Difference between revisions of "HW4.3 Nathan Najdek ECE301Fall2008mboutin" - Rhea

Latest revision as of 17:54, 25 September 2008

Part A: Unit Impulse Response and System Function

$ h(t) = 9\delta(t) $

so:

$ y(t) = \int^{\infty}_{-\infty} h(\tau) x(\tau) d\tau $

$ y(t) = \int^{\infty}_{-\infty} 9\delta(\tau)e^{-j\omega_0(t - \tau)} d\tau $

$ y(t) = e^{j\omega_0 t} \int^{\infty}_{-\infty} 9\delta(\tau)e^{-j\omega_0 \tau} d\tau $

$ H(s) = 9e^{j\omega_0} $

$ H(s) = 9 $

Part B: Response of the System

$ x(t) = 1 + sin(8\pi t) + 2cos(8\pi t) + cos(16\pi t + \frac{\pi}{4}) $

$ x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $

So with $ y(t) = H(s)e^{-st}x(t) $, the final response is:

$ y(t) = 9e^{-st} + 9e^{-st}(\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + 9e^{-st}(e^{j8 \pi t} + e^{-j8 \pi t}) + 9e^{-st}(\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $