| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant | LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant | ||
| + | |||
| + | ---- | ||
Unit Impulse Response: <math>h(t) = K \delta(t)</math> | Unit Impulse Response: <math>h(t) = K \delta(t)</math> | ||
| + | |||
| + | ---- | ||
Frequency Response: | Frequency Response: | ||
| − | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\ | + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math> |
| − | then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k* | + | then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*[h(t)*e^{j\omega_0 t}]</math> |
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition | <math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition | ||
| − | <math>H(s) = \int^{\infty}_{-\infty} K \delta( | + | <math>H(s) = \int^{\infty}_{-\infty} K \delta(t) e^{-j\omega_0 t} dt</math> |
<math>H(s) = K e^{-jw0}</math> | <math>H(s) = K e^{-jw0}</math> | ||
<math>H(s) = K</math> | <math>H(s) = K</math> | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | Response of the CT LTI system in 4.1: | ||
| + | |||
| + | <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math> | ||
| + | |||
| + | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}</math> | ||
| + | |||
| + | <math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\omega_0 t}</math> | ||
| + | |||
| + | <math>y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\omega_0 t}</math> | ||
| + | |||
| + | <math>y(t) = K\sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}</math> | ||
| + | |||
| + | <math>y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4})</math> | ||
Latest revision as of 11:33, 25 September 2008
LTI System: $ y(t) = Kx(t)\, $ where K is a constant
Unit Impulse Response: $ h(t) = K \delta(t) $
Frequency Response:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $
then $ y(t)=\sum^{\infty}_{k = -\infty}a_k*[h(t)*e^{j\omega_0 t}] $
$ H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt $ by definition
$ H(s) = \int^{\infty}_{-\infty} K \delta(t) e^{-j\omega_0 t} dt $
$ H(s) = K e^{-jw0} $
$ H(s) = K $
Response of the CT LTI system in 4.1:
$ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t} $
$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\omega_0 t} $
$ y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\omega_0 t} $
$ y(t) = K\sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t} $
$ y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4}) $
