- #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.

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 ##.

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 ##,

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.

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.