Question

Hi everyone! Can someone please show me steps to solve L{sintU(t-(pi/2))} ? We need to use 2nd Translation/Shifting Theorem. Thanks :o)

Answer 1

What is the theorem you need to use?

Answer 2

2nd Translation/Shifting Theorem

Answer 3

What does that theorem say?

Answer 4

one sec...my keyboard is out of batteries and I am clicking all these letters one by one...

Answer 5

alright

Answer 6

I looked it up and don't know it yet. Sorry.

Answer 7

L{f(t-a)U(t-a)} = e^(-as)F(s)

Answer 8

i know a=pi/2

Answer 9

I honestly don't know, but I just started watching this youtube video to see if I can figure it out https://www.youtube.com/watch?v=ns2F_WPZxvQ

Answer 10

Interesting at about 3 minutes in he says that if the two functions aren't shifted by the same amount you can't do it directly, but it looks like we can rewrite

\(\sin(t) = \cos (t- \pi/2)\)

So that looks like our first step to plug that in!

Answer 11

Hi everyone! Can someone please show me steps to solve L{sintU(t-(pi/2))} ? We need to use 2nd Translation/Shifting Theorem. Thanks :o)

$$\begin{align*}\mathscr{L}\left\{\sin(t)\mathscr{U}\left(t-\frac\pi2\right)\right\}&=\mathscr{L}\left\{\sin\left(\left(t-\frac\pi2\right)+\frac\pi2\right)\mathscr{U}\left(t-\frac\pi2\right)\right\}\\&=e^{-\frac\pi2s}\mathscr{L}\{\sin\left(t+\frac\pi2\right)\}\\&=e^{-\frac\pi2s}\mathscr{L}\{\cos(t)\}\\&=e^{-\frac\pi2s}\cdot\frac{s}{s^2+1}\end{align*}$$

Answer 12

using the fact that \(\sin(t+\pi/2)=\cos(t)\)