Meaning of the FFT of a Poynting Vector integral, reflection coefficient

  • #1
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Hello,
For calculating the mean power at a specific cross section of a waveguide, one can calculate the mean value of the temporal function of Poynting Vector, P(t), where P(t) is the ExHy-EyHx. Note that I am not talking about phasors or a sinusoidal state. If I integrate over the waveguide cross section and take the mean value over time, I simply obtain the power output. I notice that the P(t) is oscillating with double the frequency of the frequency of each field, and that's normal. If I take the fourier transform of P(t), I will notice a Peak at the double frequency. How can I relate this to power/frequency term if I want a gaussian modulated time function instead of a plain sine? Can I divide with 2 to obtain the "correct" frequency? Of course this is required to calculate the reflection coefficient of a structure as Pout/Pin as a function of frequency, but this double frequency somewhat confuses me
 

Answers and Replies

  • #2
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Is the case you're discussing ideal in that there is wave propagation only along one direction in the guide?

Also recall ##\sin(t)^2 = (1-\cos(2t))/2##
 
  • #3
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Is the case you're discussing ideal in that there is wave propagation only along one direction in the guide?

Also recall ##\sin(t)^2 = (1-\cos(2t))/2##
It is not ideal as I would have also reflections from an obstacle. My question is how Pout(w)/Pin(w) could be related with reflection coefficient?
 
  • #4
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It is not ideal as I would have also reflections from an obstacle. My question is how Pout(w)/Pin(w) could be related with reflection coefficient?
In a linear RF network the power reflected is ## P_r(\omega) = |R|^2 P_\text{in}(\omega)## Transmission power can be more tricky when the line impedance changes or is different than the input port. There are some good discussions out there on S parameters and such. Also you might look at "Waveguide Handbook" by M. Marcuvitz. Marcuvitz discusses reflection from many types of waveguide obstacles so it's worth a look.
 
  • #5
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What If I look the system as a two-port network?
In a linear RF network the power reflected is Pr(ω)=|R|2Pin(ω) P_r(\omega) = |R|^2 P_\text{in}(\omega) Transmission power can be more tricky when the line impedance changes or is different than the input port. There are some good discussions out there on S parameters and such. Also you might look at "Waveguide Handbook" by M. Marcuvitz. Marcuvitz discusses reflection from many types of waveguide obstacles so it's worth a look.
 
  • #6
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What If I look the system as a two-port network?
That's the way I was looking at it. Work a simple example. Connect a 50 ohm cable to a 100 ohm cable. The impedance mismatch will cause a reflection at the interface. The reflection and transmission coefficients are defined in terms of left and right going voltages. The power in each cable is a function of the line impedance of the cable. All fields are continuous at the boundary so,

##1 + R = T##

where 1 is the incident voltage, R the reflected voltage and T the transmitted voltage. Likewise, the current is

##Z_1 - Z_1 R = Z_2 T##

where ##Z_1=50## ohm and ##Z_2=100## ohm. Solve these for R and T.

For more complex problems like discontinuities in waveguides more complex calculations are needed to compute T and R. These include variational methods pioneered by Schwinger during WWII or more computationally intensive things like mode matching. To help more I would need to know more about the problem you're trying to solve.
 
  • #7
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That's the way I was looking at it. Work a simple example. Connect a 50 ohm cable to a 100 ohm cable. The impedance mismatch will cause a reflection at the interface. The reflection and transmission coefficients are defined in terms of left and right going voltages. The power in each cable is a function of the line impedance of the cable. All fields are continuous at the boundary so,

1+R=T1 + R = T

where 1 is the incident voltage, R the reflected voltage and T the transmitted voltage. Likewise, the current is

Z1−Z1R=Z2TZ_1 - Z_1 R = Z_2 T

where Z1=50Z_1=50 ohm and Z2=100Z_2=100 ohm. Solve these for R and T.

For more complex problems like discontinuities in waveguides more complex calculations are needed to compute T and R. These include variational methods pioneered by Schwinger during WWII or more computationally intensive things like mode matching. To help more I would need to know more about the problem you're trying to solve.
Thank you for the answer. I'm trying to calculate the reflection and transmission coefficient in a rectangular waveguide when somewhere in the waveguide there is a discontinuity along z-axis. Waveguide is excited with a specific TE or TM mode and I was thinking how can I use the power at z=0 and at a z position after the discontinuity. I'm not interested in mode conversion at all. Just raw power of the total field
 
  • #8
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What's that in the waveguide, a head? (sorry, couldn't resist: Benny Hill). Well, you can roll your own calculation or you can try to find a similar situation in Marcuvitz. For small obstructions I favor using perturbation theory with the reciprocity relation,

##\iint E_1\cdot J_2 dA = \iint E_2\cdot J_1 dA##

This relation (which is a surface integral) is true quite generally provided the problem doesn't have non-reciprocal or chiral materials. Basically, lumps of ##\mu## and ##\epsilon## and you're good to go. In this expression ##E_k, J_k## are solutions of Maxwell's equations for a driving source currents ##J_k##. The way I would apply this is ##E_1## would be just the unperturbed mode field where we may take ##J_1=\hat{n}\times H_1## for the section of guide one considers. The ##E_2## solution is the reflected and transmitted field your looking for. The surface of integration is the waveguide interior minus the obstructing thing (we'll assume a solder blob on one waveguide wall) Since both fields 1 and 2 obey the boundary conditions every surface other than the waveguide cross section and the fields on the solder blob are zero because tangent E is zero. The fields on the blob may be estimated by the physical optics calculation. Remember that the unperturbed solution is known everywhere and doesn't know or care about the blob. The two waveguide cross sections (you may only need one for reflection.) is chosen far from the obstruction. At this point one may assume,

##E_2(x,y) = R E_-(x,y)##

where ##E_-(x,y)## is a unit amplitude TE mode traveling in the direction of the reflected wave. ##R## is the reflection coefficient which is expressed as a surface integral over the obstruction by evaluating the reciprocity integral.
 
  • #9
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What's that in the waveguide, a head? (sorry, couldn't resist: Benny Hill).
A simple change in either x on y direction (or both)

I took an idea and for sure I have to look at Markuvitz book!
 

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