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 the stability of the filter, we need to check if all the poles of the transfer function H(z) lie inside the unit circle in the z-plane.
The transfer function can be rewritten as:
H(z) = (0.35z^2 + 0.70z + 0.35) / (z^2 - 1.41z + 0.55)
= (0.35(z^2 + 2z + 1) / (z^2 - 1.41z + 0.55)
= (0.35(z + 1)^2) / ((z - 0.7)(z - 0.79))
The poles of the transfer function are the values of z that make the denominator equal to zero. In this case, the poles are z=0.7 and z=0.79.
To determine stability, we need to check if the magnitude of each pole is less than 1.
|0.7| < 1, therefore, the pole at z=0.7 lies inside the unit circle.
|0.79| < 1, therefore, the pole at z=0.79 lies inside the unit circle.
Since all poles of the transfer function H(z) lie inside the unit circle, the filter is stable.
are you sure these answers are correct?
I apologize for the error in my previous response. Let me correct my calculations:
The transfer function can be rewritten as:
H(z) = (0.35z^2 + 0.70z + 0.35) / (z^2 - 1.41z + 0.55)
= (0.35z^2 + 0.70z + 0.35) / (z - 0.7)(z - 0.79)
The poles of the transfer function are the values of z that make the denominator equal to zero. In this case, the poles are z=0.7 and z=0.79.
To determine stability, we need to check if the magnitude of each pole is less than 1.
|0.7| = 0.7 < 1, therefore, the pole at z=0.7 lies inside the unit circle.
|0.79| = 0.79 < 1, therefore, the pole at z=0.79 lies inside the unit circle.
Since both poles of the transfer function H(z) lie inside the unit circle, the filter is stable.