Reference no: EM132723981
Lab Exercise: Complex Poles and Zeros
PeZ assumes real coefficients for the numerator and denominator polynomials. Therefore, if we enter a complex pole or zero, PeZ will automatically insert second root at the conjugate location. For example, if we place a root at z = 1 + j 1 , then we will also get one at z = 1 - j 1.
(a) What property of the polynomial coefficients of A(z) = 1 - a1z-1 - a2z-2 will guarantee that the roots come in conjugate pairs?
(b) Clear all the poles and zeros from PeZ. Now place a pole with magnitude 0.85 at an angle of 45?; and then two zeros at the origin. Note that PeZ automatically places a conjugate pole in the z-domain. The frequency response has a peak-record the frequency (location) of this peak.
(c) Change the angle of the pole: move the pole to 90?, then 135?. Describe the changes in |H(ejω)|. Concentrate on the location of the peak.
Next, we will put complex zeros on the unit circle to see the effect on |H(ejω)|.
(d) Clear all poles and zeros from PeZ. Now place zeros at the following locations: z1 = 1, z2 = 0 j and z3 = 0 + j (remember that conjugate pairs such as z2 and z3 will be entered simultaneously). Judging from the impulse and frequency responses what type of filter have you just implemented?
Attachment:- Signal Processing First.rar
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