- #1

yungman

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[tex]\tilde E_y=\hat y E_0 e^{-\delta z} \;\;\hbox { and } \;\; \tilde H_x = (-\hat x)\frac{E_0}{\eta} e^{-\delta z}[/tex]

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- Thread starter yungman
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- #1

yungman

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[tex]\tilde E_y=\hat y E_0 e^{-\delta z} \;\;\hbox { and } \;\; \tilde H_x = (-\hat x)\frac{E_0}{\eta} e^{-\delta z}[/tex]

- #2

yungman

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- #3

Born2bwire

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If we assume that the plates are perfect conductors and that the dielectric is lossless then no energy is lost as the wave propagates. With a PEC material, there is no energy loss in generating currents. But once we allow for the plates to have a finite conductivity then we introduce a small loss since we have to inject a certain amount of energy to keep the currents flowing.

The equations for your fields are incorrect though. The guided modes of a parallel plate waveguid bounce back and forth off of the plates. The angle of incidence upon the plates is determined by the frequency and mode of your wave. This is so that the wave will strike the plates in such a way as to satisfy the appropriate boundary conditions. So if your waveguide is setup so that the guided direction of propagation is along the z-axis and the plates are in the x-z plane, then your waves must have a dependence upon

[tex] \sim e^{ik_yy + ik_zz} [/tex]

The waves are still travelling as TEM waves but we generally decompose the solutions into those that are TE and TM to the direction of guided propagation (z in this case). So your TE_z solution is something of the form

[tex] \mathbf{E}(\mathbf{r}) = E_0 \hat{x} e^{ik_yy + ik_zz} [/tex]

[tex] \mathbf{H}(\mathbf{r}) = H_0 (\alpha\hat{y} + \beta\hat{z}) e^{ik_yy + ik_zz} [/tex]

TM_z solutions are of the form,

[tex] \mathbf{E}(\mathbf{r}) = E_0 (\alpha\hat{y} + \beta\hat{z}) e^{ik_yy + ik_zz} [/tex]

[tex] \mathbf{H}(\mathbf{r}) = H_0 \hat{x} e^{ik_yy + ik_zz} [/tex]

Further, we note that we can have a solution that travels in the +y and +z direction and also in the -y and +z direction at the same location. That is, we can have both upward and downward bouncing waves. So we generally combine the two travelling wave components as a standing wave component where the standing wave is just the superposition of two travelling waves. So that means that the E field of the TE_z solution is expressed more compactly as,

[tex] \mathbf{E}(\mathbf{r}) = E_0 \hat{x} \sin (k_yy) e^{ik_zz} [/tex]

Again, the selection of the wave vector k is dependent upon the mode and frequency of the solution (obviously in the above we note that at the locations of the plates (say y=0, y=a) the tangential electric field is zero). This, in essence, is an eigenvalue problem. So it's a simple matter to see that in our case that,

[tex] k_y = \frac{\pi m}{a} [/tex]

[tex]k_z = \sqrt{ k_0^2 - \left( \frac{\pi m}{a} \right)^2 } [/tex]

Thus the modes for our TE_z solution for a desired frequency are defined by the above eigenmodes for m=1,2,3, ...

EDIT: Ok, pfew... I think that's actually pretty much the treatment of the parallel plate waveguide in its entirety.

The equations for your fields are incorrect though. The guided modes of a parallel plate waveguid bounce back and forth off of the plates. The angle of incidence upon the plates is determined by the frequency and mode of your wave. This is so that the wave will strike the plates in such a way as to satisfy the appropriate boundary conditions. So if your waveguide is setup so that the guided direction of propagation is along the z-axis and the plates are in the x-z plane, then your waves must have a dependence upon

[tex] \sim e^{ik_yy + ik_zz} [/tex]

The waves are still travelling as TEM waves but we generally decompose the solutions into those that are TE and TM to the direction of guided propagation (z in this case). So your TE_z solution is something of the form

[tex] \mathbf{E}(\mathbf{r}) = E_0 \hat{x} e^{ik_yy + ik_zz} [/tex]

[tex] \mathbf{H}(\mathbf{r}) = H_0 (\alpha\hat{y} + \beta\hat{z}) e^{ik_yy + ik_zz} [/tex]

TM_z solutions are of the form,

[tex] \mathbf{E}(\mathbf{r}) = E_0 (\alpha\hat{y} + \beta\hat{z}) e^{ik_yy + ik_zz} [/tex]

[tex] \mathbf{H}(\mathbf{r}) = H_0 \hat{x} e^{ik_yy + ik_zz} [/tex]

Further, we note that we can have a solution that travels in the +y and +z direction and also in the -y and +z direction at the same location. That is, we can have both upward and downward bouncing waves. So we generally combine the two travelling wave components as a standing wave component where the standing wave is just the superposition of two travelling waves. So that means that the E field of the TE_z solution is expressed more compactly as,

[tex] \mathbf{E}(\mathbf{r}) = E_0 \hat{x} \sin (k_yy) e^{ik_zz} [/tex]

Again, the selection of the wave vector k is dependent upon the mode and frequency of the solution (obviously in the above we note that at the locations of the plates (say y=0, y=a) the tangential electric field is zero). This, in essence, is an eigenvalue problem. So it's a simple matter to see that in our case that,

[tex] k_y = \frac{\pi m}{a} [/tex]

[tex]k_z = \sqrt{ k_0^2 - \left( \frac{\pi m}{a} \right)^2 } [/tex]

Thus the modes for our TE_z solution for a desired frequency are defined by the above eigenmodes for m=1,2,3, ...

