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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 [[ | + | :'''Click [[2011CS-2-1|here]] to view student [[2011CS-2-1|answers and discussions]]''' |
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<math>z[n]=x[n]+jy[n]</math><br> | <math>z[n]=x[n]+jy[n]</math><br> | ||
(f) Plot <math>r_{zz}[l]</math> | (f) Plot <math>r_{zz}[l]</math> | ||
− | :'''Click [[ | + | :'''Click [[2011CS-2-2|here]] to view student [[2011CS-2-2|answers and discussions]]''' |
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[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE QE page]] |
Latest revision as of 14:34, 19 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) $.
(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.
- Click here to view student answers and discussions
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] $
- Click here to view student answers and discussions