Revision as of 07:49, 26 September 2008

Obtain the input impulse response h[n] and the system function H(z) of your system

Defining a DT LTI: $ y[n] = x[n+5] + x[n-3]\, $
So, we have the unit impulse response: $ h[n] = \delta[n-5] + \delta[n-3]\, $

Then we find the frequency response:

$ F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\, $

find m value to make the value inside the bracket zero

m = -5 for the first set and 3 for the second set

$ F(z) = e^{-5j\omega} + e^{3j\omega} \, $

Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal

Signal defined in Question 1: $ x(t) = 6\cos(2\pi t) + 8\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}\, $

From Question 1: $ x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $