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Solved Anna University Problems

(NOV/DEC 2010 - NOV/DEC 2019)

EC8651-Transmission Lines and RF

Systems

UNIT 1 - TRANSMISSION LINE THEORY

UNIT 1 - TRANSMISSION LINE THEORY

PROBLEMS

1. Find the characteristic impedance of a line at 1600Hz if ZOC = 750∠-30° Ω

and ZSC = 600 ∠-20° Ω. [EC6503-NOV/DEC 2019](2 Marks), [EC6503-

NOV/DEC 2016](2 Marks), [EC6503-APR/MAY 2019](2 Marks)

Solution:

Given: f= 1600 Hz, ZOC = 750∠-30° Ω, ZSC = 600∠-20° Ω

Z O = ZOC×Z SC

  Z (^) O = 750 ∠− 30 × 600 ∠− 20

ZO = 670.82 ∠ -

° Ω

2. The following measurement are made on a 25km line at a frequency of 796

Hz. ZSC = 3220 ∠-79.29° Ω, ZOC = 1301∠76.67° Ω. Determine the primary

constants of the line. [EC2305-NOV/DEC 2014](10 Marks)

Solution:

Given: l=25Km, f=796Km, ZSC = 3220∠-79.29° Ω, ZOC = 1301∠76.67° Ω

Z O = ZOC×Z SC

  Z (^) o = 1301 ∠ 76. 67 × 3220 ∠− 79. 29

ZO = 2046.76 ∠ -1.

°

OC

SC

Z

Z

tanh γ =





tanh γ ∠



1. 5732 77. 98 e 1

e 1 tanh γ 2 γ

2 γ

= ∠ −

1. 3276 j 1. 5387 e 1

e 1

2 γ

2 γ

= −

e 1 ( 0. 3276 j 1. 5387 )(e 1 )

2 γ 2 γ − = − +

e 1 0. 3276 e 0. 3276 j 1. 5387 e j 1. 5387

2 γ 2 γ 2 γ − = × + − × −

e [ 1 0. 3276 j 1. 5387 ] 1 0. 3276 j 1. 5387

2 γ − + = + −

e [ 0. 6724 j 1. 5387 ] 1. 3276 j 1. 5387

2 γ

• = −

98. 8 nF/Km
99. 88

C

4

=

×

−

C = 98.8 nF/Km

3. A transmission line has Zo = 745∠-12° Ω and is terminated in ZR=100Ω.

Calculate reflection factor. [EC6503-APR/MAY 2017](2 Marks)

Solution:

Given: Zo = 745∠-12° Ω, ZR=100Ω

R+^ O

R O

Z Z

2 Z Z

ReflectionFactor,k=

× ∠−





ReflectionFactor,k=





ReflectionFactor,k=

Reflection Factor = 0.6475 ∠ 4.

°

4. A transmission line has Zo = 745∠-12° Ω and is terminated in ZR=100Ω.

Calculate reflection loss in dB. [EC2305-APR/MAY 2011](2 Marks)

Solution:

Given: Zo = 745∠-12° Ω, ZR=100Ω

dB |k|

Reflection lossindB= 20 log 

R+^ O

R O

Z Z

2 Z Z

Reflection Factor,k=

× ∠−





Reflection Factor,k=





Reflection Factor,k=

Reflection Factor = 0.6475 ∠ 4.

°

dB

Reflection lossindB= 20 log  

Reflection loss in dB = 3.7752 dB

5. A transmission line has a characteristic impedance of 400 Ω and is

terminated by a load impedance of (650-j475) Ω. Determine the reflection

coefficient. [EC2305-NOV/DEC 2013](2 Marks)

Solution:

Given: ZO = 400 Ω, ZR = (650-j475) Ω

1050 j 475

250 j 475

( 650 j 475 ) 400

( 650 j 475 ) 400

Z Z

Z Z

K

R O

R O

−

 K = 0. 3675 −j 0. 2861 = 0. 4658 ∠− 37. 9

K = 0.4658 ∠ -37.9 °

6. A transmission line has a characteristic impedance of 600 Ω. Determine

magnitude of reflection coefficient if the receiving end impedance is (650-j475)

