Frequency and Impulse Response of a causal LTI system defined by a difference equation

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 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 if so inclined,

for n = 0
$ h[n] = 1\ $
for n > 0
$ h[n] = {\left(\frac{1}{2} \right)}^{n-1} $

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva