(→b) Computing the response to the system when x[n] is the input from Question 2) |
(→b) Computing the response to the system when x[n] is the input from Question 2) |
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Therefore: <font size = '4'><math>F(e^{j\pi}) = 5</math></font> | Therefore: <font size = '4'><math>F(e^{j\pi}) = 5</math></font> | ||
| + | Express <font size = '4'><math>x[n] = cos(5\pi n)</math></font> as <math>x[n] = \sum_{k = -\infty}^{\infty} a_k e^{jk\pi n} </math> | ||
| − | The response, <font size = '4'><math>y[n] = F(e^{ | + | The response, <font size = '4'><math>y[n] = \sum_{k = -\infty}^{\infty} a_k F(e^{jk\pi}) e^{jk\pi n} </math> |
Revision as of 15:58, 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_{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 $
Express $ x[n] = cos(5\pi n) $ as $ x[n] = \sum_{k = -\infty}^{\infty} a_k e^{jk\pi n} $
The response, $ y[n] = \sum_{k = -\infty}^{\infty} a_k F(e^{jk\pi}) e^{jk\pi n} $
