# 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

Question
Laplace transform
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

2020-10-29
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 :
$$L[f*g]=\frac{1}{5}\left[\frac{s}{(s^2+4)}\right]+\frac{2}{5}\left[\frac{2}{(s^2+4)}\right]+\frac{1}{5}\left[\frac{1}{(s-1)}\right]$$

### Relevant Questions

Use the table of Laplace transform and properties to obtain the Laplace transform of the following functions. Specify which transform pair or property is used and write in the simplest form.
a) $$x(t)=\cos(3t)$$
b)$$y(t)=t \cos(3t)$$
c) $$z(t)=e^{-2t}\left[t \cos (3t)\right]$$
d) $$x(t)=3 \cos(2t)+5 \sin(8t)$$
e) $$y(t)=t^3+3t^2$$
f) $$z(t)=t^4e^{-2t}$$
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
c)$$f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)$$
Find the Laplace transform of the function $$L\left\{f^{(9)}(t)\right\}$$
Use the Laplace transform to solve the following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
Use Theorem 7.4.3 to find the Laplace transform F(s) of the given periodic function.
F(s)=?
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
Let $$y(t)=\int_0^tf(t)dt$$ If the Laplace transform of y(t) is given $$Y(s)=\frac{19}{(s^2+25)}$$ , find f(t)
a) $$f(t)=19 \sin(5t)$$
b) none
c) $$f(t)=6 \sin(2t)$$
d) $$f(t)=20 \cos(6t)$$
e) $$f(t)=19 \cos(5t)$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$