DSP | Z-Domain | Impulse Response
Calculators Theory Z-Domain Filters
Z-Transform Z-Transfer Impulse Response Transfer Function
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Impulse Response
For the impulse response;
\[h[n] = \begin{cases} \frac{1}{5} & \text{for} & -2 \leq n \leq 2 \\ 0 & \text{elsewhere} \end{cases}\]Answer the following, giving answers to four decimal places;
Transfer Function
Complete the following;
\[H(\Omega) = A \times cos(\Omega) + B \times cos(2\Omega) + C \times cos(3\Omega) + D \times cos(4\Omega) + \frac{1}{5}\]where \(H(\Omega)\) is the normalised digital frequency transfer function for the system specified with the above impulse response.
\[H(\Omega) = \sum_{n=-2}^{2} h[n]e^{-j \Omega n}\]Therefore;
\[\displaylines{ H(\Omega) & = \frac{1}{5}e^{j2\Omega}+\frac{1}{5}e^{j\Omega}+e^{j\Omega}+\frac{1}{5}e^{-j\Omega}+\frac{1}{5}e^{-j2\Omega} \\ & = \frac{1}{5}(e^{j2\Omega}+e^{j\Omega}+1+e^{-j\Omega}+e^{-j2\Omega}) }\]We can convert this with Euler’s formula;
\[\frac{1}{5}(cos(2\Omega)+jsin(2\Omega)+cos(\Omega)+jsin(\Omega)+1+cos(\Omega)-jsin(\Omega)+cos(2\Omega)-jsin(2\Omega))\]Here, the \(-jsin()\) and \(+jsin()\) pairs cancel each other out, leaving us with;
\[\displaylines{ H(z) = \frac{1}{5}(cos(2\Omega)+cos(\Omega)+1+cos(\Omega)+cos(2\Omega)) \\ = \frac{1}{5}(2 \times cos(2\Omega)+2 \times cos(\Omega)+1) \\ = \frac{2}{5}cos(2\Omega)+\frac{2}{5}cos(\Omega)+\frac{1}{5} }\]Therefore, the answer is;
\[H(\Omega) = 0.4 \times cos(\Omega) + 0.4 \times cos(2\Omega) + 0 \times cos(3\Omega) + 0 \times cos(4\Omega) + \frac{1}{5}\]Difference Equation
Determine the difference equation for this filter in terms of an input \(x[n]\) and an output \(y[n]\);
\[\displaylines{ y[n] = A \times x[n-4] + B \times x[n-3] + C \times x[n-2] + D \times x[n-1] + E \times x[n] \\ + F \times x[n+1] + G \times x[n+2] + H \times x[n+3] + I \times x[n+4] }\]Recall;
\[\displaylines{ H(z) = \frac{1}{5}(e^{j2\Omega}+e^{j\Omega}+1+e^{-j\Omega}+e^{-j2\Omega}) = \frac{1}{5}(z^{2}+z+1+z^{-1}+z^{-2}) }\]This is equivelent to;
\[\displaylines{ H(z) = \frac{1}{5}\left(\frac{z^{2}+z+1+z^{-1}+z^{-2}}{1}\right) = \frac{0.2z^{2}+0.2z+0.2+0.2z^{-1}+0.2z^{-2}}{1} }\]Transforming to time domain gives;
\[\displaylines{ y[n] = 0 \times x[n-4] + 0 \times x[n-3] + 0.2 \times x[n-2] + 0.2 \times x[n-1] + 0.2 \times x[n] \\ + 0.2 \times x[n+1] + 0.2 \times x[n+2] + 0 \times x[n+3] + 0 \times x[n+4] }\]