Single stub impedance matching Impedance matching can be ...
IMPEDANCE MATCHING IN HIGH FREQUENCY LINES UNIT - III.
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Transcript of IMPEDANCE MATCHING IN HIGH FREQUENCY LINES UNIT - III.
IMPEDANCE MATCHING IN HIGH FREQUENCY LINES
UNIT - III
Impedance Matching
04/21/23 2
Maximum power is delivered when the load is matched the line and the power loss in the feed line is minimizedImpedance matching sensitive receiver components improves the signal to noise ratio of the systemImpedance matching in a power distribution network will reduce amplitude and phase errors
ComplexityBandwidthImplementation
Adjustability
Half and Quarter wave transmission lines
• The relationship of the input impedance at the input of the half-wave transmission line with its terminating impedance is got by letting L = wavelength/2in the impedance equation.
Zinput = ZL
• The relationship of the input impedance at the input of the quarter-wave transmission line with its terminating impedance is got by letting L =wavelength/4 in the impedance equation.
Zinput = (Zinput Zoutput)0.5
Series Stub
Input impedance=1/S
Voltage minimum
S
S
Xjj
dS
X
S
Sd
djjS
djSXjZ
SZ
in
in
1tan
2
tan
tan)1
1(
1
1cos
4
tan1
tan1
/1
10
0
0
10
01
01
Single Stub Tunning
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Shunt Stub
Series Stub
G=Y0=1/Z0
Single Shunt Stub Tuner Design Procedure
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1. Locate normalized load impedance and draw VSWR circle (normalized load admittance point is 180o from the normalized impedance point).
2. From the normalized load admittance point, rotate CW (toward generator) on the VSWR circle until it intersects the r = 1 circle. This rotation distance is the length d of the terminated section of t-tline. The nomalized admittance at this point is 1 + jb.
3. Beginning at the stub end (rightmost Smith chart point is the admittance of a short-circuit, leftmost Smith chart point is the admittance of an open-circuit), rotate CW (toward generator) until the point at 0 - jb is reached. This rotation distance is the stub length l.
Smith Chart
• Impedances, voltages, currents, etc. all repeat every half wavelength
• The magnitude of the reflection coefficient, the standing wave ratio (SWR) do not change, so they characterize the voltage & current patterns on the line
• If the load impedance is normalized by the characteristic impedance of the line, the voltages, currents, impedances, etc. all still have the same properties, but the results can be generalized to any line with the same normalized impedances
Smith Chart
• The Smith Chart is a clever tool for analyzing transmission lines
• The outside of the chart shows location on the line in wavelengths
• The combination of intersecting circles inside the chart allow us to locate the normalized impedance and then to find the impedance anywhere on the line
Smith ChartReal Impedance Axis
Imaginary Impedance Axis
Smith Chart Constant Imaginary Impedance Lines
Constant Real Impedance Circles
Impedance
Z=R+jX
=100+j50
Normalized
z=2+j for
Zo=50
Smith Chart
• Impedance divided by line impedance (50 Ohms)
• Z1 = 100 + j50 • Z2 = 75 -j100 • Z3 = j200 • Z4 = 150 • Z5 = infinity (an open circuit)• Z6 = 0 (a short circuit)• Z7 = 50 • Z8 = 184 -j900
• Then, normalize and plot. The points are plotted as follows:
• z1 = 2 + j• z2 = 1.5 -j2• z3 = j4• z4 = 3• z5 = infinity• z6 = 0• z7 = 1• z8 = 3.68 -j18S
Smith Chart
• Thus, the first step in analyzing a transmission line is to locate the normalized load impedance on the chart
• Next, a circle is drawn that represents the reflection coefficient or SWR. The center of the circle is the center of the chart. The circle passes through the normalized load impedance
• Any point on the line is found on this circle. Rotate clockwise to move toward the generator (away from the load)
• The distance moved on the line is indicated on the outside of the chart in wavelengths
Toward Generator
Away From Generator
Constant Reflection Coefficient Circle
Scale in Wavelengths
Full Circle is One Half Wavelength Since Everything Repeats
Single-Stub Matching
Load impedance jBYin 1
1tan
2length stub The
1
1cos
4
1
epoint wher minimum- voltage thefrom distance thebe Let
real ist coefficien reflection then thereal, is If1
1
10
10
0
S
S
S
Sd
jBY
d
Y
SY
in
L
in
Input admittance=S
Single Stub Tuning
Single-stub tuning circuits.
(a) Shunt stub. (b) Series stub.
• 2 adjustable parameters• d: from the load to the stub position.• B or X provided by the shunt or series stub.
