View source for ECE PhD QE FO2 2009 Problem1.1 - Rhea

1)

a)

<math>
P = \frac{1}{2}\int Re[\bar{E}\times\bar{H}^*]d\bar{s} =\frac{1}{2n_0}|E|^2(A) =10 \text{ TEM}
|E| = \sqrt{\frac{20n_0}{A}} = \frac{\sqrt{20n_0}}{\sqrt{(0.02)(0.001)}}= 10^3\sqrt{n_0}
</math>

b)

<math>
R_s = \frac{1}{\sigma\delta} = \frac{P}{\delta} = P \sqrt{\pi f\mu\sigma} = \sqrt{\pi f\mu P} \hspace{1cm} p=10^{-5}\Omega m
R_s = \sqrt{\pi(10^9)\mu_0(10^{-5})} =100\sqrt{\pi\mu_0}
</math>

c)

<math>
\begin{align*}
P_d &=\frac{1}{2} \int\bar{E}\cdot\bar{J}dv = \frac{1}{2} \int \frac{1}{\sigma}|\bar{J_s}|\bigg(\frac{\bar{J_s}}{\delta}\bigg)ds\\
&=\frac{1}{2} |\bar{J_s}|^2 \frac{1}{\sigma\delta}(A) =\frac{1}{2} |\bar{J_s}|^2R_s(A)\\
&=\frac{1}{2} \sigma^2|E|^2 R_s(A) =\frac{1}{2P^2} |E|^2R_s(A)\\
&=\frac{1}{2}(10^{10})(10^6n_0)(100)\sqrt{\pi\mu_0}(0.02)(0.001)\\
&=\frac{1}{2}\sqrt{\pi\mu_0}(10^{18})(2)(10^{-5})n_0\\
&= n_0\sqrt{\pi\mu_0}(10^{13})
\end{align*}
</math>

-----------------------

2)

a)

<math>
\bar{E}_i = \hat{z}E_0e^{-j\beta_0x}
\bar{H}_i = (-\hat{y})\bigg(\frac{E_0}{n_0}\bigg)e^{-j\beta_0x}
</math>

b)

<math>
\bar{P}_i = \bar{E}_i\times\bar{H}_i = \hat{x} \bigg(\frac{E_0^2}{n_0}\bigg)e^{-j2\beta_0x}
\text{time-average: } \bar{P}_i = \frac{1}{2}Re[\bar{E}_i\times\bar{H}^*] = \hat{x}\bigg(\frac{E_0^2}{2n_0}\bigg)
</math>

c)

[[Image:A1FO22009.png|Alt text|200x200px]]

<math>
Z_{TE} = \frac{n}{\cos\theta}
</math>

<math>
\begin{align*}
n_2&=n_0\frac{1}{\sqrt{2}}\\
n_1&=n_0\\
n_1\sin\theta_i &= n_2\sin\theta_t\\
\frac{1}{\sqrt{2}} = \sin\theta_i &= \sqrt{2}\sin\theta_t\\
\sin\theta_t &= \frac{1}{2}\\
\theta_t &=30^\circ
\end{align*}
</math>

<math>
\begin{align*}
\Gamma &= \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{\frac{n_2}{\cos\theta_t} - \frac{n_1}{\cos\theta_i}}{\frac{n_2}{\cos\theta_t} + \frac{n_1}{\cos\theta_i}}\\
\Gamma &= \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t} = PL\\
\Gamma &= \frac{\frac{1}{\sqrt{2}}\cos(45) - \cos(30)}{\frac{1}{\sqrt{2}}\cos(45)+ \cos(30)} \\ 
\Gamma &= \frac{\frac{1}{\sqrt{2}}\big(\frac{1}{\sqrt{2}}\big) - \frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}\big(\frac{1}{\sqrt{2}}\big)+ \big(\frac{\sqrt{3}}{2}\big)} =\frac{\frac{1-\sqrt{3}}{2}}{\frac{1+\sqrt{3}}{2}}\\ 
\Gamma &= \frac{1-\sqrt{3}}{1+\sqrt{3}} \to \bar{E}_r = \hat{z}\frac{(1-\sqrt{3})}{(1+\sqrt{3})}E_0e^{+j\beta y}
\end{align*}
</math>

\underline{Note}: Reflected wave travels along <math>-\hat{y}</math> because <math>\theta_i = \theta_r = 45^\circ</math>

d)

<math>
\tau = 1+\Gamma = \frac{(1+\sqrt{3}) + (1-\sqrt{3})}{1+\sqrt{3}} = \frac{2}{1+\sqrt{3}}
\hat{n}_t = \hat{x}\cos(15^\circ) + \hat{y}\sin(15^\circ)
</math>

e)

[[Image:A2FO22009.png|Alt text|200x200px]]

<math>
\bar{E}_t = \hat{z}\bigg(\frac{2}{1+\sqrt{3}}\bigg)e^{-j\beta_0[x\cos(15)+y\sin(15)]}
</math>
transmitted wave propogates at <math>15^\circ</math> above x-axis

-------------------------

3)

<math>
K_c^2 = \omega^2\mu\epsilon = k_x^2 + k_y^2 = \bigg(\frac{m\pi}{a}\bigg)^2 +  \bigg(\frac{n\pi}{b}\bigg)^2
f_c = \frac{c}{2}\sqrt{\bigg(\frac{m}{a}\bigg)^2 + \bigg(\frac{n}{b}\bigg)^2 }
</math>
lowest order mode: TE<math>_{01}</math>

<math>
f_c = \frac{c}{2}\bigg(\frac{1}{a}\bigg) = \frac{c}{2a}=\frac{c}{3} =10^8 = 0.1GHz
\lambda_c = \frac{c}{f_c}  = 3m
</math>

--- lowest frequency has wavelength of twice the largest dimension.<br>

---assumed rectangular waveguide cut-off frequency applies here.