Voltage pulse sketches for reflection on a lossless transmission line

In summary, voltage pulse sketches can be used to visualize and analyze reflections on a lossless transmission line. These sketches show the propagation of voltage pulses along the line and the resulting reflections when they encounter impedance changes. By studying these sketches, engineers can gain a better understanding of how to design and optimize transmission lines for efficient signal transfer with minimal reflections. This technique is particularly useful in high-frequency and high-speed applications where signal integrity is crucial.
  • #1
Master1022
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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 traveling waves which would each be of height ## V_0 / 2 ##. Therefore, the backwards traveling wave would be absorbed by the source and the forwards traveling 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.
 
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  • #2
Master1022 said:
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
berkeman said:
What end?
Apologies, I should have been more clear - the far end.

berkeman said:
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 traveling 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
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
berkeman said:
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 traveling 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 traveling wave?
 
  • #6
Master1022 said:
Thanks for your reply. However, I thought the D'Alembert solution of there being a forward and backward traveling 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 traveling 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
berkeman said:
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 realize the zero condition on the velocity might be where I am wrong... is that the case?
 
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  • #8
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.
Master1022 said:
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 realize 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
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
hutchphd said:
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
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
hutchphd said:
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.
Master1022 said:
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...

hutchphd said:
Maybe this will help.
Nice link. I like the animations! :smile:
 
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  • #13
Master1022 said:
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
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
Without looking at it again I think that must be true so the limits on the integral are what they are...
 
  • #16
Joshy said:
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
hutchphd said:
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 ## traveling 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 traveling 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
Master1022 said:
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
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).
 

Related to Voltage pulse sketches for reflection on a lossless transmission line

1. What is a voltage pulse sketch for reflection on a lossless transmission line?

A voltage pulse sketch for reflection on a lossless transmission line is a graphical representation of the voltage waveforms that occur when a pulse of electrical energy is sent down a transmission line and encounters a change in impedance. It shows the amplitude and direction of the reflected and transmitted waves as they travel along the transmission line.

2. How is a voltage pulse sketch created?

A voltage pulse sketch is created by using the principles of transmission line theory and the equations for voltage and current reflections. The sketch is typically drawn on a time domain graph, with the horizontal axis representing time and the vertical axis representing voltage amplitude.

3. What information can be obtained from a voltage pulse sketch?

A voltage pulse sketch can provide information about the impedance of the transmission line and any changes in impedance along the line. It can also show the amplitude and direction of the reflected and transmitted waves, which can be used to calculate the standing wave ratio and the amount of power that is reflected back to the source.

4. Why is a lossless transmission line used for voltage pulse sketches?

A lossless transmission line is used for voltage pulse sketches because it simplifies the calculations and allows for a more accurate representation of the voltage waveforms. In a lossless transmission line, there is no energy loss due to resistance, so the voltage and current reflections can be easily calculated using the equations for a lossless transmission line.

5. How are voltage pulse sketches used in practical applications?

Voltage pulse sketches are used in practical applications to design and analyze transmission lines, such as those used in telecommunications and power systems. They can also be used to troubleshoot and diagnose issues with transmission lines, such as impedance mismatches or faults. Additionally, voltage pulse sketches can be used to optimize the performance of transmission lines by adjusting the impedance at different points along the line.

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