Question

The transfer function of a system is given by H(s) \(= \frac{1}{{{s^2}\left( {s - 2} \right)}}\) The impulse response of the system is

1. (t² * e^-2t)u(t)
2. (t * e^2t)u(t)
3. (t² e^-2t)u(t)
4. (t e^2t) u(t)

Answer 1

Correct Answer - Option 2 : (t * e^2t)u(t)

\(H\left( s \right) = \frac{1}{{{s^2}\left( {s - 2} \right)}}\) 

\(= \frac{A}{{{s^2}}} + \frac{B}{s} + \frac{C}{{\left( {s - 2} \right)}}\) 

A = (s - 2) + Bs (s - 2) + CS² = 1      ---(1)

Putting s = 0 in (1)

A = -0.5

Putting s = 2 in (1)

C = 0.25

Putting s = 1 in (1)

-0.5 (1 – 2) B 1(1 - 2) + 0.25 = 1

-B = 0.25

∴ \(\frac{1}{{{s^2}\left( {s - 2} \right)}} = \frac{{ - 0.5}}{{{s^2}}} - \frac{{0.25}}{s} + \frac{{0.25}}{{s - 2}}\)

= -0.5t – 0.25 u(t) + 0.25 e^2t u(t)

We can also proceed from options:

∵ Multiplication in frequency domain = Convolution in time domain and vice-versa

\(\frac{1}{{{s^2}\left( {s - 2} \right)}} = {H_1}\left( s \right){H_2}\left( s \right)\) 

\({H_1}\left( s \right) = \frac{1}{{{s^2}}}\) 

∴ h₁(t) = t u(t)

\({H_2}\left( s \right) = \frac{1}{{s - 2}}\) 

∴ h₂(t) = e^2t u(t)

H₁(s) H₂(s) \(\mathop \leftrightarrow \limits^{I.L.T} \) h₁(t) * h₂(t)

∴ (t * e^2t) u(t).