# Laplace Transforms Involving: Unit-Step, and Ramp Functions

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1. Sep 19, 2015

### ConnorM

1. The problem statement, all variables and given/known data
Here is an imgur link to my assignment: http://imgur.com/N0l2Buk
I also uploaded it as a picture and attached it to this post.

2. Relevant equations

$u_c (t) = \begin{cases} 1 & \text{if } t \geq c \\ 0 & \text{if } t < c \end{cases}$

3. The attempt at a solution

Question 1.1 -

$L[tu(t)] = \int_0^∞ tu(t)e^{-st} \,dt$

Using the definition of the step function, $t \geq 0, u(t) = 1$
*Is it right to assume that $c = 0$?*

$L[tu(t)] = \int_0^∞ t(1)e^{-st} \,dt$

$L[tu(t)] = \int_0^∞ te^{-st} \,dt$

$L[tu(t)] = 1/s^2$

I'm not sure if this is correct. Should it be solved using the rule, $L[tf(t)] = -F'(s)$

Question 1.2 -

Let $r_1 (t), r_2 (t)$ be the two ramp functions

Let $u_1 (t), u_2 (t)$ be the two unit-step functions

$r_1 (t) = \begin{cases} t & \text{if } 0 \leq t < 1 \end{cases}$

$r_2 (t) = \begin{cases} t+1 & \text{if } 1 \leq t < 2 \end{cases}$

$u_2 (t) = \begin{cases} 3 & \text{if } 2 \leq t < 4 \end{cases}$

I'm not quite sure what to do for the unit-step functions. Could someone help me figure out what they should be?

#### Attached Files:

• ###### Assignment1.JPG
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Last edited: Sep 19, 2015
2. Sep 20, 2015

### Orodruin

Staff Emeritus
Your transformation is correct. For the second part, I do not think you have done what was intended. You are supposed to write the function as a linear combination, not as different expressions in different regions.

3. Sep 20, 2015

### ConnorM

So would it be something like,

Let $y(t)$ be the function on the graph that I am trying to recreate.
** Is this true, $r(t) = tu(t)$? **

So,

$y(t) = tu(t) - (t+1)u(t-1) - 3u(t-2)$

Although I still don't know what the second unit-step function should be?

Last edited: Sep 20, 2015
4. Sep 20, 2015

### Orodruin

Staff Emeritus
Have you tried plotting the function you gave? How does it differ from the one you should find?

5. Sep 20, 2015

### vela

Staff Emeritus
Try plotting the function g(t) = u(t-1)-u(t-2). You should see it's a pulse. Now consider what you'll get if you multiply g(t) by some function. Try plotting the function f(t) alone and the product f(t)g(t). For instance, try it with f(t) = t2. Do you understand the effect of multiplying by g(t) on the graph?

6. Sep 20, 2015

### ConnorM

I tried plotting it and from what I see it's not similar at all.

7. Sep 20, 2015

### ConnorM

So multiplying f(t) by g(t) gives me a point on that function?

8. Sep 20, 2015

### vela

Staff Emeritus
Which function? Which point? Can you elaborate?

9. Sep 20, 2015

### ConnorM

So for g(t) that you suggested it gave a pulse at (1,1), then when I plotted $f(t) = t^2$ I got a parabola. When I multiplied f(t) by g(t), I got a pulse again at (1,1).

10. Sep 20, 2015

### Orodruin

Staff Emeritus
I suggest that you think along the following lines:
• What happens when you pass the step of a step function? What changes? Where do these changes occur for the sought function?
• What happens when you pass the base of a function of the form $(t-a) u(t-a)$? Where do these changes occur for the sought function?

11. Sep 20, 2015

### vela

Staff Emeritus
I don't think you plotted g(t) correctly then. I'm not sure what you mean by a pulse at (1,1). I've attached a plot of what you should've gotten.

#### Attached Files:

• ###### plot.png
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12. Sep 20, 2015

### ConnorM

So I did some work with graphing the equations and I think this is the right one,

$y(t) = tu(t) + u(t-1) - (t-2)u(t-2) - 3u(t-4)$

Let me know if this latex is working or not, I'm on my phone and can't see if it's showing up correctly!

13. Sep 20, 2015

### Orodruin

Staff Emeritus
Looks reasonable.