Voltage pulse sketches for reflection on a lossless transmission line

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Homework Statement:

If we send a voltage pulse of amplitude ## V_0 ## through a lossless transmission line, what does the voltage at the output look like when the load impedance is ## \infty ## (i.e. open circuit). We can assume that the time taken by the pulse to traverse the line ## \tau ## is much greater than the duration.

Relevant Equations:

Reflection coefficient
Hi,

I was recently attempting a question about transmission lines and I don't seem to really understand how the voltages travel through the line.

Question:
If we send a voltage pulse of amplitude ## V_0 ## through a lossless transmission line, what does the voltage at the output look like when the load impedance is ## \infty ## (i.e. open circuit). We can assume that the time taken by the pulse to traverse the line ## \tau ## is much greater than the duration. The source impedance is equal to the characteristic impedance of the line ## Z_0 ##.

My approach:
We know the expression for the reflection coefficient ## \rho ## in this case at the end is given by:
$$ \rho_{end} = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{\infty - Z_0}{\infty + Z_0} = 1 $$
and at the source it is:
$$ \rho_{source} = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{Z_0 - Z_0}{Z_0 + Z_0} = 0 $$

By using the wave equation, we can derive the following solution for the voltage:
$$ V(z, t) = ( Ae^{j \beta z} + Be^{-j \beta z} ) e^{j \omega t} $$

where ## \beta ## is the propagation coefficient. I thought that this wave solution meant that there are forward and backward travelling waves which would each be of height ## V_0 / 2 ##. Therefore, the backwards travelling wave would be absorbed by the source and the forwards travelling wave would have an amplitude ## V_0 / 2 ## which would reach the end at time ## \tau ##. Then it would be reflected and sent backwards and then absorbed by the source.

However, the answer says that the voltage on the end is of amplitude ## 2 V_0 ##, but I don't really understand how this can be the case?

Would anyone be able to explain to me why this is the case? How does the voltage actually double? I think this means that my understanding of how the forward and backwards wave travel is incorrect.

Any help is greatly appreciated. Thanks.
 

Answers and Replies

  • #2
berkeman
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However, the answer says that the voltage on the end is of amplitude , but I don't really understand how this can be the case?
What end? At the far end you will get double as the outgoing pulse and reflected pulse add, but as the pulse travels back to the source after it passes itself, it should just have amplitude ##V_0## AFAIK.
 
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  • #3
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What end?
Apologies, I should have been more clear - the far end.

At the far end you will get double as the outgoing pulse and reflected pulse add, but as the pulse travels back to the source after it passes itself, it should just have amplitude ##V_0## AFAIK.
Yes, this does remind me of something I saw in a lab but I am just trying to wrap my head around the mathematics of it. This is what I am thinking:
the above solution for the voltage is like D'Alembert's solution for the wave equation. The initial pulse is an initial condition which is the sum of the forward and backwards travelling waves at ## t = 0 ##. Therefore, the two waves ought to have an amplitude of ## V_0 / 2 ## each? Then from the front of the line, we have one pulse being sent in the -z direction and another pulse in the +z direction?
 
  • #4
berkeman
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Sorry, I'm not understanding where your ##V_0/2## is coming from. The finite duration pulse injected into the line has amplitude ##V_0##, no?

1602105047251.png
 
  • #5
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Sorry, I'm not understanding where your ##V_0/2## is coming from. The finite duration pulse injected into the line has amplitude ##V_0##, no?

View attachment 270587
Thanks for your reply. However, I thought the D'Alembert solution of there being a forward and backward travelling wave apply to this situation as well. If it does, then the pulses must satisfy the initial condition at the very front of the wave and the backwards wave must have an amplitude of ## V_0 / 2 ## there as well as the forwards travelling wave?
 
