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.
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)
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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.
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.