Question

I'm trying to find the Laplace Transform of a convolution, posted below in a moment.

Answer 1

I'm looking for
\[\mathcal{L}\left\{\vphantom{}t \int\limits_{0}^{t}\sin(\tau)d \tau\right\}\]

Answer 2

I can't figure out what I'm doing wrong, but I immediately do not understand what to do with the t outside of the integral. @satellite73 , could you help possibly on this?

Answer 3

The steps I'm taking are this: \[\mathcal{L}\left\{\vphantom{}\int\limits_{0}^{t}\sin(\tau)d \tau\right\}=\mathcal{L}\left\{1\right\}\mathcal{L}\left\{\sin(\tau)\right\}=\frac{1}{s}.\frac{1}{s^2+1}\]

Answer 4

Hey, @wio . Any chance you could help me with this?

Answer 5

looks like convolution ?

Answer 6

Yeah, it does, but I'm not sure how to do it, that's what I'm asking for help on.

Answer 7

The answer is supposed to be:\[\frac{3s^2+1}{s^2(s^2+1)^2}\]

Answer 8

What I was doing was separating the integrand into the product of two laplace transforms (by the convolution theorem)-but I must be doing that wrong, because my laplace transforms are correct; the separation into two functions F(s) and G(s) must be wrong.

Answer 9

If you say that \(t\) is constant with respect to \(\tau\), then:\[
t\int_0^t\sin\tau\;d\tau = \int_0^t t\sin\tau\;d\tau
\]

Answer 10

\[F(t)=\sin(t); \ \ \ G(t) = \ ?\]\[f(\tau)=\sin(\tau); \ g(t - \tau) = \ ?\]

Answer 11

Yeah, t can be factored in or out, not really sure what to do with it; what is advantageous about factoring t into the integrand?

Answer 12

t isnt constant tho...its a variable we r gonna sub in...

Answer 13

treating it like a constant might screw stuff up..

Answer 14

I'm guessing maybe manipulating the integrand in some manner to get a second function in the from f(t-tau)?

Answer 15

We sub it into \(\tau\).

Answer 16

^

Answer 17

I factored it in because I think \(g(t-\tau)=t\).

Answer 18

I think you're right, one sec!

Answer 19

\[\mathcal{L}\left\{\vphantom{}\int\limits_{0}^{t}t \sin(\tau)d \tau\right\}=\mathcal{L}\left\{t\right\}\mathcal{L}\left\{\sin(\tau)\right\}\]

Answer 20

Whoops, sin(t).

Answer 21

@hahd \[
t\int_0^tf'(x)\;dx = t(f(t)-f(0)) = tf(t)-tf(0) = \int_0^ttf'(x)\;dx
\]

Answer 22

\[=\frac{1}{s^2}.\frac{1}{s^2+1}\]

Answer 23

No dice? Really close to the answer, I'm not sure where I'm doing something wrong.

Answer 24

Wait, maybe since:\[
\frac{d}{dt}tf(t) = tf'(t) +f(t)
\]it would be: \[
\int_0^txf'(x)+f(x)\;dx
\]

Answer 25

No, that doesn't work either.

Answer 26

^ i was thinking this

Answer 27

leibniz rule..

Answer 28

\[
\int_0^txf'(x)+f(x)\;dx = tf(t) - 0f(0) = tf(t)
\]

Answer 29

this is a dum question but is tau like 2pi? or its a variable

Answer 30

lol im really interested now too lol

Answer 31

It's a variable of integration

Answer 32

\[
0<\tau<t
\]

Answer 33

lkk

Answer 34

how do you know that tho?

Answer 35

Limits of integration...

Answer 36

oh gawd kk i get it...

Answer 37

It's just a dummy variable, it's just there to prevent ourselves from confusing up t in the limits and in the integrand.

Answer 38

kk but why doesnt leibniz work here

Answer 39

Well, I don't even understand where you guys are taking a derivative from in the first place.

Answer 40

its given in the form of an integral of a derivative ??? isnt that the natural thing to do...

Answer 41

purpose of the derivative in this case is so I can express the antiderivative.

Answer 42

The laplace transform is an integral operation, and the argument of the Laplace transform is *also* an integral, I don't see where differentiation is coming in at all other than the fact that "d tau" happens to be there, but yeah.

I dunno, I pretty much never look at integrals in the manner we are right now, unfamiliar to me.

Answer 43

@dan815

Answer 44

(Brb in 5 minutes, time on this PC is running out and need to go to the next door campus building to use another PC)

Answer 45

\[\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\mathcal{L}\left\{t\sin(t)+\int\limits_{0}^{t}\sin(\tau)d\tau\right\}=s\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{1}{s}\left[\mathcal{L}\left\{t\sin(t)\right\}+\mathcal{L}\left\{\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\right]=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{1}{s}\left[\frac{2s}{(s^2+1)^2}+\mathcal{L}\left\{\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\right]=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{1}{s}\left[\frac{2s}{(s^2+1)^2}+\frac{\mathcal{L}\left\{\sin(t)\right\}}{s}\right]=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{1}{s}\left[\frac{2s}{(s^2+1)^2}+\frac{\frac{1}{s^2+1}}{s}\right]=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{1}{s}\left[\frac{2s^2}{s(s^2+1)^2}+\frac{s^2+1}{s(s^2+1)^2}\right]=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

\[\frac{3s^2+1}{s^2(s^2+1)^2}=\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}\]

Answer 46

Alright, I'm back.

Answer 47

I'm confused about where the first term on the second LHS from the top came from; I'm confused about what operation was done there at all from the original problem.

Answer 48

@Zarkon (PS, thank you); Just wondering about that first LHS.

Answer 49

@ganeshie8 , any clue how he did this?

