A \(10 \mathrm{GHz}\) radar signal may be represented as a uniform plane wave in a sufficiently small region. Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a nonmagnetic material for which (a) \(\epsilon_{r}^{\prime}=1\) and \(\epsilon_{r}^{\prime \prime}=0 ;(b) \epsilon_{r}^{\prime}=1.04\) and \(\epsilon_{r}^{\prime \prime}=9.00 \times\) \(10^{-4} ;(c) \epsilon_{r}^{\prime}=2.5\) and \(\epsilon_{r}^{\prime \prime}=7.2\)

First, compute the wavelength using the speed of light equation: \(\lambda = \frac{c}{f \sqrt{\epsilon_{r}^{\prime}}}\) where: - \(\lambda\) is the wavelength, - \(c \approx 3\times 10^{10}\) \(\mathrm{cm/s}\) is the speed of light in vacuum, - \(f\) is the signal frequency, - and \(\epsilon_{r}^{\prime}\) is the real part of the relative complex permittivity. Once you have the wavelength, calculate the attenuation constant (\(\alpha\)) in nepers per meter: \(\alpha = \frac{\pi f\epsilon_{r}^{\prime\prime}}{c}\) where: - \(\alpha\) is the attenuation constant in nepers per meter, - \(\epsilon_{r}^{\prime\prime}\) is the imaginary part of the relative complex permittivity. Now, substitute the known values and solve for \(\lambda\) and \(\alpha\). \(\lambda = \frac{3\times 10^{10}}{(10\times 10^9)(1)} = 3\,\mathrm{cm}\) \(\alpha = \frac{\pi(10\times 10^9)(0)}{3\times 10^{10}} = 0\,\mathrm{Np/m}\)

Repeat the steps from part (a), but with the new given values of \(\epsilon_{r}^{\prime}\) and \(\epsilon_{r}^{\prime\prime}\). \(\lambda = \frac{3\times 10^{10}}{(10\times 10^9)(\sqrt{1.04})} \approx 2.94\,\mathrm{cm}\) \(\alpha = \frac{\pi(10\times 10^9)(9.00 \times 10^{-4})}{3\times 10^{10}} \approx 9.42\times 10^{-4}\,\mathrm{Np/m}\)

Repeat the steps from part (a) and (b), but with the new given values of \(\epsilon_{r}^{\prime}\) and \(\epsilon_{r}^{\prime\prime}\). \(\lambda = \frac{3\times 10^{10}}{(10\times 10^9)(\sqrt{2.5})} \approx 1.9\,\mathrm{cm}\) \(\alpha = \frac{\pi(10\times 10^9)(7.2)}{3\times 10^{10}} \approx 7.54\times 10^{-2}\,\mathrm{Np/m}\) To summarize, the wavelengths and attenuations for each case are: - Part (a): \(\lambda = 3\,\mathrm{cm}\) and \(\alpha = 0\,\mathrm{Np/m}\) - Part (b): \(\lambda = 2.94\,\mathrm{cm}\) and \(\alpha = 9.42\times 10^{-4}\,\mathrm{Np/m}\) - Part (c): \(\lambda = 1.9\,\mathrm{cm}\) and \(\alpha = 7.54\times 10^{-2}\,\mathrm{Np/m}\)