(→b) Computing the response to the system when x[n] is the input from Question 2) |
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<math>x[n] \rightarrow system \rightarrow y[n] = 5x[n]</math> | <math>x[n] \rightarrow system \rightarrow y[n] = 5x[n]</math> | ||
| − | ==a) Finding the unit impulse response h[n] and the system function | + | ==a) Finding the unit impulse response h[n] and the system function F(z).== |
<math>x[n] = \delta [n] \rightarrow system \rightarrow y[n]=5\delta [n]</math> | <math>x[n] = \delta [n] \rightarrow system \rightarrow y[n]=5\delta [n]</math> | ||
| − | Therefore the unit impulse response, <math>h[n] = 5\delta [n]</math> | + | Therefore the unit impulse response, <big><math>h[n] = 5\delta [n]</math></big> |
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| + | For a DT LTI system, | ||
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| + | <math>Z^n \rightarrow system \rightarrow F(z)Z^n</math> | ||
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| + | Output of the system, <math>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}</math> | ||
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| + | Therefore, <math>F(z) = \sum_{m = -\infty}^{\infty}h[m]Z^{-m} = \sum_{m = -\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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| + | <font size = '4'><math>x[n] = cos(5\pi n) = e^{j\pi n}</math></font> | ||
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| + | Therefore, <font size = '4'><math>Z^n = e^{j\pi n}</math></font> | ||
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| + | <font size = '4'><math>Z = e^{j\pi}</math></font> | ||
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| + | <math>F(e^{j\pi}) = \sum_{m = -\infty}^{\infty}5\delta [m] (e^{j\pi})^{-m} </math> | ||
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| + | delta[m] is 0 for all values of m except at delta[0] (m = 0) where it is 1. | ||
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| + | Therefore: <font size = '4'><math>F(e^{j\pi}) = 5</math></font> | ||
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| + | 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> | ||
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| + | The response, <font size = '4'><math>y[n] = \sum_{k = -\infty}^{\infty} a_k F(e^{jk\pi}) e^{jk\pi n} = 5\sum_{k = -\infty}^{\infty} a_k e^{jk\pi n} </math> | ||
Latest revision as of 14:59, 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} = 5\sum_{k = -\infty}^{\infty} a_k e^{jk\pi n} $
