# Thread: Laplace Transform - Determine response of the system

1. ## (Solved thanks) Laplace Transform - Determine response of the system

Hi I've an exam on system dynamics in a few weeks and as trying to understand whats going on.

Here is the problem I'm currently stuck on and it's wrecking my head.

With the assistance of laplace transform tables determine an expression for the response of the system.

T(s) = 1/(s^2+0.2s+1)

when subject to a unit ramp input. Assume initial conditions are zero.

y(t) = t-0.2-0.96((e^(-0.1t))/0.995)sin(0.995t)-0.2((e^(-0.1t))/0.995)sin(0.995t-1.471) , t > 0

I just can't find any book or web site that could explain me how to do this.

so far my working are:

1/(s^2+0.2s+1) = (1/0.995)(0.995/(s+0.1)^2+0.995^2)

which transforms into (1/0.995)((e^(-0.1t))/0.995)sin(0.995t)

But well this is not the answer ... I'm sure I'm supposed to do something about it been subject to a unit ramp input but this is where I'm getting completly lost.

Please can someone Help me??
Thanks
Bob.

2. Hi

I post an answer even though you have beaten us at rugby yesterday

OK now have a look at the link
Transformée de Laplace - Wikipédia

It is in French but see the table at the end of the page
At line 2c you have the ramp whose Laplace transform is $\displaystyle \frac{1}{s^2}$ ("p" is in French and "s" in English)

I think that you have to multiply this expression by T(s)

$\displaystyle \frac{1}{s^2} \:\frac{1}{s^2+0.2s+1}=-\frac{0.2}{s} + \frac{1}{s^2} + \frac{0.2s-0.96}{s^2+0.2s+1}$

I did not perform the whole job but :
- the first term will give -0.2
- the second term will give t
- in the third term there is 0.2 and 0.96 which are in the final solution

Therefore I think it goes in the right direction !

3. Thanks a lot for you reply ...

That makes sense, I thought I was supposed to multiply it by 1/s^2 but then I couldn't do the transformation

And I still can't do it!

I'm lost at the partial fraction you've done
How did you do it?

First I would of never guessed to use the denominators
s,s^2, s^2+0.2s+1

Second, when I try to solve it using those denominators I can't ...
I'm using the technique of A B & C but in order for me to set say B & C to zero I have to let s equal to zero and then it all equals to zero and therefore my A equals to zero!

Hope you understand where I'm coming from.
Thanks ... sorry about the match ...

4. You have to find a, b, c and d such as

$\displaystyle \frac{1}{s^2} \:\frac{1}{s^2+0.2s+1}=\frac{a}{s} + \frac{b}{s^2} + \frac{cs+d}{s^2+0.2s+1}$

One way to proceed is to use the common denominator $\displaystyle s^2 \s^2+0.2s+1)$, to expand the numerator and to set the s^3, s² and s factors to 0.

Another way is to take some values of s.

An alternative way is to multiply both sides by s²

$\displaystyle \frac{1}{s^2+0.2s+1}=as + b + \frac{s^2(cs+d)}{s^2+0.2s+1}$

and to set s to 0 which gives $\displaystyle 1= b$

5. Oh I see!!! Thanks a lot!!!

I had never come across those kind of partial fractions before and I was spending all my time researching laplace transforms when the answer was what I thought it should be in the first place lol very ironic.

I will research those partial fractions then so and the rest should work out fine.

Once again thanks alot for you help

bob.

6. Ok, it all look good but I got stuck again ...

I managed to get as far as this

Y(s) = $\displaystyle (-0.2/s) + (1/s^2) + (0.2s-0.96)/(s^2+0.2s+1)$

then I split the last term into 2 and I get this

Y(s) = $\displaystyle (-0.2/s) + (1/s^2) - (0.96)/(s^2+0.2s+1) + (0.2s)/(s^2+0.2s+1)$

Up to there it's all good ... the 1st 3 terms are as per solution previously posted ... but I can't transform the last term at all!!

$\displaystyle (0.2s)/(s^2+0.2s+1)$

7. $\displaystyle \frac{0.2s-0.96}{s^2+0.2s+1} = 0.2\: \frac{s+0.1}{(s+0.1)^2+0.995^2} - \frac{0.98}{0.995}\:\frac{0.995}{(s+0.1)^2+0.995^2 }$

Should give therefore

$\displaystyle 0.2 \:e^{-0.1t}\:\cos(0.995t) - \frac{0.98}{0.995}\:e^{-0.1t}\:\sin(0.995t)$

And

$\displaystyle \cos(0.995t) = - \sin\left(0.995t - \frac{\pi}{2}\right) = - \sin\left(0.995\left(t - \frac{\pi}{1.99}\right)\right)$

$\displaystyle \cos(0.995t) = - \sin(0.995(t - 1.579))$

8. Hi thanks a lot for all the help

Woke up fresh this morning and spotted the answer on the bus

I used the transform formula for the outstanding term

$\displaystyle s/(s^2 + 2zwns+wn^2)$

and used the values 0.1 for z(zeta) and 1 for wn

worked great and got the given solution.

Once again thanks for all your help

bob.