# The laplace transform/inverse

Hello

question is calculate the laplace transform of

F(P)= 1-P divided by P(squared) + 4P +13

I realise you have to complete the square as denominator is complex

When i complete the square i get (P+2) (all squared)+ 3(squared)

But i cant seem to get the top values of my new partial fraction and im not sure what to do after as ive never come across one of this type before.

Sorry about the notation but this website wont accept an underline

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fabsuk said:
Hello

question is calculate the laplace transform of

F(P)= 1-P divided by P(squared) + 4P +13

I realise you have to complete the square as denominator is complex

When i complete the square i get (P+2) (all squared)+ 3(squared)

But i cant seem to get the top values of my new partial fraction and im not sure what to do after as ive never come across one of this type before.

Sorry about the notation but this website wont accept an underline

Need a confirmation first. Is your function
$$F(P)=\frac{1-P}{P^2+4P+13}$$ or
$$F(P)=1-\frac{P}{P^2+4P+13}$$?

-Dan
(Left-click on one of the equations to see how to write it.)

Its the top one.

$$F(P)=\frac{1-P}{P^2+4P+13}$$

Im also having problem with a laplace differential equation where I get to this point. i dont know how to do a partial fraction with imaginary numbers.

$$Y{(P^2+p-5)}= \frac{1}{P-2} + p+ 3$$

benorin
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Gold Member
It the same as for real numbers, only that so-called irreduceable quadratics are not so, for example:

$$F(P)=\frac{1-P}{P^2+4P+13}=\frac{1-P}{(P+2+3i)(P+2-3i)}=\frac{A}{P+2+3i}+\frac{B}{P+2-3i}$$

cross-multiply to get

$$1-P=A(P+2-3i)+B(P+2+3i)$$

Now plug-in nice values for P to solve for A and B, best choices for P are $$P=-2\pm 3i$$ (the zeros of the factors):

the value $$P=-2+3i$$ gives

$$3-3i=0+6iB\Rightarrow B=\frac{3-3i}{6i}=-\frac{1}{2}(1+i)$$

and the value $$P=-2-3i$$ gives

$$3+3i=-6iA+0\Rightarrow A=\frac{3+3i}{-6i}=\frac{1}{2}(-1+i)$$

and hence

$$F(P)=\frac{1-P}{P^2+4P+13}=\frac{1-P}{(P+2+3i)(P+2-3i)}=\frac{1}{2}\left[ \frac{-1+i}{P+2+3i}-\frac{1+i}{P+2-3i}\right]$$

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HallsofIvy
Homework Helper
If you want to stay in real numbers- don't use partial fractions. Your table of Laplace transforms should have one that gives $\frac{1}{s^2+ a^2}$ and one that gives $\frac{s}{s^2+ a^2}$.

I understand now clearly what benorin has done where u put in nice values for p and it follows very clearly but is it does not seem possible to do an inverse transform at that point hence i completed the square but i cant seem to use that method either

$$F(P)=\frac{1-P}{P^2+4P+13}=\frac{?}{((P+2)^2) + (3^2)}=\frac{A}{P+2+3i}+\frac{B}{P+2-3i}$$

If u dont put it in partial fractions then what do u do?

BTW i sorted out the other problem

How do u take inverse transformations with i's in them

e,g

$$\frac{1}{2(a+p)}+\frac{1}{4(a+p-2bi)}+\frac{1}{4(a+p+2bi)}$$

Clearling up these points would i think clear up my problems in laplace.Any ideas

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benorin
Homework Helper
Gold Member
Check out what HallsofIvy wrote, for that is your solution.

i understand that you have to use the tables but i dont understand how u use them with complex components.As i said i know u have to complete the square but i just cant get it and looking at the answer just gets me angry.

AS i said ive been trying for ages now
the other example to clarify things would be helpful as well.Im stuck!!!

benorin
Homework Helper
Gold Member
Inverse Laplace transforms via rean & complex methods

The real variable method (what HallsofIvy suggested):

Notation: Suppose that $$\ell \{ f(t) \}=F(P)$$ is the Laplace transform of f(t), so that $$\ell ^{-1} \{ F(P) \}=f(t)$$ is the inverse Laplace transform of F(P).

Work: From

$$F(P)=\frac{1-P}{P^2+4P+13},$$

write

$$\ell ^{-1} \{ F(P)\} =\ell ^{-1} \left\{ \frac{1-P}{P^2+4P+13}\right\} = \ell ^{-1} \left\{ \frac{1-P}{(P+2)^2+9}\right\} = \ell ^{-1} \left\{ \frac{3-(P+2)}{(P+2)^2+3^2}\right\}$$
$$=\ell ^{-1} \left\{ \frac{3}{(P+2)^2+3^2}-\frac{P+2}{(P+2)^2+3^2}\right\} = \ell ^{-1} \left\{ \frac{3}{(P+2)^2+3^2}\right\}-\ell ^{-1} \left\{ \frac{P+2}{(P+2)^2+3^2}\right\}$$
$$= e^{-2t}\ell ^{-1} \left\{ \frac{3}{P^2+3^2}\right\}-e^{-2t}\ell ^{-1} \left\{ \frac{P}{P^2+3^2}\right\}$$

since $$\ell ^{-1} \left\{ F(P-a)\right\} = e^{at}\ell ^{-1} \left\{ F(P)\right\},$$ continuing on we have

$$\ell ^{-1} \{ F(P)\}= e^{-2t}\ell ^{-1} \left\{ \frac{3}{P^2+3^2}\right\}-e^{-2t}\ell ^{-1} \left\{ \frac{P}{P^2+3^2}\right\} = e^{-2t}\sin{3t} - e^{-2t}\cos{3t} ,$$

where the inverse transforms of sin and cos are off the table.