  • #6
berkeman
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Thanks for your reply. However, I thought the D'Alembert solution of there being a forward and backward travelling wave apply to this situation as well. If it does, then the pulses must satisfy the initial condition at the very front of the wave and the backwards wave must have an amplitude of ## V_0 / 2 ## there as well as the forwards travelling wave?
Can you provide a link where you are seeing this interpretation? So far I'm still not understanding, but I'm a little slow today... :wink:

https://mathworld.wolfram.com/dAlembertsSolution.html
 
  • #7
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Can you provide a link where you are seeing this interpretation? So far I'm still not understanding, but I'm a little slow today... :wink:

https://mathworld.wolfram.com/dAlembertsSolution.html
Apologies, so if we take the equation (19) from that link:

$$ y(x, t) = \frac{1}{2} y_0 (x - ct) + \frac{1}{2} y_0 (x - ct) + \frac{1}{2c} \int_{x - ct}^{x + ct} v_0 (s) ds $$

Then the pulse at the front of the wave (i.e. at ## t = 0 ##) must satisfy this condition. If we assume that there is a zero initial condition for the voltage (maybe this wrong to say?) so ## g(s) = 0 ##, then the equation becomes:

$$ y(x, t) = \frac{1}{2} y_0 (x - ct) + \frac{1}{2} y_0 (x - ct) $$
and therefore:
$$ y(x, 0) = \frac{1}{2} y_0 (x) + \frac{1}{2} y_0 (x ) = V_0 $$
so the pulses would each have an amplitude of ## V_0 / 2 ##. I realise the zero condition on the velocity might be where I am wrong... is that the case?
 
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  • #8
hutchphd
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Just a few observations...what you say has some insufficiencies...so it is not horribly incorrect but incomplete. The solution you present is a steady state solution and describes, not a pulse, but left and right going waves extending to infinity in time and space. Any finite disturbance will not be pure.
The General solution you have written must be made particular by choosing the complex constants A and B to match the boundary conditions for your problem on finite boundaries.
So please solve for A and B explicitly and see where you are.
Apologies, so if we take the equation (19) from that link:

$$ y(x, t) = \frac{1}{2} y_0 (x - ct) + \frac{1}{2} y_0 (x - ct) + \frac{1}{2c} \int_{x - ct}^{x + ct} v_0 (s) ds $$

Then the pulse at the front of the wave (i.e. at ## t = 0 ##) must satisfy this condition. If we assume that there is a zero initial condition for the voltage (maybe this wrong to say?) so ## g(s) = 0 ##, then the equation becomes:

$$ y(x, t) = \frac{1}{2} y_0 (x - ct) + \frac{1}{2} y_0 (x - ct) $$
and therefore:
$$ y(x, 0) = \frac{1}{2} y_0 (x) + \frac{1}{2} y_0 (x ) = V_0 $$
so the pulses would each have an amplitude of ## V_0 / 2 ##. I realise the zero condition on the velocity might be where I am wrong... is that the case?
'
You have two rightgoing waves. This is incorrect. What are you trying to show?
 
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  • #9
hutchphd
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Ok I just looked at the Wolfram treatment. The final solution is very pretty and I like his treatment.....you must have ##y_0(x+ct) ## (plus) to represent the leftgoing pulse however. The integral third terms represents the outgoing scattering from the initial pulse. It is in the spirit of a Green's function (integral equation) solution.
 
  • #10
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Ok I just looked at the Wolfram treatment. The final solution is very pretty and I like his treatment.....you must have ##y_0(x+ct) ## (plus) to represent the leftgoing pulse however. The integral third terms represents the outgoing scattering from the initial pulse. It is in the spirit of a Green's function (integral equation) solution.
Okay, so if I were to include the 'velocity initial condition' for the pulse I would find that the two waves were actually of amplitude ## V_0 ## and not ## V_0 / 2 ##?
 
  • #11
hutchphd
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On a transmission line not properly terminated, their will be (total) reflection of a pulse. The remaining incoming pulse will add to the reflecting pulse, and so the result depends upon the details of the pulses and the termination.
You have not specified your problem sufficiently for me to understand what ##V_0## represents (or at least I can't find such specification), so I can't answer.

Maybe this will help.
 
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  • #12
berkeman
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You have not specified your problem sufficiently for me to understand what ##V_0##
represents (or at least I can't find such specification), so I can't answer.
Homework Statement:: If we send a voltage pulse of amplitude ## V_0 ## through a lossless transmission
From his initial problem statement, I interpreted it as the amplitude of the pulse driven into the transmission line. But I could be wrong about that...

