# Homework Help: Inverse Laplace Transform help

1. Jul 29, 2012

Hi guys :) I'm looking to get a jump start on my uni course and have been going through some topics on my own before classes start in the Fall, I've reached the point of Laplace Transform and Fourier maths - and it's tough!

I have a small question on an inverse laplace transform equation in my maths book. Problem is, there are no examples of this type to help me in it :(

1. I want to find the inverse laplace transform of the following:

H(s) = 15/(s^3 + 6s^2 + 15s + 15)

What I've been doing so far with these questions is splitting them into smaller fractions - but I can't seem to do it with this and am having trouble trying to figure out how to go about completing it.

Any help would be very appreciated!

3. The attempt at a solution
Would I be correct in simplifying the above example into: 15/[ (s + 2)^3 + 3(s + 2) + 1 ] and then trying to find the inverse laplace transform?

Last edited: Jul 29, 2012
2. Jul 29, 2012

### uart

It's only tricky because the cubic doesn't factorize nicely. Despite that however, I'd still approach it the usual way (break into simple and quadratic factors, then partial fractions) using floating point approximations.

Since a cubic with real coefficients has at least one real root you have to be able to split it into smaller factors.

For this example you could split it up as,

$$\frac{15}{(s-\alpha)( (s+a)^2 + w^2)}$$

BTW. Are you sure it's not just a typo? s^3 + 6s^2 + 15s + 14 factorizes nicely.

Last edited: Jul 29, 2012
3. Jul 29, 2012

You're definitely right about (s^3 + 6s^2 + 15s + 14) would have made things a lot simpler! But I doubt this is a typo in the book. I think it's intentionally like this.

It's actually part of a chapter on filters and has been associated with a Bessel Function (which I didn't mention since I haven't reached this stage of the book yet).

Anyways, I tried what you recommended and is this what you were saying:

real root = 2.3222
imag. roots = 1.8389 ± 1.7544

Thus, (s + 2.3222)(0.3036s^2) but how does this = (s^3 + 6s^2 + 15s + 15)?

Apologies in advance if I misunderstood what you were saying.

Last edited: Jul 29, 2012
4. Jul 29, 2012

### LCKurtz

If you take s - each of those roots and multiply that out you get
$s^3-6.0000*s^2+15.00005973*s-15.00018720-0i$ which is approximately your polynomial.

Last edited: Jul 29, 2012
5. Jul 30, 2012

So the factor of (s3 + 6s2 + 15s + 15) would be:

(s + 2.3222)(1.8389s - 1.7544is)(1.8389s + 1.7544is)

Is this right?

Wouldn't I need it in the following form: 15/[(s + 2.3222)(s2 + s + x)] with x being some numbers. This form would allow me to split the quadratic into (s - p1)(s - p2) which I can then further break down to:

k1/(s - p1) + k2/(s - p2) + k3/( s + 2.3222) ; and then find the inverse laplace transform of this.

Last edited: Jul 30, 2012
6. Jul 30, 2012

### uart

Yes solve numerically for the real root and then use polynomial division to extract the quadratic factor. Then express the quadratic factor in a "completed square" form.

Do you have any software to aid with the numerical stuff. I used the Octave (a freeware Matlab clone) "roots" and "deconv" functions to help with finding the roots and factors.

$$\frac{15}{(s + 2.32218535462609)(s^2 + 3.67781464537391 s + 6.45943269348336)}$$

$$\frac{15}{(s + 2.32218535462609)[(s + 1.83890732268696)^2 + 3.07785255205165 ]}$$

7. Jul 30, 2012

Thanks for the confirmation I will attempt the inverse laplace of this.

So far I haven't used any software, been relying on laplace tables/google for my self-study...but I should probably consider getting some software to help with these extremely tricky parts (i.e. finding roots etc).

This is what I got:

h(t) = 15*(0.30198e-2.3222t + e(-1.8389 - 1.7544i)t((0.0416i - 0.1510)-(0.0416i + 0.1510)e(3.5088i)t)

what I did was split it into the individual partial gains (k1,k2,k3) and solved for them;

So i.e. k1 = 15/[ (p1-p2)(p1-p3) ] k3 = 15/[ (p1-p3)(p2-p3) ] etc.

Is this the correct way?

Last edited: Jul 30, 2012
8. Jul 30, 2012

### LCKurtz

I believe you would expect sines and cosines with that form. For what it's worth, Maple agrees with uart's values in post #6 and gives the following answer for the inverse:$$4.529792164e^{-2.322185354t}-4.529792162e^{-1.838907323t}\cos{(1.754380959*t)}+1.247818511e^{-1.838907323t}\sin{(1.754380959t)}$$

9. Jul 30, 2012

Ok I will go through my work again, I must have done some part wrong. I didn't get any sines/cosines the first time I attempted it. Thanks for the clarification.

10. Jul 30, 2012

### LCKurtz

What I don't understand is, if you are just reviewing concepts, why punish yourself with a problem like this with all those decimals to deal with? You can review the concepts with similar, but tractable, problems. If you can invert something like$$\phi(s) = \frac{2}{(s-3)((s+2)^2+25)}$$
you know all you need to do the current problem. And I just made up that problem but I promise it won't be as ugly as what you are messing with.

