(→System function) |
(→System function) |
||
| Line 42: | Line 42: | ||
<math>H(s) = 5e^{-2s} - 3\!</math> | <math>H(s) = 5e^{-2s} - 3\!</math> | ||
| + | |||
| + | |||
| + | ==System Response== | ||
| + | |||
| + | In this last part, you just take my signal from part (a): | ||
| + | |||
| + | |||
| + | <math>x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t}</math> | ||
| + | |||
| + | |||
| + | Then multiply that by the <math>H(s)\!</math> | ||
| + | |||
| + | |||
| + | <math>y(t) = [H(jw)][x(t)]\!</math> | ||
| + | |||
| + | |||
| + | <math>y(t) = (5e^{-2s} - 3)*[2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t}]</math> | ||
| + | |||
| + | |||
| + | <math>y(t) = 10e^{j2t - 2s} - 6e^{j2t} + 10e^{-j2t - 2s} - 6e^{-j2t} + \frac{15}{2}e^{j3t - 2s} - \frac{9}{2}e^{j3t} - \frac{15}{2}e^{-j3t - 2s} + \frac{9}{2}e^{-j3t}</math> | ||
Latest revision as of 15:10, 26 September 2008
CT LTI system
My CT LTI system is (drum roll please)...
$ y(t)=5x(t - 2) - 3x(t)\! $
Unit Impulse Response
Well, this is rather straightforward. You want the response to the unit impulse, do ya? Well, if that is what you want, that is what you will get. All you have to do is replace the $ x(t)\! $ with $ \delta(t)\! $, also known as the notorious delta function.
$ h(t) = 5\delta(t - 2) - 3\delta(t)\! $
System function
So, you want the system function? Easy. It is a simple application of something we covered way back in ECE 201, laplace transform. I personally call it DeCarlo transform, but that is another story entirely. Anyways, the following formula gets you the system function!
$ H(s) = \int_{-\infty}^{+\infty} h(t)e^{st}dt $
Note that $ s=j\omega\! $.
Here goes the math:
$ H(s) = \int_{-\infty}^{+\infty}[5\delta(t - 2) - 3\delta(t)]e^{-st}dt\! $
$ H(s) = \int_{-\infty}^{+\infty}5\delta(t - 2)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t)e^{-st}dt $
Using what is known as the sifting property, we obtain the following result:
$ H(s) = 5e^{-2s} - 3\! $
System Response
In this last part, you just take my signal from part (a):
$ x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t} $
Then multiply that by the $ H(s)\! $
$ y(t) = [H(jw)][x(t)]\! $
$ y(t) = (5e^{-2s} - 3)*[2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t}] $
$ y(t) = 10e^{j2t - 2s} - 6e^{j2t} + 10e^{-j2t - 2s} - 6e^{-j2t} + \frac{15}{2}e^{j3t - 2s} - \frac{9}{2}e^{j3t} - \frac{15}{2}e^{-j3t - 2s} + \frac{9}{2}e^{-j3t} $
