Question 3.1: CHARACTERISTICS OF A RECTANGULAR WAVEGUIDE Consider a length...
CHARACTERISTICS OF A RECTANGULAR WAVEGUIDE Consider a length of Teflon-filled, copper K-band rectangular waveguide having dimensions a = 1.07 cm and b = 0.43 cm. Find the cutoff frequencies of the first five propagating modes. If the operating frequency is 15 GHz, find the attenuation due to dielectric and conductor losses.
TABLE 3.2 Summary of Results for Rectangular Waveguide
| Quantity | \ TE_{mn} Mode | \ TM_{mn} Mode |
| k | \ \omega \sqrt{\mu \epsilon } | \ \omega \sqrt{\mu \epsilon } |
| \ k_{c} | \ \sqrt{\left(m\pi /a\right)^{2}+\left(n\pi /b\right)^{2} } | \ \sqrt{\left(m\pi /a\right)^{2}+\left(n\pi /b\right)^{2} } |
| \ \beta | \ \sqrt{k^{2}-k_{c}^{2}} | \ \sqrt{k^{2}-k_{c}^{2}} |
| \ \lambda _{c} | \ \frac{2\pi }{k_{c}} | \ \frac{2\pi }{k_{c}} |
| \ \lambda _{g} | \ \frac{2\pi }{\beta } | \ \frac{2\pi }{\beta } |
| \ v_{p} | \ \frac{\omega }{\beta } | \ \frac{\omega }{\beta } |
| \ \alpha_{d} | \ \frac{k^{2}\tan \delta }{2\beta } | \ \frac{k^{2}\tan \delta }{2\beta } |
| \ E_{z} | 0 | \ B\sin \frac{m\pi x}{a} \sin \frac{n\pi y}{b} e^{-j\beta z} |
| \ H_{z} | \ A\cos \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} | 0 |
| \ E_{x} | \ \frac{j\omega \mu n\pi }{k_{c}^{2}b}A\cos \frac{m\pi x}{a} \sin \frac{n\pi y}{b} e^{-j\beta z} | \ \frac{-j\beta m\pi }{k_{c}^{2}a} B\cos \frac{m\pi x}{a} \sin \frac{n\pi y}{b} e^{-j\beta z} |
| \ E_{y} | \ \frac{-j\omega \mu m\pi }{k_{c}^{2}a}A\sin \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} | \ \frac{-j\beta n\pi }{k_{c}^{2}b} B\sin \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} |
| \ H_{x} | \ \frac{j\beta m\pi }{k_{c}^{2}a} A\sin \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} | \ \frac{j\omega \epsilon n\pi }{k_{c}^{2}b}B\sin \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} |
| \ H_{y} | \ \frac{j\beta n\pi }{k_{c}^{2}b} A\sin \frac{m\pi x}{a} \cos \frac{n\pi y}{b} e^{-j\beta z} | \ \frac{-j\omega \epsilon m\pi }{k_{c}^{2}}B\cos \frac{m\pi x}{a} \sin \frac{n\pi y}{b} e^{-j\beta z} |
| \ Z | \ Z_{TE}=\frac{k\eta }{\beta } | \ Z_{TM}=\frac{\beta\eta }{k } |
From Appendix G, for Teflon, \ \epsilon_{r}=2.08 and tan δ= 0.0004. From (3.84)\ f_{cmn}=\frac{k_{c}}{2\pi \sqrt{\mu \epsilon } } =\frac{1}{2\pi \sqrt{\mu \epsilon } } \sqrt{\left(\frac{m\pi }{a} \right)^{2} +\left(\frac{n\pi }{b} \right)^{2}} the cutoff frequencies are given by
\ f_{cmn}=\frac{c}{2\pi \sqrt{\epsilon _{r}} } \sqrt{\left(\frac{m\pi }{a} \right)^{2} +\left(\frac{n\pi }{b} \right)^{2}}Computing \ f_{c}for the first few values of m and n gives the following results:
| Mode | m | n | \ f_{c}\left(GHz\right) |
| TE | 1 | 0 | 9.72 |
| TE | 2 | 0 | 19.44 |
| TE | 0 | 1 | 24.19 |
| TE, TM | 1 | 1 | 26.07 |
| TE, TM | 2 | 1 | 31.03 |
Thus the \ TE_{10},TE_{20},TE_{01},TE_{11} and TM_{11}modes will be the first five modes to propagate.
At 15 GHz,\ k=453.1m^{-1}, and the propagation constant for the \ TE_{10}mode is
From (3.29)\ \alpha _{d}=\frac{k^{2}\tan \delta }{2\beta } Np/m\left(TE or TM waves\right).
, the attenuation due to dielectric loss is
\ \alpha _{d}=\frac{k^{2}\tan \delta }{2\beta }= 0.119 Np/m = 1.03 dB/m.The surface resistivity of the copper walls is\ \left(\sigma =5.8\times 10^{7}S/m\right)
\ R_{s}=\sqrt{\frac{\omega \mu_{0}}{2\sigma } } = 0.032\Omega,and the attenuation due to conductor loss, from (3.96), is
\ \alpha _{c}=\frac{R_{s}}{a^{3}b\beta k\eta } \left(2b\pi ^{2}+a^{3}k^{2}\right) =0.050Np/m=0434dB/m.