Correct Answer - Option 2 : (e
-t - e
-3t) u(t)
Concept:
The transfer function is defined as the ratio of the Laplace transform of the output to that of the input with the initial conditions zero.
\(\rm TF=\dfrac{L[output]}{L[input]}\)
The Laplace transform for the integrated and differential signals are shown below:
\(\dfrac{dx}{dt} \longleftrightarrow sX(s)\)
\(\dfrac{d^2x}{dt^2} \longleftrightarrow {s^2}X(s)\)
\(\int x dt = \dfrac{1}{s} X(s)\)
The Laplace transform of the basic exponential signal is:
\(e^{-at}u(t) \longleftrightarrow \dfrac{1}{s+a}; \; ROC = \sigma > Re{(-a)}\)
\(e^{at}u(t) \longleftrightarrow \dfrac{1}{s-a}; \; ROC = \sigma > Re{(a)}\)
Calculation:
Converting the given time-domain differential equation into the frequency domain we get:
\(s^2Y(s)+4sY(s)+3Y(s)=2sX(s)+4X(s)\)
\(\frac{Y(s)}{X(s)}=\dfrac{2s+4}{s^2+4s+3}\)
\(Y(s)=\dfrac{2(s+2)}{(s+1)(s+3)}\frac{1}{s+2}\)
\(Y(s)=\dfrac{2}{(s+1)(s+3)}\)
\(H(s)=\dfrac{A}{s+1}+\dfrac{B}{s+3}\)
\(A=\left.\dfrac{2}{s+3}\right|_{s=-1}\)
\(A=\dfrac{2}{-1+3}=\dfrac{2}{2}=1\)
\(B=\left.\dfrac{2}{s+1}\right|_{s=-3}\)
\(B=\dfrac{2}{-3+1}=\dfrac{2}{-2}=-1\)
\(Y(s)=\dfrac{1}{s+1}+\dfrac{-1}{s+3}\)
\(y(t)=(e^{-t}-e^{-3t})u(t)\)