Maybe this will help.
Nice link. I like the animations! :smile:
 
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  • #13
hutchphd
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If we assume that there is a zero initial condition for the voltage (maybe this wrong to say?)
My apologies for not being careful here. You cannot make this assumption ##g(s)##is defined as the slope of the shape at t=0....no slope means no pulse i.e. voltage is constant in space and time. There will be nothing going on. Again sorry.
 
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  • #14
Joshy
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I think others have already answered the equation in a way that I would have been okay with.

Something that confuses me a bit, which is why I wanted to post is the extra comment "We can assume that the time taken by the pulse to traverse the line ##\tau## is much greater than the duration." What does this give us in a problem like this? What would other cases such as as ##\tau## is greater than or equal to the duration?
 
  • #15
hutchphd
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Without looking at it again I think that must be true so the limits on the integral are what they are....
 
  • #16
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I think others have already answered the equation in a way that I would have been okay with.

Something that confuses me a bit, which is why I wanted to post is the extra comment "We can assume that the time taken by the pulse to traverse the line ##\tau## is much greater than the duration." What does this give us in a problem like this? What would other cases such as as ##\tau## is greater than or equal to the duration?
Hi @Joshy - apologies for the confusion. I believe that detail is given to avoid confusion about the pulse simultaneously being at both the input and output and thinking about what that would cause.
 
  • #17
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My apologies for not being careful here. You cannot make this assumption ##g(s)##is defined as the slope of the shape at t=0....no slope means no pulse i.e. voltage is constant in space and time. There will be nothing going on. Again sorry.
Apologies, I was just thinking about this and upon reflection I am confused by what you mean here. If you are referring to the term in the integral, I thought that term refers to the velocity condition (as mentioned on the Wolfram Alpha link above)? Please do correct me if I am wrong.

Just to sum up where I am, I was originally incorrect by ignoring the velocity condition in the D'Alembert's solution for the wave. If I included that (whatever the correct condition would be), then that should yield the correct waves. We should find that we have a wave of amplitude ## V_0 ## travelling along the line.

However, I still have a few questions:
1) D'Alembert Solution
At the front of the line, we have a forwards and backwards travelling wave. As mentioned above, the forward wave (with amplitude ## V_0 ##) will go to the end and get reflected. However, does the backwards wave just get absorbed at the source (due to the reflection coefficient being equal to 0)?

2) Reflection at other end
I am still slightly confused about what happens at the other end... So it sounds like a 'reflected' pulse adds with the incoming pulse at the far end to create a pulse of ## 2 V_0 ##. Just to confirm, is this 'reflected' pulse just the reflection of the very same incoming pulse? I can imagine the pulse reaching the end and then changing direction and the overlap causing a pulse of ## 2 V_0 ## (with the same duration as the initial pulse) to appear at the output.

Just as a further extension to check I know what is going on (or maybe show that I have no clue): if the source impedance was also an open circuit (for the sake of argument) such that the reflection coefficient at the front was also 1, then would we observe a pulse of amplitude ## 2 V_0 ## at the output for twice the duration of the initial pulse?

Thanks
 
  • #18
hutchphd
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2) Reflection at other end
I am still slightly confused about what happens at the other end... So it sounds like a 'reflected' pulse adds with the incoming pulse at the far end to create a pulse of 2V0. Just to confirm, is this 'reflected' pulse just the reflection of the very same incoming pulse? I can imagine the pulse reaching the end and then changing direction and the overlap causing a pulse of 2V0 (with the same duration as the initial pulse) to appear at the output.
Did you look at the link for pulses on a string?
Hard boundary is shorted line soft boundary is open line. For the open line you can get a pulse that is double the incident pulse.

D'Alembert's solution is general, but its direct utility is limited. As @Joshy noted the pulses need to be separated in order to see the stru cture. For the infinite string, no driving force, initial conditions as given the solution is good. If end point boundary conditions are specified one is restricted to use sinusoidal waves for the f and g functions and their relative phase must be chosen with care. Fourier analysis is where one is led .
 
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  • #19
hutchphd
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Also I think that the Wikipedia version is a little more complete. N.B. their discussion of the inhomogeneous term (which is how you add the source).
 

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