(New page: == Define A CT LTI System == === Part A === <math> h(t) = 3u(t-4)\!</math> <math> H(j\omega) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} dt\!</math> <math> H(j\omega) = \int_{-\infty...) |
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<math> y(t) = \sum_{k = -\infty}^{\infty} a_k H(jk\omega) (5cos(2t) + 3sin(4t)) \!</math> | <math> y(t) = \sum_{k = -\infty}^{\infty} a_k H(jk\omega) (5cos(2t) + 3sin(4t)) \!</math> | ||
| − | <math> y(t) = | + | <math> y(t) = (\frac{3}{j\omega}) (\frac{5}{2}) + (\frac{3}{j\omega}) (\frac{5}{2}) + (\frac{3}{j\omega}) (\frac{3}{2j}) - (\frac{3}{j\omega}) (\frac{3}{2j}) \!</math> |
| + | |||
| + | |||
| + | <math> y(t) = \frac{15}{j\omega} \!</math> | ||
Latest revision as of 16:18, 24 September 2008
Define A CT LTI System
Part A
$ h(t) = 3u(t-4)\! $
$ H(j\omega) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} dt\! $
$ H(j\omega) = \int_{-\infty}^{\infty} 3u(t-4) e^{-j\omega t} dt\! $
$ H(j\omega) = 3\int_{4}^{\infty} e^{-j\omega t} dt\! $
$ H(j\omega) = 3(\frac{1}{-j\omega})|_{4}^{\infty} \! $
$ H(j\omega) = \frac{3}{j\omega} \! $
Part B
Input Signal: $ x(t) = 5cos(2t) + 3sin(4t) \! $
Fourier Series Coefficients:
$ a_1 = a_{-1} = \frac{5}{2} \! $
$ a_2 = a_{-2} = \frac{3}{2j} \! $
$ a_k = 0 \! $ whenever $ K \neq \pm2, \pm 4\! $
$ x(t) \rightarrow h(t) \rightarrow y(t) = H(jw)x(t) \! $
$ y(t) = \sum_{k = -\infty}^{\infty} a_k H(jk\omega) (5cos(2t) + 3sin(4t)) \! $
$ y(t) = (\frac{3}{j\omega}) (\frac{5}{2}) + (\frac{3}{j\omega}) (\frac{5}{2}) + (\frac{3}{j\omega}) (\frac{3}{2j}) - (\frac{3}{j\omega}) (\frac{3}{2j}) \! $
$ y(t) = \frac{15}{j\omega} \! $
