A CT-LTI System

$ y(t) = 10x(t)\, $

Unit Impulse Response

$ x(t) = \delta(t)\, $

$ h(t) = 10 \delta(t)\, $

Frequency Response

$ y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\, $ where $ x(t) = e^{jwt} \, $

$ y(t) = \int^{\infty}_{-\infty} 10 \delta(t) * e^{jwt} dt\, $

$ y(t) = \int^{\infty}_{-\infty} 10 \delta(r) e^{jw(t-r} dr\, $

$ y(t) = e^{jwt} \int^{\infty}_{-\infty} 10 \delta(r) e^{-jwr} dr\, $

$ H(s) = \int^{\infty}_{-\infty} 10 \delta(r) e^{-jwr} dr\, $

$ H(s) = 10 e^{-jw0}\, $

$ H(s) = 10\, $

Response of the CT system defined in Q1

CT Periodic Signal : $ x(t) = \cos(3\pi t) + \sin(4\pi t)\, $

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\, $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\pi t}\, $

$ y(t) = 10\sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $

$ y(t) = 10(\cos(3\pi t) + \sin(4\pi t))\, $

$ y(t) = 10\cos(3\pi t) + 10\sin(4\pi t)\, $

Alumni Liaison

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

Dr. Paul Garrett