Answer 50

e.g. where did\[\mathcal{L}\left\{\vphantom{}t \sin(t)+ \int\limits_{0}^{t}\sin(\tau)d \tau\right\}\] come from?

Answer 51

\[\mathcal{L}\left\{f'(t)\right\}=s\mathcal{L}\left\{f(t)\right\}-f(0)\]

Answer 52

Alright, cool. Thank you. Going to take a look at this for a little longer.

Answer 53

\[s \mathcal{L}\left\{\vphantom{}t \int\limits_{0}^{t}\sin(\tau)d \tau\right\}=s\mathcal{L}\left\{f(t)\right\}\]

Answer 54

Product rule is applied;

Answer 55

\[f(t)=t \int\limits_{0}^{t}\sin(\tau)d \tau; f'(t)=t \sin(t)+ \int\limits_{0}^{t}\sin(\tau)d \tau\]

Answer 56

Alright, cool, I got it. Thanks so much, man, that was a really unique/strange approach to apply the definition of the laplace of a derivative.

Answer 57

even easier it you evaluate the integral

\[\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}=\mathcal{L}\left\{-t\cos(t)+t\right\}=\cdots\]

Answer 58

\[\mathcal{L}\left\{-t \cos(t)+t\right\}=\frac{1}{s^2}-(-1)^{1}\frac{d}{ds}\bigg(\frac{s}{s^2+1}\bigg)\]

Answer 59

\(*\) : convolution

Answer 60

i don't see how to use convolution for the remaining integral hmm

Answer 61

** \[\begin{align}\mathcal{L}\left\{t\int\limits_{0}^{t}\sin(\tau)d\tau\right\}&=\mathcal{L} \left( (t-\tau+\tau)\int\limits_{0}^{t}\sin(\tau)d\tau \right)\\~\\
&=\mathcal{L} \left(\int\limits_{0}^{t} (t-\tau+\tau)\sin(\tau)d\tau \right) \\~\\
&=\mathcal{L} \left(t*\sin(t) \right)+ \mathcal{L}\left(\int\limits_{0}^{t} \tau\sin(\tau)d\tau \right) \\~\\

\end{align}\]

Answer 62

=]

Answer 63

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

Answer 64

Then just finding an LCD and combining like terms as appropriate:\[\frac{(s^2+1)^2}{s^2(s^2+1)^2}-\frac{(s^2)(1-s^2)}{(s^2+1)^2s^2}\]

Answer 65

(Lol how about no)

Answer 66

But seriously, I made an algebraic error somewhere, because it looks like those....yeah, it looks like those...just wait one minute.

Answer 67

nvm, right hand term is positive altogether.

Answer 68

why are you not using wolfram to do the donkey work

Answer 69

\[\frac{s^4+2s^2+1+s^2-s^4}{s^2(s^2+1)^2}=\frac{3s^2+1}{s^2(s^2+1)^2}\]

Answer 70

Cuz' need to be good at the donkey work for my test in six hours..... :'C

Answer 71

I'm pretty much having to rapid-fire since last week Monday, constantly do every problem in every section that this test covers prior to 7 PM tonight.

Answer 72

man u guys are overdoing it
its so easy and u are making it real big :(

Answer 73

i can do it in two steps

Answer 74

wanna see

Answer 75

sid can you do it in two steps without evaluating the integral ?

Answer 76

ok lets see

Answer 77

)

Answer 78

we need only two basic properties

Answer 79

yes i can do it without any integration @ganeshie8

Answer 80

:O

Answer 81

\[\mathcal{L} \left(\int_0^tu\sin u du\right)\]

Okie, i have been trying the same and got stuck at above step :O

Answer 82

first property:
\[L[\int\limits_{0}^{t}F(\tau)d \tau]=\frac{ 1 }{ s }F(s)\\\]
next property:\[L[t^n.f(t)]=(-1)^n.\frac{ d^n }{ ds^n }[F(s)]\]

Answer 83

Siddhartha about to be all like: https://www.youtube.com/watch?v=JGw8DWctAts

Answer 84

(For now I'm closing this question and moving on since I'm running out of time, heh, but feel free to keep going if you guys want. I'm in crunch-time. Thank you guys so much, again.)

Answer 85

so according to those\[L[t.\int\limits_{0}^{t}\sin \tau d \tau]=(-1)^1\frac{ d^1 }{ ds^1 }[\frac{ 1 }{ s }.\frac{ 1 }{ s^2+1 }]=(-1).\frac{ d }{ ds }\frac{ 1 }{ s(s^2+1) }\\=(-1)\frac{ -3s^2-1 }{ s^2(s^2+1)^2 }=\frac{ 3s^2+1 }{ s^2(s^2+1)^2 }\]
ta da!!!!!!!!!!
done without any integration

Answer 86

Nice :) still convolution is not used

Answer 87

but the question did'nt asked to use convolution
it asked to find the Laplace Transform of a convolution

Answer 88

(Taking a look at this now as well, the best way I think to get a grasp for how to use all techniques is to see how a single problem can be solved in multiple ways)

Answer 89

Oh right lol i misread the question and messing wid convolution :/

Answer 90

and convolution cannot be used in TIME DOMAIN it is used in FREQUENCY(S) DOMIAN right?

Answer 91

That was pretty clever, the combination of those two properties.

Answer 92

we can find inverse laplace using convolution but laplace using convolution is something complicated

Answer 93

hahaha that theme song for sidd

Answer 94

lol which one dan?

Answer 95

his youtube video up there

Answer 96

lol now i see he looks like a super saiyan :P

Answer 97

how come they make a big deal out of this property 1/s * F(s) lol/
...
like f(t)*g(t)=f(s)g(s)=int(f(tau)(G(t-tau)) tau, its easy to see if u got f(tau) only then g(t) must be a constant or 1, then g(s)= 1/s