Ω. [EC2305-NOV/DEC 2013](2 Marks)

Solution:

Given: ZO = 600 Ω, ZR = (650-j475) Ω

1250 j 475

50 j 475

( 650 j 475 ) 600

( 650 j 475 ) 600

Z Z

Z Z

K

R O

R O

−

 K = 0. 1611 −j 0. 3188 = 0. 3571 ∠− 63. 18

K = 0.3571 ∠ -63.18 °

7. A transmission line has a characteristic impedance of 300 Ω and is

terminated in a load impedance of (150-j150) Ω. Determine the reflection

coefficient. [EC2305-APR/MAY 2014](2 Marks)

Solution:

Given: ZO = 300 Ω, ZR = (150-j150) Ω

450 j 150

150 j 150

( 150 j 150 ) 300

( 150 j 150 ) 300

Z Z

Z Z

K

R O

R O

−

 K =− 0. 2 −j 0. 4 = 0. 4472 ∠− 116. 57

K = 0.4472 ∠ -116.57 °

8. Find the reflection coefficient of a 50 ohm line when it is terminated by a

load impedance of 60+j 40 ohm. [EC6503-NOV/DEC 2015](2 Marks),

[EC2305-NOV/DEC 2015](2 Marks)[EC2305-APR/MAY 2013](2 Marks)

Solution:

Given: ZO = 50 Ω, ZR = (60+j40) Ω

10. A transmission line has a characteristic impedance of (683-j138) ohm. the

propagation constant is (0.0074+j0.0356) per Km. Determine the values of R

and L of this line if the frequency is 1000Hz.

[EC2305-APR/MAY 2014](6 Marks)

Solution:

Given: Zo = 683-j138Ω, γ=P= 0.0074+j0.0356, f=1000Hz

ω = 2πf = 2 x 3.14 x 1000 = 6280

R+jωL = ZoP = (683-j138) x ( 0.0074+j0.0356)

= 9.967+j23.

Comparing real and imaginary terms,

R = 9.967 ohms/Km

ωL = 23.

3. 7092 mH/ Km 6280

L = =

L=3.7092 mH/Km

11. A Communication line has L=3.67 mH/Km, G=0.08x10-6^ mho/Km,

C=0.0083 μF/Km and R=10.4 Ω/Km. Determine the characteristic impedance,

propagation constant, phase constant, velocity of propagation, sending end

current and receiving end current for a given frequency f=1000 Hz. sending

end voltage is 1V and transmission line length is 100Kms.

[EC6503-NOV/DEC 2016] (16 Marks) [EC6503-APR/MAY 2018](13 Marks)

Solution:

Given: Es = 1V, f=1000 Hz, l=100 Km, R=10.4 Ω/Km, L = 0.00367H/Km, G

=0.08x10-6^ mho/Km, C = 0.0083 μF/Km

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL

Y

Z

Z (^) o

Z = R+jωL = 10.4 + j(6280)(.00367) = 25.285∠65.71°

Y = G+jωC = 0.08x10-6^ +j(6280)(0.0083x10-6) = 5.212x10-5∠89.

°





Z

o (^5) × ∠

−

Z

3 o

  ∠ − ×

−

Zo = 696.46 ∠ -12.

° Ω

The propagation constant is given by,

P =γ= ZY = ( R+jωL)( G+jωC)

  P γ 25. 285 65. 71 5. 212 10 89. 91

5 = = ∠ × × ∠

−

P γ 1. 319 10 ( 65. 71 89. 91 )

3   = = × ∠ +

−

P=γ = α+jβ = 0.0363 ∠77.81° (Or) 7.6649x10-3+j0.

attenuation constant, α = 7.6649x10-3^ Np/Km

and phase constant, β = 0.0355 rad/Km

The velocity of propagation is given by,

β

ω v (^) p =

1. 769 10 Km/ sec
2. 0355

v

5 p = = ×

176. 99 m
177. 0355

2 π

β

2 π λ = = =

NOTE: ZR value is not given in the question. so, assume the line is perfectly

matched line to proceed further. i.e., ZR=ZO, ZS=ZO

g S

g

g in

g S Z Z

E

Z Z

E

I

1. 4358 10 12. 1 A

I

3 S

 

= × ∠

−

IS = 1.4358x10-3^ ∠ 12.

