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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.
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} $
