3. 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 stability, we need to examine the poles of the transfer function H(z). The poles are the values of z for which the denominator of the transfer function is equal to zero.
So, we need to find the roots of the denominator polynomial z^2 - 1.41z + 0.55:
z^2 - 1.41z + 0.55 = 0
By factoring, we get:
(z - 0.55)(z - 0.86) = 0
Setting each factor to zero gives us the poles:
z - 0.55 = 0 or z - 0.86 = 0
z = 0.55 or z = 0.86
Since all poles lie inside the unit circle (|z| < 1), the filter with the given transfer function H(z) is stable.
are you sure these answers are correct?
Apologies for the confusion in my previous response. Let's correct the solution:
To determine stability, we need to examine the poles of the transfer function H(z). The poles are the values of z for which the denominator of the transfer function is equal to zero.
The denominator of the transfer function H(z) is z^2 - 1.41z + 0.55. We need to find the roots of this polynomial:
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(1.9881 - 2.2)] / 2
z = [1.41 ± sqrt(-0.2119)] / 2
The two poles are complex conjugate pairs:
z = 0.705 + j0.375
z = 0.705 - j0.375
Since the magnitude of the poles is less than 1 (|z| < 1), the filter with the given transfer function H(z) is stable. Thank you for your patience.