Syafruddin Hasan
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At any point along the transmission line, we can find
the ratio of the total voltage to the total current.
current
This ratio is known as input impedance.
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Input Impedance (Cont’d)
Looking into the line at z = - ℓ, the input impedance :
Z L + Z 0 tanh (γl )
Z in = Z 0
Z 0 + Z L tanh (γl )
For a special lossless case, it becomes:
Z L + jZ 0 tan (β l )
Z in = Z 0
Z 0 + jZ L tan (β l )
with β
=
2π
λ
See Drill 6.7 and 6.8 and 6.9
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EXAMPLE 1
A source with 50 Ω source impedance drives
a 50 Ω transmission line that is 1/8 of
wavelength long,
long terminated in a load
ZL = 50 – j25 Ω. Calculate:
(i) The reflection coefficient,
coefficient ГL
(ii) VSWR
(iii) The
Th input
i
t impedance
i
d
seen by
b the
th source.
4
SOLUTION TO EXAMPLE 1
It can be shown as:
5
SOLUTION TO EXAMPLE 1 (Cont’d)
((i)) The reflection coefficient,
Z L − Z0
ΓL =
Z L + Z0
(
50 − j 25) − 50
= 0.242e− j 76
=
(50 − j 25) + 50
0
(ii) VSWR
1 + ΓL
VSWR =
= 1.64
1 − ΓL
6
SOLUTION TO EXAMPLE 1 (Cont’d)
(iii) The input impedance seen by the source, Zin
Need to calculate
Therefore,
2π λ π
βl =
=
λ 8 4
∴ tan
π
4
=1
Z L + jZ 0 tan βl
Z in = Z 0
Z 0 + jZ L tan βl
50 − j 25 + j 50
= 50
50 + j 50 + 25
Zin = 30.8 − j 3.8Ω
7
5-6 Input Impedance of the Lossless Line
•
•
Voltage to current ratio is called input
impedance Zin
i .
The input impedance at z = −l is given as
⎛ Z L + jZ 0 tan βl ⎞
⎛ Z L cos βl + jZ 0 sin βl ⎞
⎟⎟
⎟⎟ = Z 0 ⎜⎜
Z in (− l ) = Z 0 ⎜⎜
⎝ Z 0 + jZ L tan βl ⎠
⎝ Z 0 cos βl + jZ L sin βl ⎠
and
d
~
⎛
V
g Z in
+
⎜
V0 =
⎜ Z g + Z in
⎝
⎞⎛
1
⎞
⎟⎜
⎟
⎟⎝ e jβl + Γe − jβl ⎠
⎠
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Example 5.6 Complete Solution for v(z, t) and i(z, t)
A 1.05-GHz generator circuit with series impedance
Zg = 10Ω and voltage
g source given
g
byy
vg (t ) = 10 sin (ωt + 30°) (V )
is connected to a load ZL = (100 + j50) through a
50 Ω 67-cm-long
50-Ω,
67 cm long lossless transmission line
line. The phase
velocity of the line is 0.7c, where c is the velocity of light
in a vacuum. Find v(z, t) and i(z, t) on the line.
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Solution 5.6 Complete Solution for v(z, t) and i(z, t)
We find the wavelength from
0.7 × 3 × 108
λ= =
= 0 .2 m
9
f
1.05 × 10
up
and
⎛ 2πl ⎞
⎛ 2π 0.67 ⎞
tan (βl ) = tan ⎜
⎟ = tan ⎜
⎟ = tan 126°
⎝ λ ⎠
⎝ 0. 2 ⎠
The voltage reflection coefficient at the load is
Γ=
Z L − Z 0 (100 + j 50 ) − 50
= 0.45e j 26.6°
=
Z L + Z 0 (100 + j 50 ) + 50
The input impedance of the line
⎡ Z + jjZ 0 tan β l ⎤
Z in = Z 0 ⎢ L
⎥ = (21.9 + j17.4 )Ω
⎣ Z 0 + jZ 0 tan β l ⎦
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Solution 5.6 Complete Solution for v(z, t) and i(z, t)
Rewriting the expression for the generator voltage,
vg (t ) = 10 sin (ωt + 30°)
[
= 10 cos(ωt − 60°) = ℜe 10e −60° e jwt
Thus the phasor voltage is
] (V )
~
Vg = 10e − j 60° = 10∠ − 60° (V )
The voltage on the line is
~
⎛
V
g Z in
V0+ = ⎜
⎜
⎝ Z g + Z in
⎞⎛
1
⎞
j159°
⎟⎜
=
10
.
