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1
Prof. David R. JacksonDept. of ECE
Notes 3
ECE 5317-6351 Microwave Engineering
Fall 2011
Smith ChartExamples
2
0 50
100 50L
Z
Z j
I(-d)
V(-d)+
d
ZL-
0,Z
Find Z(-d)
/ 1/ 4, 3 / 8, 1/ 2gd at
,0
2 1LL n
ZZ j
Z
/ 4
0.4 0.2
/ 4 20 10
g
n
g
d
Z j
Z j
nLZ
3 / 8 gd
/ 4gd
a
b
01/ 2 gd
or
Example 1
a
Impedance chart
3
3 / 8
0.5 0.5
3 / 8 25 25
g
n
g
d
Z j
Z j
b
/ 2
2 1
/ 2 100 50
g
n
g
d
Z j
Z j
c
nLZ
3 / 8 gd
/ 4gd
a
b
0/ 2
gd
0.087g
c0.5g
0.462g
0.212g
Example 1 (cont.) I(-d)
V(-d)+
d
ZL-
0,Z
4
0 050 20mS
8mS 4mSL
Z Y
Y j
Find Y(-d)
/ 1/ 4, 3 / 8, 1/ 2gd at
,0
0.4 0.2LL n
YY j
Y
/ 4
2 1
/ 4 40mS 20mS
g
n
g
d
Y j
Y j
nLY
3 / 8 gd
/ 4gd
a
b
01/ 2 gd
or
c
Example 2
a
I(-d)
V(-d)+
d
ZL-
0,Z
Admittance chart
5
3 / 8
1 1
3 / 8 20mS 20mS
g
n
g
d
Y j
Y j
/ 2
0.4 0.2
/ 2 8mS 4mS
g
n
g
d
Y j
Y j
Example 2 (cont.) b
c
I(-d)
V(-d)+
d
ZL-
0,Z
nLY
3 / 8 gd
/ 4gd
a
b
0
1/ 2 gd
c
Admittance chart
6
Simple answer:
* When adding elements in series use Z-chart
* When adding elements in parallel use Y-chart
A B CZ Z Z
A B CY Y Y
ZA ZB ZC
YA YB YC
Which Chart to Use?
7
Use reactively loaded section of transmission line.
most common to use open or short load
d
ZL0 ,Z
Z d
L LZ jX
LX 0LX
SC OC
dd = 0
nLX
nX d
Non-absorbing load
(RL = 0)
Using Reactive Loads
Impedance chart
Reactance
8
d
YL0 ,Z
Y d
L LY jSC OC
dd = 0
nLY
nY d
Susceptance
0
0/
n
n
Y d Y d Y
Y d Z
Using Reactive Loads (cont.)
Admittance chart
9
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.
SC
-1/2
0 .426 g
50
0 g L
25inZ j
SC50 Ω , k0
,
251/ 2
50in nZ j j
0.426 0 0.426g g gL
Example 3
L = 1.28 cm
00 0 0
2 2c
f k
0 3.0 cm
Impedance chart
10
Use an open circuited section of 75 (Y0 = 1/75 S) air-filled transmission line at f = 10 GHz to create an admittance of
j1
1/75 S
OC
L
L
1/ 75inY j
OC75 Ω , k0
, 1in nY j
0
1
75Y
1S 13.3mS
75inY j j
0.375cmL
Example 4
00.125L 0 3.0 cm
Admittance chart
11
Y0
d1
njb ,A nY
,L nY
1nG
,in inZ Y d
ZLYS
AY
0Z
0inZ Z
,,0 0 ,
1LL n
L
L nL n
YY
Y Z
ZZ
Z
, , ,in n A n S nY Y Y
Note: At d we have
Want to pick d and Ys such that Yin = Y0.
00
1Y
Z
Matching Circuit
Admittance chart
12
,S n nY jb
, ,1 1 n S nin n YY jb
We want
Choose
0
,
0
1
1 1in n n n
in
inin
Y jb b
Y
j
Y Y
Z Z
,in inZ Y d
ZLYS
AY
0Z
Y0
d1
njb ,A nY
,L nY
1nG
Matching Circuit (cont.)
Admittance chart
13
Example 5
ZL Z0
ls
Z0s
d
0 50[ ]Z
100 100 [ ]LZ j
, 2 2L nZ j
,
10.25 .25
2 2L nY jj
/6 o0.62 0.62 30jL e 0
0
1
1L Ln
LL Ln
Z Z Z
Z Z Z
In this example we will use the “usual” Smith chart, but as an admittance calculator.
14
Example 5 (cont.)
X
X
0.178 g
,Y 0.25 0. 5 2L n j
1 1.57j
1 1.57j
0.322 g
0.363 g
0.219 g
0.041 g
X
0.219 -
0.36
1.57
1.57 3
g
g
d
d
j
j
Solution :
Add at
or at
0.170 0.2.0 981 14 g 0.320 0.3.0 321 64 g
wavelengths toward loadwavelengths toward generator
Smith chart scale:
Admittance calculator
plane
SCOC
(We’ll use the first choice.)
15
Example 5 (cont.)
S / C
X0 1.57j
0.09
O / C
Admittance calculator
, cot
1.57 cot
1cot 1.57; tan
1.572
0.567 [radians]
s nB l
l
l l
l l
0.09s gl From the Smith chart:
0.0903s gl
Analytically:
16
00
0
tan
tanL T
in TT L
Z jZZ Z
Z jZ
2/ 4
4 2g
gg
20
12
0
Tin
L
T in L
ZZ
Z
Z Z Z
0
12
0 0
in
T L
Z Z
Z Z Z
ZLZ0 Z0T
Zin, inZL is real
/ 4g
Quarter-Wave Transformer
At f0 :
For in = 0
17
Match 100 load to 50 transmission Line at f0.
0 100 50
70.7TZ
0
0
0
0
2 2 2g f
r rfk k
Example 6
0 70.7TZ
1r , 50 Ω Z0T
/ 4g
100 Ω