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<math>y[n] = f(x[n]) = \sum_{k=0}^{3} a_k H(z_k) {z_k}^n</math> | <math>y[n] = f(x[n]) = \sum_{k=0}^{3} a_k H(z_k) {z_k}^n</math> | ||
| + | |||
| + | <math>{z_k}^n = e^{\frac{jk\pi}{2}}</math> | ||
| + | |||
| + | so <math> y[n] = \sum_{k=0}^{3} a_x H(z_k) e^{\frac{jk\pi n}{2}} </math> | ||
| + | |||
| + | <math> = 1/2 *H(1) + \frac{-1-j}{4} *H(j)e^{\frac{j\pi n}{2}} + \frac{j-1}{4}* H(-j) e^{\frac{3j\pi n}{2}} </math> | ||
| + | |||
| + | <math> = 1/2 * 6 + \frac{-1-j}{4} * (5 + j^{-1}) * e^{\frac{j\pi n}{2}} + \frac{j-1}{4} * (5 + (-j)^{-5}) *e^{\frac{3j\pi n}{2}}</math> | ||
Latest revision as of 16:23, 26 September 2008
DT LTI System
Lets define our system where $ y[n] = 5x[n] + x[n-5] $
What we need to do is first find h[n] and H[z] for our system.
Then we can calculate the system's response to a signal using H[z] and the fourier coefficients for the system.
Step 1:
$ h[n] = y[\delta[n]] = 5\delta[n]+\delta[n-5] $
Step 2:
$ H(z) = \sum_{k=-\infty}^{\infty} h[k] z^{-k} =\sum_{k=-\infty}^{\infty} (5\delta[n]+\delta[n-5]) z^{-k} = 5z^0+z^-5 = 5+z^{-5} $
By the shifting property
Step 3:
My signal from part 2 has fourier coefficients:
$ a_0 = 1/2, a_1 = \frac{-1-j}{4}, a_2 = 0, a_3 = \frac{j-1}{4} $
$ y[n] = f(x[n]) = \sum_{k=0}^{3} a_k H(z_k) {z_k}^n $
$ {z_k}^n = e^{\frac{jk\pi}{2}} $
so $ y[n] = \sum_{k=0}^{3} a_x H(z_k) e^{\frac{jk\pi n}{2}} $
$ = 1/2 *H(1) + \frac{-1-j}{4} *H(j)e^{\frac{j\pi n}{2}} + \frac{j-1}{4}* H(-j) e^{\frac{3j\pi n}{2}} $
$ = 1/2 * 6 + \frac{-1-j}{4} * (5 + j^{-1}) * e^{\frac{j\pi n}{2}} + \frac{j-1}{4} * (5 + (-j)^{-5}) *e^{\frac{3j\pi n}{2}} $
