(a) Finding the unit impulse response h[n] and the system function F(z).)
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Therefore, <math>F(z) = \sum_{-\infty}^{\infty}h[m]Z^{-m} = \sum_{-\infty}^{\infty}5\delta [m] Z^{-m}</math>
 
Therefore, <math>F(z) = \sum_{-\infty}^{\infty}h[m]Z^{-m} = \sum_{-\infty}^{\infty}5\delta [m] Z^{-m}</math>
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==b) Computing the response to the system when x[n] is the input from Question 2==
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x[n] = cos(5\pi n)

Revision as of 14:13, 26 September 2008

Defining the DT LTI system

$ x[n] \rightarrow system \rightarrow y[n] = 5x[n] $

a) Finding the unit impulse response h[n] and the system function F(z).

$ x[n] = \delta [n] \rightarrow system \rightarrow y[n]=5\delta [n] $

Therefore the unit impulse response, $ h[n] = 5\delta [n] $

For a DT LTI system,

$ Z^n \rightarrow system \rightarrow F(z)Z^n $

Output of the system, $ F(z)Z^n = h[n]*Z^n = \sum_{m=-\infty}^{\infty} h[m]Z^{n-m} = Z^n\sum_{-\infty}^{\infty}h[m]Z^{-m} $

Therefore, $ F(z) = \sum_{-\infty}^{\infty}h[m]Z^{-m} = \sum_{-\infty}^{\infty}5\delta [m] Z^{-m} $

b) Computing the response to the system when x[n] is the input from Question 2

x[n] = cos(5\pi n)

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood