(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Causal LTI systems defined by a linear, constant coefficient difference equation = Consider th...)
 
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=== Answer 1  ===
 
=== Answer 1  ===
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a)
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<math>\mathfrak F (y[n]-\frac{1}{2}y[n-1]) = \mathfrak F (x[n])</math>
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<math>\mathcal Y (\omega) - \frac{1}{2}\mathfrak F (y[n-1]) = \mathcal X (\omega)</math>
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<math>\mathcal Y (\omega) - \frac{1}{2}e^{-j\omega} \mathcal Y (\omega) = \mathcal X (\omega)</math>
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<math>\mathcal Y (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal X (\omega)</math>
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<math>\mathcal H (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}</math>
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b)
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<math>h[n]=\mathfrak F ^{-1} (\mathcal H (\omega))=  \mathfrak F ^{-1} \Big( \frac{1}{1-\frac{1}{2}e^{-j\omega}} \Big)</math>
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<math>use \mathfrak F (a^n u[n]) = \frac{1}{1-ae^{-j\omega}}</math>
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<math>h[n] = \Big(\frac{1}{2}\Big)^n u[n]</math>
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=== Answer 2  ===
 
=== Answer 2  ===
 
Write it here.
 
Write it here.

Revision as of 16:51, 8 March 2011


Practice Question on Causal LTI systems defined by a linear, constant coefficient difference equation

Consider the LTI system defined by the difference equation

$ y[n]-\frac{1}{2}y[n-1]=x[n]\ $

a) What is the frequency response of this system?

b) What is the unit impulse response of this system?

c) What is the system's response to the input $ x[n] = \left( \frac{1}{5}\right)^n u[n] \ $?

c) What is the system's response to the input $ x[n] =\cos \left(\frac{\pi}{2} n \right) \ $?


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Answer 1

a)

$ \mathfrak F (y[n]-\frac{1}{2}y[n-1]) = \mathfrak F (x[n]) $

$ \mathcal Y (\omega) - \frac{1}{2}\mathfrak F (y[n-1]) = \mathcal X (\omega) $

$ \mathcal Y (\omega) - \frac{1}{2}e^{-j\omega} \mathcal Y (\omega) = \mathcal X (\omega) $

$ \mathcal Y (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal X (\omega) $

$ \mathcal H (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}} $

b)

$ h[n]=\mathfrak F ^{-1} (\mathcal H (\omega))= \mathfrak F ^{-1} \Big( \frac{1}{1-\frac{1}{2}e^{-j\omega}} \Big) $

$ use \mathfrak F (a^n u[n]) = \frac{1}{1-ae^{-j\omega}} $

$ h[n] = \Big(\frac{1}{2}\Big)^n u[n] $

Answer 2

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Answer 3

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