m (Add a comment to a minor fault in the solution)
 
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</math>
 
</math>
  
−
<span style="color:green"> Lack of proof. Should mention the property of Poisson distribution to show the equivalence. See the proof in solution 2.</span>
+
b) So the impulse can be obtained by reversing Z-transform
  
−
b) So the impulse can be obtained by inverting Z-transform
 
 
<math>
 
<math>
 
h(m,n) = ba^mu[m]a^nu[n] = ba^{m+n}u[m]u[n]
 
h(m,n) = ba^mu[m]a^nu[n] = ba^{m+n}u[m]u[n]
 
</math>
 
</math>
  
−
c) Solve the differential equation in b), we have
+
c) Z-transform can be written as
−
 
+
 
<math>
 
<math>
−
\lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt}
+
X(z_1,z_2) = \sum\sum x(m,n)z_1^{-m}z_2^{-n}
 
</math>
 
</math>
  
−
d) So the integral of the density, <math>\int^T_0\mu(x)dx </math> can be written as
+
Therefore, when <math>z_1=1,z_2=1</math>, <math>X(z_1,z_2) = \sum\sum x(m,n)</math>, which is equivalent to the average of the signal. So in order to satisfy the condition, we need <math>H(1,1) = 1</math>
−
<math>
+
−
\int^T_0\mu(x)dx = -\log\left(\frac{\lambda_T}{\lambda_0}\right)
+
−
</math>
+
  
 +
<math>H(1,1) =  \frac{b}{(1-az_1^{-1})(1-az_2^{-1})} = \frac{b}{(1-a)^2} = 1</math>
  
−
e) <math>\int^T_ \mu(x)dx \simeq -\log \left( \frac{Y_T}{Y_0} \right)</math>
+
So <math>b = (1-a)^2</math>.
 +
 
 +
<span style="color:green"> Another solution is to think of DC component of the image signal, so it can be derived from <math>\omega = 0</math>.</span>
 +
 
 +
d). <math>R_x(k,l) = E[x(m,n)x(m+k,n+l)] = \sum_m\sum_n x(m,n)x(m+k,n+l)</math>
 +
 
 +
so when <math>k=l=0</math>, <math>R_x</math> is the covariance of the input signal, which is 1.
 +
 
 +
when <math>k \neq 0</math> or <math>l \neq 0</math>, according to the property of the i.i.d. random variables, we know it should be 0.
 +
 
 +
Combining those two conditions, we can get that <math>R_x(k,l) = \delta_{k,l}</math>
 +
 
 +
<span style="color:green">This can be obtained directly using the property of i.i.d.</span>
 +
 
 +
And the power spectral density can be obtained using <math>R_x</math>, and it is 1.
 +
 
 +
e) The power spectral density can be calculated from that of input signal.
 +
 
 +
<math>S_y(e^{j\mu}, e^{j\nu}) = \|H(e^{j\mu}, e^{j\nu})\|^2S_x(e^{j\mu}, e^{j\nu})</math>
 +
 
 +
So
 +
 
 +
<math>S_y(e^{j\mu}, e^{j\nu}) = \|\frac{b}{(1-ae^{-j\mu})(1-ae^{-j\nu}))}\|^2\times 1 = \frac{b^2}{\left[{(1-ae^{-j\mu})(1-ae^{-j\nu})}\right]^2}</math>
 +
<span style="color:red"> Last line is wrong. 2-norm of a complex number is product of this number with its conjugate, not the complex number squared </span>
  
 
== Solution 2: ==
 
== Solution 2: ==
  
−
a). As we know <math>P\left\{Y_x=k\right\} = \frac{e^{-\lambda_x}\lambda_x^k}{k!}</math> is a Potion distribution, it is known that the expectation of a Poisson RV is <math>\lambda_x</math>.
+
a). Make 2d Z transform of the original 2D difference equation
 +
 
 +
<math>Y(z_1,z_2) = bX(z_1,z_2)+aY(z_1,z_2)z_1^{-1}+aY(z_1,z_2)z_2^{-1}-a^2Y(z_1,z_2)z_1^{-1}z_2^{-1}</math>
  
−
Proof:
+
then
  
 
<math>
 
<math>
−
\begin{split}
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H(z_1,z_2) = \frac{Y(z_1,z_2)}{X(z_1,z_2)} = \frac{b}{1-az_1^{-1}-az_2^{-1}+a^2z_1^{-1}z_2^{-1}} = \frac{bz_1z_2}{(z_1-a)(z_2-a)}
−
E[Y_x] &= \sum^{+ \infty}_{k > 0} k \frac{e^{-\lambda_x}\lambda_x^k}{k!}\\
+
−
&= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^k}{(k-1)!}\\
+
−
&= \sum^{+ \infty}_{k = 1} \frac{e^{-\lambda_x}\lambda_x^{k-1}}{(k-1)!}\lambda_x\\
+
−
&= \lambda_xe^{-\lambda_x}\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!}\\
+
−
&= \lambda_xe^{-\lambda_x}e^{\lambda_x}\\
+
−
&= \lambda_x\\
+
−
\end{split}
+
 
