Revision as of 16:07, 24 September 2008 by Mpaganin (Talk)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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) = $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett