Question

The system given by equation y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a) is?

(a) Stable

(b) Unstable

(c) Marginally stable

(d) 0

This question was posed to me in an interview for internship.

This is a very interesting question from Laplace Transform by Properties topic in portion Laplace Transform of Engineering Mathematics

Answer 1

Right answer is (a) Stable

For explanation: Given, y(t-2a)-3y(t-a)+2y(t)=x(t-a), then

Taking Laplace transform, e^-2as Y(s)-3e^-as Y(s)+2Y(s)=e^-as X(s),

Hence, \(H(s)=\frac{e^{-as}}{(1-e^{-as})(2-e^{-as})}\)

To find the stability, we should have, \(∫_{-∞}^∞ H(s)ds>0\)

Hence, \(\int_{-\infty}^\infty \frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds\)

Let, \(e^{-as}=z=>-ae^{-as} ds=dz\)

\(\frac{1}{a} \int_0^∞ \frac{1}{(1-z)(2-z)} dz=\frac{1}{a} \int_0^∞ \left [\frac{1}{(z-1) }-\frac{1}{z-2} \right ]dz=\frac{1}{a} ln⁡\left [\frac{z-1}{z-2}\right ]\)

putting, z=0, \(\frac{1}{a} ln⁡\left [\frac{z-1}{z-2}\right ]=\frac{1}{a} ln⁡(\frac{1}{2})\)

putting, z=∞, \(\frac{1}{a} ln⁡\left [\frac{z-1}{z-2}\right ]=\frac{1}{a}  ln⁡[\frac{1-1/z}{1-2/z}]\)=0

hence, \(\int_{-\infty}^\infty \frac{e^{-as}}{(1-e^{-as})(2-e^{-as})} ds=\frac{1}{a} ln⁡(\frac{1}{2})\)  

It is stable.