Difference between revisions of "ECE PhD QE FO2 2012 Problem2.1other" - Rhea

Latest revision as of 11:55, 4 June 2017

1) $ \left\{ \begin{array}{ll} \bar{E}_1 = [e^{-j\beta_1 Z} + re^{j\beta_1 Z}]\hat{x} \hspace{1cm }\bar{H}_1 =\frac{1}{n_1} [e^{-j\beta_1 Z} - re^{j\beta_1 Z}]\hat{y}\\ \bar{E}_2 = te^{-j\beta_2 Z}\hspace{2.7cm }\bar{H}_2 = \frac{t}{n_2}e^{-j\beta_2 Z} \end{array} \right. $

$ \left\{ \begin{array}{ll} \beta_1 = \omega\sqrt{\mu_1\epsilon_1}\hspace{0.5cm} n_1=\sqrt{\frac{\mu_1}{\epsilon_1}}\\ \beta_2 = \omega\sqrt{\mu_1\epsilon_1}\hspace{0.5cm} n_2=\sqrt{\frac{\mu_2}{\epsilon_2}} \end{array} \right. $

BCs: $ \begin{align*} E_{1t} &= E_{2t}|_{Z=0}: 1+r = t\\ H_{1t} &= H_{2t}|_{Z=0}: \frac{1}{n_1}[1-r] = \frac{t}{n_2} \end{align*} $

$ 1+r = \frac{n_2}{n_1}[1+r] $

$ r\bigg[1+\frac{n_2}{n_1}\bigg] = \frac{n_2}{n_1} -1 $

$ r = \frac{n_2-n_1}{n_1+n_2} $

$ t = 1 + r = \frac{n_2+n_1+n_2-n_1}{n_2+n_1} $

$ t = \frac{2n_2}{n_1+n_2} $

2) $ \left\{ \begin{array}{ll} \bar{E}_0 = [E_0e^{-j\beta_0Z} + rE_0e^{j\beta_0Z}]\hat{x}\\ \bar{E}_1 = [Ae^{-j\beta_1Z} + Be^{j\beta_1Z}]\hat{x}\\ \bar{E}_2 = [E_0te^{-j\beta_2 Z}]\hat{x} \end{array} \right. $

$ \left\{ \begin{array}{ll} \bar{H}_0 = \frac{1}{M_0}[E_0e^{-j\beta_0Z} - rE_0e^{j\beta_0Z}]\hat{y}\\ \bar{H}_1 = \frac{1}{M_1}[Ae^{-j\beta_1Z} - Be^{j\beta_1Z}]\hat{y}\\ \bar{H}_2 = \frac{1}{M_2}[E_0te^{-j\beta_2 Z}]\hat{y} \end{array} \right. $

$ n=\sqrt{\mu_r\epsilon_r} $
$ n=\sqrt{\frac{\mu}{\epsilon}} = \frac{1}{\sqrt{\epsilon}} $
$ n = \frac{1}{n\sqrt{\epsilon_0}} = \frac{n_0}{n} $
$ \beta = \omega\sqrt{\mu\epsilon} =n\beta_0 $

BC's:

$ (1) E_{0t} = E_{1t}|_{Z=0}: E_0(1+r )= A+ B $

$ (2) H_{0t} = H_{1t}|_{Z=0}: \frac{n_1}{n_0}E_0(1-r) = A- B $

$ (3) E_{1t} = E_{2t}|_{Z=d}: Ae^{-j\beta_1 d}+Be^{j\beta_1d} = E_0te^{-j\beta_2d} $

$ (4) H_{1t} = H_{2t}|_{Z=d}: \frac{n_2}{n_1}[Ae^{-j\beta_1 d}-Be^{j\beta_1d}] = E_0te^{-j\beta_2d} $

Unknowns: $ r,t,A,B $
Equation (1) + Equation (2): $ E_0(1+r) + \frac{n_1}{n_0}E_0(1-r) = 2A $
Equation (3) - Equation (4): $ Ae^{-j\beta_1d} + Be^{j\beta_1d} = \frac{n_2}{n_1}[Ae^{-j\beta_1 d}-Be^{j\beta_1d}] $

$ \begin{align*} 1+r &= \frac{1}{E_0}(A+B)<br> -r & = \frac{n_0}{n_1}\frac{1}{E_0}(A-B)\\ 2E_0 &= [(A+B) + \frac{n_0}{n_1}(A-B)]\\ 2E_0 &= A\bigg(1+ \frac{n_0}{n_1}\bigg) + B\bigg(1- \frac{n_0}{n_1}\bigg)\\ A&=\frac{2E_0 - B(1-\frac{n_0}{n_1})}{1+n_0/n_1} \end{align*} $

$ e^{-j\beta d}\frac{2E_0}{1+n_0/n_1}+\bigg[\frac{-(1-\frac{n_0}{n_1})}{1+\frac{n_0}{n_1}}e^{-j\beta_1 d} +e^{j\beta_1 d}\bigg]B = \frac{2\frac{n_2}{n_1}E_0}{1+\frac{n_0}{n_1}}e^{-j\beta_1 d} -\bigg[\frac{\frac{n_2}{n_1}(1-\frac{n_0}{n_1})}{1+n_0/n_1}e^{-j\beta_1d}+ e^{j\beta_1d}\bigg]B $

$ B\bigg[2e^{j\beta_1d} + \frac{\frac{n_2}{n_1}(1-\frac{n_0}{n_1}) -(1-\frac{n_0}{n_1})}{1+n_0/n_1}e^{-j\beta_1d}\bigg] = e^{-j\beta_1d}\bigg[\frac{2E_0[\frac{n_2}{n_1}-1]}{1+n_0/n_1}\bigg] $

$ \to B\to A\to r\to t $

impedance transformation: $ Z_{in} = Z_1\frac{Z_2 + Z_1j\tan\beta_1d}{Z_1 + Z_2j\tan \beta_1 d} $

$ Z_{in} = n_1\frac{n_2 + n_1j\tan\beta_1d}{n_1 + n_2j\tan \beta_1 d} $

$ Z_1 = n_1 =\frac{n_0}{n_1} \hspace{0.5cm} \beta_1 = \omega\sqrt{\mu_1\epsilon_1} = n_1\beta_0 $

$ Z_2 = n_2 = \frac{n_0}{n_2} $

$ Z_0 = n_0 $

$ \begin{align*} r &= \frac{Z_{in}-Z_0}{Z_[in]+Z_0} = \frac{n_1\frac{n_2+n_1j\tan\beta_1d}{n_1+n_2j\tan\beta_1d} -n_0}{n_1\frac{n_2+n_1j\tan\beta_1d}{n_1+n_2j\tan\beta_1d} +n_0}\\ & = \frac{n_1n_2 +n_1^2j\tan\beta_1d - n_0n_1-n_0n_2j\tan\beta_1d}{n_1n_2 +n_1^2j\tan\beta_1d + n_0n_1+n_0n_2j\tan\beta_1d}\\ & = \frac{\frac{1}{n_1n_2}\cancel{n_0^2} + \frac{1}{n_1^2}\cancel{n_0^2}(j\tan n_1\beta_0d) - \frac{1}{n_1}\cancel{n_0^2} - \frac{1}{n_2}\cancel{n_0^2}j\tan(n_1\beta_0 d)}{\frac{1}{n_1n_2}\cancel{n_0^2} + \frac{1}{n_1^2}\cancel{n_0^2}(j\tan n_1\beta_0d) + \frac{1}{n_1}\cancel{n_0^2} + \frac{1}{n_2}\cancel{n_0^2}j\tan(n_1\beta_0 d)}\\ r &= \frac{n_1+n_2(j\tan n_1\beta_0d) - n_1n_2 -n_1^2(j\tan(n_1\beta_0d))}{n_1+n_2(j\tan n_1\beta_0d) +n_1n_2 +n_1^2(j\tan(n_1\beta_0d))} \end{align*} $

4) $ r = 0 = n_1 -n_1n_2 +(n_2-n_1^2)(j\tan n_1\beta_0d) $

$ n_2 = 1 =n_0 $

$ n_1\beta_0d = n\pi\hspace{0.5cm}n = 0,1,2,3,\dots $

$ d = \frac{n\pi}{n_1\beta_0}\hspace{0.5cm}\beta = \frac{2\pi}{\lambda_0} $

$ d = \frac{\lambda_0}{2n_1} = \frac{\lambda}{2} $

5) $ n_1n_3 = n_2^2 $

$ \frac{2\pi}{\lambda}\frac{\lambda}{4} = \frac{\pi}{2} $

$ \Gamma = \frac{n_2 - n_1^2}{n_2+n_1^2} = 0 $

$ n_2 = n_1^2 $

$ n_2^2 - n_1n_3 $

$ n_2l=\frac{\pi}{2} $