| Line 13: | Line 13: | ||
The system function, H(s) is: | The system function, H(s) is: | ||
| − | <math>H( | + | <math>H(j w) = \int^{\infty}_{- \infty} h(t) e^{- j w t} dt</math> |
| − | <math>H( | + | <math>H(j w) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- j w t}</math> |
| − | <math>H( | + | <math>H(j w) = 3 e^{0} + 7 e^{- j 3 t} </math> |
| − | <math>H( | + | <math>H( jw) = 3 + 7 e^{- j 3 t}</math> |
| + | |||
| + | ---- | ||
The signal used in question 1: | The signal used in question 1: | ||
<math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math> | <math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math> | ||
| + | |||
| + | <math>= \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}</math> | ||
| + | |||
| + | The response, y(t) = H(jw)*x(t) | ||
| + | |||
| + | <math>y(t) = (3 + 7 e^{- j 3 t}) * (\frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}) </math> | ||
Latest revision as of 09:55, 26 September 2008
Let us use the CT LTI system:
$ y(t) = 3x(t) + 7x(t+3) $
The impulse response, h(t), of this system is computed using the following:
$ x(t) = \delta (t) $
$ h(t) = 3\delta (t) + 7\delta (t+3) $
The system function, H(s) is:
$ H(j w) = \int^{\infty}_{- \infty} h(t) e^{- j w t} dt $
$ H(j w) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- j w t} $
$ H(j w) = 3 e^{0} + 7 e^{- j 3 t} $
$ H( jw) = 3 + 7 e^{- j 3 t} $
The signal used in question 1:
$ x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t} $
$ = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t} $
The response, y(t) = H(jw)*x(t)
$ y(t) = (3 + 7 e^{- j 3 t}) * (\frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}) $
