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;
The transfer function H[Z] for an IIR filter given the difference equation coefficients is:
H[Z] = (b0 + b1*Z^(-1) + b2*Z^(-2)) / (1 + a1*Z^(-1) + a2*Z^(-2))
Plugging in the given coefficients, we get:
H[Z] = (0.0200 + 0.0400*Z^(-1) + 0.0200*Z^(-2)) / (1 - 1.5600*Z^(-1) + 0.6420*Z^(-2))
The poles and zeros of the transfer function can be found by setting the denominator and numerator equal to zero and solving for Z:
Denominator = 1 - 1.5600*Z^(-1) + 0.6420*Z^(-2) = 0
Numerator = 0.0200 + 0.0400*Z^(-1) + 0.0200*Z^(-2) = 0
Solving the denominator equation first, we get the poles of the transfer function:
1 - 1.5600*Z^(-1) + 0.6420*Z^(-2) = 0
This can be factored as:
(1 - 0.8*Z^(-1))(1 - 0.8*Z^(-1)) = 0
The poles are at Z = 0.8 and Z = 0.8.
Solving the numerator equation, we get the zeros of the transfer function:
0.0200 + 0.0400*Z^(-1) + 0.0200*Z^(-2) = 0
This can be factored as:
(0.02)(1 + 2Z^(-1) + 1*Z^(-2)) = 0
The zeros are at Z = -1 (multiplicity 2).
Since the poles are inside the unit circle (|Z| < 1), the filter is stable. The poles at Z = 0.8 can be located on the Z-plane. The zeros at Z = -1 (multiplicity 2) can also be located on the Z-plane.
are you sure this is correct/
Let's recheck the calculations for finding the poles and zeros of the transfer function.
The transfer function H[Z] for an IIR filter given the difference equation coefficients is:
H[Z] = (b0 + b1*Z^(-1) + b2*Z^(-2)) / (1 + a1*Z^(-1) + a2*Z^(-2))
Given b0 = 0.0200, b1 = 0.0400, b2 = 0.0200, a1 = -1.5600, a2 = 0.6420, the transfer function becomes:
H[Z] = (0.0200 + 0.0400*Z^(-1) + 0.0200*Z^(-2)) / (1 - 1.5600*Z^(-1) + 0.6420*Z^(-2))
To find the poles, we need to solve the denominator equation:
1 - 1.5600*Z^(-1) + 0.6420*Z^(-2) = 0
By solving this quadratic equation, we find the poles to be at Z = 0.9759 + j0.2205 and Z = 0.9759 - j0.2205.
To find the zeros, we solve the numerator equation:
0.0200 + 0.0400*Z^(-1) + 0.0200*Z^(-2) = 0
The zeros can be found by solving this quadratic equation. However, for this specific transfer function, the zeros do not exist in the given range.
Since the poles are inside the unit circle (|Z| < 1), the filter is stable. The poles at Z = 0.9759 + j0.2205 and Z = 0.9759 - j0.2205 can be located on the Z-plane.