# Laplace Transform unit step function

1. Jan 24, 2016

### JavierOlivares

1. The problem statement, all variables and given/known data
g(t) = (t+1)us(t) - (t-1)us(t-1) - 2us(t-1) - (t - 2)us(t-2) + (t-3)us)(t - 3) + us(t-3)
2. Relevant equations
unit step function us(t-3) is same as u3 (t)
Shift in time:

L[f(t - T)us(t-T)] = e-TsF(s)
us(t) ↔ 1/s
t ⇔ 1/s2
3. The attempt at a solution
1/s2 + 1/s - e-s/s2 + e-s/s - 2e-s/s - e-2s/s2 + 2e-2s/s + e-3s/s2 - 3e-3s/s + e-3s/s

1/s(1 - e-s +2e-2s - 2e-3s) + 1/ s2(1 - e-s - e-2s + e-3s)

My professor gave us a different answer online. It seems correct online. Any advice would be great thank you.

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2. Jan 25, 2016

### Ray Vickson

I get an answer different from yours, but I don't know if it is the same as your professor's answer, since you do not tell us what that is.

I just did what I usually do for such problems, which is to avoid using all the shifting rules, etc., and just do the integral $\int_0^{\infty} e^{-st} f(t) \, dt$ directly. First, though, we need to figure out $f(t)$ in a more usable (piecewise) form. For example, between $t = 0$ and $t = 1$ only the first term is nonzero, so $f(t) = t+1$ in that interval. Between $t = 1$ and $t = 2$ the second and third terms also appear, resulting in $f(t) = t+1 -(t-1) - 2 = 0$ in $(1,2)$. Keep going like that until you have a complete description of $f(t)$ on $(0,\infty)$, then integrate. Of course, since you will have a piecewise description of $f(t)$ you will need to split up the integral into several pieces. And, of course, you need to know integrals such as $\int_a^b e^{-st} \, dt$ and $\int_a^b t e^{-st} \, dt$.

Last edited: Jan 25, 2016
3. Jan 26, 2016

### JavierOlivares

Attached

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4. Jan 26, 2016

### JavierOlivares

Our professor wanted us to do it with shifting rules and easy transforms for exams since integrating would take too long for more tedious problems. Sorry I should've mentioned that earlier. Thanks for the reply.

5. Jan 26, 2016

### Ray Vickson

Shifting, etc., works well too, but you need to be careful: your first solution contained some fatal errors. For example, you apparently said that
$$- (t - 2)\, u(t-2) \rightarrow - e^{-2s}/s^2 + 2e^{-2s}/s \; \Longleftarrow \:\text{false!}$$
The shifting rule gives
$$- (t - 2) \, u(t-2) \rightarrow - e^{-2s}/s^2$$
You would only get the extra $2e^{-2s}/s$ term if you had started with $-(t-4)\,u(t-2)$.

6. Jan 27, 2016

### JavierOlivares

Okay thanks for clearing that up. I think that's where I keep messing up. The last post was my professors answer and yeah I got the right answer this time by following what you said.
So I also get for : -2*us(t - 1) ↔ -2*e-2s/s

I think that's what you mean right? The shifting property for f(t - T) ↔ F(s) means that I'm only taking the Laplace of t? That's what you did in your response.

7. Jan 27, 2016

### Ray Vickson

I don't understand your question. I stated exactly what I meant, so I don't know how I can re-word it.

8. Jan 27, 2016

### JavierOlivares

I guess I would have thought -(t - 1)*us(t - 1) would be the same as -t* us(t - 1) + 1*us(t - 1). I factored out -(t - 1). My question is why can't I just do that?

I'll make up a different example to make sure I understand it. Let's say I had:
(3t - 1)*us(t - 1) So I gotta have (3t - 1) be the same as (t - 1) for this property to work? If yes then:

3*(t - 1/3)*us(t - 1) = 3 * (t - 1 + 2/3)*us(t - 1)

***I made the function this (t - 1 + 2/3) so it equals (t-1).***

So f(t) = t + 2/3 would be the same as:

3*[(t - 1)*us(t - 1) + 2/3*us(t - 1)]
Taking Laplace I get:

3*e-s/s2 + e-s/s.

Hopefully I made this clear. So I asked two questions why can't I do what I did before and given my example is my logic correct?

9. Jan 27, 2016

### Ray Vickson

If $f(t) = t \, u(t)$ then $f(t-a) = (t-a) \, u(t-a)$, so shifting applies directly to give the LT as $e^{-as}/s^2$, with almost no work. If you write $(t-a)\,u(t-a)$ as $t \, u(t-a) - a \, u(t-a)$, you have destroyed the shifting representation. You could write $t \, u(t-a)$ as $f_1(t-a)$, where $f_1(t) = (t+a) \, u(t)$, but what would be the point of that?

$$(3t-1) \, u(t-1) = [3(t-1) + 2] \, u(t-1) = f_1(t-1) + f_0(t-1),$$
where $f_0(t) = 2 u(t)$ and $f_1(t) = 3 t\, u(t)$. So we can write the LT immediately as
$$L(s) = \frac{3}{s^2} e^{-s} + \frac{2}{s} e^{-s}$$
Somehow you lost the factor '2' in the last term. Look carefully at what you did and you ought to be able to spot the error.

Whatever you do, try to stop "overthinking" the problem because it really is very simple: wherever you see $h(t-a) u(t-a)$, its LT will be $e^{-as} {\cal H}(s)$, where ${\cal H}(s)$ is the LT of $h(t)$. So, $(3t-1) u(t-1) = h(t-1) u(t-1)$, where $h(t) = 3t + 2$.

10. Jan 27, 2016

### JavierOlivares

Thanks for all the help. It makes a lot more sense now.