(35 intermediate revisions by 2 users not shown)
Line 16: Line 16:
 
</font size>
 
</font size>
  
August 2016 (Published on Jul 2019)
+
August 2016 (Published in Jul 2019)
 
</center>
 
</center>
 
==Problem 1==
 
==Problem 1==
  
# Calculate an expression for <math>\lambda_n^c</math>, the X-ray energy corrected for the dark current.
+
1. Calcualte an expression for <math>\lambda_n^c</math>, the X-ray energy corrected for the dark current  
  
 
<center>
 
<center>
Line 26: Line 26:
 
</center>
 
</center>
  
# Calculate an expression for <math>G_n</math>, the X-ray attenuation due to the object's presence.
+
2. Calculate an expression for <math>G_n</math>, the X-ray attenuation due to the object's presence
  
 
<center>
 
<center>
<math>
+
<math>G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n * \Delta d)\lambda_n^c</math>
G_n=-\mu(x,y_0+n*\Delta d)\lambda_n
+
</math>
+
 
</center>
 
</center>
  
# Calculate an expression for <math>\hat{P}_n</math>, an estimate of the integral intensity in terms of <math>\lambda_n</math>, <math>\lambda_n^b</math>, and <math>\lambda_b^d</math>.
+
3. Calculate an expression for <math>\hat{P}_n</math>, an estimate of the integral intensity in terms of <math>\lambda_n</math>, <math>\lambda_n^b</math>, and <math>\lambda_n^d</math>
 +
 
  
 
<center>
 
<center>
<math>
+
<math>\lambda_n = (\lambda_n^b-\lambda_n^d) e^{-\int_{0}^{x}\mu(t)dt}d)\lambda_n^c</math>
\lambda_n=(\lambda_n^b-\lambda_n^d)e^{-\int_0^x \mu(t)dt}
+
 
</math>
+
<math>\hat{P}_n = \int_{0}^{x}\mu(t)dt= -log(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d})</math>
</center>
+
<center>
+
<math>
+
\hat{P}_n=\int_0^x \mu(t)dt=-log\frac{\lambda_n}{\lambda_n^b-\lambda_n^d}
+
</math>
+
 
</center>
 
</center>
  
# For this part, assume that the object is of constant density with <math>\mu(x,y)=\mu_0</math>. Then sketch a plot of <math>\hat{P}_n</math> versus the object thickness, <math>T_n</math>, in <math>mm</math>, for the <math>n^{th}</math> detector. Label key features of the curve such as its slope and intersection.
+
4. For this part, assume that the object is of constant density with <math>\mu(x,y) = \mu_0</math>. Then sketch a plot of <math>\hat{P}_n</math> versus the object thickness, <math>T_n</math>, in mm, for the <math>n^{th}</math> detector. Label key features of the curve such as its slope and intersection.
 +
 
 +
[[Image:cs52016.jpg]]
  
 
==Problem 2==
 
==Problem 2==
  
# Specify the size of <math>YY^t</math> and <math>Y^tY</math>. Which matrix is smaller?
+
1. Specify the size of <math>YY^t</math> and <math>Y^tY</math>. Which matrix is smaller
  
 
<center>
 
<center>
<math>Y</math> is of size <math>p\times N</math>, so the size of <math>YY^t</math> is <math>p\times p</math>.
+
Y is of size <math>p \times N</math>, so the size of <math>YY^t</math> is <math>p \times p</math>
  
<math>Y</math> is of size <math>p\times N</math>, so the size of <math>Y^tY</math> is <math>N\times N</math>.
+
Y is of size <math>p \times N</math>, so the size of <math>Y^tY</math> is <math>N \times N</math>
  
Obviously, the size of <math>Y^tY</math> is much smaller, since <math>N<<p</math>.
+
Obviously, the size of <math>Y^tY</math> is much smaller, since N << p.
 
</center>
 
</center>
  
# Prove that both <math>YY^t</math> and <math>Y^tY</math> are both symmetric and positive semi-definite matrices.
+
2. Prove that both <math>YY^t</math> and <math>Y^tY</math> are both symmetric and positive semi-definite matrices.
  
 
<center>
 
<center>
 
To prove it is symmetric:
 
To prove it is symmetric:
  
<math>
+
<math>(YY^t)^t = YY^t</math>
(YY^t)^t=YY^t
+
</math>
+
  
 
To prove it is positive semi-definite:
 
To prove it is positive semi-definite:
  
Let <math>x</math> be an arbitrary vector
+
Let x be an arbitrary vector
  
<math>
+
<math>x^tYY^tx = (Y^tx)^T(Y^tx) \geq 0</math> so the matrix of <math>YY^t</math> is positive semi-definite.
x^tYY^tx=(Y^tx)^T(Y^tx)\geq 0
+
</math>
+
So the matrix <math>YY^t</math> is positive semi-definite.
+
  
The proving procedures for <math>Y^tY</math> are the same.
+
The proving procedures for <math>Y^tY</math> are the same  
 
</center>
 
</center>
  
# Derive expressions for <math>V</math> and <math>\Sigma</math> in terms of <math>T</math>, and <math>D</math>.
+
3. Derive expressions for <math>V</math> and <math>\Sigma</math> in terms of <math>T</math>, and <math>D</math>.
  
 
<center>
 
<center>
<math>
+
<math>Y^tY = (U \Sigma V^t)^T U \Sigma V^t = V\Sigma^2 V^t = TDT^t</math> therefore <math>V =T</math> and <math>\Sigma = D^{\frac{1}{2}}</math>
Y^tY=(U\Sigma V^t)^tU\Sigma V^t=V\Sigma^2V^t=TDT^t
+
</math>
+
therefore <math>V=T</math> and <math>\Sigma=D^\frac{1}{2}</math>
+
 
</center>
 
</center>
  
# Drive expressions for <math>U</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
+
4. Derive an expression for <math>U</math> in terms of <math>Y</math>, <math>T</math>, <math>D</math>.
 
+
<center>
+
<math>
+
Y=U\Sigma V^t=UD^\frac{1}{2}T^t
+
</math>
+
</center>
+
 
<center>
 
<center>
<math>
 
\therefore U=Y(D^\frac{1}{2}T^t)^{-1}
 
</math>
 
</center>
 
  
# Derive expressions for <math>E</math> in terms of <math>Y</math>, <math>T</math>, and <math>D</math>.
+
<math>Y = U\Sigma V^t= UD^{\frac{1}{2}}T^t</math>
 +
 
 +
<math>\therefore U = Y(D^{\frac{1}{2}}T^t)^{-1}</math>
 +
</center>
  
 +
5. Derive expressions for E in terms of Y, T, and D.
 
<center>
 
<center>
<math>
+
<math>YY^t = U\Sigma V^t(U\Sigma V^t)^t= U\Sigma^2 U^t = E\Gamma E^t</math>
YY^t=U\Sigma V^t(U\Sigma V^t)^t=U\Sigma^2U^t=E\Gamma E^t
+
</math>
+
  
 
therefore
 
therefore
  
<math>
+
<math>E = U = Y(D^{\frac{1}{2}}T^t)^{-1}</math>
E=U=Y(D^\frac{1}{2}T^t)^{-1}
+
</math>
+
 
</center>
 
</center>
  
# If the columns of <math>Y</math> are images from a training database, then what name do we give to the columns of <math>U</math>?
+
6. If the columns of Y are images from a training database, then what name do we give to the columns of U?
 
+
<center>
They are called eigenimages.
+
They are called '''eigenimages'''.
 +
</center>

Latest revision as of 03:07, 26 April 2020


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2016 (Published in Jul 2019)

Problem 1

1. Calcualte an expression for $ \lambda_n^c $, the X-ray energy corrected for the dark current

$ \lambda_n^c=\lambda_n^b-\lambda_n^d $

2. Calculate an expression for $ G_n $, the X-ray attenuation due to the object's presence

$ G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n * \Delta d)\lambda_n^c $

3. Calculate an expression for $ \hat{P}_n $, an estimate of the integral intensity in terms of $ \lambda_n $, $ \lambda_n^b $, and $ \lambda_n^d $


$ \lambda_n = (\lambda_n^b-\lambda_n^d) e^{-\int_{0}^{x}\mu(t)dt}d)\lambda_n^c $

$ \hat{P}_n = \int_{0}^{x}\mu(t)dt= -log(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d}) $

4. For this part, assume that the object is of constant density with $ \mu(x,y) = \mu_0 $. Then sketch a plot of $ \hat{P}_n $ versus the object thickness, $ T_n $, in mm, for the $ n^{th} $ detector. Label key features of the curve such as its slope and intersection.

Cs52016.jpg

Problem 2

1. Specify the size of $ YY^t $ and $ Y^tY $. Which matrix is smaller

Y is of size $ p \times N $, so the size of $ YY^t $ is $ p \times p $

Y is of size $ p \times N $, so the size of $ Y^tY $ is $ N \times N $

Obviously, the size of $ Y^tY $ is much smaller, since N << p.

2. Prove that both $ YY^t $ and $ Y^tY $ are both symmetric and positive semi-definite matrices.

To prove it is symmetric:

$ (YY^t)^t = YY^t $

To prove it is positive semi-definite:

Let x be an arbitrary vector

$ x^tYY^tx = (Y^tx)^T(Y^tx) \geq 0 $ so the matrix of $ YY^t $ is positive semi-definite.

The proving procedures for $ Y^tY $ are the same

3. Derive expressions for $ V $ and $ \Sigma $ in terms of $ T $, and $ D $.

$ Y^tY = (U \Sigma V^t)^T U \Sigma V^t = V\Sigma^2 V^t = TDT^t $ therefore $ V =T $ and $ \Sigma = D^{\frac{1}{2}} $

4. Derive an expression for $ U $ in terms of $ Y $, $ T $, $ D $.

$ Y = U\Sigma V^t= UD^{\frac{1}{2}}T^t $

$ \therefore U = Y(D^{\frac{1}{2}}T^t)^{-1} $

5. Derive expressions for E in terms of Y, T, and D.

$ YY^t = U\Sigma V^t(U\Sigma V^t)^t= U\Sigma^2 U^t = E\Gamma E^t $

therefore

$ E = U = Y(D^{\frac{1}{2}}T^t)^{-1} $

6. If the columns of Y are images from a training database, then what name do we give to the columns of U?

They are called eigenimages.

Alumni Liaison

Have a piece of advice for Purdue students? Share it through Rhea!

Alumni Liaison