</math>
 
</math>
  
−
So <math>E[Y_x] = \lambda_x</math>
+
b). From a). <math>H(z_1,z_2) == \frac{bz_1z_2}{(z_1-a)(z_2-a)}</math> is separable,
  
−
<span style="color:red"> Here, it used <math>\sum^{+ \infty}_{k = 0} \frac{\lambda_x^k}{k!} = e^{\lambda_x}</math> to derive the final conclusion.</span>
+
<math>H(z_1,z_2) = H(z_1)H(z_2) = \frac{\sqrt{b}}{1-az_1^{-1}}\frac{\sqrt{b}}{1-az_2^{-1}}</math>
  
−
b). Because the number of photons will decrease when increasing the depth,
+
we know that <math>a^nu(n) \Longleftrightarrow \frac{1}{1-az^{-1}}</math>
−
<math>
+
−
d\lambda_x = -\lambda_x\mu(x)dx
+
−
</math>
+
  
−
and
+
therefore <math>\frac{\sqrt{b}}{1-az_1^{-1}} \Longleftrightarrow \sqrt{b}a^mu(m)</math>
  
−
<math>
+
<math>\frac{\sqrt{b}}{1-az_2^{-1}} \Longleftrightarrow \sqrt{b}a^nu(n)</math>
−
\frac{d\lambda_x}{dx} = -\lambda_x\mu(x)
+
−
</math>
+
  
−
c). The final differential equation in b). is an ordinary differential equation. We can get the expression as
+
<math>h(m,n) = h(m)h(n) = ba^mu[m]a^nu[n] = ba^{m+n}u[m]u[n]</math>
  
−
<math>\lambda_x = \lambda_0e^{-\int^x_0\mu(t)dt}</math>
+
<span style="color:red"> In fact, by the linearity of Z-tranform, b as a scalar doesn't have to be treated this way. See solution 1. </span>
  
−
where <math>\lambda_0</math> is the initial number of photons.
+
c). if we want average values of the input and the output image remain the same, we should guarantee that the magnitude of <math>H(z_1,z_2)</math> at <math>\omega=0</math> equals to 1.
  
−
d). From part c). <math>\frac{\lambda_x}{\lambda_0} = e^{-\int^x_0\mu(t)dt}</math>, so we have
+
So we can easily get
  
−
<math>
+
<math>b\times \frac{1}{1-a}=1 \Longleftrightarrow b=1-a</math>
−
\int^x_0\mu(t)dt = -\log\left(\frac{\lambda_T}{\lambda_0}\right)
+
−
</math>
+
  
−
e). Because from a). and c)., we can get
+
<span style="color:red">This solution has the wrong result, but the right way to think of the problem: because <math>H(z_1,z_2) = \frac{bz_1z_2}{(z_1-a)(z_2-a)}</math>, it is missing another <math>1-a</math> term in the denominator. The correct answer is in solution 1.</span>
  
−
<math>
+
d). Because <math>x(m,n)</math> is i.i.d. R.V.
−
\int^x_0\mu(t)dt = -\log\left(\frac{Y_T}{Y_0}\right)
+
 
−
</math>
+
<math>R_x(k,l) = \delta(k,l)</math>
 +
 
 +
<math>S_x(e^{j\mu},e^{j\nu}) = DSFT\left\{R_x(k,l)\right\} = DSFT\left\{\delta(k,l)\right\} = 1</math>
 +
 
 +
e). Because <math>y(m,n) = x(m,n)**h(m,n)</math>
 +
 
 +
<math>S_y(e^{j\mu},e^{j\nu}) = \|H(e^{j\mu},e^{j\nu})\|^2S_x(e^{j\mu},e^{j\nu}) = \frac{b^2}{(1-ae^{-j\mu})^2(1-ae^{-j\nu})^2}</math>
  
−
<span style="color:red"> This photon attenuation question is very similar to other questions: for example 2017S-ECE637-Exam1, Problem 3. Related topics are projection problems(e.g.: 2013S-ECE637-Exam1, Problem 2; 2012S-ECE637-Exam1, Problem 3) and scan problems(e.g.: 2016QE-CS5, Problem 1). </span>
 
  
  

Latest revision as of 10:08, 13 August 2018


ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)

August 2014, Problem 2

Problem 1 , 2

Solution 1

a) Take Z-transform on both sides, we have $ \begin{split} &Y(z_1,z_2) = bX(z_1,z_2)+aY(z_1,z_2)z_1^{-1}+aY(z_1,z_2)z_2^{-1}-a^2Y(z_1,z_2)z_1^{-1}z_2^{-1}\\ &Y(z_1,z_2) = bX(z_1,z_2) +Y(z_1,z_2)\left[az_1^{-1}+az_2^{-1}-a^2z_1^{-1}z_2^{-1}\right]\\ &Y(z_1,z_2)\left[1-az_1^{-1}-az_2^{-1}+a^2z_1^{-1}z_2^{-1} \right] = bX(z_1,z_2)\\ &\frac{Y(z_1,z_2)}{X(z_1,z_2)} = \frac{b}{1-az_1^{-1}-az_2^{-1}+a^2z_1^{-1}z_2^{-1}} = \frac{b}{(1-az_1^{-1})(1-az_2^{-1})}\\ \end{split} $

b) So the impulse can be obtained by reversing Z-transform

$ h(m,n) = ba^mu[m]a^nu[n] = ba^{m+n}u[m]u[n] $

c) Z-transform can be written as $ X(z_1,z_2) = \sum\sum x(m,n)z_1^{-m}z_2^{-n} $

