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.
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;
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