Revision as of 08:56, 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

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

Signal defined in Question 1:
$X[n] = 6\cos(3 \pi n + \pi)\,$

$x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\,$

@@ Line 17: / Line 17: @@
-=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=
+=Compute the response of your system to the signal you defined in Question 2 using H(z) and the Fourier series coefficients of your signal=
-Signal defined in Question 1:
+Signal defined in Question 1:<br>
-<math>x(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br>
+<math>X[n] = 6\cos(3 \pi n + \pi)\,</math>
-<br>
-<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math>
-<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\,</math>
+<math>x[n] = \sum^{3}_{k = 0} a_k e^{jk\frac{\pi}{2} n}\,</math>
-From Question 1:
-<math>x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br>
-With this expression we can conclude:<br>
-<math>a_1 = 3\,</math><br>
-<math>a_{-1} = 3\,</math><br>
-<math>a_2 = 4\,</math><br>
-<math>a_{-2} = -4\,</math><br>