Latest revision as of 21:52, 24 April 2017

ECE Ph.D. Qualifying Exam

Fields and Optics (FO)

Question 1: Statics 1

August 2013

Question 1

Solution

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Question 2

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Question 3

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@@ Line 27: / Line 27: @@
 =Solution=
 :'''Click [[ECE_PhD_QE_FO_2013_Problem1.1|here]] to view student [[ECE_PhD_QE_FO_2013_Problem1.1|answers and discussions]]'''
-<math>
-\begin{equation*}
-\left.\begin{aligned}
-\nabla\cdot \bar{B}&=0\\
-\oint_S \bar{B}\cdot d\bar{S}&=0
-\end{aligned}\right\}
- \longrightarrow \Phi=\oint_S \bar{B}\cdot d\bar{S} \Longrightarrow  \boxed{ \Phi=\oint_S \bar{B}\cdot d\bar{S}=0}
-\end{equation*}
-</math>
-The magnetic flux through this closed surface is <math>\Phi</math>
-<math>
-\begin{equation*}
-\boxed{\Phi=0}
-\end{equation*}
-</math>
 ==Question 2==
@@ Line 49: / Line 32: @@
 =Solution=
-[[Image:cube.png|Alt text|300x300px]]
+:'''Click [[ECE_PhD_QE_FO_2013_Problem2.1|here]] to view student [[ECE_PhD_QE_FO_2013_Problem2.1|answers and discussions]]'''
-<math>
-\begin{align*}
-\nabla \cdot \bar{D} &= \rho \\
-\quad (\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z})\cdot(2\hat{x})&=\rho \\
-\frac{\partial}{\partial x}(2)&=0=\rho \quad \text{(no charge)}
-\end{align*}
-</math>
-Also:
-<math>
-\begin{align*}
-\oint \bar{D}\cdot d\bar{S}&=Q\\
-&=\int2(dS_x)+\int2(-dS_x)=2-2=\boxed{0}
-\end{align*}
-</math>
 ==Question 3==
@@ Line 74: / Line 40: @@
 =Solution=
-Using superposition <br/>
+:'''Click [[ECE_PhD_QE_FO_2013_Problem3.1|here]] to view student [[ECE_PhD_QE_FO_2013_Problem3.1|answers and discussions]]'''
-In the left cylinder <br/>
-<math>
-\begin{equation*}
-\nabla\times \bar{H}=\bar{J} \longrightarrow \oint \bar{H}\cdot d\bar{l}=\int_S\bar{J}\cdot d\bar{S}
- \qquad \left\{ \begin{aligned}
-dl&=dr\hat{r}+rd\phi\hat{\phi}+dz\hat{z}\\
-d\bar{S}_z&=rd\phi dr\hat{z}
-\end{aligned} \right.
-\end{equation*}
-</math>
-<br/>
-<math>
-\begin{align*}
-\int_0^{2\pi}H_{\phi}(rd\phi)&=\int_0^r\int_0^{2\pi} J_0(r'd\phi dr')\\
-H_{\phi}(2\pi r)&=2\pi J_0(\frac{r^2}{2}) \\
- & \boxed{\bar{H}=\frac{J_0r}{2}\hat{\phi}}
-\end{align*}
-</math>
-<br/>
-[[Image:cil.png|Alt text|300x300px]]
-<br/>
-<math>
-\begin{align*}
-\text{Transform to cartesian:}&\left\{\begin{aligned}
- r&=\sqrt{x^2+y^2}\\
-\hat{\phi}&=-sin\phi\hat{x}+cos\phi\hat{y}\\
-&=(\frac{-y}{\sqrt{x^2+y^2}})\hat{x}+(\frac{x}{\sqrt{x^2+y^2}})\hat{y}
-\end{aligned}\right. \\
-& \boxed{\bar{H}_L=\frac{J_0}{2}\left[-y\hat{x}+x\hat{y}\right]}
-\end{align*}
-</math>
-In the right cilinder <br/>
-<math>
-\begin{align*}
-&\bar{H}_R=\frac{-J_0}{2}\left[-y'\hat{x'}+x'\hat{y'}\right]&
-\left\{
-\begin{aligned}
-x'&=x-3\\
-y'&=y\\
-\hat{x}'&=\hat{x}\\
-\hat{y}'&=\hat{y}
-\end{aligned}
-\right.\\
-&\boxed{\bar{H}_R=\frac{-J_0}{2}\left[-y\hat{x}+(x-3)\hat{y}\right]}&
-\end{align*}
-</math>
-<math>
-\begin{equation*}
-\boxed{\bar{H}_T=\bar{H}_L+\bar{H}_R=\frac{3J_0}{2}\hat{y}}
-\end{equation*}
-</math>
-==Question==
-'''Part 1. '''
-Consider <math class="inline">n</math> independent flips of a coin having probability <math class="inline">p</math> of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if <math class="inline">n=5</math> and the sequence <math class="inline">HHTHT</math> is observed, then there are 3 changeovers. Find the expected number of changeovers for <math class="inline">n</math> flips. ''Hint'': Express the number of changeovers as a sum of Bernoulli random variables.
-:'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.1|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.1|answers and discussions]]'''
-----
-'''Part 2.'''
-Let <math>X_1,X_2,...</math> be a sequence of jointly Gaussian random variables with covariance
-<math>Cov(X_i,X_j) = \left\{ \begin{array}{ll}
- {\sigma}^2, & i=j\\
- \rho{\sigma}^2, & |i-j|=1\\
-, & otherwise
-  \end{array} \right.</math>
-Suppose we take 2 consecutive samples from this sequence to form a vector <math>X</math>, which is then linearly transformed to form a 2-dimensional random vector <math>Y=AX</math>. Find a matrix <math>A</math> so that the components of <math>Y</math> are independent random variables You must justify your answer.
-:'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.2|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.2|answers and discussions]]'''
-----
-'''Part 3.'''
-Let <math>X</math> be an exponential random variable with parameter <math>\lambda</math>, so that <math>f_X(x)=\lambda{exp}(-\lambda{x})u(x)</math>. Find the variance of <math>X</math>. You must show all of your work.
-:'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.3|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.3|answers and discussions]]'''
-----
-'''Part 4.'''
-Consider a sequence of independent random variables <math>X_1,X_2,...</math>, where <math>X_n</math> has pdf
-<math>\begin{align}f_n(x)=&(1-\frac{1}{n})\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\frac{n-1}{n}\sigma)^2]\\
-&+\frac{1}{n}\sigma exp(-\sigma x)u(x)\end{align}</math>.
-Does this sequence converge in the mean-square sense? ''Hint:'' Use the Cauchy criterion for mean-square convergence, which states that a sequence of random variables <math>X_1,X_2,...</math> converges in mean-square if and only if <math>E[|X_n-X_{n+m}|] \to 0</math> as <math>n \to \infty</math>, for every <math>m>0</math>.
-:'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.4|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.4|answers and discussions]]'''
 ----
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Difference between revisions of "ECE PhD QE Fields Optics 2013" - Rhea

Latest revision as of 21:52, 24 April 2017

Contents

Question 1

Solution

Question 2

Solution

Question 3

Solution

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