Laplace heaviside step function question

[imsleepy]

Homework Statement

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

Homework Equations

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.

i have the answer, but could someone please explain how to get to the answer?

 

[HallsofIvy]

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 $t\ge 1$?

 

[imsleepy]

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..

 

[imsleepy]

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?

 

[imsleepy]

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) = [30e^4s]/[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 e^4s 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{[30e^4s]/[s(s+3)(s-2)]}
= inverse L{30e^4sF(s)}

so there i noticed they made F(s) equal to 30/[s(s+3)(s-2)], and inverse L{30e^4sF(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 e^4s 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.

 

[imsleepy]

my working out from partial fractions onwards (because that's where mine and the worked solutions' answers differ):

http://i.imgur.com/YbtE8.jpg
http://i.imgur.com/Z33QW.jpg
http://i.imgur.com/EjDXU.jpg