2
e
= 10.2∠159° (V )
⎟
⎟⎝ e jβl + Γe − jβl ⎠
⎠
and phasor voltage on the line is
(
)
(
~
V ( z ) = V0+ e − jβz + Γe jβz = 10.2e j159° e − jβz + 0.45e j 26.6° e jβz
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Solution 5.6 Complete Solution for v(z, t) and i(z, t)
The instantaneous voltage and current is
[
]
~
v(z , t ) = ℜe V (z )e jωt = 10.2 cos(ωt − β z + 159°)
+ 4.55 cos(ωt + β z + 185.6°) (V )
i ( z , t ) = 0.20 cos(ωt − β z + 159°)
+ 0.091 cos(ωt + β z + 5.6°)
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5-7 Special Cases of the Lossless Line
•
Special cases
has useful properties.
properties
5-7 .1 ShortShort-Circuited Line
•
For short-circuited line at
z = − l,
~
V
sc
sc (− l )
Z in = ~
= jZ
j 0 tan βl
I sc (− l )
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Example 5.7 Equivalent Reactive Elements
Choose the length of a shorted 50- lossless transmission
line ((Fig.
g 5-16)) such that its input
p impedance
p
at 2.25
GHz is equivalent to the reactance of a capacitor with
capacitance Ceqq = 4 pF. The wave velocity on the line is
0.75c.
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Solution 5.7 Equivalent Reactive Elements
We are given
u p = 0.75c = 2.25 × 108 m/s
Z 0 = 50Ω
f = 2.25 × 10 9 Hz
Ceq = 4 × 10 −12 F
The phase constant is
2nd
quadrant is
β=
2π
λ
=
β l1 = 2.8 rad or l1 =
4th quadrant is βl2 = 5.94 rad
2πf
1
= 62.8 (rad/m ), tan β l = −
= −0.354
up
Z 0ωCeq
2 .8
β
or l2 =
=
2 .8
= 4.46 cm
62.8
5.94
= 9.46 cm
62.8
Any length l = 4.46 cm + nλ/2, where n is a positive
integer, is also a solution.
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5-7.2 OpenOpen-Circuited Line
•
With ZL = ∞, it forms an
open-circuited
open
circuited line.
line
V (− l )
Z inoc = ~oc
= − jZ 0 cot βl
I oc (− l )
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5-7.3 Application of Short
Short--Circuit and OpenOpen-Circuit Measurements
• Product and ratio of SC and OC equations give
the following results:
− Z insc
t βl =
tan
Z inoc
Z o = + Z insc Z inoc
• Radio-frequency (RF) instruments measure the
impedance of any load.
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Example 5.8 Measuring Z0 and β
Find Z0 and β of a 57-cm-long lossless transmission
line whose input
p impedance
p
was measured as Zscin =
j40.42Ω when terminated in a short circuit and as
Zocin = −j121.24Ω when terminated in an open circuit.
From other measurements, we know that the line is
between 3 and 3.25 wavelengths long.
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Solution 5.8 Measuring Z0 and β
We have,
Z 0 = + Z insc Z inoc =
( j 40.42)(− j121.24) = 70Ω
− Z insc
1
tan βl =
=
oc
3
Z in
True value of βl is
βl = 6π +
π
6
= 19.4 (rad )
and
19.4
β=
= 34 (rad/m )
0.57
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Example 5.9 QuarterQuarter-Wave Transformer
A 50-Ω lossless transmission line is to be matched to a
resistive load impedance
p
with ZL = 100Ω via a q
quarterwave section as shown, thereby eliminating reflections
along the feedline. Find the characteristic impedance of
the quarter-wave transformer.
