Question

Laplace {sint unit step (t-pi/2)}

Answer 1

By the definition of the transform,\[\mathscr{L}\left\{\sin t~u\left(t-\frac{\pi}{2}\right)\right\}=\int_0^\infty \sin t~u\left(t-\frac{\pi}{2}\right)e^{-st}~dt\]
By definition of the unit step function,
\[u\left(t-\frac{\pi}{2}\right)=\begin{cases}1&\text{for }t<\frac{\pi}{2}\\ \\
0&\text{for }t\ge\frac{\pi}{2}
\end{cases}\]
So, the integral is
\[\begin{align*}\int_0^\infty \sin t~u\left(t-\frac{\pi}{2}\right)e^{-st}~dt
&=\int_0^{\pi/2} \sin t~u\left(t-\frac{\pi}{2}\right)e^{-st}~dt\\
&~~~~~~~+\int_{\pi/2}^\infty \sin t~u\left(t-\frac{\pi}{2}\right)e^{-st}~dt\\
&=\int_0^{\pi/2} \sin t~(0)e^{-st}~dt+\int_{\pi/2}^\infty \sin t~(1)e^{-st}~dt\\
&=\int_{\pi/2}^\infty \sin t~e^{-st}~dt
\end{align*}\]
Integrate by parts!

Answer 2

but it will be better with the form {f(t-a) U (t-a)}

Answer 3

sorry

Answer 4

What I got is \[(\frac{ 1 }{ s^2+1 }+\frac{ 1 }{ s })e^{-\frac{ pis }{ 2 }}\]

Answer 5

is that correct?

Answer 6

Just a moment, I'm writing it out on paper to check.

Answer 7

Okay, so I've finally got the answer. Your answer isn't correct. The \(\dfrac{1}{s}\) shouldn't be there.

You should have
\[e^{\frac{\pi}{2}s}\frac{s}{s^2+1}\]
Notice that \(\dfrac{s}{s^2+1}\) is the transform of \(\cos t\). But, \(\sin\left(t-\dfrac{\pi}{2}\right)=\cos t\).

Answer 8

Thank you!!! But... I do not understand why (t−π 2 )=cost

Answer 9

I mean... you don't distribute t and π/2???

Answer 10

This is what i did:

Answer 11

This is what i did:\[a=\frac{ \pi }{ 2 }\] and \[f(t-a)=sint\]

Answer 12

This is what i did:\[a=\frac{ \pi }{ 2 }\] and \[f(t-a)=sint\] \[f(t)=\sin (t+a)\]

Answer 13

\[f(t)=sint + \sin \frac{ \pi }{ 2 }\]

Answer 14

\[f(t)= sint+1\]\[f(t)=sint + \sin \frac{ \pi }{ 2 }\]

Answer 15

Sorry, just split the last two answers
I continue:

Answer 16

\[F(s)=\frac{ 1 }{ s^2+1 }+\frac{ 1 }{ s }\]

Shall you please explain me what is wrong in my procedure???
In the other hand, I have another transformation:
Laplace {(Cos2t) U (t-pi)}

and my answer is

\[(\frac{ s }{ s^2+4 }+\frac{ 1 }{ s })\[e^{-\pi s}\]

I did it with the same procedure I explained you before.
Following your answer, I guess de 1/s is wronsg, but.... why???

You´re a master for this, do you mind to help me with some other transformations???
I understand most of this calculus, but saddly these last technics are not so clear for me.

Yesterday, I was trying to solve
Laplace ^-1 \[{\frac{ (1+e^{-2s})^{2}} }{ s+2 }\]

or {(1+e^-2s)^2}/(s+2)

And honestly, I do not know either how to begin.

Excuse me for annoying you with this but it seems you are the only one can help me. My apologies for my English, I´m from Colombia...

Answer 17

The problem is that
\[\sin(a+b)\not=\sin(a)+\sin(b)\]
(Distribution doesn't work.)

However,
\[\sin(a\pm b)=\sin(a)\cos(b)\pm\cos(b)\sin(a)\]
So,
\[\begin{align*}\sin\left(t\pm\frac{\pi}{2}\right)&=\sin(t)\cos\left(\frac{\pi}{2}\right)+\cos(t)\sin\left(\frac{\pi}{2}\right)\\
&=0+\cos(t)\\
&=\cos t
\end{align*}\]

Answer 18

And now your next question is
\[\mathscr{L}^{-1}\left\{\frac{1+e^{-2s}}{s+2}\right\}\]
First, break up the fraction so you have
\[\mathscr{L}^{-1}\left\{\frac{1}{s+2}+\frac{e^{-2s}}{s+2}\right\}\]
The normal and inverse transform operators are linear, which means that \(\mathscr{L}^{-1}\left\{f(t)+g(t)\right\}=\mathscr{L}^{-1}\left\{f(t)\right\}+\mathscr{L}^{-1}\left\{g(t)\right\}\)

So you then have
\[\mathscr{L}^{-1}\left\{\frac{1}{s+2}\right\}+\mathscr{L}^{-1}\left\{\frac{e^{-2s}}{s+2}\right\}\]
The first term should be fairly simple: \(e^{-2t}\). The second is somewhat different; it's the product of an exponential function and the transform of \(e^{-2t}\).

When you're given a transform of the form \(e^{-cs}F(s)\), its inverse transform would be \(u(t-c)f(t-c)\). Here, \(c=-2\) and \(f(t)=e^{-2t}\), so \(f(t-c)=e^{-2(t-2)}\). So,
\[\mathscr{L}^{-1}\left\{\frac{e^{-2s}}{s+2}\right\}=e^{-2(t-2)} u(t-2)\]
And don't worry about bothering - happy to help.