Question

show that \[\mathcal L \lbrace \cos kt \rbrace = \frac{s}{s^2+k^2}; \; \text{for} \; s >0\]

what does this mean?

Answer 1

You have to show that
\[
\int_0^\infty e^{- s t} \cos( k t)dt=\frac {s}{s^2 + k^2}
\]

Answer 2

what does s>0 mean?

Answer 3

Try to write
\[
\cos(k t) =\frac 1 2 (e^{i k t} + e^{-i kt})
\]

Answer 4

where did that come from?

Answer 5

De Moivres Formula

Answer 6

hmm...we werent taught that...i think

Answer 7

The Laplace transform of a function \(f(x)\) is defined as\[\mathcal L\{f(t)\}=\int_0^\infty e^{-st}f(t)dt~;~~~~~s>0\]if we did not put the restraint that \(s>0\) our integral would usually not converge (it would be line ignoring the negative sign)

Answer 8

or you can use
http://www.wolframalpha.com/input/?i=integrate+e^%28+a+x%29++cos%28+b+x%29

Answer 9

since all you asked was what it meant that's all I answered
@eliassaab has shown you how to find your particular transform by using the definition

Answer 10

I usually resort to using a Laplace table....

Answer 11

May be they want to see how to get this value in the table.

Answer 12

oh so s>0 doesnt really matter much in the solutionn?

Answer 13

If s <0, the integral will not converge.

Answer 14

yeah i meant it's just a condition to ensure that it converges

Answer 15

so how will i prove this? withoout the use of that de moivres thingy...coz we werent taught that

Answer 16

You have to do integration by parts twice.

Answer 17

ok i'll try

Answer 18

http://www.youtube.com/watch?v=KTerp411MuM

Answer 19

what is \[\int e^{-st}?\]is it just \[-\frac{e^{-st}}{s}\]

Answer 20

Yes

Answer 21

thanks..wanted to make sure s was a onstant

Answer 22

constant*

Answer 23

yes it is a constant.

Answer 24

so i got to \[\large \int e^{-st} \cos kt dt + \frac{1}{s^2} \int e^{-st} \cos kt dt= -\frac{e^{-st} \cos kt}{s} +\frac{e^{-st} \sin kt - 1}{s^2} \]
so how do i combine that?

Answer 25

you didn't do it quite right

Answer 26

aww

Answer 27

\[\int e^{-st}\cos(kt)dt\]\[u=e^{-st}\implies du=-se^{-st}dt\]\[dv=\cos(kt)dt\implies v=\frac1k\sin(kt)\]so\[\int udv=uv-\int vdu\]\[=\frac1ke^{-st}\sin(kt)+\frac sk\int e^{-st}\sin(kt)dt\]now just do the second integral...

Answer 28

wait...shouldnt you use trig functions instead of exponential as u? acording to LIATE

Answer 29

I don't see why it makes a difference, should work both ways...

Answer 30

just thinking about it... let's see

Answer 31

\[\int e^{-st}\sin(kt)dt=-\frac1ke^{-st}\cos(kt)+\frac sk\int e^{-st}\cos(kt)dt\]from now on let's just call the integral \(I\)
so we get\[I=\frac1ke^{-st}\sin(kt)+\frac sk\left(-\frac1ke^{-st}\cos(kt)+\frac skI\right)\]solve for \(I\)

Answer 32

\[I=\frac1ke^{-st}\sin(kt)-\frac s{k^2}e^{-st}\cos(kt)+\frac {s^2}{k^2}I\]\[(1-\frac{s^2}{k^2})I=\frac1ke^{-st}\sin(kt)-\frac s{k^2}e^{-st}\cos(kt)\]\[(k^2-s^2)I=e^{-st}[k\sin(kt)-s\cos(kt)]\]\[I={e^{-st}[k\sin(kt)-s\cos(kt)]\over(k^2-s^2)}\]hmm I seem to be missing something
let me check my notes...

Answer 33

ohh i see what i did wrong now...i didnt put a k when i did my derivatives

Answer 34

when you evaluate the limit in the improper integral as \(t\to0\) it should be what you have (clearly at least the \(e^{-st}\) will disappear) but I must have made a mistake

try again your way

Answer 35

yup i got the same as yours now

Answer 36

so what's the plan how to make it look like \[\frac{s}{s^2 + k^2}\]
substitute in the limits?

Answer 37

I should do it again, taking each limit each time, but I have to take a break first...

