Numerical1: TE Modes in a Rectangular Waveguide
Consider a rectangular waveguide having dimensions:
$ a = 2.286\text{ cm} = 2.286\times10^{-2}\text{ m} $
$ b = 1.016\text{ cm} = 1.016\times10^{-2}\text{ m} $
The cutoff frequency for any TEmn mode is given by:
$ f_c = \frac{c}{2} \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 } $
where:
- \(c = 3\times10^8\) m/s is the velocity of light
- \(a\) is the broader dimension of the waveguide
- \(b\) is the narrower dimension of the waveguide
- \(m\) and \(n\) are the mode indices
Cutoff Frequency for TE01 Mode
For TE01 mode:
$ m=0,\qquad n=1 $
Substituting into the cutoff frequency formula:
$ f_{c01} = \frac{3\times10^8}{2} \sqrt{ 0+ \left( \frac{1}{1.016\times10^{-2}} \right)^2 } $
$ f_{c01} = 14.764\text{ GHz} $
Cutoff Frequency for TE20 Mode
For TE20 mode:
$ m=2,\qquad n=0 $
Substituting into the cutoff frequency formula:
$ f_{c20} = \frac{3\times10^8}{2} \sqrt{ \left( \frac{2}{2.286\times10^{-2}} \right)^2 + 0 } $
$ f_{c20} = 13.123\text{ GHz} $
Cutoff Frequency for TE12 Mode
For TE12 mode:
$ m=1,\qquad n=2 $
Substituting into the cutoff frequency formula:
$ f_{c12} = \frac{3\times10^8}{2} \sqrt{ \left( \frac{1}{2.286\times10^{-2}} \right)^2 + \left( \frac{2}{1.016\times10^{-2}} \right)^2 } $
$ f_{c12} = 30.248\text{ GHz} $
Cutoff Frequency for TE21 Mode
For TE21 mode:
$ m=2,\qquad n=1 $
Substituting into the cutoff frequency formula:
$ f_{c21} = \frac{3\times10^8}{2} \sqrt{ \left( \frac{2}{2.286\times10^{-2}} \right)^2 + \left( \frac{1}{1.016\times10^{-2}} \right)^2 } $
$ f_{c21} = 19.753\text{ GHz} $
Final Answers
- $f_{c01}=14.764\text{ GHz}$
- $f_{c20}=13.123\text{ GHz}$
- $f_{c12}=30.248\text{ GHz}$
- $f_{c21}=19.753\text{ GHz}$
Among these modes, TE20 has the lowest cutoff frequency and would begin propagating first as the operating frequency is increased.