transmission lines and filters assign

8/18/2019 Transmission Lines and Filters Assign

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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


2/13

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,


5/13

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


7/13

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


8/13


9/13

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.

transmission lines and filters assign

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