This question was previously asked in

ESE Electronics 2012 Paper 1: Official Paper

Option 1 : C(t) = 5e^{-2t} – e^{-t} -1

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

__Concept__:

The response of a system is nothing but the convolution of the input signal with the impulse response.

For a discrete-time system, the response for an input sequence x(n) will be:

y[n] = x[n] ⊕ h[n]

\(y\left( n \right) = \mathop \sum \limits_{k = - \infty }^{k = \infty } x\left[ k \right]h\left[ {n - k} \right]\)

For a continuous-time signal, the response is given by:

\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( \tau \right)h\left( {t - \tau } \right)d\tau\)

For the continuous-time system, the step response will be:

\(s\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt\)

where h(t) is the impulse response.

**Analysis:**

\(H(s)=\frac{(3s^2-2)}{(s^2+3s+2)} \)

X(s) = 1/s

Y(s) = \(\frac{3s^2-2}{s^2+3s+2} \times \frac{1}{s}\)

Y(s) = \(\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}\)

A = -1

B = -1

C = 5

**y(t) = -1 - e ^{-t }+ 5 e^{-2t}**