Communicates & Signal Process (CS)
Question 2: Signal Processing
August 2010
Problem 1. [50 pts]
Equation below is the formula, for reconstructing the DTFT, $ X(\omega) $, from $ N $ equi-spaced samples of the DTFT over $ 0 \leq \omega \leq 2\pi $. $ X_{N}(k) = X(\frac{2\pi k}{N},k=0,1,...,N-1) $ is the N-pt DFT of x[n], which corresponds to N equi-spaced samples of the DTFT of x[n] over $ 0 \leq \omega \leq 2\pi $.
(a) Let x[n] be a discrete-time rectangular pulse of length $ L=12 $ as defined below:
(i) $ X_{N}(k) $ is computed as a 16-point DFT of x[n] and used in Eqn (1) with N=16. Write a closed-form expression for resulting reconstructed spectrum $ X_{r}(\omega) $.
(ii) $ X_{N}(k) $ is computed as a 12-point DFT of x[n] and used in Eqn (1) with N=12. Write a closed-form expression for the resulting reconstructed spectrum $ X_{r}(\omega) $.
(iii) $ X_{N}(k) $ is computed as an 8-point DFT of x[n] ans used in Eqn (1) with N=8. That is, $ X_{N}(k) $ is obtained by sampling the DTFT of x[n] at 8 equi-spaced frequencies between 0 and 2$ \pi $. Write a closed-form expression for the resulting reconstructed spectrum $ X_{r}(\omega) $.
(b) Let x[n] be a discrete-time sinewave of length L=12 as defined below. For all sub-parts of part (b), $ X_{N}(k) $ is computed as a 12-pt DFT of x[n] and used in Eqn (1) with N=12.
(i) Write a closed-form expression for the resulting reconstructed spectrum $ X_{r}(\omega) $.
(ii) What is the numerical value of $ X_{r}(\frac{\pi}{3}) $? The answer is a number and you do not need a calculator to determine the value; this also applies to the next 2 parts.
(iii) What is the numerical value of $ X_{r}(\frac{5 \pi}{3}) $?
(iv) What is the numerical value of $ X_{r}(\frac{\pi}{2}) $?
Solution :
(a)
(i)
$ x[n] = (u[n]-u[n+2]) - 2(u[n]-u[n-4]) $
$ X(w) = e^{-j \frac{12-1}{2}w } \frac{sin(\frac{12}{2}w)}{sin(\frac{1}{2}w)} - 2e^{-j \frac{4-1}{2}w } \frac{sin(\frac{4}{2}w)}{\frac{1}{2}w} $
$ X_{r}(w) = X(w) $
(ii)
same,
$ X_{r}(w)=X(w) $
(iii)
$ x_{r}[n] = \{0,0,0,0,1,1,1,1 \} = (u[n] - u[n-4])* \delta[n-4] $
$ X_{r}(w) = e^{-j \frac{4-1}{2}w} \frac{sin(\frac{4}{2}w)}{\frac{1}{2}w} e^{-j4w} $
(b)
(i)
$ X(w)=\frac{1}{2} e^{-j \frac{11}{2}(w- \frac{\pi}{3})} \frac{sin(6(w-\frac{\pi}{3}))}{sin(\frac{1}{2}(w-\frac{\pi}{3}))} + \frac{1}{2} e^{-j \frac{11}{2} (w+\frac{\pi}{3}) } \frac{sin(6(w+\frac{\pi}{3}))}{sin(\frac{1}{2}(w-\frac{\pi}{3}))} $
(ii)
$ X_{r}(\frac{\pi}{3})=\frac{1}{2} \times 12 = 6 $
(iii)
$ X_{r}(\frac{5}{3} \pi) = \frac{1}{2} \times 12 = 6 $
(iv)
$ X_{r}(\frac{\pi}{2}) = \frac{1}{2} \times 0 = 0 $
Problem 2. [50 pts]
Consider a finite-length sinewave of the form below where $ k_{o} $ is an interger in the range $ 0 \leq k_{o} \leq N-1 $.
