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Old December 20th 06, 04:25 PM posted to rec.radio.amateur.antenna
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Default Rectangular waveguide


ha escrito:

In air-filled rectangular waveguide, what is the meaning by " the
operating frequency is to be at least 25% greater than the cutoff
frequency of the dominant mode"?
Can show it in mathematically?
Thanks

Laura


Hello Laura,

Regarding your rectangular wave guide question:

Mostly a waveguide for energy transport is not used close to the
cut-off frequencies. If you do, the losses of the guide will be high.
At the cut-off frequency (and below for H10 mode), the wave vanishes
exponentially of a short distance.

To reduce the losses, you should choose the operation frequency well
above the cut-off frequency of the propagation mode that you want to
use. Minimal loss is obtained around twice the cut-off frequency for
H10 mode (H10 mode, H-field has component in direction of propagation,
E-field has not, there is one E-field maximum along the largest width
of the guide).

Many modes of propagation can exist, especially when the operating
frequency is far above the lowest cut-off frequency and other geometry
is used. The way of excitation (for example coax to waveguide
converter), together with the waveguide dimensions determine the
"dominant" propagation mode.

Besides the waveguide loss, there is also a group velocity. Close to
the cut-off frequency, the group velocity changes strongly with
frequency. This introduces envelope distortion of the propagated signal
(if you were using AM). Of course this only counts for very wide band
signals.

Best Regards,
Wim



 
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