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The origin of higher modes

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GGAPBE96

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Impedance mismatch generate higher order modes?
Please see the attached example...
 

GGAPBE96 said:
Impedance mismatch generate higher order modes?
Yep, any obstacle (imperfection) in a waveguide generetes high order modes.
These modes could be propagated or not depending on their cut-off freqs.
 

    GGAPBE96

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To navuho,

Thank you.
I confused impedance mismatch with geometrical discontinuity.
Both obstacles violate the electromagnetic field..
 

GGAPBE96 said:
To navuho,

Thank you.
I confused impedance mismatch with geometrical discontinuity.
Both obstacles violate the electromagnetic field..

Suppose you have a microstrip with a certain length. Underneath the signal line you apply material A for half of the length and material B for another half of the length. This means you will have an impedance mismatch, but does not necessarily mean it will generate higher-order mode. It may generate higher order mode if it reaches the cut-off freq depending on the εr and etc, but not always.

While, I guess discontinuity (step, bend, and etc) will always generate higher-order modes which may or may not die along line as it propagates.

Please correct me if I am wrong.

regards,
wlcsp
 

    GGAPBE96

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wlcsp said:
but does not necessarily mean it will generate higher-order mode. It may generate higher order mode if it reaches the cut-off freq depending on the er and etc, but not always.

wlcsp

To wlcsp,

Thank u for the reply.

I'm confused again...
I understood that reflection and higher order modes always arise whenever the EM fields are violated.
If signal frequency is below the cut-off frequency,higher order modes only decay exponentially.

What is the electronic difference between discontinuity and impedance mismatch?
 

wlcsp said:
This means you will have an impedance mismatch, but does not necessarily mean it will generate higher-order mode.
Yes, you are right. If the impeadnce mismatch caused only by media properties changing (εr, µr) then no high order modes will be generated near such a boundary.
My previous answer was based on the picture with kind of obstacle (hole ?) presence.

It may generate higher order mode if it reaches the cut-off freq depending on the εr and etc, but not always.
In that case I think higher order modes will be not generated at all regardless of εr.
Just because of fully ortogonality of a natural waves in the waveguide.

While, I guess discontinuity (step, bend, and etc) will always generate higher-order modes which may or may not die along line as it propagates.
Agree
 

    GGAPBE96

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