smith chart lecture
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Example 2.2The load impedance of 40+j70 Ω terminates at 100Ω tx
line that is 0.1λ long. Find Γ at the load, Γ at input terminals, zin, SWR and return loss.
Solution:Normalized input impedance is zL=ZL/Zo=0.4+0.7j
Mark the point on the chartDraw a radial line form center to plot edge through
marked point, and read the angle 104o
Open compass from center to that point (radius)Measure SWR and |Γ| from the scale below the chart
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Example 2.2
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104o
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ZL connected to λ/4 Line If an arbitrary ZL is connected to λ/4 line
Under normalized condition
Length of λ/4 corresponds to 180o rotation in Smith chart
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2tan( )
tan( )L o o
in oo L L
Z jZ l ZZ Z
Z jZ l Z
1in
L
zz
(Admittance)
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Zin when ZL at λ/4
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ZL
Zin=1/ZL
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The Admittance Smith chart
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The Slotted lineIt is a device to measure arbitrary load attached to lineIt’s a tx line configuration that is used to find E field
amplitude of standing wave on tx lineDevice is replaced by VNAsSWR and 1st null from load can be measured
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1 | |
1 | |SWR
1
1
SWR
SWR
min2 l
| | je where
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Quarter Wave TransformerCase when Zo of tx line and ZL are irreplaceable
Impedance needs to be matched to avoid reflections
(1) λ/4 transformation (2) Stub matchingInitially Zo of tx line and ZL are given
Insert a section of lossless tx line between tx line ZL
Length of segment is λ/4 and impedance Z1
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In order for Γ = 0, we put Zin = Zo
In this way SWR=1 on tx lineThe impedance will be matched only for one frequencyThe method is limited to real loads onlyComplex loads can be made real by adding proper
lengths of tx lines8
1 o LZ Z R
21
inL
ZZ
R
where βl = π/2tan( )
tan( )L o
in oo L
Z jZ lZ Z
Z jZ l
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Example 2.5Match a load of 100 Ω with a 50 Ω line using quarter
wave transformer at frequency 3GHzSolution:
Now λ for 3GHz is 0.1m, λ/4 = 0.025m Frequency response of Γ(quarter wave transformation)
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1 (50)(100) 70.71Z
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Quarter wave transformer: multiple reflection view point
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11
1
o
o
Z Z
Z Z
0 12
0 1
Z Z
Z Z
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1
L
L
R Z
R Z
11
1
21
2
2o
o
o
ZT
Z Z
ZT
Z Z
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Generator and Load MismatchWe discussed terminated tx lines irrespective of
generatorInvestigation of maximum power transfer to load
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( 2 )
( 2 )
tan( ) 1
tan( ) 1
j lL o l
in o o j lo L l
Z jZ l eZ Z Z
Z jZ l e
* 21 1 1Re{ } | | Re
2 2in in inin
P V I VZ
| | in
in gin g
ZV V
Z Z
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Let Zin = Rin+jXin and Zg=Rg+jXg then
Consider Zl = Zo, matched load with tx line
Max power is delivered under conjugate matchingRin = Rg, Xin = -Xg or Zin=Zg*
It means only half of Pin is delivered to the load
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21| |
2 ( ) ( )in
gin g in g
RP V
R R X X
21 1| |
2 4gg
P VR
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The Terminated Lossy LineIn this case γ = α + jβHere
( )
( )
z zo
z zo
o
V z V e e
VI z e e
Z
2( ) | | ll e
tan( )
tan( )L o
in oo L
Z Z lZ Z
Z Z l