Question

Inverse Laplace Transform (6 + s^2 )/( (s^2 + 4) ( 3 s - 4) )

Answer 1

do you have a book or a link on how to do this (its been a while)

Answer 2

no i dont have

Answer 3

I believe you would want to break apart that denominator. Are you familiar with partial fraction decomposition?

Answer 4

no can u show me ??

Answer 5

Your starting point is:
\( \displaystyle \frac{s^2 + 6}{(s^2 + 4)(3s - 4)} \)

Your ultimate goal is something that looks like this:
\( \displaystyle \frac{As + B}{s^2 + 6} + \frac{C}{3s - 4} \)

So we would set these equal and try to find those coefficients:
\( \displaystyle \frac{s^2 + 6}{(s^2 + 4)(3s - 4)} = \frac{As + B}{s^2 + 6} + \frac{C}{3s - 4} \)

Answer 6

I have heard of two ways of doing it, but I am most familiar with just multiplying by the common denominator and trying to set like-coefficients equal to each other.

\( \displaystyle s^2 + 6 = (As + B)(3s - 4) + C(s^2 + 4) \) multiplying through with the commmon den.

Answer 7

So
\(s^2 + 6 = 3As^2 - 4As + 3B s - 4B + Cs^2 + 4C \)
\(s^2 + \color{#555555}{0s} + 6 = (3A + C)s^2 + (3B - 4A)s + 4C - 4B \)
So you can create a three-variable system of equations to solve for A, B, and C.
3A + C = 1
3B - 4A = 0
4C - 4B = 6

May want to check my work over here, it gets a little messy. :)

Answer 8

okay i did like what u did now but how i can find the value of A ,B and C ??

Answer 9

If you have that system of linear equations, you just have to solve it. There are several ways of doing so. If you know linear algebra, maybe cramer's rule or reduced row echelon form. Otherwise, we could make this substitution:
3A + C = 1 ==> C = 1 - 3A
into equation 3: 4(1 - 3A) - 4B = 6 --> 4 - 12A - 4B = 6

-12A - 4B = 2 reduce both sides by division of 2.

-6A - 2B = 1
3B - 4A = 0 This would be an easier 2-variable system. Solve for A, B here and plug back for C.

Answer 10

3A + C = 1 ==> C = 1 - 3A into eq.3
3B - 4A = 0
4C - 4B = 6

4(1 - 3A) - 4B = 6
4 - 12A - 4B = 6
-12A - 4B = 2

-12A - 4B = 2
-4A + 3B = 0 ==> 12A - 9B = 0 add eq.1 and 2

-13B = 2
B = -2/13

-4A + 3(-2/13) = 0
-4A - 6/13 = 0
-4A = 6/13
A = -6/52 = -3/26

For C, Back to C = 1 - 3A. C = 1 - 3(-3/26) = 1 + 9/26 = 35/26
I would check over this work again! But it seems to match wolfram: http://www.wolframalpha.com/input/?i=partial+fraction+decomposition++%28s%5E2+%2B+6%29%2F%28%28s%5E2+%2B+4%29%283s+-+4%29%29

Answer 11

Once you have these values for A, B, and C, you just put them back into the original context:

\( \displaystyle \frac{s^2 + 6}{(s^2 + 4)(3s - 4)} = \frac{(-3/26) s - 2/13}{s^2 + 4} + \frac{35/26}{3s - 4} \)
and then take the inverse laplace of that new entity, which should come down to knowing the inverse of the individual fractions (usually they can be looked up on a table).

Answer 12

okaay thanks alot but can u help me in the last part how to end this context

Answer 13

Alright.
\( \displaystyle \mathcal{L} ^{-1} \left[ \frac{-3/26 \; s - 4/26}{s^2 + 4} + \frac{35/26}{3s - 4} \right] \)
Is now what we are looking at. We ought to break it up further into individual fractions:
\( \displaystyle \mathcal{L}^{-1} \left[ \frac{-3}{26} \frac{s}{s^2 + 4} + \frac{-4}{26} \frac{1}{s^2 + 4} + \frac{35}{26} \frac{1}{3s - 4} \right] \)
Because we should have a sense that these are the sine, cosine, and exponential functions' laplace transforms.

