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

u_c (t) =<br /> \begin{cases}<br /> 1 &amp; \text{if } t \geq c \\<br /> 0 &amp; \text{if } t &lt; c<br /> \end{cases}

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&#039;(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) =<br /> \begin{cases}<br /> t &amp; \text{if } 0 \leq t &lt; 1<br /> \end{cases}

r_2 (t) =<br /> \begin{cases}<br /> t+1 &amp; \text{if } 1 \leq t &lt; 2<br /> \end{cases}

u_2 (t) =<br /> \begin{cases}<br /> 3 &amp; \text{if } 2 \leq t &lt; 4<br /> \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?

 

[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 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?

 

[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 f(t) = t^2 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 f(t) = t^2 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,

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!

 

[Orodruin]

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!

Looks reasonable.