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
Answer: Option
Explanation:
Distortion less 
a = RG and Z0 = RG → 
.
Discussion:
24 comments Page 1 of 3.
Jacb said:
7 years ago
z0=√(R/G) , equate to G: G=R/(z0)^2.
atten const = √(R*G), subs. G :
atten const = √(R*(R/(z0)^2)).
atten const = √(0.1 * ( 0.1/ (50)^2)).
Hope that helps.
atten const = √(R*G), subs. G :
atten const = √(R*(R/(z0)^2)).
atten const = √(0.1 * ( 0.1/ (50)^2)).
Hope that helps.
(3)
Gopal said:
8 years ago
Hi,
Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).
Propogation constant = attenuation constant + i(phase constant).
Attenuation constant = R/2zo.
= 0.1/(2 * 50).
Shweta said:
1 decade ago
Got it by myself.
Correct formula for distortionless transmission line is:
Characteristic impedance = Square root of (resistance/Conductance).
Correct formula for distortionless transmission line is:
Characteristic impedance = Square root of (resistance/Conductance).
Ashwin said:
8 years ago
We know that in distortion less medium R/L=G/C, so alpha= R.sqrt(C/L) and Zo= sqrt(L/C), so alpha= R/Zo.
Jai2x said:
9 years ago
The real formula is Zo = sqrt(R/G) if you analyze the solution for number 2.
Tamrat Z said:
10 months ago
Anyone, please give me the clear explanation of the answer.
Shweta said:
7 years ago
How it is?
Zo= √ RG.
It is Zo= √( R/G).
Zo= √ RG.
It is Zo= √( R/G).
SAI SURESH said:
9 years ago
I can't understand can any one explain this.
Sachin kumar said:
7 years ago
How 0.1/2*50=0.1/2500 can anyone explain?
Sonal said:
8 years ago
R = α * z0. Compute by this formula.
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