Answer 14

The inverse laplace transform is linear, so you can take the inverse of each individual term and factor out any coefficients. Your only goal is to get something from this table:
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
Namely, you want expressions inside the laplace transform that look like those of sine, cosine, and exponential. You are free to multiply, divide, and factor constants to arrange it at this point.

Answer 15

For example, you would first have:
\( \displaystyle \mathcal{L}^{-1} \left[ \frac{-3}{26} \frac{s}{s^2 + 4} \right] \)
The s in the numerator and s^2 + 4 in the denominator indicate the form of cosine transform:
\( \displaystyle \mathcal{L} \left[ \cos at \right] = \frac{s}{s^2 + a^2} \)
So we can factor out the -3/26 and get that inverse laplace transform back.
\( \displaystyle \frac{-3}{26} \mathcal{L}^{-1} \left[ \frac{s}{s^2 + 4} \right] \)
Where a=4, so we have the inverse transform of -3/26 cos (2t)

Answer 16

a^2 = 4, and a = 2. Not a=4 **

Answer 17

The next part is then looking a lot like the sine transform.
\( \displaystyle \mathcal{L} \left[ \sin at \right] = \frac{a}{s^2 + a^2} \)
We just need to multiply and divide by 2 (same thing as 1=2/2) and factor out any excess to make it match the form we want. Get the picture with that? :)

Answer 18

\( \displaystyle \frac{-4}{26} \frac{1}{s^2 + 4} = \frac{-4}{26} \color{green}{\frac{2}{2}} \frac{1}{s^2 + 4} = \frac{-2}{26} \frac{2}{s^2 + 4} \)
There we now have the right form multiplied by a constant!

Answer 19

Lastly for the last term, it looks like the exponential transform:
\( \mathcal{L} \left[ e^{at} \right] = \frac{1}{s - a} \)
This one requires a slightly trickier factor of 3 from the denominator.
3s - 4 = 3(s - 4/3). But you end up with a constant multiplied to the correct transform, so you can convert it back to 35/26 * 1/3 e^(4/3 t) and you would just add the three inverses together to finish!

Answer 20

A lot of this really comes down to algebraic manipulations. At any rate, if there is anything I can go over more clearly, let me know! :)

Answer 21

last one i have not seen before

Answer 22

The exponential function transform?

Answer 23

no the way u use

Answer 24

ohhh i got it now it is okay

Answer 25

Alright! :)

Answer 26

sir for the second one u multiply and divide by 2 how u got \[-2/26\]

Answer 27

From
\( \displaystyle \frac{-4}{26} \times \frac{2}{2} \times \frac{1}{s^2 + 4} \)

The -4 in the numerator and 2 in the denominator cancel.
\( \displaystyle \frac{-2 \times \cancel{2}}{26} \times \frac{2}{\cancel{2}} \times \frac{1}{s^2 + 4} \)

The numerator's 2 goes onto the fraction, and -2/26 remains outside. You could technically just factor -4 into -2*2 and do the same.

Answer 28

Sometimes they don't quite work out that way like if it were -5 instead of -4, so multiplying and dividing can be applied more generally.

Answer 29

okayy i got it now thanks alot for everything sir

Answer 30

You're welcome! Good luck with the work! :D

Answer 31

seam to you sir

Answer 32

Oh, and one more note: I would definitely recommend looking into partial fraction decomposition because these problems sometimes love to make use of it. There are likely many resources online for it, and there are some specific cases that can become more tricky than others (like 1/(x + 1)^2 = A/(x + 1) + B/(x + 1)^2

Something like this may be of use: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/ode/laplace/pf/pf.html