1 r ji . 1 r ji r jx 1 r ji 1 r ji 1 r i 2
r
2
1 r i
1 r i 2
r
2
1 r i
x
2
2i 2
2
1 r i
2
2
2
2
2
x
2i
1 r i2
r 1 2 i r 1 r 1 r 2
2
1 1 r 1 i x x 2
2
2
Resistive Circles
Centre
r
r 1 r
;
i 0
r 1 2 r i 1 r 1 r
Radius
1 1 r
2
2
•The centers of all r-circles lie on the r axis.
r 1 2 r i 1 r 1 r
•The r = 0 circle, unity radius, centre at origin •The r-circles becomes progressively smaller as r increases from 0 toward ending at ( r = 1, i = 0) point for open circuit • All r – circles pass through the ( r = 1, i = 0) point
2
2
Reactance Circles
Centre
r 1; i
1 1 r 1 i x x 2
1
x
Radius
1 x
2
2
1 1 1 r i •The centers of all x-circles lie on the r = 1 lines; x x 2
for x > 0 (inductive reactance) lie above r axis. for x < 0 (capacitive reactance) lie below r axis. • The x = 0 circle becomes the r axis •The x-circles becomes progressively smaller as x from 0 toward ending at (r = 1, i = 0) point for open circuit • All x –circles pass through the ( r = 1, i = 0) point
2
•Smith Chart: a chart of r and x circle in r and I plane for 1 •R and x circle are everywhere orthogonal to one another •The intersection of r and x circle defines a points that represents a normalized load impedance
z L r jx
Г i x
• Actual Impedance is ZL R0 (r + jx)
1.0
5 0.
2 .
0
3 . 0
C
X / Z O =
0 .2
•r = -1 and i = 0 corresponds to r = 0 and x = 0: short circuit
2
-1
. 0 =
0.5
O
Z R
1.0
2.0
5.0
1
A
/
2 - 0.
= / O X Z
•r = 1 and i = 0 corresponds to infinite impedance: open circuit
3 . 0
. 2 0
5 .
0
-
-
0 . 1
Гr
Smith chart can also be marked as polar coordinates:
Magnitude of
Phase angle of
SC 1 OC 1
Pm
PM
Each circle intersect the real axis at two points: PM on positive real axis Pm on negative real axis Since x = 0 along the real axis: P M and Pm both represents a purely resistive load RL > R0 (r > 1) : PM RL > R0 (r < 1) : Pm
if
R L R0
S
R L R0
r
The value of the r-circle passing through the point PM is numerically equal to the standing wave ratio
if
R L R0
S
R0 R L
1
r
r
1
S
The value of the r-circle passing through the point Pm on negative real axis is numerically equal to the 1/S
Constant Circle
1. All the circles are centered at the origin and their radii vary from 0 to 1.
2. The angle measured from the positive real axis, of the line drawn from the origin through the point representing z L equal .
3. The value of the r-circle passing through the intersection of the circle and the positive real axis equals the standing wave ratio
The input impedance looking towards the load end at a distance z’ from the load is
Zi z '
V z ' I z '
I L
2
Z L Z0 e z ' 1 e2 z '
I L
2 Z 0
2 z ' z ' 1 Z Z e e L 0
1 e2 j z ' Zi z ' Z 0 2 j z ' 1 e Normalized input impedance
z i
Zi z ' Z 0
1 e
2 j z '
1 e
2 j z '
z i
Zi z ' Z 0
2 j z '
1 e
2 j z '
1 e
z i
z i
Zi z ' Z 0
Zi z ' Z 0
At z’ = 0;
2 j z '
1 e
2 j z '
1 e
1 e
1 e
2 j z ' zi z L
S
2 j z '
1 1
1 e
j
1 e
j
The magnitude of remains constant, therefore VSWR are not changed by additional length
Keeping
an angle to
constant, subtract (rotate clockwise direction) from
2 z '
4 z '
This will locate the point for
.