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Solution 5.9 QuarterQuarter-Wave Transformer
To eliminate reflections at terminal AA’, the input
impedance
p
Zin looking
g into the quarter-wave
q
line should
be equal to Z01, the characteristic impedance of the
feedline. Thus, Zin = 50 .
2
Z 02
Z in =
ZL
Z 02 = 50 × 100 = 70.7Ω
Since the lines are lossless, all the incident power will
end up getting transferred into the load ZL.
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5-8 Power Flow on a Lossless Transmission Line
•
We shall examine the flow of power carried by
incident and reflected waves.
waves
5-8.1 Instantaneous Power
•
Instantaneous power is the product of
instantaneous voltage and current
current.
5-8.2 TimeTime-Average Power
•
More interested in time-averaged power flow.
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5-8.2 TimeTime-Average Power
• There are 2 types of approach:
1) Time-Domain
Time Domain Approach
• Incident power and reflected wave power are
Pavi =
•
+ 2
V0
2Z 0
(W)
Pavr = − Γ
2
V0+
2
2Z 0
2
= − Γ Pavi
For net average power delivered
d li
d to the
h lload,
d
Pav = Pavi + Pavr =
+ 2
V0
2Z 0
[1 − Γ ]
2
(W)
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5-8.2 TimeTime-Average Power
2) Phasor-Domain Approach
• Time-average
Time average power for any propagating wave is
[
1
~ ~*
Pav = R e V ⋅ I
2
]
5-9 Smith Chart
•
The Smith Chart is used for analyzing and
d i i transmission-line
designing
t
i i li circuits.
i it
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5.9 SMITH CHART
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SMITH CHART (Cont’d)
• Graphical tool for use with transmission line circuits
and microwave circuit elements.
• Only lossless transmission line will be considered.
• Can be thought as two graphs in one ;
¾
It plots the normalized impedance at any
point along a T-line.
¾
It plots the reflection coefficient at any
point along the line.
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SMITH CHART (Cont’d)
The transmission
line calculator,
commonly
referred as the
Smith Chart
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Smith Chart
•
•
Impedances represented by normalized values, Z0.
Reflection coefficient is
zL =
•
1+ Γ
1− Γ
Normalized load
admittance is
1 1− Γ
yL =
=
(dimensionless)
zL 1 + Γ
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Example 5.11 Determining ZL using the Smith Chart
Given that the voltage standing-wave ratio is S = 3 on
a 50-Ω line,, that the first voltage
g minimum occurs at 5
cm from the load, and that the distance between
successive minima is 20 cm, find the load impedance.
Solution
The first voltage
g minimum is at
5
l min =
= 0.125λ
40
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Solution 5.11 Determining ZL using the Smith Chart
From Smith Chart,
rL = S = 3
The normalized load
impedance at point C is
z L = 0 .6 − j 0 .8
Multiplying by Z0 = 50Ω ,
we obtain
Z L = 50(0.6 − j 0.8) = (30 − j 40 )Ω
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USING THE SMITH CHART
The Smith Chart is a plot of normalized impedance.
For example, if a Z0 = 50 Ω transmission line is
terminated in a load ZL = 50 + j100 Ω as below:
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SMITH CHART (Cont’d)
To locate this p
point on Smith Chart,, you
y
would first
normalize the load impedance, zNL = ZL/ZN or
zNL = ZL/Zo to obtain zNL = 1 + j2 Ω
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SMITH CHART (Cont’d)
Th normalized
The
li d load
l d
impedance is located
att the
th intersection
i t
ti
off
the r =1 circle and the
x =+2
+2 circle.
circle
zNL = 1 + j2 Ω
Fig. 6
6-16
16
(c)
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To see how Smith Chart is also a
plot of the reflection coefficient
The reflection coefficient at any point z along line
h a magnitude
has
it d ΓL and
d an angle
l θ Γ that
th t equall tto
its angle at the load plus 2βz
Γ = ΓRe + jTIm = ΓL e j 2 βz
Γ = ΓL e
jθ Γ
The actual value of ΓL is found by taking the
distance from the center of the chart to the p
point
divided by the distance from the center of the
chart to the periphery ( ΓL = 1 ).
