Heaviside and Laplace Transforms

In summary, the student is trying to solve a homework problem that involved rearranging a function using a Heaviside step function, and then transforming it into a Laplace transform. However, they ran into a problem with the first term of the function, which they couldn't figure out. The student then asked for help, and explained that they made a mistake when they did the homework earlier. After getting some help, the student was able to solve the problem.
  • #1
nobodyuknow
64
0

Homework Statement



PehSSzM.png


Rearrange f(t) using Heaviside Step Function
Then Rearrange it so that the Laplace Transform can be written down
Then, Write the Laplace Transform of f(t)

Homework Equations


The Attempt at a Solution



So my first step is as follows...
Using the basic Piecewise Function in terms of Unit Functions formulae

f(t) = 0 -cos(2∏t)H(t-3)-cos(2∏t)H(t-6)

Which I've then rearranged into...

f(t) = 0 -cos(2∏(t-3))H(t-3) - cos(2∏(t-6))H(t-6)

Then transformed into...

Sorry, couldn't use the Math editor for this :/
http://img837.imageshack.us/img837/9594/1bff76145aea4f8f8f2dcbd.png [Broken]Is this right? All help appreciated thanks!
 
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  • #2
I think the second and third terms are correct. But why is the first term there? In the original function,
f(t) = - H(t-3) + cos(2∏(t-3))H(t-3) - cos(2∏(t-6))H(t-6)
The second and third terms make sense to me. But why the first term? Try testing out the function with several possible values of t, to make sure it has the values you want it to have.
 
  • #3
If 3< t< 6 then H(t- 3)= 1 and H(t- 6)= 0 so f(t)= -1+ cos(2pi(t- 3)).

That "-1" is not in the original function and I do not understand where the first H(t- 3) came from.
 
  • #4
Ah, it's a mistake carried over from a previous homework question I've done.

How would you go about substituting in values of 't' in order for it to validate my f(t) value?

For example...

cos(2∏(t-3))H(t-3) - cos(2∏(t-6))H(t-6)

And substitute in 0,

cos(2∏(0-3))H(0-3) - cos(2∏(0-6))H(0-6) is supposedly = 0
= H(0-3) - H(0-6)
Not sure how you'd solve since there's a stepside function in there.

EDIT:

I didn't follow off my original Piecewise Function to Unit Functions formulae, it's a mistake carried over from a previous question I've done.
 
  • #5
H(-6) is just zero, because the input to the H function is less than zero. (that's just from its definition). I'm guessing you know that, but just forgot. It is easy to forget what these 'functions' mean when they are not simple functions like x^2 +4x
 
  • #6
nobodyuknow said:

Homework Statement



PehSSzM.png


Rearrange f(t) using Heaviside Step Function
Then Rearrange it so that the Laplace Transform can be written down
Then, Write the Laplace Transform of f(t)

Homework Equations





The Attempt at a Solution



So my first step is as follows...
Using the basic Piecewise Function in terms of Unit Functions formulae

f(t) = 0 -cos(2∏t)H(t-3)-cos(2∏t)H(t-6)

Which I've then rearranged into...

f(t) = 0 -cos(2∏(t-3))H(t-3) - cos(2∏(t-6))H(t-6)

Then transformed into...

Sorry, couldn't use the Math editor for this :/
http://img837.imageshack.us/img837/9594/1bff76145aea4f8f8f2dcbd.png [Broken]


Is this right? All help appreciated thanks!

If ##2 \pi^2## means ##(2 \pi)^2## then yes, your answer is OK; otherwise it is not. If you DO mean ##(2 \pi)^2## you must either use parentheses, or expand it out to get ##4 \pi^2##.
 
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  • #7
Thanks BruceW, unfortunately, I did not know that, going through my lecture notes and slides, there's no information on double checking our Laplace Transforms. Thanks a lot for telling me this, I'll Google it later to do a little bit more research on it!

And Ray Vickson, much appreciated, I missed out this small detail, very obvious, but, didn't catch on!

Muchly appreciated all!

New Answer:

Nu0VBBj.png
 
  • #8
yeah, that's OK. happy to help. Also it was good practice for me to check your answer, since I kindof have avoided Laplace transforms in the past. I always felt that the rules were sort of arbitrary and felt like "what is the point of doing all these transforms?"

But now that I know a bit about contour integration, I see some of the rules are not so arbitrary. (like where the 'region of convergence' is). And now that I have experience in doing differential equations, I see that it can be useful to do transforms, so we have an algebraic equation instead of a differential equation.

There is still loads of stuff I don't fully know. Like how there are several possible formal definitions, depending on if you want to talk about Lebesgue integrals or conditionally convergent improper integrals, or in the sense of a distribution. And how the two-sided Laplace transform relates to the Fourier transform - which is probably my favourite transform :)

Anyway, I guess this is all deeper stuff, and most of the time it is fine to just know the 'shallower' stuff. I have come to terms with the fact that I don't know most of the deeper reasons behind it all. As I guess all of us have had to do at some time. (except maybe the 'pure' mathematicians).

Sorry for the massive tangent I have just gone off on.
 
