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Homework Help: Laplace Transform

  1. Jul 17, 2013 #1
    I attached the problem as a word document. I'm stuck trying to determine the laplace transform for t-tU(t-1). I know I'm supposed to work with 1/s^2(s+2) and solve for A, B,C. I got B=1/2, A=-1/4, and C=1/4 when 1=(As+B)(s+2)+Cs^2. The answer to the problem is
    y= 1/4 + 1/2t +1/4 e^-2t -[1/4 +1/2(t-1) -1/4 e^-2(t-1)] U(t-1). I understand where 1/4 + 1/2t +1/4 e^-2t comes from but I don't understand where -[1/4 +1/2(t-1) -1/4 e^-2(t-1)] U(t-1) comes from.

    Attached Files:

  2. jcsd
  3. Jul 17, 2013 #2


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    You are trying to find the Laplace transform for f(t)= t for [tex]t \le 1[/tex], f(t)= 0 for t>1?

    I would just use the definition of "Laplace transform":
    [tex]\int_0^\infty f(t)e^{-st}dt= \int_0^1 te^{-st}dt[/tex]
    Integrate that "by parts". Let u= t, [itex]dv= e^{-st}dt[/itex], du= dt, [itex]v= -(1/s)e^{-st}[/itex] so this is
    [tex]\left[-(t/s)e^{-st}\right]_0^1+ (1/s)\int e^{-st}dt[/tex][tex]= -e^{-s}/s+ \left[(-1/s^2)e^{-st}\right]_0^1[/tex]
    [tex]= -e^{-s}/s- e^{-s}/s^2+ 1/s^2[/tex]
  4. Jul 20, 2013 #3
    I still don't understand where -[1/4 +1/2(t-1) -1/4 e^-2(t-1)] U(t-1) comes from. The answer is y= 1/4 + 1/2t +1/4 e^-2t -[1/4 +1/2(t-1) -1/4 e^-2(t-1)] U(t-1). Where does the second part of the answer come from.
  5. Jul 20, 2013 #4


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    Do you not know what "U(t-1)" means? U(t), the Heaviside function or "unit step function" is 0 for x< 0, 1 for x> 0 (generally it is defined to be 1 at x= 1 but that is not important). The "t- 1" argument just shifst the step to x= 1: when x= 1, x- 1= 0 so U(1-1)= U(0). That is, U(t- 1) is 0 for x< 1, 1 for x> 1.

    If x< 1 then -[1/4 +1/2(t-1) -1/4 e^-2(t-1)] U(t-1)= -[1/4+ (1/2)(t- 1)]+ 1/4e^{-2(t-1)}(0)= -1/4- (1/2)t- 1/2= -[(1/2)t+ 3/4]. If [itex]x\ge 1[/itex], this is -[1/4+ (1/2)(t- 1)]+ 1/4e^{-2(t-1)}(1)= -[(1/2)t+ 3/4- 1/4e^{-2(t-1)}]
  6. Jul 20, 2013 #5

    Ray Vickson

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    When the author writes ##t - t u(t-1)## he/she is essentially saying that
    [tex] \int_0^1 e^{-st} t \, dt = \int_0^{\infty} e^{-st} t \, dt - \int_1^{\infty} e^{-st} t \, dt.[/tex] You could put ##t = x+1## in the second term to get
    [tex] \int_1^{\infty} e^{-st} t \, dt= \int_0^{\infty} (x+1) e^{-s(x+1)} \, dx[/tex]
    and then do the integral on the right as a sum of two already-known integrals.
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