DSP | Z-Domain | Transfer Function
Calculators Theory Z-Domain Filters
Z-Transform Z-Transfer Impulse Response Transfer Function
← Back to Digital Signal Processing
Transfer Function
The transfer function for a digital system is given by;
\[H(z) = \frac{0.23z(z-0.46)}{(z-0.59)(z-0.5)}\]Answer the following, giving answers to 4 decimal places;
Factors
Determine the factors in the following recursion for the system;
\[h[k] + A \times h[k-1] + B \times h[k-2] = C \times \delta[k] + D \times \delta[k-1]\]where \(\delta[k-v]\) is the unit impulse function.
\[\displaylines{ H(z) = \frac{0.23z^{2} - 0.23 \times 0.46z}{z^{2}-0.5z-0.59z+0.295} \\ = \frac{0.23z^{2} - 0.1058z}{z^{2}-1.09z+0.295} }\]Divide through by \(z^{2}\);
\[H(z) = \frac{0.23 - 0.1058z^{-1}}{1-1.09z^{-1}+0.295z^{-2}}\]Results in;
\[h[k] + -1.09 \times h[k-1] + 0.295 \times h[k-2] = 0.23 \times \delta[k] + -0.1058 \times \delta[k-1]\]MATLAB
z1 = 0.23; z2 = -0.46; p1 = -0.59; p2 = -0.5;
A = p1 + p2
B = p1 * p2
C = z1
D = z1 * z2
Difference Equation
If the input sequence to the system is \(x[n]\) and its output sequence is \(y[n]\), derive the difference equation which specifies the current output in terms of previous outputs and current and previous inputs.
To answer this question, complete the following;
\[y[n] = A \times x[n] + B \times x[n-1] + C \times y[n-1] + D \times y[n-2]\]Take the function after dividing by \(z^{2}\) under the Factors heading;
Cross multiply;
\[y(z) - 1.09 y(z)z^{-1} + 0.295y(z)z^{-2} = 0.23x(z) - 0.1058x(z)z^{-1}\]Invert and fit to required answer by changing signs that cross;
\[y[n] = 0.23 \times x[n] - 0.1058 \times x[n-1] + 1.09 \times y[n-1] - 0.295 \times y[n-2]\]Partial Fraction Expansion
Perform partial fraction expansion with;
\[\frac{A}{z-0.59} + \frac{B}{z-0.5}\]Thus;
\[H(z) = \frac{0.23z(z-0.46)}{(z-0.59)(z-0.5)} = \frac{A}{z-0.59} + \frac{B}{z-0.5}\]Times through;
\[H(z) = \frac{0.23(z-0.46)}{(z-0.59)(z-0.5)} = \frac{A(z-0.5) + B(z-0.59)}{(z-0.59)(z-0.5)}\]Divide through the denominator;
\[H(z) = 0.23(z-0.46) = A(z-0.5) + B(z-0.59)\]Determine \(A\) by making \(z = 0.59\);
\[\displaylines{ A = 0.23(0.59-0.46) = A(0.59-0.5) + (B \times 0) \\ A = 0.23(0.13) = A(0.09) + (B \times 0) \\ A = 0.0299 = A(0.09) + (B \times 0) \\ A = \frac{0.0299}{0.09} = 0.332222222 = 0.3322 }\]Determine \(B\) by making \(z = 0.5\);
\[\displaylines{ B = 0.23(0.5-0.46) = (A \times 0) + B(0.5-0.59) \\ B = 0.23(0.04) = (A \times 0) + B(-0.09) \\ B = 0.0092 = (A \times 0) + B(-0.09) \\ B = \frac{0.0092}{-0.09} = -0.10222222 = -0.1022 }\]MATLAB
z1 = 0.23; z2 = -0.46;
p1 = -0.59; p2 = -0.5;
A = (z1 * (abs(p1) + z2))/(abs(p1) + p2)
B = (z1 * (abs(p2) + z2))/(abs(p2) + p1)
%
% Note, if the numerator is of the form 0.23z^2, then;
%
z1 = 0.23;
p1 = -0.59; p2 = -0.5;
A = (z1 * abs(p1))/(abs(p1) + p2)
B = (z1 * abs(p2))/(abs(p2) + p1)
Pulse Response
Determine the pulse response of the system in the form;
\[h[k] = A \times (0.59^{k}) + B \times (0.5^{k})\]Here, we simply populate the values with those determined above;
\[h[k] = 0.3322 \times (0.59^{k}) + -0.1022 \times (0.5^{k})\]