°

We know that,

− ZYs

R o

ZYs R o

0

R R o e Z Z

Z Z

e Z

Z Z

I

I

At sending end, s=l, and put K=0, because ZR = ZO.

ZY l IS =IRe

ZY l IR ISe

−

( )  e e e e 203. 4

ZYl γl -7.6649 10 +j0.0355 100 0. 76649

= = = ∠

− − × × −

 e 0. 4646 203. 4

ZYl = ∠

−

  I 1. 4358 10 12. 1 0. 4646 203. 4

3 ∴ (^) R = × ∠ × ∠

−

IR = 6.671x10-4 ∠ -144.

° A

13. A 2 meter long transmission line with characteristics impedance of 60+j 40

ohm is operating at ω = 10^6 rad / sec has attenuation constant of 0.921 Np/m

and phase shift constant of 0 rad /m. If the line is terminated by a load of

20+j50 ohm, determine the input impedance of this line.

[EC6503-NOV/DEC 2015](6 Marks), [EC6503-APR/MAY 2017](8 Marks),

[EC2305-NOV/DEC 2015](6 Marks), [EC2305-APR/MAY 2019](8 Marks)

Solution:

Given: l = 2m, ZO = 60+j 40Ω, ω = 10^6 rad/sec, α= 0.921 Np/m, β= 0 rad /m,

ZR = 20+j50 Ω

Reflection Coefficient, K

R O

R O

Z Z

Z Z

K



1. 1586 j 0. 30345 0. 3424 117. 6 ( 20 j 50 ) ( 60 j 40 )

( 20 j 50 ) ( 60 j 40 ) K = − + = ∠

K=0.3424 ∠ 117.6 °

e e e βl e ( 0 )( 2 )

γl (α jβ)l αl ( 0. 921 )( 2 ) = = ∠ = ∠

γl 1. 842 e =e

e

γ l = 6.3091 ∠ 0 °

e e e βl e ( 0 )( 2 )

γl (α jβ)l αl ( 0. 921 )( 2 ) = = ∠− = ∠−

− − + − −

γl 1. 842 e e

− −

e-

γ l = 0.1585 ∠ 0 °

−

−

γl γ l

γl γ l

s o e Ke

e Ke Z Z

Z (^) s 72. 11 33. 69 

 



 

Z (^) s 72. 11 33. 69



 

Z (^) s 72. 11 33. 69

{ }

  Z (^) s = 72. 11 ∠ 33. 69 0. 9920 ∠ 0. 8740

ZS = 71.54 ∠ 34.

° Ω

14. A generator of 1V, 1000 Hz, supplies power to a 100 Km open wire line

terminated in ZO and having following parameters R = 10.4 ohm per Km, L =

0.00367 Henry per Km, G =

6

1. 8 10

− × mho per Km, C = 0.00835 μF per Km.

Calculate Zo, α, β, λ, v. also find the received power.

[EC6503-MAY/JUN 2016](16 Marks), [EC6503-APR/MAY 2016](16 Marks),

[EC6503-APR/MAY 2015](16 Marks)

Solution:

Given: Eg = 1V, f=1000 Hz, l=100 Km, when ZR = ZO then ZS = ZO, R=10.

Ω/Km, L = 0.00367H/Km, G =0.8x10-6^ mho/Km, C = 0.00835 μF/Km

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL

Y

Z

Z (^) o

Z = R+jωL = 10.4 + j(6280)(.00367) = 25.285∠65.71°

Y = G+jωC = 0.8x10-6^ +j(6280)(0.00835x10-6) = 5.244x10-5∠89.

°





Z

5 o × ∠

−

Z

3 o

  ∠ − ×

−

Zo = 694.48 ∠ -11.

° Ω

The propagation constant is given by,

P =γ= ZY = ( R+jωL)( G+jωC)

  P γ 25. 285 65. 71 5. 244 10 89. 13

5 = = ∠ × × ∠

−

P γ 1. 326 10 ( 65. 71 89. 13 )

3   = = × ∠ +

−

P=γ = α+jβ = 0.0364 ∠77.42° (Or) 7.928x10-3+j0.

attenuation constant, α = 7.928x10-3^ Np/Km

and phase constant, β = 0.0355 rad/Km

The velocity of propagation is given by,

β

ω v (^) p =

1. 769 10 Km/ sec
2. 0355

v

5 p = = ×

176. 99 m
177. 0355

2 π

β

2 π λ = = =

15. A generator of 1V, 1000 cycles, supplies power to a 100 mile open wire line

terminated in ZO and having following parameters: Series resistance R = 10.

ohm/ mile, Series inductance L = 0.00367 H/mile, Shunt conductance G = 0.

x 10-6^ mho/ mile, and capacitance between conductors C = 0.00835 μF/mile.

Find the characteristic impedance, Propagation constant, attenuation

constant, phase shift constant, velocity of propagation and wavelength.

[EC6503-APR/MAY 2017](6 Marks)

Solution:

Given: Eg = 1V, f=1000 cycles = 1000 Hz, l=100 mile, when ZR = ZO then ZS =

ZO, R=10.4 Ω/ mile, L = 0.00367H/ mile, G =0.8x10-6^ mho/ mile, C = 0.

μF/ mile

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL

Y

Z

Z (^) o

Z = R+jωL = 10.4 + j(6280)(0.00367) = 25.285∠65.71°

Y = G+jωC = 0.8x10-6^ +j(6280)(0.00835x10-6) = 5.244x10-5∠89.

°





Z

o (^5) × ∠

−

Z

o (^3)

  ∠ − ×

−

Zo = 694.48 ∠ -11.

° Ω

The propagation constant is given by,

P =γ= ZY = ( R+jωL)( G+jωC)

  P γ 25. 285 65. 71 5. 244 10 89. 13

5 = = ∠ × × ∠

−

P γ 1. 326 10 ( 65. 71 89. 13 )

3   = = × ∠ +

−

P=γ = α+jβ = 0.0364 ∠77.42° (Or) 7.928x10-3+j0.

attenuation constant, α = 7.928x10-3^ Np/Km

and phase constant, β = 0.0355 rad/Km

The velocity of propagation is given by,

β

ω v (^) p =

1. 769 10 Km/ sec
2. 0355

v

5 p = = ×

176. 99 m
177. 0355

2 π

β

2 π λ = = =

16. A generator of 1V, 1KHz, supplies power to a 100 Km open wire line

terminated in 200 Ω resistance. The line parameters are R = 10 Ω/Km, L =

3.8mH/Km, G =1x10-6^ mho/Km, C = 0.0085 μF/Km. Calculate Zo, α, β, λ, v.

also find the received power. [EC2305-NOV/DEC 2018](16 Marks)

Solution:

Given: Eg = 1V, f=1000 Hz, l=100 Km, ZR =200 Ω, R=10 Ω/Km, L =

3.8mH/Km, G =1x10-6^ mho/Km, C = 0.0085 μF/Km

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL

Y

Z

Z (^) o

Z = R+jωL = 10 + j(6280)(3.8x10-3) = 25.87∠67.26°

Y = G+jωC = 1x10-6^ +j(6280)(0.0085x10-6) = 5.339x10-5∠88.

°

Zo

∠

×

Zo

×

Zo = 696.06 ∠ -10.