Answer 38

i kknow e^-infty is jjust 0 so i guess i'll just jump down to the good stuff
\[\huge -\frac{e^0 (k\sin (0) - s\cos(0))}{s^2 + k^2}\]
\[\huge \implies -\frac{-s}{s^2+k^2}\]
\[\huge \implies \frac{s}{s^2 + k^2}\]
nice

Answer 39

yep, I somehow had s^2-k^2 in mine
I must have made a sign error somewhere
nice job

Answer 40

if you just added a negative to turn \(s^2-k^2\) into \(s^2+k^2\) that's a mistake though you realize?

Answer 41

prolly in the second integral

Answer 42

nahh i did what i did earlier...i just snuck in a k

Answer 43

really you should redo the problem, taking the limit every time you find v in the integration by parts steps, not waiting until the end

Answer 44

ok I trust your ability to check your own work

Answer 45

what do you mean "taking the limit everytime you find v" and "waiting until the end"

Answer 46

well my solution was something like\[\int e^{-st} \cos kt = -\frac{e^{-st}\cos kt}{s} - \frac ks \left[-\frac{e^{-st}\sin kt}{s} + \frac ks\int e^{-st} \cos ktdt\right]\]
\[\int e^{-st} \cos kt dt + \frac{k^2}{s^2} \int e^{-st} \cos kt dt = -\frac{e^{-st} \cos kt}{s} + \frac{ke^{-st}\sin kt }{s^2}\]
etc.

Answer 47

you should really evaluate the integral every time you take it along the way\[\int e^{-st} \cos kt dt= -\left.\frac{e^{-st}\cos kt}{s}\right|_0^\infty - \frac ks \left.\left[\left(-\frac{e^{-st}\sin kt}{s}\right)\right|_0^\infty + \frac ks\int e^{-st} \cos ktdt\right]\]this is what I mean by that...

Answer 48

oh..but isnt it same if you just do the whole integral then simplify then evaluate integral?

Answer 49

not always, though you may be able to get away with it sometimes (maybe even most if the time)

Answer 50

it's one of those little details that mathematicians like Zarkon can make sound like a crime, but probably don't affect us more engineer-minded types

Answer 51

evaluating you get\[I= -\lim_{n\to\infty}\left.\frac{e^{-st}\cos kt}{s}\right|_0^n - \frac ks \left.\left[\lim_{n\to\infty}\left(-\frac{e^{-st}\sin kt}{s}\right)\right|_0^n + \frac ksI\right]\]

Answer 52

evaluating you get\[I= \lim_{n\to\infty}\left.\frac{e^{-st}\cos kt}{s}\right|_n^0 - \frac ks \left.\left[\lim_{n\to\infty}\left(\frac{e^{-st}\sin kt}{s}\right)\right|_n^0 + \frac ksI\right]\]

Answer 53

\[I= \frac1s-\frac1s\lim_{n\to\infty}e^{-sn}\cos kn- \frac k{s^2} \left[0-\lim_{n\to\infty}\left(e^{-sn}\sin kn\right) + kI\right]\]

Answer 54

are you following this still?

Answer 55

Using the method I suggested before:

\[
e^{ - s t}\cos(k t) =\frac 1 2 (e^{ (i k-s) t} + e^{- (i k+s)t})\\
\int_0^\infty e^{-i s t}\cos(k t) =\frac 1 2 (\int_0^\infty e^{ (i k-s) t}dt + \int_0^\infty e^{{- (i k+s)t}})dt=\\
\frac 1 2\left[ \frac{e^{ (i k-s) t}}{i k-s } \right]_{t=0}^{\infty}-\frac 1 2\left[ \frac{e^{- (i k+s)t}}{i k+s)} \right]_{t=0}^{\infty}=\\
-\frac 1 2 \frac 1 {i k-s} + \frac 1 2 \frac 1 {i k+s} =\\
\frac{s}{(k-i s) (k+i s)}=
\frac s{k^2+s^2}


\]

Answer 56

...or continuing from my last post, notice that the limits are zero (this should be obvious if you consider that sin and cos can only vary between -1 and 1 while e^(-st) gets smaller and smaller, but can be easily proven with either de moivre or l'hospital)\[I= \frac1s-\frac1s\cancel{\lim_{n\to\infty}e^{-sn}\cos (kn)}^{\huge0}- \frac k{s^2} \left[0-\cancel{\lim_{n\to\infty}\left(e^{-sn}\sin (kn)\right)}^{\huge0} + kI\right]\]\[I=\frac1s-\frac{k^2}{s^2}I\implies(1+\frac{k^2}{s^2})I=\frac1s\implies\left(s^2+k^2\over s^2\right)I=\frac1s\]\[I=\int_0^\infty e^{-st} \cos( kt)dt={s\over s^2+k^2}=\mathcal L\{\cos (kt)\} \]