Completing the Square.

Ok. This is embarassing because I am a senior in a EE program and I can’t for the life of me remember how to complete the square. I am trying to do some Laplace and think I need to complete the square but am unsure how because I haven’t done it since Freshman year!


I need to Laplace this (for my circuits class). I may have constructed the transfer function wrong, but if I did it correctly I should be able to invlaplace this bad boy.

Do I need to complete the square on this to make it laplaceable? If so how? And if not, how the hell DO I complete the square?

I’ve never actually tried to learn Laplace transforms, so I can’t help you with that, but I can help you with completing the square.

Notice that:

(x+a)[sup]2[/sup] = x[sup]2[/sup] + 2ax + a[sup]2[/sup].

In particular, you can get the constant term by taking the coefficient of x, divide it by 2, then square it.

You have s[sup]2[/sup] + 40s + 2000.

Take 40, divide it by 2, then square it (gives you 400):

(s[sup]2[/sup] + 40s + 400) + 1600

(s+20)[sup]2[/sup] + 1600.

Thanks for the refresher. But, that doesnt solve my problem as the entire equation is this:

(40s)/(s+40s+2000). If I complete the square on the bottom it will look like this: (40s)/((s+20)^2 +(40^2)) which I cannot find in any of my Laplace tables.

Anyone who knows thier DE I have to Partial Frac this thing dont I? When I do I get (s+20-40i) and (s+20+40i) for my roots. If I plug and chug into the original equation I get K=20+10i and K*=20-10i .

Does this sound correct to anyone? The reason I am asking is because I am pretty sure that I set up my s domain circuit correctly and this seems to produce a someone strange impulse function when I inverse laplace this bad boy.

For anyone who is curious what my original circuit looks like it is very simple. it is one loop. in series there is a 5H inductor, a 100 micro F cap, and a 200 ohm resistor. The output is over the 200 ohm resistor. I am 99.99% sure I set up my transfer function correctly. The original equation is my transfer function.

If any EE guys want to double check my transfer function if they have time that would be super. I have an exam on Monday and I need to know how to do this for the test!!!

:mad: :mad: :mad:

If you have a reasonably complete table of transforms you should find that form directly. Your table might have the constants rearranged so that the quadradic ends in 1. This is done by dividing both numerator and denominator by 2000. Or you can find the roots of the denominator using the quadratic formula and use partial fractions to get two terms of the form:

A/(s + a) + B/(s + b). In any case your transform will be of the form:

Ae[sup]-at[/sup] + Be[sup]-bt[/sup] which looks to me like A, B, a, and b will be such that it will turn out as either a sin or cos.

I feel like I am pounding my head on the desk here. I pumped the function into mathcad and it does result in exactly the form you guessed, but I cannot find a plug and chug for the function as it sits. When I partial frac it I get 2 distinct complex roots so my answer is all jacked. I must be doing something wrong here. I cannot get it into A/(s+b) form!!!

If anyone would take the time to explain that Laplace all the way thru or tell me which form it can fit into I would apprecitate it. I have spent 6 hours on this one piece of this shitty problem and I cannot stand it any more!!!

There’s nothing wrong with complex roots! If you just work out the problem like you would with real roots, everything should work out fine.

You’ve got (40s)/((s+20)[sup]2[/sup]+1600), which equals A/(s+20+40i) + B/(s+20-40i) for some A and B. Cross-multiply and solve: We need A+B=40, and (20-40i)A+(20+40i)B=0. Solve for A and B by whatever method you like; you’ll get complex answers, but that’s okay. Just plug the values of A and B into Aesupt[/sup]+Besupt[/sup], which is the inverse laplace transform. Then use Euler’s formula to replace esupt[/sup] with e[sup]-20t/sup, and similarly for the other term. If you’ve solved for A and B correctly, the imaginary terms will cancel out at the end.

But if you really, really want to avoid complex numbers, you can do that too. You’ll need to find the laplace transforms of e[sup]at[/sup]cos(bt) and e[sup]at[/sup]sin(bt) in a table somewhere and use those.