
Single-Stub Matching (cont.)
Next, we find the length of the short-circuited stub: BsN  1.57
Rotate clockwise from S/C to desired Bs value.
 Im   z 
0+j1
Assume Z0s = Z0
Otherwise, we have to be
careful with the normalization
(see the note below).
0+j2
0+j0.5
S/C
Yn  
0+j0
 Re   z 
0-j0.5
Note: In general,
0-j2
0-j1
BsN    BinN Y0  / Y0 s
 1.57 Y0 / Y0 s 
YsN  0  j1.57
18
Single-Stub Matching (cont.)
From the Smith chart:
ls  0.340  0.250
ls  0.09
Analytically:
0
S / C 0.250
O/C
Z s  jZ 0 s tan   ls 
 Ys   jY0 s cot   ls 
0.09
 BsN   cot   ls 
0  j1.57
Hence :
0.340
1.57   cot  ls
cot  ls  1.57; tan  ls 
 ls 
2

1
 0.637
1.57
ls  tan 1  0.637   0.567 [radians]
ls  0.0903
19
Single-Stub Matching (cont.)
UNMATCHED
1+  L  1.62
1.62
1.55
1.0
ZL
V z
0.78
V  1
1-  L  0.38
0.38
z
0
 z
0
0.178
Z LN
1.55
0.292 0.219
0.042
z
0.042 (0.25  0.178 )
Crank diagram
1.62
0.78
L  0.62
0.219
0.397
20
Single-Stub Matching (cont.)
1+  L  1.62
1.62

V 1
UNMATCHED
1.55
1.0
V z
ZL
0.78
SWR = 4.26
0
1-  L  0.38
0.38
z
0.292 0.219
z
0.042
MATCHED
1.62
ZL
jBs
1.55
SWR = 1.0
V z
0.78
0.219
0
z
0.219
0.042
z
21
```