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Problem 1. [60 pts] <br>
 
Problem 1. [60 pts] <br>
 
In the system below, the two analysis filters, <math>h_0[n]</math> and <math>h_1[n]</math>, and the two synthesis filters, <math>f_0[n]</math> and <math>f_1[n]</math>,form a Quadrature Mirror Filter (QMF). Specially, <br>
 
In the system below, the two analysis filters, <math>h_0[n]</math> and <math>h_1[n]</math>, and the two synthesis filters, <math>f_0[n]</math> and <math>f_1[n]</math>,form a Quadrature Mirror Filter (QMF). Specially, <br>
<math>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</math> <br>
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<math>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</math> with <math>\beta=0.5</math> <br>
 
<math>h_1[n]=(-1)^n h_0[n]</math>        <math>f_0[n]=h_0[n]</math>        <math>f_1[n]=-h_1[n]</math> <br>
 
<math>h_1[n]=(-1)^n h_0[n]</math>        <math>f_0[n]=h_0[n]</math>        <math>f_1[n]=-h_1[n]</math> <br>
 
The DTFT of the halfband filter <math>h_0[n]</math> above may be expressed as follows:<br>
 
The DTFT of the halfband filter <math>h_0[n]</math> above may be expressed as follows:<br>
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(e) Plot the magnitude of the DTFT of <math>y_1[n]</math>, <math>Y_1(\omega)</math>, over <math>-\pi<\omega<\pi</math>. Show all work.<br>
 
(e) Plot the magnitude of the DTFT of <math>y_1[n]</math>, <math>Y_1(\omega)</math>, over <math>-\pi<\omega<\pi</math>. Show all work.<br>
 
(f) Plot the magnitude of the DTFT of the final output <math>y[n][n]</math>, <math>Y(\omega)</math>, over <math>-\pi<\omega<\pi</math>. Show all work.<br>
 
(f) Plot the magnitude of the DTFT of the final output <math>y[n][n]</math>, <math>Y(\omega)</math>, over <math>-\pi<\omega<\pi</math>. Show all work.<br>
:'''Click [[ECE_PhD_QE_CNSIP_2015_Problem1.1|here]] to view student [[ECE_PhD_QE_CNSIP_2015_Problem1.1|answers and discussions]]'''
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:'''Click [[2011CS-2-1|here]] to view student [[2011CS-2-1|answers and discussions]]'''
 
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Problem 2. [50 pts] <br>
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Problem 2. [40 pts] <br>
Consider a finite-length sinewave of the form below where <math> k_{o} </math> is an interger in the range <math> 0 \leq k_{o} \leq N-1 </math>.<br/>
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(a) Let <math>x[n]</math> and <math>y[n]</math> be real-valued sequences both of which are even-symmetric: <math>x[n]=x[-n]</math> and <math>y[n]=y[-n]</math>. Under these conditions, prove that
<center><math> x[n] = e^{j 2 \pi \frac{k_{o}}{N} n} \{ u[n] - u[n-N] \} </math>   (2)</center><br/>
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<math>r_{xy}[l]=r_{yx}[l]</math> for all <math>l</math>.<br>
In addition, h[n] is a causal FIR filter of length L, where L < N. In this problem <math> y[n]=x[n] \star h[n] </math> 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.<br/>
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(b) Express the autocorrelation sequence r_{zz}[l] for the complex-valued signal <math>z[n]=x[n]+jy[n]</math> where <math>x[n]</math> and <math>y[n]</math> are real-valued sequences, in terms of <math>r_{xx}[l]</math>, <math>r_{xy}[l]</math>, <math>r_yx[l]</math> and <math>r_{yy}[l]</math>.<br>
<center><math> y[n]=x[n] \star h[n] </math></center><br/>
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(c) Determine a closed-form expression for the autocorrelation sequence <math>r_{xx}[l]</math> for the signal <math>x[n]</math> below.<br>
(a) The region <math> 0 \leq n \leq L-1 </math> corresponds to partial overlap. The covolution sum can be written as:<br/>
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<math>x[n]=({\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}})({1+(-1)^n})</math><br>
<center><math> y[n]=\sum_{k=??}^{??} h[k] x[n-k] \ \ partial \ \ overlap:\ 0 \leq n \leq L-1  </math> (3) </center><br/>
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(d) Determine a closed-form expression for the autocorrelation sequence <math>r_yy[l]</math> for the signal <math>y[n]</math> below <br>
Determine the upper and lower limits in the convolution sum above for <math> 0 \leq n \leq L-1 </math><br/>
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<math>y[n]=(\dfrac{sin(\dfrac{\pi}{4})n}{\pi n})cos(\dfrac{\pi}{2}n)</math><br>
(b) The region <math> L \leq n \leq N-1 </math> corresponds to full overlap. The convolution sum is: <br/>
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(e) Determine a closed-form expression for the autocorrelation sequence <math>r_{zz}[l]</math> for the complex-valued signal <math>z[n]</math> formed with <math>x[n]</math> and <math>y[n]</math> defined above as the real and imaginary parts, respectively, as defined below. You must show all work and simplify as much as possible.<br>
<center><math> y[n]=\sum_{k=??}^{??} h[k]x[n-k] \ \ full \ \ overlap: \ L \leq n \leq N-1 </math> (4)</center><br/>
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<math>z[n]=x[n]+jy[n]</math><br>
(i) Determine the upper and lower limits in the convolution sum for <math> L \leq n \leq N-1 </math>.<br/>
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(f) Plot <math>r_{zz}[l]</math>
(ii) Substituting x[n] in Eqn (1), show that for this range y[n] simplifies to:<br/>
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:'''Click [[2011CS-2-2|here]] to view student [[2011CS-2-2|answers and discussions]]'''
<center><math> y[n]=H_{N}(k_{o}) e^{j 2 \pi \frac{k_{o}}{N} n} \ \ for L \leq n \leq N-1  </math>(5)</center> <br/>
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where <math> H_{N}(k) </math> is the N-point DFT of h[n] evaluated at <math> k = k_{o} </math>. To get the points, you must show all work and explain all details. <br/>
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(c) The region <math> N \leq n \leq N+L-2 </math> corresponds to partial overlap. The convolution sum: <br/>
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<center><math> y[n] = \sum_{k=??}^{??} h[k]x[n-k] \ \ partial \ \ overlap: \ N \leq n \leq N+L-2 </math>(6)</center>  <br/>
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Determine the upper and lower limits in the convolution sum for <math> N \leq n \leq N+L-2 </math>. <br/>
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(d) Add the two regions of partial overlap at the beginning and end to form: <br/>
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<center><math> z[n] =y[n]+y[n+N]=\sum_{k=??}^{??}h[n]x[n-k] \ \ for: \ 0 \leq n \leq L-1 </math>(7)</center> <br/>
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(i) Determine the upper and lower limits in the convolution sum above. <br/>
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(ii) Substituting x[n] in Eqn (1), show that for this range z[n] simplifies to: <br/>
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<center> <math> z[n] = y[n]+y[n+N]=H_{N}(k_{o})e^{j2 \pi \frac{k_{o}}{N} n} \ \ for \ 0 \leq n \leq L-1 </math>  (8)</center><br/>
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where <math> H_{N}(k) </math> is the N-point DFT of h[n] evaluated at <math> k = k_{o}</math> as defined previously. <br/>
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(e) <math> y_{N}[n] </math> is formed by computing <math> X_{N}(k) </math> as an N-pt DFT of x[n] in Enq (2), <math> H_{N}(k) </math> as an N-pt DFT of h[n], and then <math> y_{N}[n] </math> as the N-pt inverse DFT of <math> Y_{N}(k) = X_{N}(K)H_{N}(k) </math>. Write a simple, closed-form expression for <math> y_{N}(k) </math>. Is <math> z[n]=y_{N}[n] + y[n+N] \ \ for \ 0 \leq n \leq N-1 </math>? <br/>
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Latest revision as of 14:34, 19 February 2019


