Latest revision as of 09:04, 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:

$X[n] = 6\cos(3 \pi n + \pi)\,$

$x[n] = \sum^{2}_{k = -1} a_k e^{jk\frac{\pi}{2} n}\,$

$X[0] = -6 \,$

$X[1] = 6 \,$

$X[2] = -6 \,$

$X[-1] = 6 \,$

The pattern of k can be seen since it forms a wave.

$y[n] = \sum^{2}_{k = -1} a_k F(z) e^{jk\frac{\pi}{2} n}\,$

$y[n] = \sum^{2}_{k = -1} a_k (e^{-5j\omega} + e^{3j\omega}) e^{jk\frac{\pi}{2} n}\,$

$y[n] = 6 (e^{-5j\omega} + e^{3j\omega}) e^{j(-1)\frac{\pi}{2} n} + (-6) (e^{-5j\omega} + e^{3j\omega}) e^{0}\,$

@@ Line 42: / Line 42: @@
 <math>y[n] = \sum^{2}_{k = -1} a_k (e^{-5j\omega} + e^{3j\omega}) e^{jk\frac{\pi}{2} n}\,</math>
-<math>y[n] = 2j(1 + e^{-j\frac{\pi}{2}})e^{j\frac{\pi}{2} n} + 2(1 + e^{-j\frac{\pi}{2}})e^{j\pi n} - 2j(1 + e^{-j\frac{\pi}{2}})e^{jk\frac{3\pi}{2} n}\,</math>
+<math>y[n] = 6 (e^{-5j\omega} + e^{3j\omega}) e^{j(-1)\frac{\pi}{2} n} + (-6) (e^{-5j\omega} + e^{3j\omega}) e^{0}\,</math>
-<math>y[n] = 2je^{j\frac{\pi}{2} n}(1 + e^{-j\frac{\pi}{2}}) + 2e^{j\pi n}(1 + e^{-j\frac{\pi}{2}}) - 2je^{jk\frac{3\pi}{2} n}(1 + e^{-j\frac{\pi}{2}})\,</math>