Question

Consider an LTI system whose response to the input x(t) = [e^-t + e^-3t] u(t) is y(t) = [2e^-t – 2e^-4t]. The system’s impulse response will be
1. \(\frac{3}{2}\left[ {{e^{ - 2t}} + {e^{ - 4t}}} \right]u\left( t \right)\)
2. \(\frac{3}{2}\left[ {{e^{ - 2t}} - {e^{ - 4t}}} \right]u\left( t \right)\)
3. \(\frac{1}{2}\left[ {{e^{ - 2t}} + {e^{ - 4t}}} \right]u\left( t \right)\)
4. \(\frac{1}{2}\left[ {{e^{ - 2t}} - {e^{ - 4t}}} \right]u\left( t \right)\)

Answer 1

Correct Answer - Option 1 : \(\frac{3}{2}\left[ {{e^{ - 2t}} + {e^{ - 4t}}} \right]u\left( t \right)\)

Concept:

The transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L^-1 [TF]

Calculation:

Input x(t) = [e^-t + e^-3t]

By applying the Laplace transform,

\(X\left( s \right) = \frac{1}{{s + 1}} + \frac{1}{{s + 3}} = \frac{{2s + 4}}{{\left( {s + 1} \right)\left( {s + 3} \right)}}\)

y(t) = [2e^-t – 2e^-4t]

By applying the Laplace transform,

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

Transfer function, \(H\left( s \right) = \frac{{Y\left( s \right)}}{{X\left( s \right)}}\)

\(H\left( s \right) = \frac{{3\left( {s + 3} \right)}}{{\left( {s + 2} \right)\left( {s + 4} \right)}}\)

By applying the partial fraction method,

\(H\left( s \right) = \frac{{1.5}}{{s + 2}} + \frac{{1.5}}{{s + 4}}\)

The impulse response is the inverse Laplace transform of the transfer function.

By applying inverse Laplace transform,

\(h\left( t \right) = \frac{3}{2}\left[ {{e^{ - 2t}} + {e^{ - 4t}}} \right]u\left( t \right)\)