Laplace Transforms Involving: Unit-Step, and Ramp Functions

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Homework Help Overview

The discussion revolves around Laplace transforms involving unit-step and ramp functions. The original poster presents a problem that includes defining these functions and applying the Laplace transform to them, particularly focusing on the transformation of a function involving a ramp and unit-step functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the Laplace transform to a function involving unit-step and ramp functions, questioning the assumptions about the parameters involved. Participants discuss the need for a linear combination of functions rather than separate expressions for different intervals. There are inquiries about the correct formulation of unit-step functions and the implications of multiplying functions by step functions.

Discussion Status

Some participants have provided guidance on how to approach the transformation and suggested plotting functions to visualize the relationships. There is ongoing exploration of the correct formulation of the functions involved, with multiple interpretations being discussed. The conversation reflects a collaborative effort to clarify the problem without reaching a definitive conclusion.

Contextual Notes

There are indications of confusion regarding the definitions and applications of unit-step functions in the context of the problem. The original poster expresses uncertainty about the correct formulation and the implications of their transformations, highlighting the need for further clarification.

ConnorM
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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) =<br /> \begin{cases}<br /> 1 & \text{if } t \geq c \\<br /> 0 & \text{if } t < c<br /> \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) =<br /> \begin{cases}<br /> t & \text{if } 0 \leq t < 1<br /> \end{cases}[/itex]

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

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

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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.
 
Orodruin said:
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?
 
Last edited:
ConnorM said:
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?
 
ConnorM said:
$$y(t) = tu(t) - (t+1)u(t-1) - 3u(t-2)$$
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?
 
Orodruin said:
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.
 
vela said:
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?
 
Which function? Which point? Can you elaborate?
 
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).
 
  • #10
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
ConnorM said:
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.
 

Attachments

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  • #12
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!
 
  • #13
ConnorM said:
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.
 

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