# Laplace Transform

1. Jul 17, 2013

### jtruth914

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.

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2. Jul 17, 2013

### HallsofIvy

Staff Emeritus
You are trying to find the Laplace transform for f(t)= t for $$t \le 1$$, f(t)= 0 for t>1?

I would just use the definition of "Laplace transform":
$$\int_0^\infty f(t)e^{-st}dt= \int_0^1 te^{-st}dt$$
Integrate that "by parts". Let u= t, $dv= e^{-st}dt$, du= dt, $v= -(1/s)e^{-st}$ so this is
$$\left[-(t/s)e^{-st}\right]_0^1+ (1/s)\int e^{-st}dt$$$$= -e^{-s}/s+ \left[(-1/s^2)e^{-st}\right]_0^1$$
$$= -e^{-s}/s- e^{-s}/s^2+ 1/s^2$$

3. Jul 20, 2013

### jtruth914

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.

4. Jul 20, 2013

### HallsofIvy

Staff Emeritus
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 $x\ge 1$, this is -[1/4+ (1/2)(t- 1)]+ 1/4e^{-2(t-1)}(1)= -[(1/2)t+ 3/4- 1/4e^{-2(t-1)}]

5. Jul 20, 2013

### Ray Vickson

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