Question

Let H₁(z) = (1 – pz^-1)^-1, H₂(z) = (1 – qz^-1)^-1, H(z) = H₁(z) + r H₂(z). The quantities p, q, r are real numbers. Consider \(p = \frac{1}{2},\;q = - \frac{1}{4},\;\left| r \right| < 1\). If the zero of H(z) lies on the unit circle, then r = ______

Answer 1

Given:

H₁(z) = (1 – pz^-1)^-1

H₂(z) = H₁(z) + r H₂(z)

\(p = \frac{1}{2},\;q = - \frac{1}{4},\;\left| r \right| < 1\)

and zero of H(z) lies on the unit circle.

\(H\left( z \right) = \frac{1}{{1 - P{z^{ - 1}}}} + \frac{r}{{1 - q{z^{ - 1}}}}\)

\(H\left( z \right) = \frac{{\left( {1 - q{z^{ - 1}}} \right) + r\left( {1 - P{z^{ - 1}}} \right)}}{{\left( {1 - P{z^{ - 1}}} \right)\left( {1 - q{z^{ - 1}}} \right)}}\)

\(H\left( z \right) = \frac{{1 - q{z^{ - 1}} + r - pr{h^{ - 1}}}}{{\left( {1 - P{z^{ - 1}}} \right)\left( {1 - q{z^{ - 1}}} \right)}}\)

\(H\left( z \right) = \frac{{\left( {\left( {1 + r} \right) - \left( {pr + q} \right){z^{ - 1}}} \right)}}{{\left( {1 - p{z^{ - 1}}} \right)\left( {1 - q{z^{ - 1}}} \right)}}\)

Zero of H(z) ⇒ (1 + r) – (pr + q) z^-1 = 0

(1 + r) = (pr + q) z^-1

\(z = \frac{{pr + q}}{{1 + r}}\)

\(z = \frac{{\frac{1}{2}r- \frac{1}{4}}}{{1 + r}}\)

And given |z| = 1 [Zero lies on unit circle]

z = ± 1

Taking z = 1

\(1 = \frac{{\frac{1}{2}r - \frac{1}{4}}}{{1 + r}}\)

 

\(r = \frac{5}{2}\)

Taking z = -1

\(- 1 = \frac{{\frac{1}{2}r - \frac{1}{4}}}{{1 + r}}\)

Solving the above we get,

\(r = -0.5\)

And Given that |r| < 1

Hence \(r = \frac{5}{2}\) cannot be possible

∴ \(r = -\frac{1}{2}\;;\left| r \right| < 1\) is satisfied.