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 all your work
b0 = 0.0168;
b1 = 0.0336;
b2 = 0.0168;
a1 = –1.6000;
a2 = 0.6683;
First, we need to determine the transfer function H(z). For the given IIR filter with difference equation:
y[n] = b0 * x[n] + b1 * x[n-1] + b2 * x[n-2] - a1 * y[n-1] - a2 * y[n-2]
The transfer function H(z) can be determined as follows:
H[z] = Y[z]/X[z] = (b0 + b1 * z^(-1) + b2 * z^(-2))/(1 + a1 * z^(-1) + a2 * z^(-2))
Plugging in the coefficients, we get:
H[z] = (0.0168 + 0.0336 * z^(-1) + 0.0168 * z^(-2))/(1 - 1.6000 * z^(-1) + 0.6683 * z^(-2))
We'll now find the zeros and poles of this transfer function.
Zeros are the values of z that make the numerator of the transfer function equal to zero:
0.0168 + 0.0336 * z^(-1) + 0.0168 * z^(-2) = 0
To solve this equation, first multiply through by z^2:
0.0168 * z^2 + 0.0336 * z + 0.0168 = 0
We can find the zeros:
z1, z2 = (-0.0336 ± √(0.0336^2 - 4 * 0.0168 * 0.0168))/(2 * 0.0168)
z1 = 0.5
z2 = 2
The zeros are located at z = 0.5 and z = 2 in the Z-plane.
Now, we'll find the poles of the transfer function. Poles are the values of z that make the denominator of the transfer function equal to zero:
1 - 1.6000 * z^(-1) + 0.6683 * z^(-2) = 0
To solve this equation, first multiply through by z^2:
z^2 - 1.6000 * z + 0.6683 = 0
We can find the poles:
p1, p2 = (1.6000 ± √(1.6000^2 - 4 * 0.6683))/(2)
p1 = 0.8358
p2 = 0.7642
The poles are located at p1 = 0.8358 and p2 = 0.7642 in the Z-plane.
Since both poles are inside the unit circle (|p1| < 1 and |p2| < 1), the filter is considered stable.