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} \! $