• For the shunt-stub case, • Select d so that Y seen looking into the line at
d from the load is Y0+jB
• Then the stub susceptance is chosen as –jB.• For the series-stub case,
• Select d so that Z seen looking into the line at d from the load is Z0+jX
• Then the stub reactance is chosen as –jX.
Shunt Stubs• Single-Stub Shunt Tuning
ZL=60-j80
.
(b) The two shunt-stub tuning solutions. (c) Reflection coefficient magnitudes versus frequency
for the tuning circuits of (b).
• To derive formulas for d and l, let ZL= 1/YL= RL+ jXL.
• Now d is chosen so that G = Y0=1/Z0,
00
0
( ) tan
( ) tanL L
L L
R jX jZ dZ Z
Z j R jX d
2
2 20
20 02 2
0 0
1
(1 tan )where
( tan )
tan ( tan )( tan )
[ ( tan ) ]
L
L L
L L L
L L
Y G jBZ
R dG
R X Z d
R d Z X d X Z dB
Z R X Z d
2 2 20 0 0 0
2 20 0
00
( ) tan 2 tan ( ) 0
[( ) ] /tan , for
L L L L L
L L L LL
L
Z R Z d X Z d R Z R X
X R Z R X Zd R Z
R Z
• If RL = Z0, then tanβd = -XL/2Z0. 2 principal solutions are
• To find the required stub length, BS = -B.
for open stub
for short stub
1
0 0
1
0 0
1tan for - 0
2 2 2
1tan for - 0
2 2 2
L L
L L
X X
Z Zd
X X
Z Z
1 10
0 0
1 1tan tan
2 2Sl B B
Y Y
1 10 0 01 1tan tan
2 2S
l Y Y
B B
Series Stubs• Single Stub Series Tuning
ZL = 100+j80
(a) Smith chart for the series-stub
tuners.
(b) The two series-stub tuning
solutions. (c) Reflection coefficient
magnitudes versus frequency for the
tuning circuits of (b).
• To derive formulas for d and l, let YL= 1/ZL= GL+ jBL.
• Now d is chosen so that R = Z0=1/Y0,
00
0
( ) tan
( ) tanL L
L L
G jB jY dY Y
Y j G jB d
2
2 20
20 02 2
0 0
1
(1 tan )where
( tan )
tan ( tan )( tan )
[ ( tan ) ]
L
L L
L L L
L L
Z R jXY
G dR
G B Y d
G d Y B d B Y dX
Y G B Y d
2 2 20 0 0 0
2 20 0
00
( ) tan 2 tan ( ) 0
[( ) ] /tan , for
L L L L L
L L L LL
L
Y G Y d B Y d G Y G B
B G Y G B Yd G Y
G Y
• If GL = Y0, then tanβd = -BL/2Y0. 2 principal solutions are
• To find the required stub length, XS = -X.
for short stub
for open stub
1
0 0
1
0 0
1tan for - 0
2 2 2
1tan for - 0
2 2 2
L L
L L
B B
Y Yd
B B
Y Y
1 10
0 0
1 1tan tan
2 2Sl X X
Z Z
1 10 0 01 1tan tan
2 2S
l Z Z
X X
Analytic Solution• To the left of the first stub in Fig. 5.7b,
Y1 = GL + j(BL+B1) where YL = GL + jBL
• To the right of the 2nd stub,
• At this point, Re{Y2} = Y0
1 02 0
0 1
( ) where tan
( )L L
L L
G j B B Y tY Y t d
Y jt G jB jB
222 0 1
0 2 2
2 220 1
0 2 2 2 20
( )10
4 ( )11 1
2 (1 )
LL L
LL
Y B t B ttG G Y
t t
t Y B t B ttG Y
t Y t
• Since GL is real,
• After d has been fixed, the 1st stub susceptance can be determined as
• The 2nd stub susceptance can be found from the negative of the imaginary part of (5.18)
2 20 12 2 2
0
4 ( )0 1
(1 )Lt Y B t B t
Y t
2
00 2 2
10
sinL
YtG Y
t d
2 2 20 0
1
(1 ) L LL
Y t G Y G tB B
t
• B2 =
• The open-circuited stub length is
• The short-circuited stub length is
2 2 20 0 0(1 )L L L
L
Y Y G t G t G Y
G t
10
0
1tan
2
l B
Y
10 01tan
2
l Y
B
For a load impedance ZL=60-j80Ω, design two single-stub (short circuit) shunt tunning networks to matching this load to a 50 Ω line. Assuming that the load is matched at 2GHz and that load consists of a resistor and capacitor in series.
