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== Define a DT LTI system == | == Define a DT LTI system == | ||
<math>y[n] = x[n+1] + x[n]\,</math> | <math>y[n] = x[n+1] + x[n]\,</math> | ||
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== Obtain the Unit Impulse Response h[n] == | == Obtain the Unit Impulse Response h[n] == | ||
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<math>h[n] = \delta[n+1] + \delta[n]\,</math> | <math>h[n] = \delta[n+1] + \delta[n]\,</math> | ||
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| + | == Obtain the System Function <math>F(z)\,</math> of the System == | ||
| + | <math>F(z) = \sum^{\infty}_{m=-\infty} h[m]e^{jm\omega} \,</math> | ||
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| + | <math>F(z) = \sum^{\infty}_{m=-\infty} (\delta[m+1] + \delta[m])e^{jm\omega} \,</math> | ||
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| + | <math>F(z) = \sum^{\infty}_{m=-\infty} \delta[m+1]e^{jm\omega} + \delta[m]e^{jm\omega} \,</math> | ||
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| + | Since the delta function is only valid when its input is zero, | ||
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| + | <math>F(z) = e^{-j\omega} + e^{0j\omega} \,</math> | ||
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| + | <math>F(z) = 1 + e^{-j\omega} \,</math> | ||
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| + | == Compute the Response of My Signal from Question 2 == | ||
| + | Signal: <math>x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\,</math> | ||
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| + | <math>x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\,</math>, where | ||
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| + | <math>a_0 = 0 , k = ..., -8, -4, 0, 4, 8, ... \,</math> | ||
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| + | <math>a_1 = 2j , k = ..., -7, -3, 1, 5, 9, ... \,</math> | ||
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| + | <math>a_2 = 2 , k = ..., -6, -2, 2, 6, 10, ... \,</math> | ||
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| + | <math>a_3 = -2j , k = ..., -5, -1, 3, 7, 11, ... \,</math> | ||
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| + | <math>y[n] = \sum^{3}_{k = 0} a_k F(z) e^{jk\frac{\pi}{2} n}\,</math> | ||
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| + | <math>y[n] = \sum^{3}_{k = 0} a_k (1 + e^{-j\frac{\pi}{2}}) e^{jk\frac{\pi}{2} n}\,</math> | ||
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| + | <math>y[n] = 2j(1 + e^{-j\frac{\pi}{2}})e^{j\frac{\pi}{2} n} + 2(1 + e^{-j\frac{\pi}{2}})e^{j\pi n} - 2j(1 + e^{-j\frac{\pi}{2}})e^{jk\frac{3\pi}{2} n}\,</math> | ||
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| + | <math>y[n] = 2je^{j\frac{\pi}{2} n}(1 + e^{-j\frac{\pi}{2}}) + 2e^{j\pi n}(1 + e^{-j\frac{\pi}{2}}) - 2je^{jk\frac{3\pi}{2} n}(1 + e^{-j\frac{\pi}{2}})\,</math> | ||
Latest revision as of 13:13, 25 September 2008
Contents
Define a DT LTI system
$ y[n] = x[n+1] + x[n]\, $
Obtain the Unit Impulse Response h[n]
By definition, to obtain the unit impulse response from a system defined by $ y[n] = x[n]\, $, simply replace the $ x[n]\, $ by $ \delta[n]\, $.
$ h[n] = \delta[n+1] + \delta[n]\, $
Obtain the System Function $ F(z)\, $ of the System
$ F(z) = \sum^{\infty}_{m=-\infty} h[m]e^{jm\omega} \, $
$ F(z) = \sum^{\infty}_{m=-\infty} (\delta[m+1] + \delta[m])e^{jm\omega} \, $
$ F(z) = \sum^{\infty}_{m=-\infty} \delta[m+1]e^{jm\omega} + \delta[m]e^{jm\omega} \, $
Since the delta function is only valid when its input is zero,
$ F(z) = e^{-j\omega} + e^{0j\omega} \, $
$ F(z) = 1 + e^{-j\omega} \, $
Compute the Response of My Signal from Question 2
Signal: $ x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\, $
$ x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\, $, where
$ a_0 = 0 , k = ..., -8, -4, 0, 4, 8, ... \, $
$ a_1 = 2j , k = ..., -7, -3, 1, 5, 9, ... \, $
$ a_2 = 2 , k = ..., -6, -2, 2, 6, 10, ... \, $
$ a_3 = -2j , k = ..., -5, -1, 3, 7, 11, ... \, $
$ y[n] = \sum^{3}_{k = 0} a_k F(z) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = \sum^{3}_{k = 0} a_k (1 + e^{-j\frac{\pi}{2}}) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = 2j(1 + e^{-j\frac{\pi}{2}})e^{j\frac{\pi}{2} n} + 2(1 + e^{-j\frac{\pi}{2}})e^{j\pi n} - 2j(1 + e^{-j\frac{\pi}{2}})e^{jk\frac{3\pi}{2} n}\, $
$ y[n] = 2je^{j\frac{\pi}{2} n}(1 + e^{-j\frac{\pi}{2}}) + 2e^{j\pi n}(1 + e^{-j\frac{\pi}{2}}) - 2je^{jk\frac{3\pi}{2} n}(1 + e^{-j\frac{\pi}{2}})\, $
