The origin of higher 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.

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

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?

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.

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.

Agree