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TI-84 Plus CE: Rectangular Waveguide – Cutoff Frequency

TI-84 Plus CE:  Rectangular Waveguide – Cutoff Frequency 

Introduction

The program WAVEGDE computes the cutoff frequency of a rectangular guide.  The waveguide has energy propagating in the TE_m,n mode with dimensions a x b (in meters) with relative permittivity E.  The cutoff frequencies is calculated as:

V = speed of light / √E
where speed of light = 299,792,458 m/s

Cutoff frequency (in Hz):
f_c = V/2 * √( (m/a)^2 + (n/b)^2 )

The second part of the calculation will depend on the target frequency, either the attenuation or phase and group velocities will be calculated:

If f < f_c, attenuation (in nerpers) is calculated:

attenuation = (2 π f_c)/V * √( 1 – (f/f_c)^2 )

If f > f_c, the phase and group velocities are calculated:

phase velocity = V / √( 1 – (f_c/f)^2 )

group velocity = V * √(1 – (f_c/f)^2 )

TI-84 Plus CE Program WAVEGDE

“EWS 2019-07-14”
Disp “RECTANGULAR WAVEGUIDE”
Input “WIDE (M): “,A
Input “NARROW (M): “,B
Input “PERMITTIVITY: “,E
Disp “HALF-WAVE VARIATIONS”, “T-M,N”
Input “M: “,M
Input “N: “,N
Input “FREQUENCY (HZ):”,F
299792458/√(E)→V
V/2*√((M/A)²+(N/B)²)→C
Disp “CUTOFF FREQ: (HZ)”,C
Pause 
If F
Then
(2πC)/V*√(1-(F/C)²)→U
Disp “ATTENUATION:”,U,”NEPERS”
Else
V/√(1-(C/F)²)→P
V*√(1-(C/F)²)→G
Disp “PHASE VELOCITY: M/S”,P
Disp “GROUP VELOCITY: M/S”,G
End

Examples

Example 1:
A rectangular waveguide has the dimensions 2.7 cm x 1.09 cm with permittivity of 1.    (a = 2.7/100, b = 1.09/100, E = 1).  The waves propagate in the TE_2,1 mode  (m = 2, n = 1).  The target frequency is 10 GHZ (10E9 Hz).

Results:
Cutoff Frequency = 1.767490017E10 Hz
Attenuation = 305.4488993 nepers

Example 2: 
Same as Example 1, but set target frequency as 20 GHZ.

Results:
Cutoff Frequency = 1.767490017E10 Hz
Phase Velocity = 640,624,938.6 m/s
Group Velocity = 140,293,504.8 m/s

Source:

“Rectangular Waveguide Calculations”  HP-65 E.E. Pac 2  Hewlett Packard  Cupertino, CA  1977

Eddie

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