Parallel Plate Waveguide Basic Functionality Issue

AI Thread Summary
The discussion centers on the behavior of the electric field (E-field) in a parallel plate waveguide or transmission line when an oscillating voltage source is applied. It highlights that the E-field intensity varies across the waveguide due to the time it takes for voltage changes to propagate along the conductor, which is constrained by the speed of light. This propagation delay means that the assumption of equipotentiality in circuit theory breaks down for longer transmission lines. The conversation emphasizes the importance of understanding wave propagation in microwave engineering, especially when the length of the conductor approaches the wavelength of the signal. Overall, the nature of the E-field in such systems is a crucial aspect of waveguide and transmission line theory.
fred3142
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Hi,

I think I'm missing something with how a parallel plate waveguide works. In the picture I've shown below, there intensity of the E-field change depending on how far across the waveguide it is. While I know that this has to happen in order for anything to propagate, I don't quite understand why it is happening. If the top plate is connected to an oscillating voltage source, and the bottom plate is grounded, shouldn't the bottom plate always be at 0 and shouldn't the E-field of the whole top plate float up and down with the voltage source? I would've thought that at an instant in time, the E-filed would have the same magnitude everywhere in dielectric (as the bottom plate is constant 0V, and the whole top plate would float up and down together).

Why, at a particular instant in time, does the E-field magnitude change through the dielectric?



Thanks.
 

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Are you sure that picture is of a waveguide? It's labeled as a transmission line.
 
It's a parallel plate transmission line. Is that different to a parallel plate waveguide?
 
fred3142 said:
Hi,
I think I'm missing something with how a parallel plate waveguide works. In the picture I've shown below, there intensity of the E-field change depending on how far across the waveguide it is. While I know that this has to happen in order for anything to propagate, I don't quite understand why it is happening. If the top plate is connected to an oscillating voltage source, and the bottom plate is grounded, shouldn't the bottom plate always be at 0 and shouldn't the E-field of the whole top plate float up and down with the voltage source? I would've thought that at an instant in time, the E-filed would have the same magnitude everywhere in dielectric (as the bottom plate is constant 0V, and the whole top plate would float up and down together).
When you have the oscillating voltage source hooked up to one end of the parallel plate system, it takes time for the voltage to travel along the conductor. Think of the (absurd) case of a parallel plate waveguide that is long enough to stretch from here to the sun. You know that sunlight takes about 8 minutes to propagate from the sun to the earth; why should a voltage pulse take any less time, let alone exactly zero time? This would violate relativity theory.

The above argument is simply showing that in the limit of a very long transmission line that the "conductor = equipotential" approximation used in circuit theory must break down. The issue is that the voltage travels along the transmission line as a wave propagating at some velocity. The group velocity must be no greater than the speed of light in a vacuum. That is what the wave equation for the parallel plate guide is telling you.

Note that as long as a conductor is much shorter than one wavelength then the standard circuit theory "conductor = equipotential" approximation is good. Otherwise it isn't. Analyzing and designing circuits in the regime where the full wave nature of the signals must be taken into account is the domain of microwave engineering, but all EEs should learn about transmission lines at some point in their education.

Jason
 
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