A rectangular waveguide has dimensions \(a=6 \mathrm{~cm}\) and \(b=4 \mathrm{~cm} .(a)\) Over what range of frequencies will the guide operate single mode? \((b)\) Over what frequency range will the guide support both \(\mathrm{TE}_{10}\) and \(\mathrm{TE}_{01}\) modes and no others?

For the TE\(_{10}\) mode, \(m=1\) and \(n=0\). Using the cut-off frequency formula, we calculate the cut-off frequency: $$f_{c_{10}} = \dfrac{c}{2\pi}\sqrt{\left(\dfrac{1\pi}{a}\right)^2 + \left(\dfrac{0\pi}{b}\right)^2} = \dfrac{c}{2a}$$ Plug in the values for \(c\) and \(a\): $$f_{c_{10}} = \dfrac{3\times10^8\,\text{m/s}}{2\times(6\times10^{-2}\,\text{m})} \approx 2.5\times10^9\,\text{Hz}$$

For the TE\(_{01}\) mode, \(m=0\) and \(n=1\). Using the cut-off frequency formula, we calculate the cut-off frequency: $$f_{c_{01}} = \dfrac{c}{2\pi}\sqrt{\left(\dfrac{0\pi}{a}\right)^2 + \left(\dfrac{1\pi}{b}\right)^2} = \dfrac{c}{2b}$$ Plug in the values for \(c\) and \(b\): $$f_{c_{01}} = \dfrac{3\times10^8\,\text{m/s}}{2\times(4\times10^{-2}\,\text{m})} \approx 3.75\times10^9\,\text{Hz}$$

For single-mode operation, the operating frequency must be higher than the cut-off frequency of the TE\(_{10}\) mode and lower than the cut-off frequency of the TE\(_{01}\) mode. This can be written as: $$f_{c_{10}} \lt f \lt f_{c_{01}}$$ So the frequency range for single-mode operation is given by: $$2.5\times10^9\,\text{Hz} \lt f \lt 3.75\times10^9\,\text{Hz}$$

For both \(\mathrm{TE}_{10}\) and \(\mathrm{TE}_{01}\) modes and no others, the operating frequency must be higher than the cut-off frequencies of both modes and lower than the next higher cut-off frequency of the waveguide (TE\(_{11}\)). Therefore, we need to first determine the cut-off frequency for TE\(_{11}\) mode. For TE\(_{11}\) mode, \(m=1\) and \(n=1\). Using the cut-off frequency formula: $$f_{c_{11}} = \dfrac{c}{2\pi}\sqrt{\left(\dfrac{1\pi}{a}\right)^2 + \left(\dfrac{1\pi}{b}\right)^2}$$ Plug in the values for \(c\), \(a\), and \(b\): $$f_{c_{11}} = \dfrac{3\times10^8\,\text{m/s}}{2\pi}\sqrt{\left(\dfrac{1\pi}{6\times10^{-2}\,\text{m}}\right)^2 + \left(\dfrac{1\pi}{4\times10^{-2}\,\text{m}}\right)^2} \approx 4.34\times10^9\,\text{Hz}$$ Now, the operating frequency must be higher than the cut-off frequencies of TE\(_{10}\) and TE\(_{01}\) modes and lower than the cut-off frequency of TE\(_{11}\) mode: $$\max(f_{c_{10}}, f_{c_{01}}) \lt f \lt f_{c_{11}}$$ So the frequency range for both TE\(_{10}\) and TE\(_{01}\) modes and no others is given by: $$3.75\times10^9\,\text{Hz} \lt f \lt 4.34\times10^9\,\text{Hz}$$