The unit impulse response of a system is -4e^-t + 6e^-2t. The step response of the same system for t ≥ 0 is Ae^-t + Be^-2t + C, where A, B and C are respectively 

Concept:

• Differentiation of parabolic input response is ramp input response.
• Differentiation of the ramp input response is step input response.
• Differentiation of step input response is impulse input response.
• Laplace transform of the impulse response is known as the transfer function of the system.

Analysis:

Unit impulse response is given as -4e-t + 6e-2t 

Hence, the transfer function of the system is

L [Impulse response] will be:

\(H\left( s \right) = \frac{{ - 4}}{{s + 1}} + \frac{6}{{s + 2}}=\frac{{Y\left( s \right)}}{{X\left( s \right)}}\)

\( = \frac{{6s + 6 - 4s - 8}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)  \(=\frac{{2\left( {s - 1} \right)}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\)

For step input u(t), X(s) = \(\frac{1}{s}\)

Y(s) = H(s).X(s)

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

By partial fraction,

Y(s) \( = \frac{A}{{\left( {s + 1} \right)}} + \frac{B}{{\left( {s + 2} \right)}} + \frac{c}{s}\)

On solving,

A = 4 , B = - 3 , C = -1