In addition, h[n] is a causal FIR filter of length L, where L < N. In this problem $ y[n]=x[n] \star h[n] $ is the linear convolution of the causal sinewave of length N in Equation (1) with a causal FIR filter of length L, where L < N.
(a) The region $ 0 \leq n \leq L-1 $ corresponds to partial overlap. The covolution sum can be written as:
Determine the upper and lower limits in the convolution sum above for $ 0 \leq n \leq L-1 $
(b) The region $ L \leq n \leq N-1 $ corresponds to full overlap. The convolution sum is:
(i) Determine the upper and lower limits in the convolution sum for $ L \leq n \leq N-1 $.
(ii) Substituting x[n] in Eqn (1), show that for this range y[n] simplifies to:
where $ H_{N}(k) $ is the N-point DFT of h[n] evaluated at $ k = k_{o} $. To get the points, you must show all work and explain all details.
(c) The region $ N \leq n \leq N+L-2 $ corresponds to partial overlap. The convolution sum:
Determine the upper and lower limits in the convolution sum for $ N \leq n \leq N+L-2 $.
(d) Add the two regions of partial overlap at the beginning and end to form:
(i) Determine the upper and lower limits in the convolution sum above.
(ii) Substituting x[n] in Eqn (1), show that for this range z[n] simplifies to:
where $ H_{N}(k) $ is the N-point DFT of h[n] evaluated at $ k = k_{o} $ as defined previously.
(e) $ y_{N}[n] $ is formed by computing $ X_{N}(k) $ as an N-pt DFT of x[n] in Enq (2), $ H_{N}(k) $ as an N-pt DFT of h[n], and then $ y_{N}[n] $ as the N-pt inverse DFT of $ Y_{N}(k) = X_{N}(K)H_{N}(k) $. Write a simple, closed-form expression for $ y_{N}(k) $. Is $ z[n]=y_{N}[n] + y[n+N] \ \ for \ 0 \leq n \leq N-1 $?
Solution:
(a) $ y[n] = \sum_{k=0}^{n} h[k] \cdot x[n-k] $
So, the lower and upper limits are 0 ~ n.
(b) $ y[n] = \sum_{k=0}^{L-1} h[k] \cdot x[n-k] $
So, the lower and upper limits are 0 ~ L-1.
$ x[n] = e^{2 \pi j (\frac{K_{0}}{N}) n } (u[n] - u[n-N]) $
$ y[n] = \sum_{k=-\infty}^{+ \infty} h_{N}[k] \cdot x[n-k] $
$ = \sum_{k=-\infty}^{+\infty} h_{N}[k] \cdot e^{2 \pi j (\frac{k_{0}}{N})(n-k)} (u[n-k]-u[n-N-k]) $
$ = e^{2 \pi jn \frac{k_{0}}{N}} \sum_{k=0}^{L-1} h_{N}[k] e^{-2 \pi j \frac{k_{0}}{N}} k $
$ = e^{2 \pi jn \frac{k_{0}}{N}} \cdot H_{N}(k_{0}) $
(c)
$ y[n] = \sum_{k=n-N+1}^{L-1} h[k] \cdot x[n-k] $
(d)
upper and lower limits are 0 ~ L-1.
$ z[n] = \sum_{k=-\infty}^{+ \infty} h_{N}[k] \cdot x[n-k] $
$ = \sum_{k=-\infty}^{+\infty} h_{N}[k] \cdot e^{2 \pi j (\frac{k_{0}}{N})(n-k)} (u[n-k]-u[n-N-k]) $
$ = e^{2 \pi jn \frac{k_{0}}{N}} \sum_{k=0}^{L-1} h_{N}[k] e^{-2 \pi j \frac{k_{0}}{N}} k $
$ = e^{2 \pi jn \frac{k_{0}}{N}} \cdot H_{N}(k_{0}) $
(e)
$ y_{N}[n] = H_{N}(k_{0}) e^{j 2\pi \frac{k_{0}}{N}} n $
.
Yes, the equation is correct.
