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(→Part a) |
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===Define a CT LTI system=== | ===Define a CT LTI system=== | ||
==Part a== | ==Part a== | ||
| − | Obtain the unit impulse response h(t) and the system function H(s) | + | Obtain the unit impulse response h(t) and the system function H(s)<br><br> |
:<math>\, y(t)=2x(t) \rightarrow h(t)=2\delta (t)</math><br><br> | :<math>\, y(t)=2x(t) \rightarrow h(t)=2\delta (t)</math><br><br> | ||
:<math>\begin{align} \, H(s)&=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau | :<math>\begin{align} \, H(s)&=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau | ||
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\\&=2 | \\&=2 | ||
\end{align}</math> | \end{align}</math> | ||
| + | |||
==Part b== | ==Part b== | ||
Compute the response of the system to the signal using H(s) and the Fourier series coefficients of the signal.<br><br> | Compute the response of the system to the signal using H(s) and the Fourier series coefficients of the signal.<br><br> | ||
Latest revision as of 18:11, 26 September 2008
Define a CT LTI system
Part a
Obtain the unit impulse response h(t) and the system function H(s)
- $ \, y(t)=2x(t) \rightarrow h(t)=2\delta (t) $
- $ \begin{align} \, 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\cdot 0} \\&=2 \end{align} $
Part b
Compute the response of the system to the signal using H(s) and the Fourier series coefficients of the signal.
- $ \, x(t)=6sin(6t)=\frac{3e^{j6t}}{j}-\frac{3e^{-j6t}}{j} $
- $ \begin{align} \, \therefore y(t)&=\frac{3e^{j6t}}jH(j6)-\frac{3e^{-j6t}}{j}H(-j6)\\&= \frac{6e^{j6t}}{j}-\frac{6e^{-j6t}}{j} \\ &= 2\cdot (\frac{3e^{j6t}}{j}-\frac{3e^{-j6t}}{j}) \\ &=2\cdot 6sin(6t) \\ &= 12sin(6t) \\ a_1 &= \frac{6}{j} \\ a_{-1}&=\frac{-6}{j}\end{align} $
