- #1

- 57

- 8

## Homework Statement

(a) State what is meant by a ‘distortionless’ and a ‘lossless’ transmission line.

(b) A transmission line has the primary coefficients as given below. Determine the line’s secondary coefficients

**,**

*Z*o**α**and

**β**at a frequency of 1 GHz.

R =

*2 Ω/m*

L =

*8 nH/m*

G=

*0.5 mS/m*

C=0

*.23 pF/m*

## The Attempt at a Solution

I've seen some old threads with this question, but it's more the workings, methodology and definitions - if someone could be so kind to have a look over.

a)

Distortionless:

The transmission line in order to be 'distortionless' must

**both**attenuate all signal frequencies in the same proportion

**and**shift them in time by the same amount.

Loseless:

A transmission line is known as loseless when there are no energy loses along the line due to coeffecients R & G. This occurs when both of the aforementioned coefficients equal 0.

b)

So:

R &= 2 Ω/m

L &= 8 nH/m = 8x10^{-9} H/m

G=0.5 mS/m = 0.0005 S/m

C &=0.23 pF/m = 0.23*10^{-12} F/m

ω=2πf = 1x10^{9} *2π = 6.283*10^{9)

Due to high frequency the formulas:

β = ω√(LC)

= 6.283*10^{9} * √ ( 8*10^{-9) * 0.23*10^{-12}

= 0.26951 Radm-1

α = R/2 * √(C/L)+G/*√(L/C)

= 2/2 * √(0.23*10^{-12}/8*10^{-9})+0.0005/2*√(8*10^{-9}/0.23*10^{-12})

= 1 * 0.00536+0.00025*186.501

= 0.05198714308

= 51.987 mNepers m-1

Zo = √((R+jωL)/ (G+jωC))

= √((2+j6.283*10^{9}*8*10^{-9}) / (0.0005+j6.283*10^{9}*0.23*10^{-12}))

= √ ( (2+j16π) / (0.0005+j(0.00046π) )

= √ (31491.630+j9511.802)

= 179.427+j26.506Ω

**I appreciate any and all help.**If anyone else is doing this question and wants advice I can try to help also.