e
j
, which determine
z i
2 j z ' The outer scale on smith chart is marked “wavelength toward generator” in clockwise direction (increasing z’)
The inner scale is marked “wavelength toward load” in counter clockwise direction (decreasing z’)
2 z ' if
z ' / 2;
4 z '
2 z '
4 z '
2
Therefore complete revolution gives the z’ of /2
Example:1 Find L if the load impedance Z L is 25+j100 and characteristic impedance of transmission line is 50
z L 0.5 j 2.0
Example:2: ZL = 25 + j 100; z’ = length of transmission line d = 0.18 ; Find Zin and (d)
L 0.8246 L 50.906 0.1793
z L 0.5 j 2.0
0.3593
Example:3: ZL = 25 + j 100; Find the location of first voltage maximum (d max) and first voltage minimum from load end (d min)
z L 0.5 j 2.0
Example:4: ZL = 25 - j 100; Find the location of first voltage maximum (d max) and first voltage minimum from load end (d min)
z L 0.5 j 2.0
0.3207
Example:5: Find the VSWR on transmission line (i) if ZL1 = 25 + j 100 and Z0 = 50 ; (ii) if ZL1 = 25 - j 100 and Z0 = 50
Circle of Constant resistance r = 10.4
Example:5: Given: R0 = 50 , S = 3.0, = 0.4 m, First voltage minima z m’ = 0.05 m: Find (i) , (ii) ZL
0.05
z m '
0.4
0.125
z L 0.6 j 0.8 Z L 50 0.6 j 0.8
Z L 30 j 40
d e
j
d / 4 e
j 2 / 4
e zn d
1 d 1 d
;
j
e e j
j
d
yn d
1 d 1 d
1 d 4 1 d zn d yn d 1 d 4 1 d 4
zn d yn d 4 Actual Impedance
Actual Admittance
Z d Z0 .zn d 4 4 Y d Y0 . yn d
The input admittance of short circuited stub is purely susceptive y s
1 y B yS
y B 1 jbB y s jbB
B’
R0
l
Example: A 50 transmission line is connected to a load impedance Z L = 35 – j 47.5 . Find the position and length of a short-circuited stub required to match the line. Solution:
Solution for location of stub For P3: (from P2’ to P3’) d1 = 0.168 - 0.109 = 0.059 For P4: (from P2’ to P4’) d2 = 0.332 - 0.109 = 0.223 Solution for length of stub yS = - jbB For P3: (P SC on the extreme right of chart to P3’’ which represents – jbB1 = -j 1.2 l 1 =
0.361 - 0.250 = 0.111
For P4: (from PSC to P4’’: + j 1.2)
l 2 =
0.139 + 0.250 = 0.389
First Solution: d1 = 0.059 and l 1 = 0.111 Second Solution: d2 = 0.223 and l 2 = 0.389
Double Stub Matching: d0
B
yB
yi
A y A ZL
ySB R0
ySA B’
A’
R0
y A = ySA + yL yi = ySB + yB
l A
l B
d0 is constant and can be arbitrary chosen as /8 or 3/8
R0
d0
B
yB
yi
A y A ZL
ySB R0
Yi YB Y SB
ySA B’
A’
R0
y A = ySA + yL yi = ySB + yB
For matching
Yi Y0
Y B Y0
Yi Y 0
l A
l B
1
R0
Y SB
yi yB ySB
Y 0
1 y B ySB
R0
The input admittance of short circuited stub is purely susceptive y sB
1 y B ySB
y B 1 jbB y sB jbB
Example: A 50 transmission line is connected to a load impedance Z L = 60 + j 80 . A double stub tuner spaced /8 apart is used to matched the load as shown below. Find the required lengths of short circuited stubs. d0
B
Solution:
yB
yi
A y A ZL
ySB R0
ySA B’
A’
R0
y A = ySA + yL yi = ySB + yB
y L
1
z L
R0 ZL
50 60 j80
0.30 j 0.40
l A
l B
R0
Draw g0 = 1 circle Rotate this circle by /8 towards load yL = 0.30 – j 0.40 as PL Move on constant g circle (g = 0.3) which intersects rotated g circle at PA1 and PA2. PA1: y A1 = 0.30 + j 0.29 PA2: y A2 = 0.30 + j 1.75 Move /8 on Constant | | circle from PA1 or PA2 and reaches PB1 or PB2 respectively PB1: yB1 = 1 + j 1.38 PB2: y = 1 j 3.5
PA1: y A1 = 0.30 + j 0.29 PA2: y A2 = 0.30 + j 1.75 yL = 0.30 – j 0.40 as PL
First stub length (ySA)1 = y A1 – yL = j 0.69 (ySA)2 = y A2 – yL = j 2.15
