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Math Help - Express f(t) in terms of unit step functions, then find the Laplace

  1. #1
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    Express f(t) in terms of unit step functions, then find the Laplace

    I've been trying to do this problem forever. I'm not sure how to type this so I took a screenshot of the problem:



    I have no idea how to do this. From what I could gather in my notes, I started off like this:

    For the unit step functions of f(t):

    t^2 - t^2U(t-1)
    (t^2 - 1)U(t-1) - (t^2 - 1)U(t-2)
    (t^2 - 1)U(t-2)

    These may not be right... and even if they are right, can someone briefly explain why? I still don't have a thorough understanding of this.
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  2. #2
    Senior Member yeKciM's Avatar
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    it's like this i think u wrote that there

    t^2[u(t)-u(t-1)]

    t^2[u(t+11)-u(t-2)] or it's (t^2-1)[u(t-1)-u(t-2)] don't know if u wrote there ( t^2) for t goes from -11 to 2 or its  (t^2-1) for t goes from 1 to 2

    (t^2 - 2)[u(t-2)]

    Step function u(t) is defined :  1 for  t\ge0 , and  0 for  t<0
    but when u need to make let's say square signal with amplitude 1 from 0 to 1 u'll do it like this... u put step function and then add one more but negative one that begins at 1 so it looks like this :

    [u(t)-u(t-1)]


    and for Laplace , just put it in integral and solve it it's quite simple do u know how it's done ?

    Laplace transform (bilateral):

    \displaystyle X(S) =  \int _{-\infty} ^{+\infty} x(t) e^{-st} \,dt

    unilateral, just goes from 0- not from -\infty ....

    P.S. if needed i can write u detailed explanation


    here are few signals with step functions, hope it will help u understand how to present signals by using step functions

    signal x(t)=u(t)



    signal x(t)=(u(t)-u(t-5))



    signal x(t)=t(u(t)-u(t-1))



    signal x(t)=(u(t+1)-u(t-1))-2(u(t-1)+u(t-4))




    to see signals bigger just click on them


    and here u are a few basic Laplace transformations

    \displaystyle\begin{array}[b]{||c||c||c||}\hline<br />
Signal&Laplace\; transformation&Region\; of\; convergence\\\hline\hline<br />
\delta(t)&1&ROC: \forall S\\\hline\hline<br />
u(t)&\frac{1}{S}&ROC: Re(S)>0\\\hline\hline<br />
-u(-t)&\frac{1}{S}&ROC: Re(S)<0\\\hline\hline<br />
tu(t)&\frac{1}{S^2}&ROC: Re(S)>0\\\hline\hline<br />
e^{-at}u(t)&\frac{1}{S+a}&ROC: Re(S)>-Re(a)\\\hline\hline<br />
te^{-at}u(t)&\frac{1}{(S+a)^2}&ROC: Re(S)>-Re(a)\\\hline\hline<br />
\cos(\omega_0 t)u(t)&\frac{S}{S^2+\omega_0 ^2}&ROC: Re(S)>0\\\hline\hline<br />
\sin(\omega_0 t)u(t)&\frac{\omega_0}{S^2+\omega_0 ^2}&ROC: Re(S)>0\\\hline<br /> <br />
\end{array}

    well that's not all of "basic" but if u need more.. just say
    Last edited by yeKciM; July 25th 2010 at 05:04 PM. Reason: added few signals
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  3. #3
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    In the first line, I don't get why you put the original u(t) in there. Also, just to clarify, the 2nd part of f(t) is t^2 - 1, 1<t<2, there is no 11 in there (its hard to tell since the function and boundaries are written so close together).

