Revision as of 14:32, 26 September 2008 by Cztan (Talk)

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_{m = -\infty}^{\infty}h[m]Z^{-m} $

Therefore, $ F(z) = \sum_{m = -\infty}^{\infty}h[m]Z^{-m} = \sum_{m = -\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) = e^{j\pi n} $

Therefore, $ Z^n = e^{j\pi n} $

$ Z = e^{j\pi} $

$ F(e^{j\pi}) = \sum_{m = -\infty}^{\infty}5\delta [m] (e^{j\pi})^{-m} $

delta[m] is 0 for all values of m except at delta[0] (m = 0) where it is 1.

Therefore: $ F(e^{j\pi}) = 5 $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang