Correct Answer - Option 3 : (1.5 – e
-t – 0.5 e
-2t) u(t)
Concept:
The impulse response of a system is the response of the system to an impulse function.
The response of the system to any signal x(t), is given by the convolution of the input signal with the impulse response, i.e.
y(t) = x(t) ⊗ h(t)
Also, the multiplication is s-domain property states that:
\(y\left( t \right)\mathop \leftrightarrow \limits^{\;\;\;L.T.\;\;\;} \;X\left( s \right) \cdot H\left( s \right)\)
X(s) = Laplace Transform of x(t)
H(s) = Laplace Transform of h(t)
The above property is used to evaluate the response of a signal.
Application:
Given, x(t) = e-t + e-2t
x(t) = u(t)
y(t) = x(t) ⊗ u(t)
Taking the Laplace Transform, we get:
Y(s) = X(s) ⋅ H(s)
Given,
H(s) = L [e-t + e-2t]
\(H\left( s \right) = \frac{1}{{s + 1}} + \frac{1}{{s + 2}}\)
x(t) = u(t)
\(X\left( s \right) = \frac{1}{s}\)
So, Y(s) = X(s) ⋅ H(s)
\(Y\left( s \right) = \left( {\frac{1}{{s + 1}} + \frac{1}{{s + 2}}} \right)\frac{1}{s}\)
\(= \frac{{s + 2 + s + 1}}{{\left( {s + 1} \right)\left( {s + 2} \right)\left( s \right)}}\)
\(= \frac{{2s + 3}}{{s\left( {s + 1} \right)\left( {s + 2} \right)}}\)
y(t) is obtained by taking the inverse Laplace transform of the above.
We can partial fraction to simplify the above:
\(\frac{{2s + 3}}{{s\left( {s + 1} \right)\left( {s + 2} \right)}} = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{s + 2}}\)
This can be written as:
⇒ 2s + 3 = A (s + 1) (s + 2) + B(s) (s + 2) + C(s) (s + 1)
For s = -1
⇒ 1 = B (-1) (1) + 0
⇒ B = -1 ---(1)
For s = 0
3 = A(2) (1) + 0
\(A = \frac{3}{2}\) ---(2)
For s = -2
⇒ 2(-2) + 3 = C (-2) (-2 + 1) + 0
⇒ -1 = 2C
\(C = - \frac{1}{2}\)
\(Y\left( s \right) = \frac{3}{2} \times \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{2} \cdot \frac{1}{{\left( {s + 2} \right)}}\)
Taking the inverse Laplace transform we get:
\(y\left( t \right) = \frac{3}{2}\;u\left( t \right) - {e^{ - t}}u\left( t \right) - \frac{1}{2}\;{e^{ - 2t}}u\left( t \right)\)
= (1.5 – e
-t – 0.5 e
-2t) u(t)