    With the laplace, we never learned it using integrals, we were given the transformations of common ones. for Laplaces of things with U, its supposed to be in the form of f(t-a)U(t-a), the laplace of which would be e^-as * L{f(t)}. But these aren't in that form =\
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  4. #4
    Senior Member yeKciM's Avatar
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    let's see

    in the first line [u(t)-u(t-1)], u must do it like that. that first signal of yours (or function) is x(t) = t^2 \;, for  t\;, goes from  0 to  1. well if u look at definition of step function that I wrote in the last post u'll see that step function is infinity function but with values 0 for t<0 and 1 for t\ge0.

    now, let's first look at simple function (not that first of yours, but simple one) let's look first and second one that i posted with picture x(t)=u(t) and x(t)=[u(t)-u(t-5)] meaning that your question why do i put original u(t) (or i didn't get u right) is because your signal is bounded from 0 to infinity for the first and from 0 to 5 in second one.... Now do u concur that in fact that x(t)=[u(t)-u(t-1)] (part of your first function) we use step functions to bound that signal from 0 to 1... and for every another value out from that interval it's 0 ... Now U multiply that with t^2 and U will get your first signal (function) looking like this :

    signal (function) x(t)=t^2[u(t)-u(t-1)]



    i hope that I understand your question with that "original u(t) there"


    now for Laplace, it's not much harder (if not easier) to do Laplace transformation with integral, (we where strictly forbid to use Laplace tables )... Juts put your let's say step function in the integral and u'll see strait away that it's 1 now Laplace transformation is procedure that "moves" signals from time to S- region where it's easier to solve and do a lot of things and then u use inverse Laplace to get back in time region

    just try using integral, and maybe u should put some work that is correctly done by your teachers and we'll see where is misunderstanding


    P.S. please check that Laplace that U wrote there....
    Last edited by yeKciM; July 26th 2010 at 11:13 AM.
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  5. #5
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    the f(t) will be:

    t^2U(t) - t^2U(t-1) + (t^2 - 1)U(t-1) - (t^2 - 1)U(t-2) + (t^-2)U(t-2)
    which simplifies to:
    t^2U(t) - U(t-1) - U(t-2)

    ^can you please double check if that is correct?

    Assuming thats correct, then the laplace of it would be L{t^2U(t)} - e^-s - e^-2s, right?

    Is there some way (an identity or something) to find the first laplace L{t^2 * U(t)}? and did I do the 2nd two terms correctly?
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  6. #6
    Senior Member yeKciM's Avatar
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    yes it's correct

    your signal represented with step functions is :

    \displaystyle t^2[u(t)-u(t-1)]+(t^2-1)[u(t-1)-u(t-2)]+(t^2-2)[u(t-2)] = t^2u(t) -u(t-1) -u(t-2)

    and for your Laplace transformation, i truly don't know how do anyone can teach u to use Laplace without using Laplace integral... Your solution is hmmmm it's missing S it can't be done like that that i'm very sure

    when u do Laplace transformation on signals :

    \displaystyle u(t-1) it will become \displaystyle \int _{-\infty} ^{+\infty}u(t-1) e^{-st} \,dt=\int _1 ^{+\infty} e^{-st} \,dt = \frac {1}{Se^S}

    \displaystyle u(t-2) it will become \displaystyle \int _{-\infty} ^{+\infty}u(t-2) e^{-st} \,dt= \int _2 ^{+\infty} e^{-st} \,dt = \frac {1}{Se^{2S}}

    and first one u can put in integral and solve it with maybe one substitution ... i can do it but, latter (I have some important work to finish)
    I'll try to do it as quickly as i can



    here u go and first one

    \displaystyle  t^2u(t) it will become \displaystyle \int _{-\infty} ^{+\infty}t^2u(t) e^{-st} \,dt=\int _0 ^{+\infty} t^2e^{-st} \,dt = \frac {2}{S^3}

    hope i helped if not i'll try again
    hehehehe



    P.S. one question do u like for exams have to memorize table of Laplace transformations or they even let u use it written on the paper on exam ? there's to much of it to memorize it's easier to just use Laplace integral or with Furrier transformation ? I'm just curious
    Last edited by yeKciM; July 26th 2010 at 12:49 PM. Reason: added first one
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