GATE 2014 EC (EC3) – Question 53
$H_1(z)=(1-pz^{-1})^{-1}$, $H_2(z)=(1-qz^{-1})^{-1}$ and $H(z)=H_1(z)+rH_2(z)$, with real p,q,r. For p = 1/2, q = −1/4 and |r| < 1, a zero of H lies on the unit circle. Find r.
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Correct answer: -0.6 to -0.4
Explanation
Combining H1+rH2 gives numerator(1+r)−(q+rp)z⁻¹. Thus z0=(q+rp)/(1+r), with p=1/2and q=−1/4. A real unit-circle zero must be±1. Setting z0=−1 gives−1/4+r/2=−1−r, so **r=−1/2**. The+1case gives r=−2.5, excluded by |r|<1.