Difference between revisions of "Frequency Response Example Old Kiwi" - Rhea

Revision as of 13:18, 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     <math> h[n] = {\left(\frac{1}{2} \right)}^{n-1} $

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