° Ω

The propagation constant is given by,

P =γ= ZY = ( R+jωL)( G+jωC)

P γ 25. 87 67. 26 5. 339 10 ∠

= = ∠ × ×

P γ 1. 3812 10

= = ×

P=γ = α+jβ = 0.03716 ∠78.095° (Or) 7.6657x10-3+j0.

attenuation constant, α = 7.6657x10-3^ Np/Km

and phase constant, β = 0.03636 rad/Km

The velocity of propagation is given by,

β

ω v (^) p =

Km/ sec

v (^) p= = ×

172. 72 m
173. 03636

2 π

β

2 π λ = = =

Reflection Coefficient, K

R O

R O

Z Z

Z Z

K

[ ]

ZYl ZYl

0

R R o S e Ke Z

Z Z

I

I

− − 

ZR + ZO = 893.28∠-8.

°

Ke

√ZYl = 0.5609∠173.

° x 2.1524 ∠208.

°

Ke

√ZYl = 1.2073∠21.

°

[ ]

  

 

2. 1524 208. 3 1. 2073 21. 58 696.06 -10.

I

1.5992x10 - 1.

• 3 R ∠ − ∠ 

( )[ ]

  

1. 2833 2. 413 3. 354 154. 11 2

I

1.5992x10 - 1.

• (^3) R ∠ = ∠ ∠−

  1.5992x10 - 1.548 IR 2. 1523 151. 70

• ∠ = × ∠−

IR = 7.43x10-4 ∠ 150.

°

ER = IR x ZR = 7.43x10-4∠150.

° x 200 = 0.1486∠150.

°

ER = 0.1486 ∠ 150.

° V

PS = |ES|. |IS|.cos (ES^ IS)

PS = 1 x 1.5992 x 10-3^ x cos (-1.5477)

PS = 1.5986 x 10-3^ W

and PR = |ER|. |IR|.cos (ER^ IR)

PR = 0.1486 x 7.43x10-4^ cos(0)

PR = 0.1486 x 7.43x10-4^ W

PR = 1.1041x10-4^ W

17. A generator of 1V, 1KHz, supplies power to a 100 Km open wire line

terminated in 200 Ω resistance. The line parameters are R = 10 Ω/Km, L =

3.8mH/Km, G =1x10-6^ mho/Km, C = 0.0085 μF/Km. Calculate the

impedance, reflection coefficient, power and transmission efficiency.

[EC2305-APR/MAY 2011](16 Marks)

Solution:

Given: Eg = 1V, f=1000 Hz, l=100 Km, ZR =200 Ω, R=10 Ω/Km, L =

3.8mH/Km, G =1x10-6^ mho/Km, C = 0.0085 μF/Km

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL

Y

Z

Z (^) o

Z = R+jωL = 10 + j(6280)(3.8x10-3) = 25.87∠67.26°

Y = G+jωC = 1x10-6^ +j(6280)(0.0085x10-6) = 5.339x10-5∠88.

°

Zo

∠

×

Zo

×

Zo = 696.06 ∠ -10.

° Ω

The propagation constant is given by,

P =γ= ZY = ( R+jωL)( G+jωC)

P γ 25. 87 67. 26 5. 339 10 ∠

= = ∠ × ×

P γ 1. 3812 10

= = ×

P=γ = α+jβ = 0.03716 ∠78.095° (Or) 7.6657x10-3+j0.

attenuation constant, α = 7.6657x10-3^ Np/Km

and phase constant, β = 0.03636 rad/Km

Reflection Coefficient, K

R O

R O

Z Z

Z Z

K

1. 557 j 0. 0656 0. 5609 173. 28

200 696. 06 10. 84

K = − + = ∠

K=0.5609 ∠ 173.28 °

e e e βl e ( 0. 03636 )( 100 )

γl (α jβ)l αl ( 7. 665710 )( 100 )

3 = = ∠ = ∠

−

• ×

e e ( 0. 03636 )( 100 )

γl ( 7. 665710 )( 100 )

3 = ∠

− × rad

e

γl = e0.76657^ ∠208.3°

e

γ l = 2.1524 ∠ 208.3 °

e e e βl e ( 0. 03636 )( 100 )

