adapted from notes by ece 5317-6351 prof. jeffery …courses.egr.uh.edu/ece/ece5317/class...
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Prof. David R. JacksonDept. of ECE
ECE 5317-6351 Microwave Engineering
Fall 2019
1
Notes 5Smith Charts
Adapted from notes by Prof. Jeffery T. Williams
Recall: ( ) ( ) ( )( )
( ) ( ) ( )( )
( ) ( )( )
( )( )
20 0
20 0
0 0
2
0 02
1 1
1 1
111 1
z z zL
z z zL
zL
zL
V z V e e V e z
V VI z e e e zZ Z
V z zeZ z Z ZI z e z
γ γ γ
γ γ γ
γ
γ
+ − + + −
+ +− + −
+
+
= + Γ = + Γ
= − Γ = − Γ
+ Γ + Γ= = = − Γ − Γ
Generalized reflection Coefficient: ( ) 2 zLz e γ+Γ = Γ
Generalized Reflection Coefficient
2
0
0
LL
L
Z ZZ Z
−Γ =
+
0,Z β ( )V z
( )I z
0z =z
LZ+
−
( )
( ) ( )
2
2L
zL
j zL
R I
z e
e e
z j z
γ
φ γ
+
+
Γ = Γ
= Γ
= Γ + Γ
Lossless transmission line (α = 0)
( ) ( )2Lj zLz e φ β+Γ = Γ
Generalized Reflection Coefficient (cont.)
Re 0 1L LZ ≥ ⇒ Γ ≤
( )( )( )( )
0
0
0
0
L LL
L L
L L
L L
R jX ZR jX Z
R Z jXR Z jX
+ −Γ =
+ +
− +=
+ +
( )( )
2 22 0
2 20
1L LL
L L
R Z XR Z X
− +⇒ Γ = ≤
+ +
Proof:
3
Different forms for Γ(z) Magnitude property of Γ(z)
Complex Γ Plane( )
( ) ( )( )
( )
( )
( )
2
2
2
2
L
L
R I
j zL
j zL
j dL
j dL
z
z j z
e
e
e
e
β
φ β
φ β
β
+
+
−
−
Γ = Γ
= Γ + Γ
= Γ
= Γ
= Γ
= Γ
Increasing d(moving towards
generator)
4
ReΓ
Im Γ
Lφ
LΓ
LΓ
2L dφ β−
Γ
Lossless line
z d= −d = distance from load
1
Note: Going λ/2 on the line
corresponds to going all the way around the Smith chart.
2 dβ
Clockwise movement!
( ) ( )( )0
11
zZ z Z
z + Γ
= − Γ
( ) ( ) ( )( )0
11n
Z z zZ z
Z z + Γ
≡ = − Γ Define
n n nZ R jX= +
Hence we have:
Z Chart
( )( )
11
R In n
R I
jR jX
j + Γ + Γ
+ = − Γ + Γ
Next, multiply both sides by the RHS denominator term and equate real and imaginary parts. Then solve the resulting equations for ΓR and ΓI in terms of Rn or Xn. This gives two equations.
5
Note: The z dependence is being
suppressed here.
The Z chart is the “usual” Smith chart.
Start with
1) Equation #1:2 2
2 11 1
nR I
n n
RR R
Γ − + Γ = + +
Equation for a circle in the Γ plane
,011
1
n
n
n
RR
R
= +
=+
Center
Radius
Z Chart (cont.)
6
1
ΓR
ΓI
( )2 2
2 1 11R In nX X
Γ − + Γ − =
Equation for a circle in the Γ plane
11,
1n
n
X
X
=
=
Center
Radius
2) Equation #2:
7
1
ΓR
ΓI
Z Chart (cont.)
9
Open ckt. (Γ=1)
Imag. (reactive)impedance
Match pt. (Γ=0)
Realimpedance
Short ckt. (Γ= −1)
Inductive (Xn > 0)
Capacitive (Xn < 0)
Γ planeRn = 1
Xn = 1
Xn = -1
Γ plane
Z Chart (cont.)
Note: ( ) ( )( )( )0
11 11
zY z
Z z Z z − Γ
= = + Γ
( )( )( )( )0
11
zY
z + −Γ
= − −Γ
( ) ( ) ( )( )( )( ) ( ) ( )
0
11n n n
zY zY z G z jB z
Y z + −Γ
⇒ ≡ = = + − −Γ
( )0 01 /Y Z≡
Admittance Calculations with the Z Chart
Define:
( ) ( )z z′Γ ≡ − Γ( ) 1
1nY z′+ Γ = ′− Γ
Same mathematical form as for Zn:
Conclusion: The same Smith chart can be
used as an admittance calculator.
