# Laplace heaviside step function question

• imsleepy
In summary, the student is having trouble solving part (ii) of the homework and is seeking help from a friend. The basic Heaviside step function, H(t), is defined to be 0 for t< 0, 1 for t\ge 0. Since the "break" in this function is at t= 1, you want H(t-1) which is 0 for t< 1, 1 for t\ge 1. You want the function to equal t for t< 1 so you want g(t)= t+ (***)H(t-1). Now, what should (***) be so that g(t)= 2- t for t\ge 1? i'm
imsleepy

## Homework Statement

(quick sketch of part i. yes i know the graph should specifically end at t=4 lol)

u(t-a)
= 0 for t>a
= 1/2 for t=a
= 1 for t>a

## The Attempt at a Solution

i'm having trouble solving part (ii).
i remember a friend of mine saying something like "start with whatever the function is at the initial point, then minus what it was, and add what it will become at that particular point 'a' on the t-axis".
i'm not too sure how to put this into an equation.

The basic Heaviside step function, H(t), is defined to be 0 for t< 0, 1 for $t\ge 0$. Since the "break" in this function is at t= 1, you want H(t- 1) which is 0 for t< 1, 1 for $t\ge 1$.

You want the function to equal t for t< 1 so you want g(t)= t+ (***)H(t-1).
Now, what should (***) be so that g(t)= 2- t for $t\ge 1$?

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i'm not sure i follow.

you start off with t, then... you want to turn "off" what it is and turn "on" what it becomes, at point a on the t-axis, yes?
so... g(t) = t + (-t)H(t-1), your *** = -t to turn "off" what it was?

then how do you add on what it will become... you add on + (2-t)H(###), I'm not sure what goes in the brackets..

so overall equation so far is

g(t) = t + (-t)H(t-1) + (2-t)H(###)

yes?

the solution says
g(t) = t(1 - u(t-1)) + (2-t)u(t-1)
g(t) = t + (2-2t)u(t-1)

so i worked backwards from there to try to understand it and i expanded it to
g(t) = t - (t)u(t-1) + (2-t)u(t-1)

which would mean the ### is t-1
but i don't understand why..

OH I THINK I GET IT!

ok so in it's basic form, g(t) = (what the function is) MINUS (what it was)u(t-a) PLUS (what it will become)u(t-a)
where u(t-a) basically turns a function 'on' or 'off', where a is it's position on the horizontal t-axis.

and since the change takes place at t=1, so you make a = 1 because in the formula u(t-a), a is the position along the axis it's turned on/off at?

g(t) = t - (t)u(t-1) + (2-1)u(t-1) ?

if this is right then i think i understand it now!

i just don't get
u(t-a)
= 0 for t>a
= 1/2 for t=a
= 1 for t>a
that we're given in the equation sheet, it's probably some standard or something?
what does it mean?

another laplace heaviside question, i figured i'd just ask here since i don't have a big thing to ask.

here's the question:

i've done as much as i can (basically solved it *almost*) but i went a different route to what the provided worked solutions have.

after applying laplacing everything and making Y(s) the subject, you get
Y(s) = [30e4s]/[s(s+3)(s-2)]

now apparently the proper way to do the question (not the way i did it, i didnt see this method but after looking at the working out it makes more sense to me, it's just sort of tricky to spot) is that you take out e4s and you find the partial fractions of 30/etc, then inverse laplace THAT.

then they wrote
Now, y(t) = inverse L{Y(s)} = inverse L{[30e4s]/[s(s+3)(s-2)]}
= inverse L{30e4sF(s)}

so there i noticed they made F(s) equal to 30/[s(s+3)(s-2)], and inverse L{30e4sF(s)} is one of the categories from the laplace table we get in the exam.

the rest is then simply worked out as per usual, inverse laplacing 30/etc which equals f(t), therefore f(t-4) is equal to blah blah.

the final answer is y(t) = (-5 + 2e-3(t-4) + 3e-2(t-4))u(t-4)
which i understand.

NOW my way of doing this (before checking the solutions)... i did NOT separate the e4s from the fraction, i found the partial fraction of the entire thing.
i did not use the f(t-a)u(t-a) line from the laplace table.
my final answer was y(t) = -5 + 2e-3(t+4) + 3e-2(t+4)

now my question is... Since i did not apply any heaviside function to my method of solving this question, and since the question specifies that it involves a heaviside step function, can i simply just work out the answer as i did, then multiply the final answer by u(t-4)? (because it's a heaviside function)

i just noticed I'm gettingn t+4 in the brackets of my answer, not sure if I've just done an error, or my method is just incorrect.

## 1. What is the Laplace Heaviside Step Function?

The Laplace Heaviside Step Function, also known as the Heaviside function or the unit step function, is a mathematical function that has a value of 0 for all negative inputs and a value of 1 for all positive inputs.

## 2. What is the purpose of the Laplace Heaviside Step Function?

The Laplace Heaviside Step Function is often used in engineering and physics to represent a sudden change or "step" in a system. It can also be used to model a switch turning on or off at a specific time.

## 3. How is the Laplace Heaviside Step Function different from other step functions?

The Laplace Heaviside Step Function is continuous, meaning that it has a smooth, unbroken line. Other step functions, such as the unit step function, have a discontinuity at the step point.

## 4. Can the Laplace Heaviside Step Function be expressed in terms of other mathematical functions?

Yes, the Laplace Heaviside Step Function can be expressed as a piecewise function using other mathematical functions such as the sign function or the unit step function.

## 5. How is the Laplace Heaviside Step Function used in real-world applications?

The Laplace Heaviside Step Function is commonly used in signal processing, control systems, and circuit analysis. It is also used in differential equations to model phenomena such as population growth and chemical reactions.

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