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To avoid this calculation, the magnitude ΓL
can be
measured
d using
i a scale
l for
f magnitude
it d off reflection
fl ti
coefficient provided below the Smith Chart.
θ Γ is indicated
The
h angle
l off the
h reflection
fl
coefficient
ff
d
d on
the angle of reflection coefficient scale shown outside the
ΓL = 1 circle on chart.
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SMITH CHART (Cont’d)
Scale for magnitude
g
off reflection
f
coefficient
ff
Scale for angle
g off
reflection coefficient
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SMITH CHART (Cont’d)
The normalized load
zL=1+j2
1 j2 corresponds
d to
t
ΓL = 0.7 andθ Γ =
o.
45
For this example,
example
Γ = ΓL e jθ
Γ
= 0 . 7 e j 45
Verify:
Z −Z
ΓL = L o
ZL + Z0
ΓL =
0
50+ j100− 50
50+ j100+ 50
ΓL = 0.5 + j0.5 = 0.707e j 45
o
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ΓL as Function of Position along the
T-line
The value of normalized impedance ZNL and reflection
coefficient ΓL are function of position along the T
T-line
line (Fig.6.17)
Fig.6.
17
After locating the normalized impedance point, draw the
jθ
constant ΓL circle. Since Γ = ΓL e
, we let θΓ change and
as we ΓL hold constant,
constant and this traces out a circle of
constant ΓL on the chart. For example, the line is 0.3λ
length (Figure 6.17).
Γ
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SMITH CHART (Cont’d)
• Move along the constant ΓL e jθ
transmission line
line.
Γ
circle is akin to moving along the
Moving away from the load (towards
generator)) corresponds
d
to moving
i
i
in
the
h
clockwise direction on the Smith Chart.
¾
Moving towards the load corresponds to
moving in the anti-clockwise direction on the
Smith Chart.
¾
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• Move towards the generator by:
¾Drawing
a line from the center of chart to
outside Wavelengths Toward Generator (WTG)
scale, to get starting point “a” at 0.188λ
0 3λ moves along the constant ΓL e jθ
0.3λ
circle to 0.488λ on the WTG scale.
¾Adding
¾Read
Γ
the corresponding normalized input
impedance point “c”, zNIN = 0.175 - j0.08Ω
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SMITH CHART (Cont’d)
Denormalizing, to find
an input impedance,
impedance
Z IN = z NIN xZ 0
ZIN = 8.75 – j4 Ω
voltage standing wave ratio
(VSWR) can be determined by
reading the value r at the θr
= 0o crossing the constant
circle (p
(point “b” in Figure
g
6.17).
VSWR = 5.9
Figure 6
6.17
17
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Locating the load
impedances terminating a 50 Ω T-line
a Æ ZL = 0 (short cct)
b Æ ZL = ∞ (open cct)
c Æ ZL = 100 + j100 Ω
d Æ ZL = 100 - j100 Ω
e Æ ZL = 50 Ω
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Take out your Smith Chart,
pencil and compass!
LETS TRY!!
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EXAMPLE 2
Repeat Example 1 using the Smith Chart.
ZO =
50 Ω
ZL = 50 –
j25 Ω
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SOLUTION TO EXAMPLE 2
(i) Locate the normalized load, and label it as point a,
where
h
it corresponds
d to
t
zNL = 1 – j0.5 Ω
(ii) Draw constant ΓL e jθ
Γ
circle.
(iii) It can be
b seen that
th t
ΓL = 0.245e
j − 76 0
and
VSWR = 1.66
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SOLUTION TO EXAMPLE 2 (Cont’d)
(iv) Move from point a (at 0.356 λ) on the WTG
scale, clockwise toward generator a distance
λ/8 or 0.125 λ to point b, which is at 0.481
λ.