γl (α jβ)l αl ( 7. 665710 )( 100 ) 3 = = ∠− = ∠ −

− − − + − − ×

e e ( 0. 03636 )( 100 )

γl ( 7. 665710 )( 100 )

3 = ∠ −

− − − × rad

e-

γl = e-0.76657^ ∠-208.3°

e-

γ l = 0.4646 ∠ -208.3 °

−

−

γl γ l

γl γ l

s o e Ke

e Ke Z Z

Z (^) s 696. 06 10. 84   

   

 

  

Z (^) s 696. 06 10. 84

PR = 0.1486 x 7.43x10-4^ W

PR = 1.1041x10-4^ W

Transmission Efficiency, 100 P

P

η S

R = ×

1. 1041 x 10 η 3

4

× ×

−

−

η = 6.907 %

18. A telephone cable 64 km long has a resistance of 13 Ω/km and a

capacitance of 0.008 μf /km. Calculate the attenuation constant, velocity and

wavelength of the line at 1000 Hz.

[EC2305-NOV/DEC 2016](6 Marks), [EC2305-MAY/JUN 2015](6 Marks),

[EC2305-APR/MAY/JUN 2016](4 Marks)

Solution:

Given: l=64Km, R=13 Ω/km, C=0.008 μf /km, f=1000 Hz

ω = 2πf = 2x3.14x1000 = 6280

The characteristic impedance is given by,

G jω C

R jωL Z (^) o

j( 6280 )( 0. 008 10 )

Z

o (^) − 6 ×





j 5. 024 10

Z

o (^55) × ∠

×

− −

Z

o 3

  ∠ − ×

−

Zo = 509.31 ∠ -

°

The propagation constant is given by,

P =γ= (R +jωL)(G +jωC)

P γ ( 13 ) ( j( 6280 )( 0. 008 10 ))

− 6 = = × ×

5 P γ 13 j 5. 024 10

− = = × ×

 P γ 6. 5312 10 90

4 = = × ∠

−

P=γ = α+jβ = 25.56x10-3^ ∠ 45 ° (Or) 0.01807+j0.

attenuation constant, α = 0.01807 Np/Km

and phase constant, β = 0.01807 rad/Km

The velocity of propagation is given by,

β

ω v (^) p =

3. 48 10 Km/ sec
4. 01807

v

5 p = = ×

347. 54 m

. 01807

2 π

β

2 π λ = = =

19. A transmission line has L = 10 mH/m, C = 10 -7^ F/m, R = 20 Ω/m and

G =10 -5^ mho/m. Find the input impedance at a frequency of (^) 



2 π

Hz, if the

line is very long. [EC2305-NOV/DEC 2013] (6 Marks)

Solution:

Given: L = 10 mH/m, C = 10 -7^ F/m, R = 20 Ω/m and G =10 -5^ mho/m,

2 π

f

ω = 2πf = 2π x (^) 



2 π

The characteristic impedance is given by,

G jω C

R jωL Z (^) o

10 j( 5000 )( 10 10 )

20 j( 5000 )( 10 10 ) Z 5 7

3

o (^) − −

−

+ ×

+ ×





10 j 5 10

20 j 50 Z o (^544) × ∠

+ ×

− − −

Z (^) o

  = ∠ −

Zo = 327.59 ∠ -10.

° Ω

20. The characteristic impedance of a uniform transmission line is 2309.

ohms at a frequency of 800 Hz. At this frequency, the propagation constant is

0.054(0.0366+j0.99). Determine R and L.

[EC2305-NOV/DEC 2013] (6 Marks), [EC2305-NOV/DEC 2010] (6 Marks)

Solution:

Given: Zo=2309.6 Ω , f=800Hz, P=γ=0.054(0.0366+j0.99) = 0.054∠87.

°

ω = 2πf = 2 x 3.14 x 800 = 5024

R=jωL = ZoP = 2309.6 x 0.054∠87.

°

= 4.57+j124.