1. For the IIR filter difference equation with the specified coefficients determine the transfer function H[Z] and its poles and zeros. Indicate locations of poles and zeros on the Z-plane and determine if the filter is stable. Show your work.
b0 = 0.0200;
b1 = 0.0400;
b2 = 0.0200;
a1 = –1.5600;
a2 = 0.6420;
H |z| = ?
Poles = ?
Zeros = ?
To determine the transfer function H[Z], we write the difference equation as:
H[Z] = (b0 + b1 * Z^(-1) + b2 * Z^(-2)) / (1 + a1 * Z^(-1) + a2 * Z^(-2))
Substitute the given coefficient values into the equation:
H[Z] = (0.0200 + 0.0400 * Z^(-1) + 0.0200 * Z^(-2)) / (1 - 1.5600 * Z^(-1) + 0.6420 * Z^(-2))
Next, simplify the transfer function by multiplying both numerator and denominator by Z^2:
H[Z] = (0.0200 * Z^2 + 0.0400 * Z + 0.0200) / (Z^2 - 1.5600 * Z + 0.6420)
Therefore, the transfer function H[Z] is:
H[Z] = (0.0200 * Z^2 + 0.0400 * Z + 0.0200) / (Z^2 - 1.5600 * Z + 0.6420)
The poles of the system are the roots of the denominator polynomial:
Z^2 - 1.5600 * Z + 0.6420 = 0
Solving the quadratic equation for Z, we get:
Z = (1.5600 ± sqrt(1.5600^2 - 4*0.6420)) / 2
Z = (1.5600 ± sqrt(2.4336 - 2.568)) / 2
Z = (1.5600 ± sqrt(0.1336)) / 2
Z ≈ 1.2790 or Z ≈ 0.5010
So, the poles of the system are located at Z = 1.2790 and Z = 0.5010.
Since the numerator is a simple polynomial, there are no zeros in this transfer function.
To determine if the filter is stable, we need to check if the poles lie within the unit circle on the Z-plane. Since both poles Z = 1.2790 and Z = 0.5010 have magnitudes less than 1, the filter is stable.