# The laplace transform/inverse

1. Mar 5, 2006

### fabsuk

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

Last edited: Mar 5, 2006
2. Mar 5, 2006

### topsquark

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.)

3. Mar 5, 2006

### fabsuk

Its the top one.

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

4. Mar 5, 2006

### fabsuk

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.

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

5. Mar 5, 2006

### 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

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

Last edited: Mar 6, 2006
6. Mar 6, 2006

### HallsofIvy

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

7. Mar 6, 2006

### 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 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

Last edited: Mar 6, 2006
8. Mar 6, 2006

9. Mar 6, 2006

### benorin

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

10. Mar 6, 2006

### fabsuk

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!!!

11. Mar 7, 2006

### 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

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

Last edited: Mar 7, 2006
12. Mar 7, 2006

### 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 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

13. Mar 7, 2006

### benorin

Last edited: Mar 7, 2006
14. Mar 7, 2006

### 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:

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

Last edited: Mar 7, 2006
15. Mar 7, 2006

### 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.

16. Mar 7, 2006

### benorin

Thanks, I found it and fixed it.