| Line 22: | Line 22: | ||
== Part B == | == Part B == | ||
| − | <math>x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\,</math> | + | <math>x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)]\,</math> |
<math>y(t) = H(jw)x(t)\,</math> | <math>y(t) = H(jw)x(t)\,</math> | ||
<math>y(t) = 10e^{-jw}\times[4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\,</math> | <math>y(t) = 10e^{-jw}\times[4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\,</math> | ||
Revision as of 12:04, 25 September 2008
Part A
CT LTI system:
y(t) = 10x(t-1)
plugging in delta(t) into the system we get:
h(t) = 10delta(t-1)
$ s = j\omega $
$ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st} $
$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $
$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $
$ H(s) = 10e^{-s}\, $
Part B
$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)]\, $
$ y(t) = H(jw)x(t)\, $
$ y(t) = 10e^{-jw}\times[4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $
