Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 2)
2.
A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m, if the line is distortion less, the attenuation constant (in Np/m) is
500
5
0.014
0.002
Answer: Option
Explanation:

Distortion less

a = RG and Z0 = RG

.

Discussion:
24 comments Page 2 of 3.

Sonal said:   8 years ago
R = α * z0. Compute by this formula.

Fadhil said:   8 years ago
Now I undestand it thanks.

Gopal said:   8 years ago
Hi,

Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).

Jitendra said:   8 years ago
Anyone, please explain this.

Rajesh said:   8 years ago
I understood the explanation.

Sowmya said:   8 years ago
Can any 1 explain how 0.1/2500 came?

Jai2x said:   9 years ago
The real formula is Zo = sqrt(R/G) if you analyze the solution for number 2.

SAI SURESH said:   9 years ago
I can't understand can any one explain this.

PAVI said:   9 years ago
Z0 = sqrt of RG.

Dilip N said:   10 years ago
G/C = R/L.
Zo = sqrt(L/C).
Zo = sqrt(R/G).


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