Transfer function of transmission line

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Bromio
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Homework Statement


Calculate the output vo(t) if the input is the signal vin(t) shown in attached figures.

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.
 

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Bromio said:

Homework Statement


Calculate the output vo(t) if the input is the signal vin(t) shown in attached figures.

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.

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

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?
 
berkeman said:
You need the output impedance of the voltage source

That information is unknown.

berkeman said:
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?

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.
 
Bromio said:
The delay is tau = 10 ns, but I don't know what happens when transmitted wave rises the load.

What delay is 10ns? It takes a lot less than 10ns for the waveform to propagate down 0.5m of TL cable...
 
Why? tau = l*sqrt(LC) = 0.5*sqrt(10^(-6)*400^(-12)) = 10 ns.
 
Bromio said:
Why? tau = l*sqrt(LC) = 0.5*sqrt(10^(-6)*400^(-12)) = 10 ns.

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.