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Given the system <math>y(t) = 2x(t+3)\,</math> | Given the system <math>y(t) = 2x(t+3)\,</math> | ||
| − | <math> x(t)=\ | + | <math> x(t)= 2\delta(t+3) </math> |
Then find the response to <math>x(t) = cos(4t) + sin(2t)\,</math> | Then find the response to <math>x(t) = cos(4t) + sin(2t)\,</math> | ||
| + | |||
| + | <math>x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t}</math> | ||
| + | |||
| + | <math>H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | ||
| + | |||
| + | <math>H(j\omega)=\int_{-\infty}^\infty (2\delta(\tau+3))e^{-j\omega\tau}d\tau</math> | ||
| + | |||
| + | y(t) = x(t)*H(t) | ||
| + | |||
| + | <math>y(t) = (2e^{3j\omega})*(\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t})</math> | ||
Latest revision as of 16:08, 26 September 2008
Given the system $ y(t) = 2x(t+3)\, $
$ x(t)= 2\delta(t+3) $
Then find the response to $ x(t) = cos(4t) + sin(2t)\, $
$ x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t} $
$ H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
$ H(j\omega)=\int_{-\infty}^\infty (2\delta(\tau+3))e^{-j\omega\tau}d\tau $
y(t) = x(t)*H(t)
$ y(t) = (2e^{3j\omega})*(\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t}) $
