Question

A telephone line has r = 30 0/km, l = 100 mh/km, g = 0, and c = 20 jtf/km. At / = 1 khz, obtain: (a) the characteristic impedance of the line (b) the propagation constant (c) the phase velocity

Answer 1

Answer:

Propagation constant:

γ

=

√

(

R

+

j

ω

L

)

(

G

+

j

ω

C

)

=

√

(

30

+

j

2

π

×

10

3

×

100

×

10

−

3

)

(

j

2

π

×

10

3

×

20

×

10

−

6

)

γ

==

√

(

30

+

j

200

π

)

(

j

0.04

π

)

=

√

(

−

78.95

+

j

3.76

)

=

8.889

∠

88.63

o

γ

=

α

+

j

β

=

[

0.212

+

j

8.887

]

k

m

−

1

β

=

8.887

rad/km

Phase velocity,  

ν

p

=

ω

β

=

2

π

×

10

3

(

8.887

)

10

3

ν

p

=

0.707

×

10

6

∴

ν

p

=

0.707

Mm/sec

Explanation:

Answer 2

The value of propagation constant is νp = 0.707 Mm/sec

Explanation:

Given data:

• Radius r = 30 Km
• L = 100 mh / km
• g = 0

Solution:

Propagation constant:

γ = √ (R + j ωL)

( G + j ω C ) = √( 30 + j 2 π × 10^3 × 100 × 10^−3

(j 2 π × 10^3 × 20 × 10^−6) γ = √( 30 + j 200 π)

(j 0.04 π ) = √( − 78.95 + j 3.76 )

= 8.889 ∠ 88.63o

γ = α + j

β = [ 0.212 + j 8.887 ] km^− 1

β = 8.887 rad/km

Phase velocity:

νp = ω

β =2π × 10^3 (8.887)10^3

νp= 0.707×10^6

∴νp = 0.707 Mm/sec