ECE Ph.D. Qualifying Exam

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} $
Wan82_ECE538_problem1.PNG 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) $.
(a) Plot the magnitude of the DTFT of $ x[n] $, $ X(\omega) $, over $ -\pi<\omega<\pi $. Show all work.
(b) Plot the magnitude of the DTFT of $ x_0[n] $, $ X_0(\omega) $, over $ -\pi<\omega<\pi $. Show all work.
(c) Plot the magnitude of the DTFT of $ x_1[n] $, $ X_1(\omega) $, over $ -\pi<\omega<\pi $. Show all work.
(d) Plot the magnitude of the DTFT of $ y_0[n] $, $ Y_0(\omega) $, over $ -\pi<\omega<\pi $. Show all work.
(e) Plot the magnitude of the DTFT of $ y_1[n] $, $ Y_1(\omega) $, over $ -\pi<\omega<\pi $. Show all work.
(f) Plot the magnitude of the DTFT of the final output $ y[n][n] $, $ Y(\omega) $, over $ -\pi<\omega<\pi $. Show all work.

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Problem 2. [40 pts]
(a) Let $ x[n] $ and $ y[n] $ be real-valued sequences both of which are even-symmetric: $ x[n]=x[-n] $ and $ y[n]=y[-n] $. Under these conditions, prove that $ r_{xy}[l]=r_{yx}[l] $ for all $ l $.
(b) Express the autocorrelation sequence r_{zz}[l] for the complex-valued signal $ z[n]=x[n]+jy[n] $ where $ x[n] $ and $ y[n] $ are real-valued sequences, in terms of $ r_{xx}[l] $, $ r_{xy}[l] $, $ r_yx[l] $ and $ r_{yy}[l] $.
(c) Determine a closed-form expression for the autocorrelation sequence $ r_{xx}[l] $ for the signal $ x[n] $ below.
$ x[n]=({\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}})({1+(-1)^n}) $
(d) Determine a closed-form expression for the autocorrelation sequence $ r_yy[l] $ for the signal $ y[n] $ below
$ y[n]=(\dfrac{sin(\dfrac{\pi}{4})n}{\pi n})cos(\dfrac{\pi}{2}n) $
(e) Determine a closed-form expression for the autocorrelation sequence $ r_{zz}[l] $ for the complex-valued signal $ z[n] $ formed with $ x[n] $ and $ y[n] $ defined above as the real and imaginary parts, respectively, as defined below. You must show all work and simplify as much as possible.
$ z[n]=x[n]+jy[n] $
(f) Plot $ r_{zz}[l] $

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