Laplace Transform, impulse response

Suppose a differential equation, \(a\ddot{w}+b\dot{w}+cw=\delta(t)\), has the following unit impulse response:

\[w(t)=e^{-t}-e^{-3t}\]

What would its response to the unit step function be?

\[a\ddot{v}+b\dot{v}+cv=u(t)\]

Using convolution we get

\[v(t)=\int_0^t w(t-\tau)*u(\tau)d\tau=\int_0^t w(\tau)*u(t-\tau)d\tau=\int_0^te^{-t}-e^{-3t} d\tau\]

So, \(v(t)=\left[-e^{-\tau}+\frac{1}{3}e^{-3\tau}\right]_0^t=-e^{-t}+1+\frac{1}{3}e^{-3t}-\frac{1}{3}=\frac{2}{3}-e^{-t}+\frac{1}{3}e^{-3t}\)

Alternatively, we know that \(w(t)=\mathop{\mathscr{L^{-1}}}\left\{\frac{1}{as^2+bs+c}\right\}\).

Comparing that to \(\mathop{\mathscr{L}}\left\{w(t)\right\}=\mathop{\mathscr{L}}\left\{e^{-t}-e^{-3t}\right\}=\frac{1}{s+1}-\frac{1}{s+3}\).

\[\frac{1}{s+1}-\frac{1}{s+3}=\frac{2}{(s+1)(s+3)}=\frac{1}{\frac{1}{2}(s^2+4s+3)}=\frac{1}{\frac{1}{2}s^2+2s+\frac{3}{2}}\]

Cool! \(a=\frac{1}{2}\) , \(b=2\) , and \(a=\frac{3}{2}\).

For fun we could solve \(\frac{1}{2}\ddot{v}+2\dot{v}+\frac{3}{2}s=u(t)\)

Using Laplace transforms with rest conditions \(v(0)=0\) and \(\dot{v}(0)=0\)

\[\hat{v}(s)=\frac{1}{s}\frac{1}{\frac{1}{2}s^2+2s+\frac{3}{2}}=\frac{2}{s(s+1)(s+3)}\]

Applying partial fractions

\[\hat{v}(s)=\frac{2/3}{s}+\frac{-1}{s+1}+\frac{1/3}{s+3}\]

Finally

\[v(t)=\frac{2}{3}-e^{-t}+\frac{1}{3}e^{-3t}\]

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