How to Calculate Laplace Transform of a Complex Fraction

[fabsuk]

Hello
can someone please help I've been struggling on this problem for 2 days now.

question is calculate the laplace transform of

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

I realize 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 can't seem to get the top values of my new partial fraction and I am not sure what to do after as I've never come across one of this type before.

Sorry about the notation but this website won't accept an underline

 

[topsquark]

Hello
can someone please help I've been struggling on this problem for 2 days now.

question is calculate the laplace transform of

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

I realize 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 can't seem to get the top values of my new partial fraction and I am not sure what to do after as I've never come across one of this type before.

Sorry about the notation but this website won't 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.)

 

[fabsuk]

Its the top one.

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

 

[fabsuk]

Im also having problem with a laplace differential equation where I get to this point. i don't know how to do a partial fraction with imaginary numbers.

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

 

[benorin]

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

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

 

[HallsofIvy]

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 $\frac{s}{s^2+ a^2}$.

 

[fabsuk]

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 can't 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 don't 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

 

[benorin]

Check out the http://www.eecircle.com/applets/007/ILaplace.html

 

[benorin]

Check out what HallsofIvy wrote, for that is your solution.

 

[fabsuk]

i understand that you have to use the tables but i don't understand how u use them with complex components.As i said i know u have to complete the square but i just can't get it and looking at the answer just gets me angry.

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

 

[benorin]

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

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

 

[fabsuk]

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 I am 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 can't find any other examples like it.PLEASE

 

[benorin]

PF resources for learning LaTeX Math Typesetting

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

An extremely useful excerpt:

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://amath.colorado.edu/documentation/LaTeX/Symbols.pdf

A bit more information on the amsmath package is available here:

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

 

[benorin]

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:

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

 

[benorin]

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]

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

Thanks, I found it and fixed it.