We could find that at this point, it corresponds
to
z NIN = 0.62 − j 0.07
Denormalizing it
ZIN = 50 (0.62 –
j0 07)
j0.07)
Z IN = 31 − j 3.5Ω
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EXAMPLE 3
The input impedance for a 100 Ω lossless
transmission line of length 1.162 λ is
measured as 12 + j42Ω.
j42Ω Determine the
load impedance.
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SOLUTION TO EXAMPLE 3
(i) Normalize the input impedance:
Z in 12 + j 42
zin =
=
= 0.12 + j 0.42
100
Z0
(ii) Locate the normalized input impedance and
label it as point a
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SOLUTION TO EXAMPLE 3 (Cont’d)
(iii) Take note the value of wavelength Wavelengths Toward
Load (WTL) scale, for point a at WTL scale.
At point a, WTL = 0.436λ
(iv) Move a distance 1.162λ towards the load to point b
WTL = 0.436λ + 1.162λ
= 1.598λ
But, to plot point b, 1.598λ – 1.500λ = 0.098λ
Note: One complete rotation of WTL/WTG = 0.5λ
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SOLUTION TO EXAMPLE 3 (Cont’d)
(v) Read the point “b” as normalized load
i
impedance
d
z NL = 0.15 − j 0.7
Denormalized it:
ZL = zNL Zo = 100 (0.15 – j0.7)
ZL = 15 – j70 Ω
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EXAMPLE 4
On a 50 Ω lossless transmission line,
the VSWR is measured as 3.4.
3 4 A voltage
maximum is located 0.079λ away from
th load
the
l d (towards
(t
d generator).
t ) D
Determine
t
i
the load.
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SOLUTION TO EXAMPLE 4
jθ
(i) Use the given VSWR to draw a constant ΓL e
Γ
circle.
(ii) Then move from maximum voltage at
WTG = 0.250λ (towards the load) to point a
at WTG = 0.250λ - 0.079λ = 0.171λ.
(iii) At this point we have ZNL = 1 + j1.3 Ω,
or ZL = 50 + j65 Ω.
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EXAMPLE 5 (TRY THIS!)
Use Smith Chart to determine the input impedance
Zin of the two line configuration shown as below:
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ANSWER FOR EXAMPLE 5
ZIN = 65.7 – j 124.7Ω
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We often desired that all of the power propagating along a T-line be
dropped across the terminating load impedance.
impedance
An impedance mismatch can result in the reflection of much of this
power
• The transmission line is said to be matched when
Z0 = ZL which no reflection occurs.
• The purpose of matching network is to
transform the load impedance ZL such that the
input impedance Zin looking into the network is
equal to Z0 of the transmission line.
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5-10 Impedance Matching
•
•
Transmission line is matched to the load when
Z0 = ZL.
Alternatively, place an impedance-matching
network between load and transmission line.
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IMPEDANCE MATCHING (Cont’d)
Adding an impedance matching networks
ensures that all power make it or delivered to
the load.
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IMPEDANCE MATCHING (Cont’d)
z
Techniques of impedance matching :
¾
Quarter wave transformer
Quarter-wave
¾
Single / double stub tuner
¾
Lumped element tuner
¾
Multi section transformer
Multi-section
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QUARTER WAVE TRANSFORMER
The
quarter
q
wave
transformer
matching
g
network only can be constructed if the load
impedance
p
is all real ((no reactive component)
p
)
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QUARTER WAVE TRANSF. (Cont’d)
To find the impedance looking into the quarter
wave long section of lossless ZS impedance line
terminated in a resistive load RL:
RL + jZ S tan βl
Zin = Z S
Z S + jRL tan βl
2π λ π
= ,
But, for quarter wavelength, β l =
λ 4 2
∴ tan β l = ∞
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QUARTER WAVE TRANSF. (Cont’d)
So,
2
ZS
Z in =
= Z0
RL
Rearrange to get impedance matched line,
Z S = Z 0 RL
See Drill 6.14
Suppose a 50 Ω T-line is terminated in a 100
load. Determine the required impedance of a
quarter-wave matching
hi section
i off T-line.
