transmission lines and filters assign
TRANSCRIPT
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MARKS BREAKDOWN
ASSESSOR’S FEEDBACK:
MAXIMUM AWARDE
TOTAL MARKS
STRUCTIONS: If any coursework assessment is found to be copied from other candidates using unacceptab
COURSE WORK TASK SHEETart A (To be fe! b" STUDENT#
SEMESTER: SPRIN$ %&'
Date o) S*b+,--,o.: /0&10%
Part B (To be fe! b" ASSESSOR#
De2art+e.t o) Ee3tro.,3- 4 Co++*.,3at,o.
MODERATOR SI$NATURE:
ASSESSOR’S
SI$NATURE
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5':
k =Z L−Z 0
Z L+Z 0
For inductor$
k I = jX −Z
0
jX +Z 0
For capacitor$
k C =− jX −Z
0
− jX +Z 0
k I k C =( jX −Z 0)(− jX −Z 0)( jX +Z
0)(− jX +Z
0)
k 2=
X 2− jX Z 0+ jX Z 0+Z 0
2
X 2+ jX Z
0− jX Z
0+Z
0
2
k 2=
X 2+Z 0
2
X 2+Z
0
2
k 2=1
|k | &
5%:
'hen a transmission line which attenuation constant is fre(uency
independent and the phase shift constant is linearly dependent on fre(uency
is called distortionless line. )shiva, *+&*
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#ondition for ditortionless line is$
R
L=
G
C
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5/:
Z T =√ Z ¿× Z L
Z ¿=Z T
2
Z L=150
2
150 =150Ω
|k |=Z L−Z 0Z L+Z 0
=150−300150+300
=0.333
VSWR=1+|k |1−|k |
=1+0.333
1−0.333=1.99
Z 0=√ Z ¿× Z L=√ 300×150=212.13Ω
56:
Z ¿=Z 0 [ Z L+ jZ 0 tan ( βl )Z 0+ jZ L tan ( βl ) ]
40=Z 0 l [ 200+ j 100+ jZ 0 l tan ( βl )Z
0 l+(200+ j 100) j tan ( βl ) ]
40=Z 0 l [ 200+ j 100+ jZ 0 l tan ( βl )Z
0 l+(200+ j 100) j tan ( βl ) ]
-ssuming that tan ( βl )= x
200Z 0 l+ j 100Z 0 l+ j Z 0 l2
x=40 Z ol+40 (200+ j100) jx
200Z 0 l+ j 100Z 0 l+ j Z 0 l2
x=40 Z ol−4000 x+ j8000 x
Now by taking all the real part imaginary parts separately,
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200Z 0 l=40Z ol−4000 x
200Z 0 l−40Z ol=−4000 x
Z 0 l=−25 x e(n &
j 100Z 0 l+ j Z 0 l2
x= j 8000 x
j(100 Z 0l+Z 0 l2
x )= j 8000 x
100Z 0 l+Z 0 l2
x=8000 x
Z 0 l2 x−8000 x=−100Z 0 l
x (Z 02−8000 )=−100Z 0
x= −100Z
0
( Z 02−8000)
/his is e(n *
No substitute e(n * in e(n &
Z 0 l=−25
−100 Z 0
(Z 02−8000 )
Z 02−8000=2500
Z 02=2500+8000
Z 02=10500
Z 0=102.46
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tan ( βl )= x
tan ( βl )= −100×102.46
(102.462−8000 )
tan ( βl )=−4.10
No take the tan inverse on both sides,
βl=−1.33
βl=−1.33+π
βl=1.81
2 π
λ l=1.81
l=1.81 λ
2 π
l=0.288 λ
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5:
Z L=100+ j 60Ω , Z 0=50Ω
z L=
Z L
Z 0=100+ j 60
50 =2+ j 1.2
a. k L=
A
B=
4
8.2=0.487∠29
0
b. Z L isat 0.21 λ+0.3 λ=0.51 λ−0.5 λ=0.01 λ
Z ¿ isat 0.01 λ
k ¿= AB = 4
8.3=0.481∠1730
c. SVWR=2.85
d. V max
lmax=0.04 λ
+¿=V s×( Z
0
Z 0+Z s )V
0
¿
+¿=10×( 5050+0 )V
0
¿
+¿=10V
V 0¿
+¿
V max=(1+k ¿ )×V 0¿
V max=(1+0.481 )×10=14.81V
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51:
Z L=150+ j 200Ω , Z 0=400Ω
z L=
Z L
Z 0=150+ j 200
400 =0.375+ j 0.5
l=37cm λ=15cm
l
λ=37 cm
15cm=2.46
l=2.46 λ
a. 0istance of 1st
V max=(0.25−0.08 ) λ=0.17 λ
0istance of 1st
V min=(0.17+0.25 ) λ=0.42 λ
b. Number of voltage ma1imas2, number of voltage minimas5.
c. 0istance of stub (0.172−0.08 ) λ=0.09 λ
3ength of open stub (0.352−0.25 ) λ=0.102 λ
d. 0istance 0.09 λ=0.09×0.15=0.0135m
3ength 0.102 λ=0.102×0.15=0.0153m
ln!t"=0.0153 m #istanc=0.0135m
Z V s
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4eferencesshiva, *+&*. DISTORTIONLESS LINE. 6nline7
-vailable at$ http$88www.winnerscience.com8transmission9lines8distortionless9line8
-ccessed *: may *+&57.