j. j. −=−Ω λ - University Of Illinoisemlab.uiuc.edu/ece451/appnotes/app_smith.pdf · Smith...
Transcript of j. j. −=−Ω λ - University Of Illinoisemlab.uiuc.edu/ece451/appnotes/app_smith.pdf · Smith...
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Smith Chart
Find
1) Reflection coefficient at load
2270 3 0 4 0 6 jR Rz . j . . e °= − ⇒ Γ =
2) SWR on the line
SWR=4.0
3) dmin
( )0 5 0 435 0 065mind . . .= − λ = λ
4) Line impedance at 0.05λ to the left
( )50 0 26 0 09 13 4 5. j . j .− = − Ω
5) Line admittance at 0.05λ
( )3 5 1 2 50 0 068 0 025. j . / . j .+ = +
6) Location nearest to load where Real[y]=1
0 14 0 325 0 185 0 14. . j . .λ = λ − λ = λ
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VSW minimum occurs at 0.30λ from the termination of a lossless 50-Ω line. Find angle
of reflection coefficient
Answer: 36o
If the VSWR is 2.0 what is zR?
Answer: zR =1.57+j0.7
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Single Stub
Find location and length of stub
zR = ZR/50=0.6-j0.8
yR = 0.6+j0.8
First Method
Rotate on constant VSWR circle from 0.125λ to 0.1665λ until intersection with unit
conductance circle at 1 1 1 16'y j .= + . Distance ( )0 1665 0 125 0 0415sd . . .= λ − λ = λ Move
toward center of chart. Change in susceptance: j(0-0.16)=-j1.16
The length of the stub is such that
1 1 16s
.tan l
= −β
or ( )0 363 0 25 0 113sl . . .= − λ = λ
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Second Method
yR = 0.6+j0.8
Rotate on constant VSWR circle from until intersection with unit conductance circle at
0.335λ at 1 1 1 16'y j .= − . ( )0 335 0 125 0 210sd . . .= λ − λ = λ Move toward center of chart.
Change in susceptance: is +j1.16. Therefore, the length of the stub is such that
1 1 16.tan l
= +β
or ( )0 632 0 25 0 382sl . . .= − λ = λ
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Analysis of double stub
1) Get yR from ZR
2) Rotate on constant VSWR circle by d1 to get y1'
3) Move on constant g circle by susceptance of stub1 to get to y1.
4) Rotate on constant VSWR circle by d2 until intersection with g =1 circle.
5) Move toward center of chart.1
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Double Stub – Example 1
Find ls1 and ls2
( )30 40 50 0 6 0 8Rz j / . j .= − = −
1st Solution 2nd Solution
1) Convert to yR by rotating by 180 degrees.
0 6 0 8Ry . j .= + (yR=y1’ since d1=0)
2) Draw auxiliary circle (g=1 circle rotated by 3λ/8)
3) Auxiliary circle intersects g=0.6 circle at 0.6-j0.1=y1.
4) Rotate by 3λ/8 until intersection at g=1 circle; y2’=1-j0.5.
5) Move along constant conductance circle until center.
Calculate ls1:
( )1
1 0 1 0 8 0 9s
j . . j .j tan l
= − − = −β
( )1 0 384 0 25 0 134sl . . .= − λ = λ
( )2
1 0 0 5 0 5s
j . j .j tan l
= − − = +⎡ ⎤⎣ ⎦β
( )2 0 574 0 25 0 324sl . . .= − λ = λ
1) Convert to yR by rotating by 180 degrees.
0 6 0 8Ry . j .= + (yR=y1’ since d1=0)
2) Draw auxiliary circle (g=1 circle rotated by 3λ/8)
3) Auxiliary circle intersects g=0.6 circle at 0.6-j1.9=y1.
4) Rotate by 3λ/8 until intersection at g=1 circle; y2’=1+j2.5.
5) Move along constant conductance circle until center.
Calculate ls1:
( )1
1 1 9 0 8 2 7s
j . . j .j tan l
= − − = −β
( )1 0 307 0 25 0 057sl . . .= − λ = λ
[ ]2
1 0 2 5 2 5s
j . j .j tan l
= − = −β
( )2 0 31 0 25 0 06sl . . .= − λ = λ
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Double Stub – Example 2
Find ls1 and ls2
( )50 100 100 0 5 1 0Rz j / . j .= + = +
1) Normalize admittance, 0 4 0 8Ry . j .= −
2) Rotate by λ/4 toward source 1 0 5 1 0'y . j .= +
3) Draw auxiliary circle (g=1 circle rotated by λ/8)
4) Auxiliary circle intersects g = 0.5 circle at y1 = 0.5+j0.14. ( )0 14 1 0 0 86change j . . j .= − = − ,
5) Add stub ( )0 388 0 25 0 138. . .− λ = λ
6) Rotation by λ/8 must end on g = 1 circle, 2 1 0 73'y j .= +
7) Add stub2 such that 2
1 0 73s
j .j tan l
= −β
8) ( )2 0 4 0 25 0 15sl . . .− λ = λ
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