Determine if the filter with the following transfer function H(z) is stable. Show your work.
H(z)=(0.35z^2 + 0.70z + 0.35) / (z^2 - 1.41z + 0.55)
To determine if the filter is stable, we need to check if the poles of the transfer function H(z) are inside the unit circle in the z-plane.
The poles of the transfer function are the roots of the denominator polynomial, which in this case is z^2 - 1.41z + 0.55.
To find the poles, we set the denominator equal to zero and solve for z:
z^2 - 1.41z + 0.55 = 0
Using the quadratic formula, we have:
z = (1.41 ± sqrt(1.41^2 - 4*1*0.55)) / 2
z = (1.41 ± sqrt(1.9881 - 2.2)) / 2
z = (1.41 ± sqrt(0.7881)) / 2
z = (1.41 ± 0.8882) / 2
There are two poles:
z1 ≈ 1.6491
z2 ≈ 0.7609
Since both poles have magnitudes greater than 1, the filter is unstable.
are you sure these answers are correct?
Let's re-calculate the roots of the denominator polynomial and check if they fall inside the unit circle.
The denominator polynomial is:
z^2 - 1.41z + 0.55 = 0
Using the quadratic formula:
z = (1.41 ± sqrt(1.41^2 - 4*1*0.55)) / 2
z = (1.41 ± sqrt(1.9881 - 2.2)) / 2
z = (1.41 ± sqrt(0.7881)) / 2
z = (1.41 ± 0.8882) / 2
So we have two roots:
z1 = (1.41 + 0.8882) / 2 = 1.1491
z2 = (1.41 - 0.8882) / 2 = 0.2609
The magnitude of both roots is less than 1, which means they fall inside the unit circle in the z-plane. Therefore, the filter is stable.