In conclusion,

$$\boxed{ \ell ^{-1} \{ F(P)\}= e^{-2t}\left( \sin{3t} - \cos{3t}\right) }$$​

The complex variable/partial fraction decomposition method (continuing my eariler post): From

$$F(P) = \frac{1-P}{P^2+4P+13} = \frac{1-P}{(P+2+3i)(P+2-3i)} = \frac{1}{2}\left( \frac{-1+i}{P+2+3i}-\frac{1+i}{P+2-3i}\right)$$

write

$$\ell ^{-1} \{ F(P)\} = \ell ^{-1} \left\{ \frac{1}{2}\left[ \frac{-1+i}{P-(-2-3i)}-\frac{1+i}{P-(-2+3i)}\right] \right\}$$
$$= \frac{1}{2} \ell ^{-1} \left\{ \frac{-1+i}{P-(-2-3i)} \right\} -\frac{1}{2} \ell ^{-1} \left\{ \frac{1+i}{P-(-2+3i)}\right\}$$
$$= \frac{-1+i}{2} \ell ^{-1} \left\{ \frac{1}{P-(-2-3i)} \right\} -\frac{1+i}{2} \ell ^{-1} \left\{ \frac{1}{P-(-2+3i)}\right\}$$
$$= \frac{-1+i}{2} e^{(-2-3i)t} -\frac{1+i}{2} e^{(-2+3i)t} = \frac{-1+i}{2} e^{-2t}e^{-3it} -\frac{1+i}{2} e^{-2t}e^{3it}$$
$$= \frac{e^{-2t}}{2}\left[ (-1+i) e^{-3it} -(1+i)e^{3it}\right] = \frac{1}{2}e^{-2t}\left[ -(e^{3it} +e^{-3it}) -i(e^{3it} -e^{-3it}) \right]$$
$$= e^{-2t}\left[ -\frac{e^{3it} +e^{-3it}}{2} +\frac{e^{3it} -e^{-3it}}{2i} \right] = e^{-2t}\left( -\cos{3t} +\sin{3t}\right)$$

So, in conclusion, except for the sign error I'm hoping someone will spot for me, we have

$$\boxed{ \ell ^{-1} \{ F(P)\} = e^{-2t}\left( -\cos{3t} +\sin{3t}\right) }$$​

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I will spot you,
It should be -cos and i now see how you bracket things

Does anybody know about step functions with laplace as again im lost

let LT of the function g(t) is L(G) = G(p).
Calculate the Lt of the function

f(t)= g(t-a) t>a>0
0, t<a

Help me get started i cant find any other examples like it.PLEASE

benorin
Homework Helper
Gold Member
PF resources for learning LaTeX Math Typesetting

This thread will help: PF's Introducing LaTeX Math Typesetting

An extremely useful excerpt:

chroot said:
A pdf file of the most useful LaTeX commands, symbols, and constructs is provided here:

https://www.physicsforums.com/misc/howtolatex.pdf

More symbol reference:

http://www.cds.caltech.edu/~dunbar/docs/amsldoc.pdf [Broken]

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benorin
Homework Helper
Gold Member
Do you mean:

for $$g(t):=\left\{\begin{array}{cc}0,&\mbox{ if }t\leq 0\\1, & \mbox{ if } t>0\end{array}\right.$$

put $$f(t)=g(t-a)=\left\{\begin{array}{cc}0,&\mbox{ if }t\leq a\\1, & \mbox{ if } t>a\end{array}\right.$$ ?

BTW, It looks better like this:

$$\mbox{For } g(t):=\left\{\begin{array}{cc}0,&\mbox{ if }t\leq 0\\1, & \mbox{ if } t>0\end{array}\right. \mbox{, put }f(t):=g(t-a)=\left\{\begin{array}{cc}0,&\mbox{ if }t\leq a\\1, & \mbox{ if } t>a\end{array}\right. ?$$

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benorin
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Gold Member
If so, here it goes...

For a>0,

$$\mathfrak{L}\{ f(t)\} = \mathfrak{L}\{ g(t-a)\} = e^{-ap}\mathfrak{L}\{ g(t)\}$$

oops, gives up to much.

benorin
Homework Helper
Gold Member
fabsuk said:
I will spot you,
It should be -cos and i now see how you bracket things

Thanks, I found it and fixed it.