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==Obtain the input impulse response h(t) and the system function H(s) of your system== | ==Obtain the input impulse response h(t) and the system function H(s) of your system== | ||
| − | A very simple | + | A very simple system: |
| + | <br><br> | ||
| + | <math>y(t)=x(t)\,</math> and <math>x(t)=\delta(t)\,</math> | ||
| + | <br><br> | ||
| + | We can get <math>h(t)=\delta(t)\,</math> | ||
| + | <br><br> | ||
| + | <math>y(t) = \int^{\infty}_{-\infty} \delta(t) dt\,</math><br> | ||
<br> | <br> | ||
| − | <math> | + | <math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau</math><br> |
<br> | <br> | ||
| − | + | <math>H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}d\tau</math><br> | |
<br> | <br> | ||
| + | <math>H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau</math><br> | ||
| + | <br> | ||
| + | <math>H(s)=-se^{-s\tau}|_0^\infty \,</math><br> | ||
| + | <br> | ||
| + | <math>H(s)=-s(e^{-\infty} - e^{0})\,</math><br> | ||
| + | <br> | ||
| + | <math>H(s)=s\,</math><br> | ||
==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal== | ==Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal== | ||
| + | |||
| + | Signal defined in Question 1: | ||
| + | <math>x(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br> | ||
| + | <br> | ||
| + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math> | ||
| + | |||
| + | <math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math> | ||
| + | |||
| + | From Question 1: | ||
| + | <math>x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br> | ||
| + | With this expression we can conclude:<br> | ||
| + | <math>a_1 = 3\,</math><br> | ||
| + | <math>a_{-1} = 3\,</math><br> | ||
| + | <math>a_2 = 4\,</math><br> | ||
| + | <math>a_{-2} = -4\,</math><br> | ||
| + | |||
| + | <math>x(t) = 3H(s)e^{j2\pi t}+3H(s)e^{-j2\pi t} + 4H(s)e^{j4\pi t}-4H(s)e^{-j4\pi t}\,</math><br> | ||
| + | |||
| + | <math>x(t) = 3j\omega_0e^{j2\pi t}+3j\omega_0e^{-j2\pi t} + 4j\omega_0e^{j4\pi t}-4j\omega_0e^{-j4\pi t}\,</math><br> | ||
| + | |||
| + | <math>\omega_0\,</math> value as the base frequency is 2 | ||
| + | |||
| + | <math>x(t) = 6je^{j2\pi t}+6je^{-j2\pi t} + 8je^{j4\pi t}-8je^{-j4\pi t}\,</math><br> | ||
Latest revision as of 08:25, 26 September 2008
Obtain the input impulse response h(t) and the system function H(s) of your system
A very simple system:
$ y(t)=x(t)\, $ and $ x(t)=\delta(t)\, $
We can get $ h(t)=\delta(t)\, $
$ y(t) = \int^{\infty}_{-\infty} \delta(t) dt\, $
$ H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau $
$ H(s)=\int_{-\infty}^{\infty}u(\tau)e^{-s\tau}d\tau $
$ H(s)=\int_{0}^{\infty}e^{-s\tau}d\tau $
$ H(s)=-se^{-s\tau}|_0^\infty \, $
$ H(s)=-s(e^{-\infty} - e^{0})\, $
$ H(s)=s\, $
Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal
Signal defined in Question 1:
$ x(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\, $
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\, $
From Question 1:
$ x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $
$ x(t) = 3H(s)e^{j2\pi t}+3H(s)e^{-j2\pi t} + 4H(s)e^{j4\pi t}-4H(s)e^{-j4\pi t}\, $
$ x(t) = 3j\omega_0e^{j2\pi t}+3j\omega_0e^{-j2\pi t} + 4j\omega_0e^{j4\pi t}-4j\omega_0e^{-j4\pi t}\, $
$ \omega_0\, $ value as the base frequency is 2
$ x(t) = 6je^{j2\pi t}+6je^{-j2\pi t} + 8je^{j4\pi t}-8je^{-j4\pi t}\, $