10
( ) 11nZ z + Γ = − Γ
11
Short ckt. (Γ′ =1)
Imag. (reactive)admittance
Match pt. (Γ′ =0)
Realadmittance
Open ckt. (Γ′ = −1)
Capacitive (Bn > 0)
Inductive (Bn < 0)
Γ′ planeBn = 1
Bn = -1
Gn = 1
Γ′ plane
Admittance Calculations with the Z Chart (cont.)
Impedance or Admittance Calculations with the Z Chart
12
The Smith chart can be used for either impedance or
admittance calculations, as long as we are consistent.
The complex plane is either the Γ plane or the Γ′ plane.
Normalized impedance or admittance coordinates
As an alternative way to do admittance calculations, we can continue to use the original Γ plane, and add admittancecurves to the chart.
( ) ( )( )( )( ) ( ) ( )
11n n n
zY z G z jB z
z + −Γ
= = + − −Γ
Y Chart
( ) ( )( )( )( ) ( ) ( )
11n n n
zZ z R z jX z
z + Γ
= = + − Γ
Compare with previous Smith chart derivation, which started with this equation:
Side note: A 180o rotation on a Smith chart makes a normalized impedance become its reciprocal. 13
Rn = 1 circle, rotated 180o, becomes Gn = 1 circle.Xn = 1 circle, rotated 180o, becomes Bn = 1 circle.
( ) ( )( ) ( )
n n
n n
R z G z
X z B z
→
→( ) ( )z zΓ → −Γ (rotation of 180o)
Examples:
Y Chart (cont.)
14
Open ckt.
Match pt.
Gn = 0
Short ckt.
Inductive (Bn < 0)
Capacitive (Bn > 0)
Gn = 1
Bn = +1
Bn = -1
Bn = 0
Γ planeΓ plane
The Y chart is the “mirror image” of the usual Smith chart.
All Four Possibilities for Smith Charts
16
Z chart, used for admittance
Γ′ plane
Z chart, used for impedance
Γ plane
The first two are the most common.
The third is sometimes convenient.
The fourth is almost never used.
1 2
Γ′ plane
4Y chart, used for impedance
Γ plane
3Y chart, used for admittance
ZY Chart
17
This is convenient for doing matching problems that involve both series and shunt elements (done later).
Short-hand version
Γ plane
Gn = 1 Rn = 1
Xn = 1
Xn = -1
Bn = -1
Bn = 1
Inductive
Capacitive
The SWR is given by the value of Rn on the positive real axis of
the Smith chart (Rnmax).
Standing Wave Ratio
Proof:1
SWR1
L
L
+ Γ=
− Γ
( )( )
2
2
2
2
11
1111
L
L
n
j zL
j zL
j j zL
j j zL
zZ
z
eee ee e
β
β
φ β
φ β
+
+
+
+
+ Γ=
− Γ
+ Γ=
− Γ
+ Γ=
− Γ
18
max 12 0
1L
L nL
z Rφ β+ Γ
+ = ⇒ =− Γ
LΓ
maxn nR R=
2 0L zφ β+ =2L zφ β+
Γ plane
( )zΓ
At this link:
http://www.sss-mag.com/topten5.html
Download the following zip file:smith_v191.zip
Extract the following files:
smith.exe mith.hlp smith.pdf
This is the application file
Electronic Smith Chart
19
0 50100 50L
ZZ j
= Ω= + Ω
,0
2 1LL n
ZZ jZ
= = +
( )
/ 40.4 0.2
/ 4 20 10
g
n
g
dZ j
Z j
λ
λ
=
≈ −
⇒ − ≈ − Ω
Example 1
a
20
( )I d−
( )V d−
z d= −
+
−
0z =
zLZ
(- ) / 1 / 4, 3 / 8, 1 / 2gZ d d λ =Find at
Use the Z chart.
a
b
Z chart
Γ plane
,L nZ
1nR =
1nX =
1nX = −
( )
3 / 80.5 0.5
3 / 8 25 25
g
n
g
dZ j
Z j
λ
λ
=
≈ +
⇒ − ≈ + Ω
b
( )
/ 22 1
/ 2 100 50
g
n
g
dZ j
Z j
λ
λ
=
≈ +
⇒ − = + Ω
c
Example 1 (cont.)