11. Jul 30, 2012

The question's listed as a tough one in the book i.e. If I can solve this then I should have no problem in my Electronics Signals Analysis module. Which is why I figured I would give it a shot after I went through some of the other examples in the book.

It's part of a big question with further parts asking for system response of a unit step signal [ y(t) ] and then onto designing an ISIS circuit for this transfer function. So it pretty much covers a big chunk of the module and what I can expect.

12. Jul 30, 2012

### uart

I still think the question is probably a typo. There is no essential difference in how we approach the problem for one that factors nicely compared with an ugly one like the present.

In the case of a cubic with just one real root we do the following steps.

1. Solve for the real root, either numerically of sometimes guess and verify (with the aid of rational root theorem) for ones that factorize nicely.

2. Use polynomial division to take out the simple factor found in step 1.

3. If the remaining quadratic factor has no real roots then express it as a completed square as above.

4. Write in partial fractions form. For example in this case write as, (with known values for $\alpha, a ,w$)

$$\frac{15}{(s-\alpha)[(s+a)^2 + w^2]} = \frac{A}{s-\alpha} + \frac{B(s+a) + C w}{(s+a)^2 + w^2}$$

5. a) Use the "residue" method (mult both sides by $s-\alpha$ and then subst $s=\alpha$) to solve directly for A.

5. b) Subst in s=-a leaving C as the only unknown. Solve for C.

5. c) Subst in any other convenient number for "s" (eg $s=0$) leaving B as the only unknown. Hence solve for B.

6. With the function in it's now decomposed form, the inverse Laplace transform can be read directly from Laplace transform tables.

That is,

$$f(t) = A e^{\alpha t} + e^{-a t} \left( B \cos(w t) + C \sin(w t) \right)$$

Last edited: Jul 30, 2012
13. Jul 31, 2012

Thanks! This simplifies things a lot...I initially did it another way, but I'll try it this method. Definitely looks more organized.

14. Jul 31, 2012

### vela

Staff Emeritus
In case you want to torture yourself a bit more, I'll point out that the denominator is a depressed cubic, and you can show that the real root is given by
$$s+2 = \varphi^{-1/3} - \varphi^{1/3}$$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.

15. Jul 31, 2012

Thanks, read most of that article (still trying to get some parts of it pinned down).

16. Aug 2, 2012

Apologies if double-posting is against the forum rules, but I had a question related to this transfer function.

I've made an ISIS circuit for this transfer funtion as in this website: http://sim.okawa-denshi.jp/en/Sallenkey3Lowkeisan.htm and the question asks for a differential equation relating the output voltage and the input voltage.

What I've done is split the circuit into 3 nodes - A,B,C

Then I've found voltage equations for the 3 nodes using kirchoffs circuit laws - but the thing is, wouldn't this only give me a transfer function?

Firstly, are the 3 following equations right?

Va = (Va-Vin)/Z1 + (Va-Vb)/R2 + (Va-Vout)/Z2 = 0

Vb = (Vb-Va)/R2 + (Vb-Vout)/Z2 + (Vb-Vc)/R3 = 0

Vc = (Vc-Vb)/R3 + (Vc-0)/Z3 = 0

where Vc = Vout.

I'm not entirely sure since this is the first 3rd order circuit I'm using kirchoffs laws on.

But from here how do I proceed to find the differential equation for the relationship between Vin and Vout?

17. Aug 2, 2012

### uart

1. At this point you might get better response to this question in the Electrical Engineering forums.

2. Those nodal equations look correct.

3. The 3rd order sallen-key implementation on the page you link is a little smaller (in terms of component count) but harder to analyze compared with a "buffered" implementation that is often used. That is, you sometimes see a unity gain amplifier after the first R1-C1 stage so as to isolate it from the input impedance of the second stage. This allows us to analyze the two stages independently. The buffered version is quite a lot easier to analyze, but in practice if the first resistor of the second stage is large (R2 >> R1) then you can treat the two stages as approximately independent.

4. You can work out the differential equations directly from the transfer function by simply replacing "s" by the "d/dt" operator. (Assuming zero initial conditions).

Last edited: Aug 2, 2012
18. Aug 2, 2012

Will do

So in other words: 15Vin(s) = (s3 + 6s2 + 15s + 15) Vout(s) ?

19. Aug 2, 2012

### uart

That's correct. So the resulting DE is,

$$\frac{d^3 v_o}{dt^3} + 6 \frac{d^2 v_o}{dt^2} + 15 \frac{d v_o}{dt} + 15 v_o = v_i$$

20. Aug 13, 2012

thanks

I'm onto the next chapter now, and this same question is repeated but this time the question asks me to find the characteristic modes of the above equation.

What I did:

Vout = Ceλt

so for the differential equation: (λ3 + 6λ2 + 15) * Ceλt = 0

then to determine λ: λ3 + 6λ2 + 15 = 0

Then I need to solve for λ, but how do I do that?