EDIT: Ok, pfew... I think that's actually pretty much the treatment of the parallel plate waveguide in its entirety.

Last edited:

- #4

yungman

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Are your equations without the same requirement so you have to worry about bouncing between plates. Therefore your propagation constant

Why there would be no energy spent to assemble the charges and surface current on the surfaces of the plates like we deal with the energy spent to assemble charges in electrostatic study? Is it because it is sinodal and the assemble and disassemble cancel out or what?

Thanks for your time.

- #5

Born2bwire

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Are your equations without the same requirement so you have to worry about bouncing between plates. Therefore your propagation constantkhas both y and z components?

Why there would be no energy spent to assemble the charges and surface current on the surfaces of the plates like we deal with the energy spent to assemble charges in electrostatic study? Is it because it is sinodal and the assemble and disassemble cancel out or what?

Thanks for your time.

If that's the case then it is an evanescent mode, which bears with your original equations assuming that \delta is real (in which case the solution is decaying exponentially). We can see that they come out easily enough. Assuming that the width is 1/10th of a wavelength,

[tex]k_z = \sqrt{ k_0^2 - \left( \frac{\pi m}{a} \right)^2 } = \sqrt{ \left( \frac{2\pi}{\lambda} \right)^2 - 100 m^2 \left( \frac{\pi}{\lambda} \right)^2} = \frac{\pi}{\lambda} \sqrt{ 4-100m^2} = i\delta [/tex]

But I expect that this is not the same as the case you are thinking about with electronic circuits as this is an infinite parallel plate waveguide. Such a configuration is not useful since it does not support propagation modes with such small separations. Those would be microstrip or striplines which have a different set of governing equations and behaviors. The dispersion relations for microstrip lines allow solutions down to DC. (EDIT: This does not work with a parallel plate waveguide since we cannot have a TEM_z solution that satisfies the boundary conditions. This is probably why the given expressions in your text do not have the sinusoidal portion. They are basically enforcing an unsupported mode.)

It doesn't take any energy because it's a PEC. The voltage difference between any two points on or inside the PEC is zero. So there is no work needed to be expended to move a charge about or on the PEC. So inducing currents on the PEC does not expend energy.

Last edited:

- #6

yungman

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What is PEC?

- #7

Born2bwire

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Perfect electrical conductor.

- #8

yungman

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Perfect electrical conductor.

Thanks for the help. I think the book use parallel plates as the first step to show how to calculate capacitance, inductance and impedance. I don't even see a lot of use with parallel plate tx line, later in the book, it just lead into microstrip and stripline which are the most common in circuit boards.

Another question is why there is not x component in the

Again, thanks for your time.

Alan

- #9

Born2bwire

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Thanks for the help. I think the book use parallel plates as the first step to show how to calculate capacitance, inductance and impedance. I don't even see a lot of use with parallel plate tx line, later in the book, it just lead into microstrip and stripline which are the most common in circuit boards.

Another question is why there is not x component in thek, ie the propagation has no x component in your equations in #3? That is one question I don't have also. What prevent the EM wave from skipping out from the side of a true parallel plate tx line? Your equation still assume the wave propagate in y and z direction only.

Again, thanks for your time.

Alan

The analysis of a parallel plate waveguide assumes that the plates are infinite. Since the problem is invariant in two directions, we can solve the problem as a 2D problem. The choice of coordinates for the 2D representation is arbitrary.

- #10

yungman

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The analysis of a parallel plate waveguide assumes that the plates are infinite. Since the problem is invariant in two directions, we can solve the problem as a 2D problem. The choice of coordinates for the 2D representation is arbitrary.

What is the reason the EM wave stay inside between the parallel plate even when the guide line turn corner?

I want to verify that the way to launch the TEM is from the

Then the next question is why terminating the line electrically can terminate the EM wave? Like you said, it does not take energy to generate current and charges in PEC. At the end of the line, only the voltage or the current see the termination. But the EM wave is still there going. How dose the EM wave get terminated?

Again thanks for your help.

Alan

- #11

Born2bwire

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The EM wave gets "terminated" as you put it by being absorbed. This can be done by gradual loss of the wave by leaking out of the waveguide. It can be absorbed due to the lossy materials (finitely conductive metals, lossy dielectrics) that make up the waveguide. And it can be absorbed by the receiving object like an antenna or chip. Without some kind of mechanism to absorb or leak the energy of the wave the wave will always just travel down, back and forth, the waveguide.

The voltage and currents are consequences of the electromagnetic wave. When the wave impinges on a conductor, it induces currents. In addition, the electric field and magnetic fields induce a voltage difference across sections of the waveguide. When we want to excite an electromagnetic wave we can do this by replicating the currents or voltages that the wave would induce on the line by virtue of reciprocity.

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