(New page: For the discrete time L.T.I. system described by <math>y[n]-\frac{1}{2}y[n-1]=x[n]+\frac{1}{2}x[n-1] </math> Find the frequency response H(|omega|) and the impulse response h[n] of the ...) |
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<math> h[n] = {\left(\frac{1}{2} \right)}^{n-1}</math> | <math> h[n] = {\left(\frac{1}{2} \right)}^{n-1}</math> | ||
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| + | Next : [[Geometric Series Note_Old Kiwi]] | ||
Latest revision as of 18:08, 4 April 2008
For the discrete time L.T.I. system described by
$ y[n]-\frac{1}{2}y[n-1]=x[n]+\frac{1}{2}x[n-1] $
Find the frequency response H(|omega|) and the impulse response h[n] of the system.
- Frequency Response:**
1: Take the Fourier transform ([Meaning of Fourier Transform]) of the equation,
$ Y(\omega)-\frac{1}{2}{e}^{-j\omega}Y(\omega)=X(\omega)+\frac{1}{2}{e}^{-j\omega}X(\omega) $
2: Solve for Y(|omega|)/X(|omega|) which is the frequency response H(|omega|),
.. $ H(\omega)=\frac{Y(\omega)}{X(\omega)}=\frac{1+\frac{1}{2}{e}^{-j\omega}}{1-\frac{1}{2}{e}^{-j\omega}} $
- Impulse Response:**
1: Expand into two terms using partial fraction expansion ([Guide to Partial Fraction Expansion]) to facilitate use of inverse Fourier transform,
$ H(\omega)=\frac{1}{1-\frac{1}{2}{e}^{-j\omega}}+\frac{1}{2}\frac{{e}^{-j\omega}}{1-\frac{1}{2}{e}^{-j\omega}} $
2: Take the inverse Fourier transform of H(|omega|) ([Fourier Transform Table]),
$ h[n]={\left(\frac{1}{2} \right)}^{n}u[n]+\frac{1}{2}{\left(\frac{1}{2} \right)}^{n-1}u[n-1] $
3: Simplify is so inclined,
for n = 0
$ h[n] = 1 $
for n > 0
$ h[n] = {\left(\frac{1}{2} \right)}^{n-1} $
