(New page: Suppose we have a LTI CT signal y(t)=2x(t) ==Unit Impulse Response h(t) and System Function H(s)== <math>y(t)=2x(t)=>h(t)=2\delta(t)</math> <math>H(j\omega)=\int_{-\infty}^\infty h(\tau)...) |
(→Response of the Signal and Fourier Series Coefficients) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 2: | Line 2: | ||
==Unit Impulse Response h(t) and System Function H(s)== | ==Unit Impulse Response h(t) and System Function H(s)== | ||
| − | |||
| − | <math> | + | i) |
| + | <math>y(t)=2x(t)=> h(t)=2\delta(t)</math> | ||
| + | |||
| + | ii) | ||
| + | <math>H(s)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | ||
| + | |||
| + | <math>=\int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau</math> | ||
| + | |||
| + | <math>=\int_{-\infty}^\infty 2\delta(\tau)e^{-s\tau}d\tau</math> | ||
| + | |||
| + | <math>=2e^{-s*0}</math> | ||
| + | |||
| + | <math>=2</math> | ||
| + | |||
| + | ==Response of the Signal and Fourier Series Coefficients== | ||
| + | |||
| + | <math>x(t)=3cos(3t)</math> | ||
| + | |||
| + | <math>=\frac{3}{2}[(e^{j3t})+(e^{-j3t})]</math> | ||
| + | |||
| + | <math>=\frac{3}{2}(e^{j3t})+\frac{3}{2}(e^{-j3t})</math> | ||
| + | |||
| + | since we have <math>e^{st}</math> has a response of <math>H(s)e^{st}</math> | ||
| + | |||
| + | so we can get | ||
| + | |||
| + | <math>y(t)=\frac{3}{2}[(e^{j3t})+(e^{-j3t})]H(s)</math> | ||
| + | |||
| + | <math>=\frac{3}{2}(e^{j3t})H(j3)+\frac{3}{2}(e^{-j3t})H(-3j)</math> | ||
| + | |||
| + | <math>=3e^{j3t}+3e^{-j3t} = 6cos(3t)</math> | ||
| + | |||
| + | |||
| + | Coefficients: | ||
| + | |||
| + | we observe that the output is just double the input, | ||
| + | so the coefficients should be doubleed,too | ||
| + | |||
| + | From Q1 we can get that when k=1, <math>a_1=3</math>, and when k=-1,<math>a_{-1}=3</math> | ||
| + | |||
| + | others are all zero | ||
Latest revision as of 16:41, 26 September 2008
Suppose we have a LTI CT signal y(t)=2x(t)
Unit Impulse Response h(t) and System Function H(s)
i) $ y(t)=2x(t)=> h(t)=2\delta(t) $
ii) $ H(s)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
$ =\int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $
$ =\int_{-\infty}^\infty 2\delta(\tau)e^{-s\tau}d\tau $
$ =2e^{-s*0} $
$ =2 $
Response of the Signal and Fourier Series Coefficients
$ x(t)=3cos(3t) $
$ =\frac{3}{2}[(e^{j3t})+(e^{-j3t})] $
$ =\frac{3}{2}(e^{j3t})+\frac{3}{2}(e^{-j3t}) $
since we have $ e^{st} $ has a response of $ H(s)e^{st} $
so we can get
$ y(t)=\frac{3}{2}[(e^{j3t})+(e^{-j3t})]H(s) $
$ =\frac{3}{2}(e^{j3t})H(j3)+\frac{3}{2}(e^{-j3t})H(-3j) $
$ =3e^{j3t}+3e^{-j3t} = 6cos(3t) $
Coefficients:
we observe that the output is just double the input, so the coefficients should be doubleed,too
From Q1 we can get that when k=1, $ a_1=3 $, and when k=-1,$ a_{-1}=3 $
others are all zero
