# Given the function begin{cases}e^{-t}& text{if } 0leq t<2 0&text{if } 2leq tend{cases} Express f(t) in terms of the shifted unit step function u(t -a) F(t) - ? Now find the Laplace transform F(s) of f(t) F(s) - ?

Question
Laplace transform
Given the function $$\begin{cases}e^{-t}& \text{if } 0\leq t<2\\ 0&\text{if } 2\leq t\end{cases}$$</span>
Express f(t) in terms of the shifted unit step function u(t -a)
F(t) - ?
Now find the Laplace transform F(s) of f(t)
F(s) - ?

2020-11-09
Step 1
Since f(t)=0 for $$2\leq t \text{ and } t<0$$</span> so create step function: $$u(t)-u(t-2)$$ which will give those zero values.
Thus, $$f(t)=e^{-t}[u(t)-u(t-2)] \text{ or } f(t)=e^{-t}u(t)-e^{-t}u(t-2)$$
Step 2
Calculate Laplace transform:
$$F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt$$
$$=\int_{-\infty}^\infty(e^{-t}u(t)-e^{-t}u(t-2))e^{-st}dt$$
$$=\int_{-\infty}^\infty(e^{-t}u(t))e^{-st}dt - \int_{-\infty}^\infty(e^{-t}u(t-2))e^{-st}dt$$
$$=\int_0^\infty e^{-(s+1)t}dt - \int_2^\infty e^{-(s+1)t}dt$$
$$=\frac{1}{(s+1)}- \frac{e^{-2(s+1)}}{(s+1)}$$
$$=\frac{1-e^{-2(s+1)}}{(s+1)}$$
Step 3
Thus, Laplace transform is $$F(s)=\frac{1-e^{-2(s+1)}}{(s+1)}$$

### Relevant Questions

Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
$$x"+8x'+20x=f(t)$$
$$x(0)=-3$$
$$x'(0)=5$$
$$\text{where the forcing } f(t) \text{ is given by }$$
$$f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}$$
Solve the following differential equations using the Laplace transform and the unit step function
$$y"+4y=g(t)$$
$$y(0)=-1$$
$$y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}$$
$$y"-y=g(t)$$
$$y(0)=1$$
$$y'(0)=2 , \text{ where } g(t)=\begin{cases}1 &, t\leq 3\\t & ,t > 3\end{cases}$$
The function
$$\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}$$
has the following Laplace transform,
$$L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt$$
True or False
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$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}$$
Use the definition of Laplace Transforms to find $$L\left\{f(t)\right\}$$
$$f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}$$
Then rewrite f(t) as a sum of step functions, $$u_c(t)$$, and show that by taking Laplace transforms, this yields the same answer as your direct computation.
Part II
29.[Poles] (a) For each of the pole diagrams below:
(i) Describe common features of all functions f(t) whose Laplace transforms have the given pole diagram.
(ii) Write down two examples of such f(t) and F(s).
The diagrams are: $$(1) {1,i,-i}. (2) {-1+4i,-1-4i}. (3) {-1}. (4)$$ The empty diagram.
(b) A mechanical system is discovered during an archaeological dig in Ethiopia. Rather than break it open, the investigators subjected it to a unit impulse. It was found that the motion of the system in response to the unit impulse is given by $$w(t) = u(t)e^{-\frac{t}{2}} \sin(\frac{3t}{2})$$
(i) What is the characteristic polynomial of the system? What is the transfer function W(s)?
(ii) Sketch the pole diagram of the system.
(ii) The team wants to transport this artifact to a museum. They know that vibrations from the truck that moves it result in vibrations of the system. They hope to avoid circular frequencies to which the system response has the greatest amplitude. What frequency should they avoid?
Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$$F(s)=\int_0^\infty e^{-st}f(t)dt$$
where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^{-t}, the following improper integral is evaluated using integration by parts:
$$F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{(s+1)}$$
Verify the following Laplace transforms, where u is a real number.
$$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$
$$\begin{cases}t & 0,4\leq t<\infty \\0 & 4\leq t<\infty \end{cases}$$
$$L\left\{f(t)\right\} - ?$$
$$f(t)=\sin \pi t \text{ if } 2\leq t\leq3 ,$$
$$f(t)=0 \text{ if } t<2 \text{ or if } t>3$$
$$g(t)=\begin{cases}0 & 0 \(g(t)=\prod_{1,3}(t)+6\prod_{3,5}(t)+4u(t-5)$$