(New page: =Obtain the input impulse response h[n] and the system function H(z) of your system= Defining a DT LTI: <math>y[n] = x[n+5] + x[n-3]\,</math><br> So, we have the unit impulse response: <ma...) |
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=Obtain the input impulse response h[n] and the system function H(z) of your system= | =Obtain the input impulse response h[n] and the system function H(z) of your system= | ||
Defining a DT LTI: | Defining a DT LTI: | ||
| − | <math>y[n] = x[n+5] + x[n-3]\,</math> | + | <math>y[n] = x[n+5] + x[n-3]\,</math> |
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So, we have the unit impulse response: | So, we have the unit impulse response: | ||
| + | |||
<math>h[n] = \delta[n-5] + \delta[n-3]\,</math> | <math>h[n] = \delta[n-5] + \delta[n-3]\,</math> | ||
Then we find the frequency response: | Then we find the frequency response: | ||
| + | |||
<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math> | <math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math> | ||
| − | <math>F(z) = \sum^{\ | + | find m value to make the value inside the bracket zero |
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| + | m = -5 for the first set and 3 for the second set | ||
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| + | <math>F(z) = e^{-5j\omega} + e^{3j\omega} \,</math> | ||
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| + | =Compute the response of your system to the signal you defined in Question 2 using H(z) and the Fourier series coefficients of your signal= | ||
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| + | Signal defined in Question 1: | ||
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| + | <math>X[n] = 6\cos(3 \pi n + \pi)\,</math> | ||
| + | |||
| + | <math>x[n] = \sum^{2}_{k = -1} a_k e^{jk\frac{\pi}{2} n}\,</math> | ||
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| + | <math>X[0] = -6 \,</math> | ||
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| + | <math>X[1] = 6 \,</math> | ||
| + | |||
| + | <math>X[2] = -6 \,</math> | ||
| + | |||
| + | <math>X[-1] = 6 \,</math> | ||
| + | |||
| + | The pattern of k can be seen since it forms a wave. | ||
| + | |||
| + | |||
| + | <math>y[n] = \sum^{2}_{k = -1} a_k F(z) e^{jk\frac{\pi}{2} n}\,</math> | ||
| + | |||
| + | <math>y[n] = \sum^{2}_{k = -1} a_k (e^{-5j\omega} + e^{3j\omega}) e^{jk\frac{\pi}{2} n}\,</math> | ||
| − | = | + | <math>y[n] = 6 (e^{-5j\omega} + e^{3j\omega}) e^{j(-1)\frac{\pi}{2} n} + (-6) (e^{-5j\omega} + e^{3j\omega}) e^{0}\,</math> |
Latest revision as of 09:04, 26 September 2008
Obtain the input impulse response h[n] and the system function H(z) of your system
Defining a DT LTI: $ y[n] = x[n+5] + x[n-3]\, $
So, we have the unit impulse response:
$ h[n] = \delta[n-5] + \delta[n-3]\, $
Then we find the frequency response:
$ F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\, $
find m value to make the value inside the bracket zero
m = -5 for the first set and 3 for the second set
$ F(z) = e^{-5j\omega} + e^{3j\omega} \, $
Compute the response of your system to the signal you defined in Question 2 using H(z) and the Fourier series coefficients of your signal
Signal defined in Question 1:
$ X[n] = 6\cos(3 \pi n + \pi)\, $
$ x[n] = \sum^{2}_{k = -1} a_k e^{jk\frac{\pi}{2} n}\, $
$ X[0] = -6 \, $
$ X[1] = 6 \, $
$ X[2] = -6 \, $
$ X[-1] = 6 \, $
The pattern of k can be seen since it forms a wave.
$ y[n] = \sum^{2}_{k = -1} a_k F(z) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = \sum^{2}_{k = -1} a_k (e^{-5j\omega} + e^{3j\omega}) e^{jk\frac{\pi}{2} n}\, $
$ y[n] = 6 (e^{-5j\omega} + e^{3j\omega}) e^{j(-1)\frac{\pi}{2} n} + (-6) (e^{-5j\omega} + e^{3j\omega}) e^{0}\, $
