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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} $
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.
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).
