Question: Establish
From definition f\left ( t\right ) \circledast g\left ( t\right ) =\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g\left ( \tau \right ) d\tau
From definition f\left ( t\right ) \circledast \left ( g_{1}\left ( t\right ) +g_{2}\left ( t\right ) \right ) =\int _{-\infty }^{\infty }f\left ( t-\tau \right ) \left ( g_{1}\left ( \tau \right ) +g_{1}\left ( \tau \right ) \right ) d\tau
From definition\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }\left ( f\circledast g\right ) \left ( \tau \right ) h\left ( t-\tau \right ) d\tau \\ & =\int _{\Re }\left [ \int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) d\tau _{1}\right ] h\left ( t-\tau \right ) d\tau \\ & =\int _{\Re }\int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau _{1}d\tau \end{align*}
By Fubini, we can change order of integration\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }\int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau d\tau _{1}\\ & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau \right ] d\tau _{1} \end{align*}
By translation, if we add \tau _{1} to \tau for both functions in the inner integral above, we obtain\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \left ( \tau +\tau _{1}\right ) -\tau _{1}\right ) h\left ( t-\left ( \tau +\tau _{1}\right ) \right ) d\tau \right ] d\tau _{1}\\ & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \tau \right ) h\left ( \left ( t-\tau _{1}\right ) -\tau \right ) d\tau \right ] d\tau _{1} \end{align*}
But now we see that inner integral is \int _{\Re }g\left ( \tau \right ) h\left ( \left ( t-\tau _{1}\right ) -\tau \right ) d\tau =\left ( g\circledast h\right ) \left ( t-\tau _{1}\right ) , hence the above becomes\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }f\left ( \tau _{1}\right ) \left ( g\circledast h\right ) \left ( t-\tau _{1}\right ) d\tau _{1}\\ & =\left ( f\circledast \left ( g\circledast h\right ) \right ) \left ( t\right ) \end{align*}
QED
Find an example showing \left ( f\circledast 1\right ) \left ( t\right ) need not be equal to f\left ( t\right )
Solution Let f\left ( t\right ) =e^{t}, hence\begin{align*} \left ( f\circledast 1\right ) \left ( t\right ) & =\int _{0}^{t}f\left ( t-\tau \right ) \times 1d\tau \\ & =\int _{0}^{t}e^{\left ( t-\tau \right ) }d\tau \\ & =\left [ \frac{e^{\left ( t-\tau \right ) }}{-1}\right ] _{\tau =0}^{\tau =t}\\ & =-\left [ e^{\left ( t-t\right ) }-e^{\left ( t-0\right ) }\right ] \\ & =-\left [ e^{0}-e^{t}\right ] \\ & =-\left ( 1-e^{t}\right ) \\ & =e^{t}-1 \end{align*}
Which is not the same as e^{t}. QED
Show that \left ( f\circledast f\right ) \left ( t\right ) is not necessarily non-negative, using f\left ( t\right ) =\sin \left ( t\right )
Solution From definition \left ( f\circledast f\right ) \left ( t\right ) =\int _{0}^{t}\sin \left ( \tau \right ) \sin \left ( t-\tau \right ) d\tau
For the second integral above, since it is w.r.t \tau , then we can pull \cos \left ( t\right ) outside, which gives\begin{align*} \left ( f\circledast f\right ) \left ( t\right ) & =\frac{1}{2}\left ( \frac{\sin \left ( 2\tau -t\right ) }{2}\right ) _{\tau =0}^{\tau =t}-\frac{1}{2}\cos \left ( t\right ) \int _{0}^{t}d\tau \\ & =\frac{1}{4}\left ( \sin \left ( 2t-t\right ) -\sin \left ( -t\right ) \right ) -\frac{1}{2}t\cos t\\ & =\frac{1}{4}\left ( \sin \left ( t\right ) +\sin \left ( t\right ) \right ) -\frac{1}{2}t\cos t\\ & =\frac{1}{2}\sin t-\frac{1}{2}t\cos t \end{align*}
Let t=2\pi then\begin{align*} \left ( f\circledast f\right ) \left ( t\right ) & =0-\frac{1}{2}\left ( 2\pi \right ) \\ & =-\pi \end{align*}
Which is negative. Hence we showed that \left ( f\circledast f\right ) \left ( t\right ) can be negative at some t. QED.
Find Laplace transform of f\left ( t\right ) =\int _{0}^{t}\left ( t-\tau \right ) ^{2}\cos \left ( 2\tau \right ) d\tau
Solution We see that f\left ( t\right ) =t^{2}\circledast \cos \left ( 2t\right )
Find Laplace transform of f\left ( t\right ) =\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin \left ( \tau \right ) d\tau
Solution We see that f\left ( t\right ) =e^{-t}\circledast \sin \left ( t\right )
Find Laplace transform of f\left ( t\right ) =\int _{0}^{t}\left ( t-\tau \right ) e^{\tau }d\tau
Solution We see that f\left ( t\right ) =t\circledast e^{t}
Find Laplace transform of f\left ( t\right ) =\int _{0}^{t}\sin \left ( t-\tau \right ) \cos \tau d\tau
Solution We see that f\left ( t\right ) =\sin \left ( t\right ) \circledast \cos \left ( t\right )
Find the inverse Laplace transform of F\left ( s\right ) =\frac{1}{s^{4}\left ( s^{2}+1\right ) } using convolution theorem.
Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{s^{4}}\frac{1}{s^{2}+1}\\ & =\mathcal{L}\left ( \frac{t^{3}}{6}\right ) \mathcal{L}\left ( \sin t\right ) \end{align*}
Hence, using convolution theorem\begin{align*} f\left ( t\right ) & =\frac{t^{3}}{6}\circledast \sin t\\ & =\frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau \end{align*}
Integrate by parts. \int udv=uv-\int vdu. Let u=\left ( t-\tau \right ) ^{3},dv=\sin \tau \rightarrow du=-3\left ( t-\tau \right ) ^{2},v=-\cos \tau , hence\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( -\left [ \left ( t-\tau \right ) ^{3}\cos \tau \right ] _{0}^{t}-\int _{0}^{t}-3\left ( t-\tau \right ) ^{2}\left ( -\cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( -\left [ \left ( t-t\right ) ^{3}\cos t-\left ( t-0\right ) ^{3}\cos 0\right ] -3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( -\left [ 0-t^{3}\right ] -3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \end{align*}
Integrate by parts. Let u=\left ( t-\tau \right ) ^{2},dv=\cos \tau \rightarrow du=-2\left ( t-\tau \right ) ,v=\sin \tau , hence\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( t^{3}-3\left [ \left ( \left ( t-\tau \right ) ^{2}\sin \tau \right ) _{0}^{t}-\int _{0}^{t}-2\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\left [ \left ( \left ( t-t\right ) ^{2}\sin t-\left ( t-0\right ) ^{2}\sin 0\right ) _{0}^{t}+2\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\left [ 0+2\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ) \end{align*}
Integrate by parts. Let u=\left ( t-\tau \right ) ,dv=\sin \tau \rightarrow du=-1,v=-\cos \tau , hence above becomes\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( t^{3}-6\left [ \left ( -\left ( t-\tau \right ) \cos \tau \right ) _{0}^{t}-\int _{0}^{t}\cos \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left [ -\left ( \left ( t-t\right ) \cos t-\left ( t-0\right ) \cos 0\right ) -\left ( \sin \tau \right ) _{0}^{t}\right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left [ -\left ( 0-t\right ) -\sin t\right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left ( t-\sin t\right ) \right ) \\ & =\frac{1}{6}\left ( t^{3}-6t+6\sin t\right ) \end{align*}
Hence f\left ( t\right ) =\frac{1}{6}\left ( t^{3}-6t+6\sin t\right )
Find the inverse Laplace transform of F\left ( s\right ) =\frac{s}{\left ( s+1\right ) \left ( s^{2}+4\right ) } using convolution theorem.
Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{s+1}\frac{s}{s^{2}+4}\\ & =\mathcal{L}\left ( e^{-t}\right ) \mathcal{L}\left ( \cos 2t\right ) \end{align*}
Hence, using convolution theorem\begin{align*} f\left ( t\right ) & =e^{-t}\circledast \cos 2t\\ & =\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau \end{align*}
Integrate by parts. \int udv=uv-\int vdu. Let u=\cos 2\tau ,dv=e^{-\left ( t-\tau \right ) }\rightarrow du=-2\sin 2\tau ,v=e^{-\left ( t-\tau \right ) }, hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau & =\left ( \cos 2\tau e^{-\left ( t-\tau \right ) }\right ) _{0}^{t}-\int _{0}^{t}e^{-\left ( t-\tau \right ) }\left ( -2\sin 2\tau \right ) d\tau \\ & =\left ( \cos 2te^{-\left ( t-t\right ) }-\cos 0e^{-\left ( t-0\right ) }\right ) +2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau d\tau \\ & =\left ( \cos 2t-e^{-t}\right ) +2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau d\tau \end{align*}
Integrate by parts. Let u=\sin 2\tau ,dv=e^{-\left ( t-\tau \right ) }\rightarrow du=2\cos 2\tau ,v=e^{-\left ( t-\tau \right ) }, hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \left ( \sin 2\tau e^{-\left ( t-\tau \right ) }\right ) _{0}^{t}-\int _{0}^{t}e^{-\left ( t-\tau \right ) }2\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \left ( \sin 2te^{-\left ( t-t\right ) }-0\right ) -2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \sin 2t-2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\sin 2t-4\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \end{align*}
Hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau +4\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\cos 2t-e^{-t}+2\sin 2t\\ 5\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\cos 2t-e^{-t}+2\sin 2t\\ \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\frac{1}{5}\left ( \cos 2t-e^{-t}+2\sin 2t\right ) \end{align*}
Therefore f\left ( t\right ) =\frac{1}{5}\left ( \cos 2t-e^{-t}+2\sin 2t\right )
Find the inverse Laplace transform of F\left ( s\right ) =\frac{1}{\left ( s+1\right ) ^{2}\left ( s^{2}+4\right ) } using convolution theorem.
Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{\left ( s+1\right ) ^{2}}\frac{1}{s^{2}+4}\\ & =\mathcal{L}\left ( te^{-t}\right ) \mathcal{L}\left ( \frac{1}{2}\sin 2t\right ) \end{align*}
Hence, using convolution theorem \begin{align*} f\left ( t\right ) & =te^{-t}\circledast \frac{1}{2}\sin 2t\\ & =\frac{1}{2}\int _{0}^{t}\left ( t-\tau \right ) e^{-\left ( t-\tau \right ) }\sin 2\tau \\ & =\frac{1}{2}\int _{0}^{t}te^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau -\frac{1}{2}\int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau \end{align*}
The first integral is \int _{0}^{t}te^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau =t\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau
Now we need to evaluate the second integral \int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau . This can also be done using integration by part. But I used CAS here, the result is \int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau =\frac{1}{25}\left ( -4e^{-t}+\left ( 4-10t\right ) \cos 2t+\left ( 3+5t\right ) \sin 2t\right )
Find the inverse Laplace transform of F\left ( s\right ) =\frac{G\left ( s\right ) }{s^{2}+1} using convolution theorem.
Solution We see that F\left ( s\right ) =G\left ( s\right ) \frac{1}{s^{2}+1}=G\left ( s\right ) \mathcal{L}\left ( \sin t\right )