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

## Alumni Liaison

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

Francisco Blanco-Silva