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----
 
----
 
----
 
----
==Question==
+
==Question 1==
 
[[Image:Q1FO12013.png|Alt text|500x500px]]
 
[[Image:Q1FO12013.png|Alt text|500x500px]]
  
[[Image:Q1FO12013D.png|Alt text|300x300px]]
+
[[Image:Q1FO12013D.png|Alt text|200x200px]]
  
 
=Solution=
 
=Solution=
 +
:'''Click [[ECE_PhD_QE_FO_2013_Problem1.1|here]] to view student [[ECE_PhD_QE_FO_2013_Problem1.1|answers and discussions]]'''
 
<math>
 
<math>
 
\begin{equation*}
 
\begin{equation*}
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</math>
 
</math>
  
==Question==
+
==Question 2==
 
[[Image:Q2FO12013.png|Alt text|500x500px]]
 
[[Image:Q2FO12013.png|Alt text|500x500px]]
  
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<math>
 
<math>
 
\begin{align*}
 
\begin{align*}
\nabla\dot \bar{D} &= \rho \\
+
\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 \\
 
\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)}
 
\frac{\partial}{\partial x}(2)&=0=\rho \quad \text{(no charge)}
 
\end{align*}
 
\end{align*}
 
</math>
 
</math>
 +
 +
Also:
  
 
<math>
 
<math>
\underline{Also}:
 
 
\begin{align*}
 
\begin{align*}
 
\oint \bar{D}\cdot d\bar{S}&=Q\\
 
\oint \bar{D}\cdot d\bar{S}&=Q\\
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\end{align*}
 
\end{align*}
 
</math>
 
</math>
 +
 +
==Question 3==
 +
[[Image:Q3FO12013.png|Alt text|500x500px]]
 +
 +
[[Image:Q3FO12013D.png|Alt text|500x500px]]
 +
 +
=Solution=
 +
Using superposition <br/>
 +
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>
 +
 +
 +
  
  

Revision as of 21:50, 24 April 2017


ECE Ph.D. Qualifying Exam

Fields and Optics (FO)

Question 1: Statics 1

August 2013



Question 1

Alt text

Alt text

Solution

Click here to view student answers and discussions

$ \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*} $

The magnetic flux through this closed surface is $ \Phi $

$ \begin{equation*} \boxed{\Phi=0} \end{equation*} $

Question 2

Alt text

Solution

Alt text

$ \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*} $

Also:

$ \begin{align*} \oint \bar{D}\cdot d\bar{S}&=Q\\ &=\int2(dS_x)+\int2(-dS_x)=2-2=\boxed{0} \end{align*} $

Question 3

Alt text

Alt text

Solution

Using superposition
In the left cylinder
$ \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*} $
$ \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*} $
Alt text
$ \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*} $

In the right cilinder
$ \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*} $

$ \begin{equation*} \boxed{\bar{H}_T=\bar{H}_L+\bar{H}_R=\frac{3J_0}{2}\hat{y}} \end{equation*} $



Question

Part 1.

Consider $ n $ independent flips of a coin having probability $ p $ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $ n=5 $ and the sequence $ HHTHT $ is observed, then there are 3 changeovers. Find the expected number of changeovers for $ n $ flips. Hint: Express the number of changeovers as a sum of Bernoulli random variables.

Click here to view student answers and discussions

Part 2.

Let $ X_1,X_2,... $ be a sequence of jointly Gaussian random variables with covariance

$ Cov(X_i,X_j) = \left\{ \begin{array}{ll} {\sigma}^2, & i=j\\ \rho{\sigma}^2, & |i-j|=1\\ 0, & otherwise \end{array} \right. $

Suppose we take 2 consecutive samples from this sequence to form a vector $ X $, which is then linearly transformed to form a 2-dimensional random vector $ Y=AX $. Find a matrix $ A $ so that the components of $ Y $ are independent random variables You must justify your answer.

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Part 3.

Let $ X $ be an exponential random variable with parameter $ \lambda $, so that $ f_X(x)=\lambda{exp}(-\lambda{x})u(x) $. Find the variance of $ X $. You must show all of your work.

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Part 4.

Consider a sequence of independent random variables $ X_1,X_2,... $, where $ X_n $ has pdf

$ \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} $.

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 $ X_1,X_2,... $ converges in mean-square if and only if $ E[|X_n-X_{n+m}|] \to 0 $ as $ n \to \infty $, for every $ m>0 $.

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