# Homework Help: Transfer function of transmission line

1. Sep 9, 2011

### Bromio

1. The problem statement, all variables and given/known data
Calculate the output vo(t) if the input is the signal vin(t) shown in attached figures.

3. The attempt at a solution
I've calculated that ransfer function transmission line: H(s) = Vo(s)/Vin(s) = exp(-j*gamma*l), where gamma = -j*omega*sqrt(LC). However, tau = sqrt(LC)*l, H(j*omega) = exp(-j*tau*omega), where tau is the delay.

I know that Vo(j*omega) = H(j*omega)*Vin(j*omega). How can I continue?

Thank you.

File size:
6 KB
Views:
226
File size:
3.7 KB
Views:
173
2. Sep 9, 2011

### Staff: Mentor

I think this problem is a bit simpler than the approach you are taking.

What is the characteristic impedance Zo for this cable? Does that simplify things at all?

Are you given the source impedance for the signal source?

3. Sep 9, 2011

### Bromio

Zo = 50 ohm = RL. So, there is no reflection. However, I don't see how does this fact affect.

4. Sep 9, 2011

### Staff: Mentor

You need the output impedance of the voltage source, but other than that, the waveform just propagates down the TL and shows up across the load resistor. How long does it take to propagate? What is its peak amplitude as it passes into the load?

5. Sep 9, 2011

### Bromio

That information is unknown.

The delay is tau = 10 ns, but I don't know what happens when transmitted wave rises the load. How can I continue?

Thank you.

6. Sep 9, 2011

### Staff: Mentor

What delay is 10ns? It takes a lot less than 10ns for the waveform to propagate down 0.5m of TL cable...

7. Sep 9, 2011

### Bromio

Why? tau = l*sqrt(LC) = 0.5*sqrt(10^(-6)*400^(-12)) = 10 ns.

8. Sep 9, 2011

### Staff: Mentor

Hmm, I guess you are right. It's just a slower cable than I usually work with. The speed of light in air is about 3ns per meter, so the velocity of propagation in this cable is 20ns per meter or only 0.15c. Usually I see closer to 0.5c (but I mostly work with twisted pair TLs), but whatever. I think your calculation is correct.

So, since you are given the voltage wavecform seen at the entry point of the cable (after the voltage source and its output resistance, presumably), you only need to worry about the propagation delay in getting through the cable to the load. So the voltage waveform at the load resistor is the time delayed version of the input voltage waveform.