  • #9
Aha, that's not a problem at all, I'm assuming a lot of what you've just mentioned is something I'll be coming across in my course :)
 
  • #10
This was my first homework question, I assumed it was correct since the tutorer looked at it and ticked me off, but, this is the question where I dragged my error from... Attempt #2

WBmBgbf.png


(t²-1)H(t-2) + (0-t²)H(t-3)
(t²-1) = (t-2)²+4(t-2)+4-1
H(t-2) [(t-2)²+4(t-2)+3]

and

t²H(t-3) = (t² - 3 + 3)²H(t-3)
[(t-3)²+6(t-3)+9]H(t-3) = H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)
into this
H(t-2)(t-2)² + 4H(t-2)² + 3H(t-2) - H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)

Laplace Transform into this...
http://img24.imageshack.us/img24/9443/2a5314ca4f9c4d748b92907.png [Broken]
 
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  • #11
nobodyuknow said:
This was my first homework question, I assumed it was correct since the tutorer looked at it and ticked me off, but, this is the question where I dragged my error from... Attempt #2

WBmBgbf.png


(t²-1)H(t-2) + (0-t²)H(t-3)
(t²-1) = (t-2)²+4(t-2)+4-1
H(t-2) [(t-2)²+4(t-2)+3]

and

t²H(t-3) = (t² - 3 + 3)²H(t-3)
[(t-3)²+6(t-3)+9]H(t-3) = H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)
into this
H(t-2)(t-2)² + 4H(t-2)² + 3H(t-2) - H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)

Laplace Transform into this...
http://img24.imageshack.us/img24/9443/2a5314ca4f9c4d748b92907.png [Broken]

In questions of this type I, personally, find the widespread use of H to be not worth the bother: it often just complicates things and gets in the way. IMHO it is much easier to just use the original definition:
[tex] \cal{L}(f)(s) = \int_0^{\infty} e^{-st} f(t) \, dt = \int_1^2 e^{-st} 1 \, dt
+ \int_2^3 e^{-st} t^2 \, dt. [/tex]
However, do whatever seems easiest to you; I just wanted to draw your attention to an alternative.
 
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  • #12
Fair enough then, I have a Chemistry Prac in a bit so I don't have much time to calculate it, but, would the end result come out to as what I have gotten?

Or is there a mistake that needs to be corrected here?

Thanks, I appreciate the feedback.
 
  • #13
nobodyuknow said:
This was my first homework question, I assumed it was correct since the tutorer looked at it and ticked me off, but, this is the question where I dragged my error from... Attempt #2

WBmBgbf.png


(t²-1)H(t-2) + (0-t²)H(t-3)
(t²-1) = (t-2)²+4(t-2)+4-1
H(t-2) [(t-2)²+4(t-2)+3]

and

t²H(t-3) = (t² - 3 + 3)²H(t-3)
[(t-3)²+6(t-3)+9]H(t-3) = H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)
into this
H(t-2)(t-2)² + 4H(t-2)² + 3H(t-2) - H(t-3)(t-3)² + 6H(t-3) + 9H(t-3)

Laplace Transform into this...
http://img24.imageshack.us/img24/9443/2a5314ca4f9c4d748b92907.png [Broken]
You made a few sign errors on the H(t-3) terms because you didn't use parentheses where you were supposed to. Also, you left out the H(t) term.
 
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  • #14
nobodyuknow said:
Fair enough then, I have a Chemistry Prac in a bit so I don't have much time to calculate it, but, would the end result come out to as what I have gotten?

Or is there a mistake that needs to be corrected here?

Thanks, I appreciate the feedback.

Yes, there are some serious errors. You have
[tex] \frac{9+3}{s} e^{-3s} + \frac{6+4}{s^2} e^{-2s} +\frac{2}{s^3}e^{-2s}
- \frac{2}{s^3} e^{-3s}
= \frac{12}{s}e^{-3s} + \frac{10}{s^2} e^{-2s} + \frac{2}{s^3}e^{-2s} -
\frac{2}{s^3}e^{-3s}\\
= \left(\frac{10}{s^2} + \frac{2}{s^3}\right) e^{-2s}
+ \left( \frac{12}{s} - \frac{2}{s^3}\right) e^{-3s} .[/tex]
However, the correct result is
[tex]\frac{1}{s}e^{-s}+\left( \frac{3}{s}+\frac{2}{s^3}+\frac{4}{s^2}\right) e^{-2s}
-\left( \frac{9}{s}+\frac{6}{s^2}+\frac{2}{s^3}\right) e^{-3s} [/tex]
 

1. What are Heaviside and Laplace Transforms?

Heaviside and Laplace Transforms are mathematical tools used to simplify and solve differential equations. They transform a function from the time domain to the frequency domain, making it easier to analyze and solve complex systems.

2. What is the difference between Heaviside and Laplace Transforms?

Heaviside Transform is a unilateral transform, meaning it only considers the values of the function for positive time values. Laplace Transform is a bilateral transform, considering both positive and negative time values. This makes Laplace Transform more versatile in solving a wider range of problems.

3. How are Heaviside and Laplace Transforms used in science?

Heaviside and Laplace Transforms are commonly used in engineering and physics to solve differential equations in electrical circuits, mechanical systems, and control systems. They are also used in signal processing, communication systems, and other fields of science and mathematics.

4. What are the benefits of using Heaviside and Laplace Transforms?

The main benefit of using Heaviside and Laplace Transforms is that they simplify the process of solving complex differential equations. They also allow for the analysis of systems in the frequency domain, providing insights into the behavior and stability of a system. Additionally, these transforms can be used to solve problems that are difficult or impossible to solve using traditional methods.

5. Are there any limitations to using Heaviside and Laplace Transforms?

One limitation of using Heaviside and Laplace Transforms is that they only work for linear systems, meaning the input and output are proportional. They also require the system to have a well-defined initial state, which may not always be the case in practical applications. Additionally, the inverse transform can be challenging and time-consuming to calculate, especially for more complex functions.

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