21
3 / 8 0.212 0.5 0.087+ − =Note :
( )I d−
( )V d−
z d= −
+
−
0z =
zLZ
a
b
0.087λg
c0.5λg
0.462λg
0.212λgZ chart
Γ plane
0.0 gλ0.5 gλ
( )0 050 20mS8mS 4 mSL
Z YY j
= Ω =
= −
,0
0.4 0.2LL n
YY jY
= = −
( )
/ 42 1
/ 4 40mS 20mS
g
n
g
dY j
Y j
λ
λ
=
≈ +
⇒ − ≈ +
Example 2
a
22
( )I d−
( )V d−
z d= −
+
−
0z =
zLZ
(- ) / 1 / 4, 3 / 8, 1 / 2gY d d λ =Find at
Use the Y chart.
a
bc
Y chart
Γ plane
,L nY
1nB = −
1nB =
1nG =
( )
3 / 81 1
3 / 8 20mS 20mS
g
n
g
dY j
Y j
λ
λ
=
≈ −
⇒ − ≈ −
( )
/ 20.4 0.2
/ 2 8mS 4mS
g
n
g
dY j
Y j
λ
λ
=
≈ −
⇒ − = −
Example 2 (cont.) b
c
23
( )I d−
( )V d−
z d= −
+
−
0z =
zLZ
a
bc
Y chart
Γ plane
,L nY
1nB = −
1nB =
1nG =
Use a short-circuited section of air-filled TEM, 50 Ω transmission line(β = k0, λg = λd =λ0) to create an impedance of Zin = -j25 Ω at f = 10 GHz.
( ),25 0.550in nZ j j= − = −
00.426 0 0.426 0.426g g gL λ λ λ λ= − = =
Example 3
24
00 0 0
2 2cf k
π πλω µ ε
= = =0 3.0 cmλ =
25inZ j= − Ω
050 , kΩ SC
L
1.28cmL =
Z chart
Γ plane
SC
-1/2
50Ω
Use the Z chart.
Use an open-circuited section of 75 Ω (Y0 = 1/75 S) air-filled transmission line at f = 10 GHz to create an admittance of Yin = j1/75 S:
0.375cmL =
Example 4
25
00.125 0.125gL λ λ= = 0 3.0 cmλ =
( ) ( ),1 / 75 S 1in in nY j Y j= ⇒ =
075 , kΩ OC
L
Y chart
Γ plane
j1
1/75 S
OC
L
1nB = −
1nB =
Use the Y chart.
Example 5
0 50Z = Ω
100 100LZ j= + Ω
( ),1 0.25 .25
2 2L nY jj
= = −+
/6 o0.62 0.62 30jL e πΓ = = ∠0
0
11
L LnL
L Ln
Z Z ZZ Z Z
− −Γ = =
+ +26
In this example we will use the “usual”
Smith chart (Z chart), but as an admittance
calculator.
Single-stub matching
, 2 2L nZ j= +
0Z
0sZ
LZ
d
slWe want Gin = Y0 (Gin,n = 1)
0 0sZ Z=Choose
Example 5 (cont.)
0.178 gλ
, 0.25 0.25L nY j= −
1 1.57j+
1 1.57j−
0.322 gλ
0.041 gλ
0.219- 1.57
= 1.57 0.363gn
gn
dY
Y j d
j λ
λ
= =
= +
Add ator at
Solution:0.170 0.2.0 981 14 gλ λλ =+0.320 0.3.0 321 64 gλ λλ =+
wavelengths toward loadwavelengths toward generator
Smith chart scale:
Z chart
27
Γ′ plane
(We’ll use the first choice.)
SCOC
1nG =
Example 5 (cont.)
( )( )
( )
( )
0
0
,
1
tan
cot
cot
1.57 cot1cot 1.57; tan 0.637
1.572 tan 0.637 0.567 [rad]
scin s
scin s
s n s
s
s s
s sg
Z jZ l
Y jY l
B l
l
l l
l l
β
β
β
β
β βπβ
λ−
=
= −
= −
− = −
= = =
= = =
0.09s gl λ=From the Smith chart:
0.0903s gl λ=
Analytically:
28
S / C
0 1.57j−
0.09 gλ
O / C
Z chart
Γ′ plane
0.25 gλ
0.34 gλ
wavelengths toward generator
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