Determine if the filter described by the following difference equation is stable:
y[n] = b0x[n] + b1x[n –1] + b2x[n – 2] – a1y[n – 1] – a2y[n – 2]
b0 = 0.035; b1 = 0.070; b2 = 0.035;
a1 = –1.41; a2 = 0.55;
To determine if the filter is stable, we need to check whether the poles of the filter transfer function lie within the unit circle in the z-plane.
The filter transfer function can be written as:
H(z) = (b0 + b1z^-1 + b2z^-2) / (1 + a1z^-1 + a2z^-2)
Setting the denominator equal to zero gives us the characteristic equation:
1 + a1z^-1 + a2z^-2 = 0
z^2 - a1z - a2 = 0
Using the quadratic formula to solve for the poles gives:
z = (a1 ± sqrt(a1^2 - 4*a2)) / 2
Plugging in the given values of a1 and a2 :
z = (1.41 ± sqrt((-1.41)^2 - 4*0.55)) / 2
z = (1.41 ± sqrt(1.9881 - 2.2)) / 2
z = (1.41 ± sqrt(-0.2119)) / 2
Since the square root of a negative number results in an imaginary number, the poles of the filter transfer function have a real part and an imaginary part. Therefore, we need to check if the magnitude of the poles is less than 1 for stability.
Calculating the magnitude of the poles:
|z| = |1.41/2 ± 0.3076i|
Since the denominator is less than 1, the filter is stable.