Question

SOLVE WHAT IS BELOW FULLY AND CORRECTLY SHOWING THE WORK AND I WILL GIVE YOU THUMBS UP.

a) sketch the graph of the given function on the interval t ≥ 0.

8(t) = uy(t) + 2из(t) — бид(t)

b) Sketch the graph of the given function and express f ( t ) in terms of the unit step function

c)Sketch the graph of the given function and express f ( t ) in terms of the unit step function

d) find the Laplace transform of the given function

e) find the Laplace transform of the given function

Answer 1

(a) Let us first understand what each component of g(t) means. We have

$\small u_1(t)=\begin{cases} 0 & \text{ if } t<1 \\ 1 & \text{ if } t\geq 1 \end{cases}$

2u3(t) if t <3 if t 2 تت

$\small -6u_3(t)=\begin{cases} 0 & \text{ if } t<4 \\ -6 & \text{ if } t\geq 4 \end{cases}$

Thus, g(t) becomes

$\small g(t)=\begin{cases} 0+0+0 & \text{ if } 0\leq t<1 \\ 1+0+0 & \text{ if } 1\leq t <3 \\ 1+2+0 & \text{ if } 3\leq t < 4 \\ 1+2-6 & \text{ if } t\geq 4 \end{cases}=\begin{cases} 0& \text{ if } 0\leq t<1 \\ 1 & \text{ if } 1\leq t <3 \\ 3 & \text{ if } 3\leq t < 4 \\ -3 & \text{ if } t\geq 4 \end{cases}$

Now, we can simply sketch g(t) as a piecewise function, and we get:

44 24 0 -2+ -44

this completes part (a)

(b) We have the function

$\small f(t)}=\begin{cases} 0& \text{ if } 0\leq t<3 \\ -2 & \text{ if } 3\leq t <5 \\ 2 & \text{ if } 5\leq t < 7 \\ 1 & \text{ if } t\geq 7 \end{cases}$

Sketching the graph of the function we have

0 -24 24 - 100

Now, we write f(t) using the unit step function u_c(t) as

$\small f(t)=0 [u_0(t)-u_3(t)]-2[u_3(t)-u_5(t)]+2[u_5(t)-u_7(t)]+1[u_7(t)]$

Opening the brackets and simplifying we get

$\small \Rightarrow f(t)=-2u_3(t)+4u_5(t)-u_7(t)$

which is the required answer.

(c) Now we have

$\small f(t)=\begin{cases} 1 & \text{ if } 0\leq t <2 \\ e^{-(t-2)} & \text{ if } t\geq 2 \end{cases}$

Sketching the graph of the function we have

N 1 0 3 N

Now, we write f(t) using the unit step function u_c(t) as

$\small f(t)=1[u_0(t)-u_2(t)]+e^{-(t-2)}[u_2(t)]$

Opening brackets and simplifying we have

$\small \Rightarrow f(t)=u_0(t)+(e^{-(t-2)}-1)u_2(t)$

which is the required answer

(d) To find the Laplace transform, first we need to write the function using the unit step function, as

$\small f(t)=(t-2)^2u_2(t)$

Now, we use the following formula to calculate the Laplace transform:

$\small \frak L \{u_c(t)f(t)\}=e^{-cs}\frak L \{f(t+c)\}$

So, we have

$\small \frak L \{(t-2)^2u_2(t)\}=e^{-2s}\frak L \{((t+2)-2)^2\}=e^{-2s}\frak L \{t^2\}=e^{-2s}\cdot {2\over s^3}$

Therefore our answer is

$\small \frak L \{f(t)\}={2\over s^3}e^{-2s}$

(e) We find the Laplace transform in a similar way, so

$\small \frak L \{f(t)\}\\ \\ =\frak L \{(t-3)u_2(t)\}-\frak L \{(t-2)u_3(t)\} \\ \\ =e^{-2s}\frak L \{t+2-3\}-e^{-3s}\frak {L} \{t+3-2\} \\ \\ =e^{-2s}\frak L \{t-1\}-e^{-3s}\frak {L} \{t+1\} \\ \\= e^{-2s}\left ( {1\over s^2}-{1\over s} \right )-e^{-3s}\left ({1\over s^2}+{1\over s} \right ) \\ \\ \\ = {e^{-2s}-e^{-3s}\over s^2}-{e^{-2s}+e^{-3s}\over s}$

which is the required answer