Laplace Transforms Involving: Unit-Step, and Ramp Functions

[ConnorM]

Homework Statement

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.

Homework Equations

[itex]u_c (t) =
\begin{cases}
1 & \text{if } t \geq c \\
0 & \text{if } t < c
\end{cases}[/itex]

The Attempt at a Solution

Question 1.1 -

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

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

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

[itex]L[tu(t)] = \int_0^∞ te^{-st} \,dt[/itex]

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

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

Question 1.2 -

Let [itex] r_1 (t), r_2 (t) [/itex] be the two ramp functions

Let [itex] u_1 (t), u_2 (t) [/itex] be the two unit-step functions

[itex]r_1 (t) =
\begin{cases}
t & \text{if } 0 \leq t < 1
\end{cases}[/itex]

[itex]r_2 (t) =
\begin{cases}
t+1 & \text{if } 1 \leq t < 2
\end{cases}[/itex]

[itex]u_2 (t) =
\begin{cases}
3 & \text{if } 2 \leq t < 4
\end{cases}[/itex]

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

 

[Orodruin]

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.

 

[ConnorM]

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.

So would it be something like,

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

So,

[itex]y(t) = tu(t) - (t+1)u(t-1) - 3u(t-2)[/itex]

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

 

[Orodruin]

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

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

 

[vela]

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

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?

 

[ConnorM]

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

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

 

[ConnorM]

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?

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

 

[vela]

Which function? Which point? Can you elaborate?

 

[ConnorM]

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

 

[Orodruin]

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?

 

[vela]

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

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.

 

[ConnorM]

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

[itex] y(t) = tu(t) + u(t-1) - (t-2)u(t-2) - 3u(t-4) [/itex]

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!

 

[Orodruin]

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

[itex] y(t) = tu(t) + u(t-1) - (t-2)u(t-2) - 3u(t-4) [/itex]

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!

Looks reasonable.