Question

Solve both
a)using the integral definition , find the convolution
\(f*g \text{ of } f(t)=\cos 2t , g(t)=e^t\)
b) Using above answer , find the Laplace Transform of f*g

Answer 1

Step 1
Given:
The functions \(f(t)=\cos 2t , g(t)=e^t\)
Step 2
a) Definition of convolution:
The convolution of piecewise continuous functions f, g : \(\mathbb{R}\rightarrow\mathbb{R}\) is the function f * g : \(\mathbb{R}\rightarrow\mathbb{R}\) given by:
\((f*g)(t)=\int_0^tf(\tau)g(t-\tau)d \tau\)
Therefore, by definition
\((f*g)(t)=\int_0^t\cos(2\tau)e^{t-\tau}d \tau\) Integrate using integration by parts:
\(u=\cos(2\tau) , dv=e^(t-\tau)d \tau\)
\(du =-2\sin(2\tau)d \tau ,v = \int e^{t-\tau}d \tau = -e^{t-\tau} \int u dv= uv- \int v du\)
\(\Rightarrow I= \int \cos(2\tau)e^{t-\tau}d \tau = -cos(2\tau)e^{t-\tau} - \int (-e^{t-\tau})(-2\sin(2\tau)d \tau)\)
\(\Rightarrow I= -\cos(2\tau)e^{t-\tau}-2\int(e^{t-\tau})(\sin(2\tau))d \tau \dots(1)\)
Step 3
To simplify further, use parts of integration for the second term Compute the integral
\(\int e^{t-\tau}\sin(2\tau)d \tau\)
\(u=\sin(2\tau) , dv=e^{t-\tau}d \tau\)
\(du=2\cos(2\tau) , v=-e^{t-\tau}\)
\(\Rightarrow \int e^{t-\tau} \sin(2\tau)d \tau = -e^{t-\tau}\sin(2\tau)-\int(-e^{t-\tau})(2\cos(2\tau))d \tau\)
\(= -e^{t-\tau}\sin(2\tau)+2\int(e^{t-\tau})(\cos(2\tau))d \tau\)
\(\Rightarrow \int e^{t-\tau} \sin(2\tau)d \tau = -e^{t-\tau}\sin(2\tau) +2I\)
\(\left\{\text{Because } I=\int(e^{t-\tau})(\cos(2\tau))d \tau\right\}\)
Equation (1) becomes:
\(\Rightarrow I=-\cos(2\tau)e^{t-\tau}-2[-e^{t-\tau}\sin(2\tau)+2I]\)
\(\Rightarrow I=-\cos(2\tau)e^{t-\tau}+2e^{t-\tau}\sin(2\tau)-4I\)
\(\Rightarrow 5I=-\cos(2\tau)e^{t-\tau}+2e^{t-\tau}\sin(2\tau)\)
\(\Rightarrow I=-\frac{1}{5}\cos(2\tau)e^{t-\tau}+\frac{2}{5}e^{t-\tau}\sin(2\tau)\)
\(\int \cos(2\tau)e^{t-\tau}d \tau = -\frac{1}{5} \cos(2\tau)e^{t-\tau} + \frac{2}{5} e^{t-\tau}\sin(2\tau)\)
Step 4
Substitute the upper and lower limit and simplify
\(\int_0^t \cos(2\tau)e^{t-\tau}d \tau = [-\frac{1}{5}\cos(2\tau)e^{t-\tau}+\frac{2}{5}e^{t-\tau}\sin(2\tau)]_0^t\)
\(=[-\frac{1}{5}\cos(2t)e^{t-t}+\frac{2}{5}e^{t-t}\sin(2t)+\frac{1}{5}\cos(2(0))e^{t-0}-\frac{2}{5}e^{t-0}-\frac{2}{5}e^{t-0}\sin(2(0))]\)
\(=[-\frac{1}{5}\cos(2t)+\frac{2}{5}\sin(2t)+\frac{1}{5}e^t]\)
\(\text{Therefore, }\)
\(\text{The convolution } f * g is:\)
\((f*g)(t)=-\frac{1}{5}\cos(2t)+\frac{2}{5}\sin(2t)+\frac{1}{5}e^t\)
Step 5
b) To find the Laplace transformation of convolution, use the formula:
\(L[f*g]=L[f]L[g]=L[g*f]\)
\(L[f*g]=\int_0^\infty e^(-s \tau) (f*g)(\tau)d \tau\)
\(L[f*g]=L[-\frac{1}{5}\cos(2t)+\frac{2}{5}\sin(2t)+\frac{1}{5}e^t]\)
Use Laplace transformation properties of sum and constant : \(L[f*g]=-\frac{1}{5}L[\cos(2t)]+\frac{2}{5}L[\sin(2t)]+\frac{1}{5}L[e^t]\)
Now, use the basic Laplace transformation formula:
\(L[\cos(at)]=\frac{s}{s^2+a^2} , L[\sin(at)]=\frac{a}{s^2+a^2} , L[e^{at}]=\frac{1}{(s-a)}\)
\(L[f*g]=\frac{1}{5}\left[\frac{s}{(s^2+4)}\right]+\frac{2}{5}\left[\frac{2}{(s^2+4)}\right]+\frac{1}{5}\left[\frac{1}{(s-1)}\right]\)
Therefore, The Laplace transformation of f*g is :
\(L[f*g]=\frac{1}{5}\left[\frac{s}{(s^2+4)}\right]+\frac{2}{5}\left[\frac{2}{(s^2+4)}\right]+\frac{1}{5}\left[\frac{1}{(s-1)}\right]\)