Determine the transfer function, the impulse response function and give a formula for solving the following initial value problem.

Question

$$y''-4y'+5y=g(t);\quad y(0)=0,~y'(0)=1$$ Knowing the initial values of $y(x)$ and $y'(x)$ it's easy to take the $\mathcal{L}$ on both sides of the equation: $$\mathcal{L}\left\lbrace y''-4y'+5y\right\rbrace=\mathcal{L}\left\lbrace g(t) \right\rbrace$$ And this results on: $$\mathcal{L}\left\lbrace y\right\rbrace=\frac{\mathcal{L}\left\lbrace g(t) \right\rbrace+1}{(s^{2}-4s+5)}$$ without knowing $g(t)$, how should I proceed? Also, how exactly is the transfer function and the impulse response function calculated?

Answer 1

From the general context of exercises of this kind you know that $g(t)=0$ for $t<0$, as all dynamical components are considered "switched off" for $t<0$ and activated at $t=0$.

Set $u(t)=e^{-2t}y(t)$ to find $$ u''(t)+u(t)=e^{-2t}(y''(t)-4y'(t)+5y(t))=e^{-2t}g(t)=h(t). $$ The general solution formula for the harmonic oscillator $$ u(t)=u(0)\cos(t)+u'(0)\sin(t)+\int_0^t\sin(t-s)h(s)\,ds $$ then translates to $$ y(t)=e^{2t}\sin(t)+\int_0^te^{2(t-s)}\sin(t-s)g(s)\,ds. $$ You could also read this off the Laplace transform formula by completing the square in $s^2-4s+5=(s-2)^2+1$.