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Consider the following input signal <br> | Consider the following input signal <br> | ||
<math>x[n]=16\dfrac{sin(\dfrac{3\pi}{8}n)}{\pi n}\dfrac{sin(\dfrac{\pi}{8}n)}{\pi n}cos(\dfrac{\pi}{2}n)</math><br> | <math>x[n]=16\dfrac{sin(\dfrac{3\pi}{8}n)}{\pi n}\dfrac{sin(\dfrac{\pi}{8}n)}{\pi n}cos(\dfrac{\pi}{2}n)</math><br> | ||
| + | HINT: The solution to problem is greatly simplified if you exploit the fact that the DTFT of the input signal <math>x[n]</math> is such that <math>X(\omega)=X(\omega-\pi)</math>.<br> | ||
| + | |||
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Revision as of 22:44, 17 February 2019
Communicates & Signal Process (CS)
Question 2: Signal Processing
August 2011
Problem 1. [60 pts]
In the system below, the two analysis filters, $ h_0[n] $ and $ h_1[n] $, and the two synthesis filters, $ f_0[n] $ and $ f_1[n] $,form a Quadrature Mirror Filter (QMF). Specially,
$ h_0[n]=\dfrac{2\beta cos[(1+\beta)\pi(n+5)/2]}{\pi[1-4\beta^2(n+5)^2]}+\dfrac{sin[(1-\beta)\pi(n+0.5)/2]}{\pi[(n+.5)-4\beta^2(n+.5)^3]},-\infty<n<\infty with \beta=0.5 $
$ h_1[n]=(-1)^n h_0[n] $ $ f_0[n]=h_0[n] $ $ f_1[n]=-h_1[n] $
The DTFT of the halfband filter $ h_0[n] $ above may be expressed as follows:
$ H_0(\omega)= \begin{cases} e^{j\dfrac{\omega}{2}} |\omega|<\dfrac{\pi}{4},\\ e^{j\dfrac{\omega}{2}} cos[(|\omega|-\dfrac{\pi}{4})], \dfrac{\pi}{4}<|\omega|<\dfrac{3\pi}{4} \\ 0 \dfrac{3\pi}{4}<|\omega|<\pi \end{cases} $
Consider the following input signal
$ x[n]=16\dfrac{sin(\dfrac{3\pi}{8}n)}{\pi n}\dfrac{sin(\dfrac{\pi}{8}n)}{\pi n}cos(\dfrac{\pi}{2}n) $
HINT: The solution to problem is greatly simplified if you exploit the fact that the DTFT of the input signal $ x[n] $ is such that $ X(\omega)=X(\omega-\pi) $.
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 $?
