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 :

\(L[f*g]=\frac{1}{5}\left[\frac{s}{(s^2+4)}\right]+\frac{2}{5}\left[\frac{2}{(s^2+4)}\right]+\frac{1}{5}\left[\frac{1}{(s-1)}\right]\)

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 :

\(L[f*g]=\frac{1}{5}\left[\frac{s}{(s^2+4)}\right]+\frac{2}{5}\left[\frac{2}{(s^2+4)}\right]+\frac{1}{5}\left[\frac{1}{(s-1)}\right]\)