# Stability of a cicuit: Laplace Transform

#### JohanEricson

Joined Aug 30, 2017
36
Hi!
I'm having a hard time getting when a system is stable or not.. I've read that the poles should have negative real part because that would give a negative power for the exponential e^(-at) term in the time domain. That makes sense!
But now i have a problem hat is really confusing me..
I have the circuit with a capacitor, inductance and a resistor.

Source: 12v
Capacitor: 1F
Inductance: 1H
Resistor: R
And the question is: For which values of R will we get an oscillating behavior for the current?

The answer the book gives: R>0.5.

But values for R above that still gives me a negative real part for the poles..? And in my mind the real part of the poles would get close to zero from the negative side but never actually get to zero or positive, because that would require us to have negative resistance. Or am i totally wrong?

So what makes the current unstable for R>0.5?

Also, the way i've done it i can show that the poles are all real part for R<0.5 but never actually going to be positive or zero.

Can someone please shed some light on this?

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#### MrAl

Joined Jun 17, 2014
11,448
Hi!
I'm having a hard time getting when a system is stable or not.. I've read that the poles should have negative real part because that would give a negative power for the exponential e^(-at) term in the time domain. That makes sense!
But now i have a problem hat is really confusing me..
I have the circuit with a capacitor, inductance and a resistor.

Source: 12v
Capacitor: 1F
Inductance: 1H
Resistor: R
And the question is: For which values of R will we get an oscillating behavior for the current?

The answer the book gives: R>0.5.

But values for R above that still gives me a negative real part for the poles..? And in my mind the real part of the poles would get close to zero from the negative side but never actually get to zero or positive, because that would require us to have negative resistance. Or am i totally wrong?

So what makes the current unstable for R>0.5?

Also, the way i've done it i can show that the poles are all real part for R<0.5 but never actually going to be positive or zero.

Can someone please shed some light on this?

Hello,

I think you may be confusing "when a system is stable or not" with "oscillating behavior". They are sometimes two different things because sometimes a damped oscillation can be called "oscillating behavior" and this is especially true if the damping is minimal so the although the oscillations are lengthy they do end after some very very long time, while an unstable system never stabilizes.

This is probably what is happening here because a value of R=1/2 will cause critical damping, so anything under that is exponential and anything over that is oscillatory (but damped nonetheless) and when R=1/2 it is a sum of exponential and ramp and constant.

So take another look at this and see if you understand this better now.

Also, it's a passive network so the only way it might be called unstable is if R goes to infinity.

#### JohanEricson

Joined Aug 30, 2017
36
Okay yeah that makes a lot of sense!
So how would you go about solving the "limit" for R?

Thank you!!

#### MrAl

Joined Jun 17, 2014
11,448
Okay yeah that makes a lot of sense!
So how would you go about solving the "limit" for R?

Thank you!!
Hi,

You're welcome

The sinusoidal response occurs when one of the roots becomes complex. That means the pole leaves the real axis.
Since this is second order we can look for a certain damping factor, or if we solve for 's' in the denominator of your:
s^2*R+s+R

and solving for s we get:
s=-(sqrt(1-4*R^2)+1)/(2*R)
s=(sqrt(1-4*R^2)-1)/(2*R)

Now we look and ask which one of these (or both) will become complex with R. Since they both have the same discriminant and when that becomes imaginary we get a complex result, we can use one of them:
d=1-4*R^2

since 'd' is under the square root, when d becomes negative we get an imaginary number. But just before that, we get zero which means it is critically damped and everything after that is sinusoidal, so we solve that:
1-4*R^2=0

and we get:
R=1/2

Now when R is even a tiny tiny bit above 0.5 (like say 0.5+1e-99) we get a sinusoidal response because then that makes the root complex. Anything under 0.5 (like say 0.5-1e-99) then the response is purely exponential.
A simpler example is R=0.49 causes a real solution, R=0.51 causes a complex solution, R=0.50 causes the critically damped real solution. So we can see that anything over 0.50 causes a complex solution, except for R approaching infinity which causes a true unending oscillation which may or may not be deemed unstable depending on the application when the amplitude is stable and less than infinite (if the amplitude becomes infinite than the circuit is unstable for every application that can not deal with that condition when that infinity is actually clamped to some finite value).

So the short answer is find out when one of the roots becomes complex, and that can sometimes be boiled down to some part of it becoming imaginary. The catch is that if two parts become imaginary and they are multiplied then they may come out to a real number, so you have to try to simplify it to be sure it is really complex. In your case here it was simpler though.

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#### JohanEricson

Joined Aug 30, 2017
36
Hi,

You're welcome

The sinusoidal response occurs when one of the roots becomes complex. That means the pole leaves the real axis.
Since this is second order we can look for a certain damping factor, or if we solve for 's' in the denominator of your:
s^2*R+s+R

and solving for s we get:
s=-(sqrt(1-4*R^2)+1)/(2*R)
s=(sqrt(1-4*R^2)-1)/(2*R)

Now we look and ask which one of these (or both) will become complex with R. Since they both have the same discriminant and when that becomes imaginary we get a complex result, we can use one of them:
d=1-4*R^2

since 'd' is under the square root, when d becomes negative we get an imaginary number. But just before that, we get zero which means it is critically damped and everything after that is sinusoidal, so we solve that:
1-4*R^2=0

and we get:
R=1/2

Now when R is even a tiny tiny bit above 0.5 (like say 0.5+1e-99) we get a sinusoidal response because then that makes the root complex. Anything under 0.5 (like say 0.5-1e-99) then the response is purely exponential.
A simpler example is R=0.49 causes a real solution, R=0.51 causes a complex solution, R=0.50 causes the critically damped real solution. So we can see that anything over 0.50 causes a complex solution, except for R approaching infinity which causes a true unending oscillation which may or may not be deemed unstable depending on the application when the amplitude is stable and less than infinite (if the amplitude becomes infinite than the circuit is unstable for every application that can not deal with that condition when that infinity is actually clamped to some finite value).

So the short answer is find out when one of the roots becomes complex, and that can sometimes be boiled down to some part of it becoming imaginary. The catch is that if two parts become imaginary and they are multiplied then they may come out to a real number, so you have to try to simplify it to be sure it is really complex. In your case here it was simpler though.
Ah okay!
And that sinusoidal component can be seen when i do the inverse transform back to time domain where the complex pole give a sinusoidal term in the expression right?
Thank you very very much for the awesome teaching!

#### MrAl

Joined Jun 17, 2014
11,448
Hi again,

You're welcome

Yes, the sine component can be seen when you do the inverse transform into the time domain, and it is usually damped so as time goes on the sine amplitude gets smaller and smaller.

I should add that again this only means a damped sinusoidal which is not in itself unstable. It starts to become application specific if the sinusoidal goes for too long though even though it eventually damps out.

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