# Challenging Laplace from diagram

#### ihaveaquestion

Joined May 1, 2009
314
http://img132.imageshack.us/img132/9986/2551copy.jpg

There's my work... this one looks tough.

Does this look right/wrong to anyone? I tried doing it in parts... the first slope using the idea of y=mx+b and same with the end slope... the middle was just 'normal' stuff, I'm not sure how to explain it... still learning this stuff.. thanks

#### notxjack

Joined Sep 7, 2009
20
I have no idea what to make of your work. The way you should approach a question which is piecewise smooth is by integrating it piecewise inside each interval where the function behaves uniformally. Basically treat the function like:

2x for x in [0,1]
2 for x in [1,3]
-2x+8 for x in [3,4]

This yields a three term sum of integrals (L = e^(-st)):

$$\int_0^1 \! L*2t \, dt.$$
plus
$$\int_1^3 \! L*2 \, dt.$$
plus
$$\int_3^4 \! (-2t+8)*L \, dt.$$

LaTeX in a forum is just cruel. Can we have a clipart drawer app instead?

#### notxjack

Joined Sep 7, 2009
20
I'm not sure whether my post is editing, but I see you're doing delay/shift theorems and that u is the step function. You should still be doing this in three parts, and your third part is still -2t+8 and not 2t as you have in your notes. When you have a broken up domain like this, you want to set it up as follows:

2t[u(t)-u(t-1)]+2[u(t-1)-u(t-3)]+(-2t+8)[u(t-3)-u(t-4)]

Idea being that as you go past each of the intermediate bounds (1, 2, and 4) your u(t)-u(t-a) terms will go to zero for the 'useless' parts of our piecewise definition.

#### t_n_k

Joined Mar 6, 2009
5,447
This was my approach - needs double checking.

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#### notxjack

Joined Sep 7, 2009
20
That 'looks' more correct, though I don't understand your graphical notation. The easiest way to break up problem into something more legible would just be to put dotted lines at each 'shift' and then write a roman numeral in each section, then list out each partial Laplace for each roman numeral, then sum them together. Re-drawing the graph at each stage both bloats your work as well as introduces opportunities for omission (such as mislabeling a point in the domain or range). Your graders will appreciate it.

Ex:

I. L(the u(t)-u(t-1) part)
II. L(the u(t-1)-u(t-3) part)
III. L(the u(t-3)-u(t-4) part)

L(h(t)) = I + II + III = some expression in terms of s

h(t) = the inverse Laplace transform of our L(h(t)) expression

#### t_n_k

Joined Mar 6, 2009
5,447
That 'looks' more correct, though I don't understand your graphical notation. The easiest way to break up problem into something more legible would just be to put dotted lines at each 'shift' and then write a roman numeral in each section, then list out each partial Laplace for each roman numeral, then sum them together. Re-drawing the graph at each stage both bloats your work as well as introduces opportunities for omission (such as mislabeling a point in the domain or range). Your graders will appreciate it.

Ex:

I. L(the u(t)-u(t-1) part)
II. L(the u(t-1)-u(t-3) part)
III. L(the u(t-3)-u(t-4) part)

L(h(t)) = I + II + III = some expression in terms of s

h(t) = the inverse Laplace transform of our L(h(t)) expression
Hi notxjack,

I'm not sure if your comments were addressed to me. Thanks for your feedback if they were. If so they brought a wry smile to my face - at the age of 60 years it's unlikely you're going to get this old dog to learn any new tricks or (alternatively) "make a silk purse from a sow's ear". Fortunately I don't have to suffer the strictures of grading by any professors these days. I'm also not sure what you mean by "more correct" - presumably one is or isn't correct.

Apologies if you weren't commenting on my contribution.

Rgds,

t_n_k

#### t_n_k

Joined Mar 6, 2009
5,447
And I do see where I am incorrect BTW. I don't need the two step bits I included.

So the correct answer is (presumably) something like ...

F(s)=(2/s^2)*(1-exp(-s)-exp(-3s)+exp(-4s))

What do you think?