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

Here's a start....

http://www.microwaves101.com/encyclo...eguidemath.cfm

Scott
N0EDV

wrote:

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

wrote in message
ups.com...
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


As frequency decreases wavelength increases, Cutoff is the longest
wavelength that will fit in the pipe.
Fcutoff = speed of light/2 times the largest crosssectional dimension of the
waveguide

or Fc= C/2d

However the wavlength of a frequency in free space is significanty
shorter than the wavelength inside the waveguide so the actual Fc will
be greater than that calculated usng freespace values. My slide rule says
that if the angle of incidence inside the waveguide is 45 degrees and the
free space wavelenghth is .05 meters wavelength inside the waveguide is
about .075 meters. Dont hold me to these numbers but they should be
reasonably close. Im not very good at writting equations using a computer
but the Practical Antenna Handbook by Joeseph Carr has a pretty good
discription of whats going on. I am sure others will give you more
references and probably more precise numbers.

Jimmie

wrote in message
ups.com...
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


Try this:

http://www.microwaves101.com/encyclo...ath.cfm#cutoff

Regards,

Frank

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