Question

The value of \(Z^{-1}\Big\{\frac{z^2}{(z-a)(z-b)}\Big\}\) is ____________

(a) \(\frac{a^{n+1} – b^{n+1}}{a+b}\)

(b) \(\frac{a^{n+1} – b^{n+1}}{a-b}\)

(c) \(\frac{a^{n+1} + b^{n+1}}{a-b}\)

(d) \(\frac{a^{n+1} + b^{n+1}}{a+b}\)

This question was addressed to me in class test.

This intriguing question originated from Inverse Z-Transform in chapter Z-Transform and Digital Filtering of Signals and Systems

Answer 1

Correct choice is (b) \(\frac{a^{n+1} – b^{n+1}}{a-b}\)

Best explanation: We know that, \(Z^{-1}{\frac{z}{z-a}} = a^n\) and\(Z^{-1}{\frac{z}{z-b}} = b^n\)

∴\(Z^{-1}\Big\{\frac{z^2}{(z-a)(z-b)}\Big\} = Z^{-1}{\frac{z}{z-a} . \frac{z}{z-b}} = a^n * b^n\)

= \(∑_{m=0}^n a^m.b^{n-m}\)

 = \(b^n ∑_{m=0}^n \frac{a^m}{b}\)

= \(b^n . \frac{\frac{a^{n+1}}{b}-1}{\frac{a}{b}-1}\)

 = \(\frac{a^{n+1} – b^{n+1}}{a-b}\).