For the IIR filter difference equation with the specified coefficients determine the transfer function H[Z] and its poles and zeros.

Question

For the IIR filter difference equation with the specified coefficients determine the transfer function H[Z] and its poles and zeros.

b0 = 0.0233;
b1 = 0.0466;
b2 = 0.0233;
a1 = –1.5230;
a2 = 0.6167;

Answer 1

First, let's represent the IIR filter difference equation in the following form:

y[n] - a1 * y[n-1] - a2 * y[n-2] = b0 * x[n] + b1 * x[n-1] + b2 * x[n-2]

Now, we take the z-transform of both sides of this equation:

(H(z) * X(z)) - a1 * z^(-1) * (H(z) * X(z)) - a2 * z^(-2) * (H(z) * X(z)) = b0 * X(z) + b1 * z^(-1) * X(z) + b2 * z^(-2) * X(z)

Next, we can factor out the H(z) * X(z) term:

H(z) * X(z) * (1 - a1 * z^(-1) - a2 * z^(-2)) = X(z) * (b0 + b1 * z^(-1) + b2 * z^(-2))

Now, we find the transfer function H(z) by dividing both sides by X(z):

H(z) = (b0 + b1 * z^(-1) + b2 * z^(-2)) / (1 - a1 * z^(-1) - a2 * z^(-2))

Replacing the coefficients with the given values:

H(z) = (0.0233 + 0.0466 * z^(-1) + 0.0233 * z^(-2)) / (1 + 1.5230 * z^(-1) - 0.6167 * z^(-2))

For better readability, let's multiply the numerator and the denominator by z^2:

H(z) = (0.0233 * z^2 + 0.0466 * z + 0.0233) / (z^2 + 1.5230 * z - 0.6167)

Now, we will find the zeros of H(z) by finding the roots of the numerator:

0.0233 * z^2 + 0.0466 * z + 0.0233 = 0

We can use the quadratic formula to find the roots:

z = (-B ± √(B^2 - 4AC)) / 2A

A = 0.0233
B = 0.0466
C = 0.0233

z = (-0.0466 ± √(0.0466^2 - 4 * 0.0233 * 0.0233)) / (2 * 0.0233)
z = (-0.0466 ± √(0.00217236 - 0.00129729)) / 0.0466
z = (-0.0466 ± √0.00087507) / 0.0466

The zeros of H(z) are:

z1 = (-0.0466 + √0.00087507) / 0.0466 = 0.5
z2 = (-0.0466 - √0.00087507) / 0.0466 = -1

To find the poles of H(z), we need to find the roots of the denominator:

z^2 + 1.5230 * z - 0.6167 = 0

Using the quadratic formula, we have:

z = (-1.5230 ± √(1.5230^2 + 4 * 0.6167)) / 2

The poles of H(z) are:

z1 = (-1.5230 + √(2.320329)) / 2 = 0.3613
z2 = (-1.5230 - √(2.320329)) / 2 = -1.8843

So, the transfer function H(z) is:

H(z) = (0.0233 * z^2 + 0.0466 * z + 0.0233) / (z^2 + 1.5230 * z - 0.6167)

The zeros are 0.5 and -1, and the poles are 0.3613 and -1.8843.