(New page: == Define a DT LTI system == <math>\,y[n]=2x[n-1]+3x[n]</math> The unit impluse response <math>\,h[n]</math> is just <math>h[n]=2\delta [n-1] +3\delta[n]\,</math> So take laplace trans...) |
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<math>\,H(\pi \omega) = 3+2e^{-j\pi}=3-2=1</math> | <math>\,H(\pi \omega) = 3+2e^{-j\pi}=3-2=1</math> | ||
| − | So if my transfer function <math>\,H(\pi \omega)=1</math>, then my <math>\,y(t)=x(t)</math> | + | So if my transfer function <math>\,H(\pi \omega)=1</math>, then my <math>\,y(t)=x(t)</math>, but just for the first 2 terms. The next 2 terms my <math>\,H(2\pi \omega)=2</math>, so my final result is |
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| + | <math>y(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{j}e^{j2\pi t}+\frac{-1-j}{j}e^{-j2\pi t}</math> | ||
Revision as of 14:28, 25 September 2008
Define a DT LTI system
$ \,y[n]=2x[n-1]+3x[n] $
The unit impluse response $ \,h[n] $ is just
$ h[n]=2\delta [n-1] +3\delta[n]\, $
So take laplace transform of this function to get $ H(s)\, $, the system function.
$ H(z) = \sum_{n=-\infty}^{\infty}h[n] e^{-sn} =\sum_{n=-\infty}^{\infty}2\delta [n-1] +\sum_{n=-\infty}^{\infty}3\delta[n] e^{-sn} =2e^{-s}+3 $
$ \,H[z]=3+2e^{-j\omega} $
Compute the response
My original function:
$ x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t} $
I have to multiply each of these fourier coefficients by my $ H[z]\, $, also noting that my $ \,\omega = \pi $
amazingly enough with my $ \,\omega=\pi $, my function turns to
$ \,H(\pi \omega) = 3+2e^{-j\pi}=3-2=1 $
So if my transfer function $ \,H(\pi \omega)=1 $, then my $ \,y(t)=x(t) $, but just for the first 2 terms. The next 2 terms my $ \,H(2\pi \omega)=2 $, so my final result is
$ y(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{j}e^{j2\pi t}+\frac{-1-j}{j}e^{-j2\pi t} $