T li
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Admittance of Shunt Stubs
• It much more convenient to add shunt elements
rather than series elements Æ Easier to work in
terms of admittances.
• Admittance
Ad itt
iis th
the iinverse off impedance:
i
d
Fig. 6-25a shows those relationship
Y=
1
Z
Fig. 6
6-25a
25a
The characteristic admittance Yo = 1/Zo and
the load admittance YL = 1/Z
/ L.
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Admittance of Shunt Stubs (cont’d)
The convenient of admittances is the shunt values
may be
b added:
dd d YTOT = YL1 + YL2
The Smith Chart is also a chart of normalized admittance.
YL 1
yL = =
The normalized load admittance is
Y0 Z L
Drill 6.16
6 16
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Consider the normalized impedance zL = 2 + j1.
We can calculate the normalized admittance as
yL = 1/zL = 0.040 – j0.20
With Smith chart, it is easy to find normalized
admittance – move to a p
point on the opposite
pp
side of
the constant ΓL e jθ circle.
Γ
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Example 5.12 SingleSingle-Stub Matching
50-Ω transmission line is connected to an antenna
with load impedance
p
ZL = ((25 − j50).
j ) Find the position
p
and length of the short-circuited stub required
to match the line.
Solution
The normalized load impedance is
Z L 25 − j 50
zL =
=
= 0.5 − j
Z0
50
Located at point A.
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Solution 5.12 Single
Single--Stub Matching
Value of yL at B is yL = 0.4 +
0.115λ on the WTG scale.
At C,
yd = 1 + j1.58
j 0.8
which locates at position
located at 0.178λ on the WTG scale.
Distant B and C is d = (0.178 − 0.155)λ = 0.063λ
Normalized input admittance at
the juncture is
yin = ys + y d
1 + j 0 = ys + 1 + j1.58
ys = − j1.58
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Solution 5.12 Single
Single--Stub Matching
Normalized admittance of −j 1.58 at F and position
0.34λ on the WTG scale g
gives l1 = (0.34 − 0.25)λ = 0.09λ
At point D, yd = 1 − j1.58
Distant B and C is d 2 = (0.322 − 0.115)λ = 0.207λ
Normalized input admittance
ys = + j1.58
at G.
Rotating from point E to point G, we get
l2 = (0.25 + 0.16 )λ = 0.41λ
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Solution 5.12 Single
Single--Stub Matching
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SHUNT STUB MATCHING NETWORK
In Smith Chart term, the objective of SSM network as
shown in Figure 6.28a
6 28a is to move to the center of the
chart.
Fig.6-28a The generic layout
of the shorter SSM network
Since a shunt stub will be added, we will work in
admittance chart.
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SHUNT STUB MATCHING NETWORK
The matching network has to transform the real part
off load
l d impedance,
i
d
RL to
t Z0 and
d reactive
ti part,
t XL to
t
zero Æ Use two adjustable parameters – e.g. shuntstub.
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SHUNT STUB MATCHING NET. (Cont’d)
Thus, the main idea of shunt stub matching
network is to:
(i) Find the stub length d and the line length l in
order to get yd and yl .
(ii) Ensure total admittance ytot = yd + yl = 1 for
complete matching network.
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Procedure for Constructing a SSM
Network
• Locate the normalized load impedance zNL.
• Draw constant SWR ΓL circle and use it to locate yNL.
• Move clockwise (WTG) along ΓL e jθ circle to intersect
with 1 ± jb Æ value of yd.
Γ
• The length moved from yNL towards yd is the through
line length,
g
d.
• Locate yl at the point m jb.
• Depends
p
on the shorted/open
/ p
stub,, move along
g the
periphery of the chart towards yl (WTG).
• The distance traveled is the length of stub, l .
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SHORTED SHUNT STUB MATCHING
Generic layout of the shorted shunt stub
matching network:
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EXAMPLE 6 (TRY THIS!) = Ex. 6-8
Construct the shorted shunt stub
matching network for a 50Ω line
terminated in a load ZL = 20 – j55Ω
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SOLUTION TO EXAMPLE 6
1.
Locate the normalized load impedance,
ZNL = ZL/Z0 = 0.4 – j1.1Ω
2.
Draw constant ΓL e jθ circle.
3.
Locate YNL. (0.112λ at WTG)
4.
Moving to the first intersection with the
Γ
1 ± jb circle, which is at 1 + j2.0 Æ yd
5 Get the value of through line length
5.
length, d
Æ from 0.112λ to 0.187λ on the WTG scale,
so d = 0.075λ
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SOLUTION TO EXAMPLE 6 (Cont’d)
6. Locate the location of short on the Smith Chart
(note: when short circuit, ZL = 0, hence YL = ∞)
Æ
Æon
the
h right
i h side
id off the
h chart
h
with
i h WTG=0.25λ
T
5λ
7. Move clockwise (WTG) until point m jb, which
is at 0 - j2.0, located at WTG= 0.324λ Æ yl
8. Determine the stub length, l
Æ 0.324λ – 0.25λ = 0.074 λ
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SOLUTION TO EXAMPLE 6 (Cont’d)
Thus, the values are:
d = 0.075 λ
l = 0.074 λ
yd = 1 + j2.0 Ω
yl = -j2.0 Ω
Where YTOT = yd + yl = (1 + j2.0) + (-j2.0) = 1
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OPEN END SHUNT STUB MATCHING
Generic layout of the open ended shunt stub
matching network:
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EXAMPLE 7 (TRY THIS!) = Ex. 6-9
Construct an open ended shunt stub
matching network for a 50Ω line
terminated in a load ZL = 150 + j100 Ω
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SOLUTION TO EXAMPLE 7
1.
Locate the normalized load impedance,
ZNL = ZL/Z0 = 3.0 + j2.0Ω
2.
Draw constant ΓL e jθ
3.
Locate YNL. (0.474λ at WTG)
4.
Moving to the first intersection with the
Γ
circle.
1 ± jb circle, which is at 1 + j1.6 Æ yd
5 Get the value of through line length
5.
length, d
Æ from 0.474λ to 0.178λ on the WTG scale,
so d = 0.204λ
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SOLUTION TO EXAMPLE 7 (Cont’d)
6. Locate the location of open end on the Smith
Chart (note: when short circuit, ZL = ∞, hence
YL = 0) Æon the left side with WTG = 0.00λ
7. Move clockwise (WTG) until point m jB, which
is at 0 – j1.6, located at WTG= 0.339λ Æ yl
8. Determine the stub length, l
Æ 0.339λ – 0.00λ = 0.339 λ
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SOLUTION TO EXAMPLE 7 (Cont’d)
Thus, the values are:
d = 0.204 λ
l = 0.339 λ
yd = 1 + j1.6 Ω
yl = -j1.6 Ω
Where YTOT = yd + yl = (1 + j1.6) + (-j1.6) = 1
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04)
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IMPORTANT!!
I both
In
b th previous
i
example,
l we chose
h
the
th first
fi t
intersection with the1 ± jB circle in designing our
matching
t hi
network.
t
k We
W could
ld also
l have
h
continued
ti
d
on to the second intersection.
Thus, try both intersection to determine which
solution p
produces max/min
/
length
g of through
g
line, d or length of stub, l.
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EXERCISE (TRY THIS!)
Determine the through line length and stub
length for both example above by using second
intersection.
F shorted
For
h t d shunt
h t stub
t b (example
(
l 6):
6)
d = 0.2 λ and l = 0.426 λ
For open ended shunt stub (example 7):
d = 0.348 λ and l = 0.161 λ
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5-11 Transients on Transmission Lines
•
•
Transient response is a time record of voltage
pulse.
pulse
An example of step function is shown below.
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5-11.1 Transient Response
•
Steady-state voltage V∞ for d-c analysis of the
circuit is
V∞ =
Vg Z L
Rg + Z L
where Vg = DC voltage source
•
St d t t currentt iis
Steady-state
Vg
V∞
I∞ =
=
Z L Rg + Z L
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CHAPTER 5
END
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