Therefore, when $ z_1=1,z_2=1 $, $ X(z_1,z_2) = \sum\sum x(m,n) $, which is equivalent to the average of the signal. So in order to satisfy the condition, we need $ H(1,1) = 1 $

$ H(1,1) = \frac{b}{(1-az_1^{-1})(1-az_2^{-1})} = \frac{b}{(1-a)^2} = 1 $

So $ b = (1-a)^2 $.

Another solution is to think of DC component of the image signal, so it can be derived from $ \omega = 0 $.

d). $ R_x(k,l) = E[x(m,n)x(m+k,n+l)] = \sum_m\sum_n x(m,n)x(m+k,n+l) $

so when $ k=l=0 $, $ R_x $ is the covariance of the input signal, which is 1.

when $ k \neq 0 $ or $ l \neq 0 $, according to the property of the i.i.d. random variables, we know it should be 0.

Combining those two conditions, we can get that $ R_x(k,l) = \delta_{k,l} $

This can be obtained directly using the property of i.i.d.

And the power spectral density can be obtained using $ R_x $, and it is 1.

e) The power spectral density can be calculated from that of input signal.

$ S_y(e^{j\mu}, e^{j\nu}) = \|H(e^{j\mu}, e^{j\nu})\|^2S_x(e^{j\mu}, e^{j\nu}) $

So

$ S_y(e^{j\mu}, e^{j\nu}) = \|\frac{b}{(1-ae^{-j\mu})(1-ae^{-j\nu}))}\|^2\times 1 = \frac{b^2}{\left[{(1-ae^{-j\mu})(1-ae^{-j\nu})}\right]^2} $ Last line is wrong. 2-norm of a complex number is product of this number with its conjugate, not the complex number squared

Solution 2:

a). Make 2d Z transform of the original 2D difference equation

$ Y(z_1,z_2) = bX(z_1,z_2)+aY(z_1,z_2)z_1^{-1}+aY(z_1,z_2)z_2^{-1}-a^2Y(z_1,z_2)z_1^{-1}z_2^{-1} $

then

$ H(z_1,z_2) = \frac{Y(z_1,z_2)}{X(z_1,z_2)} = \frac{b}{1-az_1^{-1}-az_2^{-1}+a^2z_1^{-1}z_2^{-1}} = \frac{bz_1z_2}{(z_1-a)(z_2-a)} $

b). From a). $ H(z_1,z_2) == \frac{bz_1z_2}{(z_1-a)(z_2-a)} $ is separable,

$ H(z_1,z_2) = H(z_1)H(z_2) = \frac{\sqrt{b}}{1-az_1^{-1}}\frac{\sqrt{b}}{1-az_2^{-1}} $

we know that $ a^nu(n) \Longleftrightarrow \frac{1}{1-az^{-1}} $

therefore $ \frac{\sqrt{b}}{1-az_1^{-1}} \Longleftrightarrow \sqrt{b}a^mu(m) $

$ \frac{\sqrt{b}}{1-az_2^{-1}} \Longleftrightarrow \sqrt{b}a^nu(n) $

$ h(m,n) = h(m)h(n) = ba^mu[m]a^nu[n] = ba^{m+n}u[m]u[n] $

In fact, by the linearity of Z-tranform, b as a scalar doesn't have to be treated this way. See solution 1.

c). if we want average values of the input and the output image remain the same, we should guarantee that the magnitude of $ H(z_1,z_2) $ at $ \omega=0 $ equals to 1.

So we can easily get

$ b\times \frac{1}{1-a}=1 \Longleftrightarrow b=1-a $

This solution has the wrong result, but the right way to think of the problem: because $ H(z_1,z_2) = \frac{bz_1z_2}{(z_1-a)(z_2-a)} $, it is missing another $ 1-a $ term in the denominator. The correct answer is in solution 1.

d). Because $ x(m,n) $ is i.i.d. R.V.

$ R_x(k,l) = \delta(k,l) $

$ S_x(e^{j\mu},e^{j\nu}) = DSFT\left\{R_x(k,l)\right\} = DSFT\left\{\delta(k,l)\right\} = 1 $

e). Because $ y(m,n) = x(m,n)**h(m,n) $

$ S_y(e^{j\mu},e^{j\nu}) = \|H(e^{j\mu},e^{j\nu})\|^2S_x(e^{j\mu},e^{j\nu}) = \frac{b^2}{(1-ae^{-j\mu})^2(1-ae^{-j\nu})^2} $


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