Single Stub Tunning
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yL=0.3+j0.4
d1=0.176-0.065=0.110λ
d2=0.325-0.065=0.260λ
y1=1+j1.47
y2=1-j1.47
l1=0.095λl1=0.405λ
Single Stub Tunning
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Results
For a load impedance ZL=25-j50Ω, design two single-stub (short circuit) shunt tunning networks to matching this load to a 50 Ω line.
Single Stub tunning
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yL=0.4+j0.8
d1=0.178-0.115=0.063λ
d2=0.325-0.065=0.260λ
y1=1+j1.67
y2=1-j1.6
l1=0.09λl1=0.41λ
Single Series Stub Tuner Design Procedure
04/21/23 34ELCT564
1. Locate normalized load impedance and draw VSWR circle
2. From the normalized load impedance point, rotate CW (toward generator) on the VSWR circle until it intersects the r = 1 circle. This rotation distance is the length d of the terminated section of t-tline. The nomalized impedance at this point is 1 + jx.
3. Beginning at the stub end (leftmost Smith chart point is the impedance of a short-circuit, rightmost Smith chart point is the impedance of an open-circuit), rotate CW (toward generator) until the point at 0 ! jx is reached. This rotation distance is the stub length l.
For a load impedance ZL=100+j80Ω, design single series open-circuit stub tunning networks to matching this load to a 50 Ω line. Assuming that the load is matched at 2GHz and that load consists of a resistor and inductor in series.
Single Stub Tunning
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zL=2+j1.6
d1=0.328-0.208=0.120λ
d2=0.5-0.208+0.172=0.463λ
z1=1-j1.33
z2=1+j1.33
l1=0.397λl1=0.103λ
Single Stub Tunning
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Single Stub Tunning
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Double Stub Matching Network
jB1jB2YL
ab
2
1
P tocircle econductancconstant along
point themoves which Bj esusceptanc a adds stubfirst The
aa plane at the
into ed transformis
LLL
LL
BjGY
YY
b a
Double-Stub Tuning• If an adjustable tuner was desired, single-tuner
would probably pose some difficulty.
Smith Chart Solution
• yL add jb1 (on the rotated 1+jb circle) rotate by d thru SWR circle (WTG) y1 add jb2 Matched
• Avoid the forbidden region.
Double Stub Tunning
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The susceptance of the first stub, b1, moves the load admittance to y1, which lies on the rotated 1+jb circle; the amount of rotation is de wavelengths toward the load. Then transforming y1 toward the generator through a length d of line to get point y2, which is on the 1+jb circle. The second stub then adds a susceptance b2.
Design a double-stub shunt tuner to match a load impedance ZL=60-j80 Ω to a 50 Ω line. The stubs are to be open-circuited stubs and are spaced λ/8 apart. Assuming that this load consists of a series resistor and capacitor and that the match frequency is 2GHz, plot the reflection coefficient magnitude versus frequency from 1 to 3GHz.
Double Stub Tunning
04/21/23 43ELCT564
yL=0.3+j0.4
b1=1.314
b1’=-0.114
y2=1-j3.38
l1=0.46λ
l2=0.204λ
Double Stub Tunning
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Double-stub tuning.
(a) Original circuit with the
load an arbitrary
distance from the first stub. (b) Equivalent-circuit with load at the first stub.
Smith chart diagram for the operation of a double-stub
tuner.
Solution to Example 5.4. (a) Smith chart for the double-
stub tuners.
ZL = 60-j80
Open stubs, d = λ/8
(b) The two double-stub tuning solutions. (c) Reflection coefficient magnitudes versus
frequency for the tuning circuits of (b).
0
r=1
x=1
x=-1
Real part ofRefl. Coeff.
Pshort circuit Popen circuit
r=0.5
Smith Chart
YL
. cancel willstub The circle. 1G on the liemust P The
.Y is admittanceinput theb-b plane At the
d2
anglean through circle radiusconstant a along P toP from Move
3
b
32
b
bb
Bj
BjG
0
r=1
x=1
x=-1
Real part ofRefl. Coeff.
Pshort circuit Popen circuit
r=0.5
Smith Chart
YL
Rotate the the G=1 circle through an angle -
The intersection of G=1 and the GL circle determine the point P2
P2
P3
G1=1
0
r=1
x=1
x=-1
Real part ofRefl. Coeff.
Pshort circuit Popen circuit
r=0.5
Smith Chart
YL
The shaded range is for the load impedance whichcannot be matched when d=1/8 wavelength
04/21/23 53ELCT564