# Solve for the Differential Characteristic Roots

#### watersnick

Joined May 26, 2016
4
Hey gang, I had this circuit for T > 0 and I keep getting a Critically Damped condition when using Nodal Analysis at the node with the Inductor, resistor, and capacitor. Likewise, I did a KVL aroud the outside and ended up with the same Crit Damped Cond.

Can someone please deny or confirm this!

Cheers,

Nick

#### watersnick

Joined May 26, 2016
4

#### MrAl

Joined Jun 17, 2014
11,578
Hello there,

I do not get the same result as you do, and since critical damping is a condition which is exactly defined, i can only say that it is not critically damped.

Did you try using a second method to find out this information? That's the best way to check your results.

#### watersnick

Joined May 26, 2016
4
Hey Mr AI thanks for taking the time to check this out. Would mind sharing your process, likewise, can you tell if anything in my process is out of place?

Also, I have done a KVL around the entire outside loop and ended up with the same roots.

Cheers,

Nick

#### MrAl

Joined Jun 17, 2014
11,578
Hello again,

I'll have to check out your math, but in the mean time here are some more ideas.

First, as you probably know, a critically damped response has no overshoot, and neither does an overdamped response. The critically and over damped responses are monotonic, so if you see any hint of oscillation even if it is damped, you know it must be underdamped. If we do a simulation of this circuit, we see a slight overshoot, which right there says that it must be underdamped. It's small, but even one nano volt overshoot tells us it must be underdamped. If you want to set your own limit as to what constitutes a real life critically damped circuit, that's up to you, but the theoretical limit is the true limit which is even a tiny tiny bit of overshoot, even that which cant be measured.

Second, an analysis of the circuit shows an underdamped response. Transforming to the time domain the overshoot shows up again, and in the frequency domain looking at the damping coefficient we see it is less than 1 so it is underdamped.

Working in the frequency domain is pretty simple so i did it that way. Using the transformed impedances:
zC=1/(s*C)
zL=s*L

we can lump all the impedances into one single impedance at the node of interest to ground. If this unforced circuit was driven with a step current source it would show a voltage response at that node.
But even without doing that if we just look at the denominator of the transfer function we can get it into the form:
denominator=s^2+2*d*w*s+w^2

and here the variable 'd' is the damping factor. If that factor, once all the known values are substituted, comes out to anything other than 1, then the response is NOT critically damped. This is a simple test, as long as we get the right transfer function.
The way i did it was to first lump L and R2, then lump that with R1, then lump that with C. That gave me the single impedance from the node to ground, then i could 'pretend' to excite it with a constant current step function or just leave it alone, and solve for that variable 'd'. If 'd' comes out to less than 1, then it is not critically damped but is really underdamped. The value i got was around 0.9 which means there is a little overshoot and that it is really underdamped.

For your reference, the transfer function i got was:
(R1*(R2+s*L1))/(s*C*R1*R2+R2+s^2*C*L1*R1+R1+s*L1)

where R1=2 ohms and R2=8 ohms as per your schematic.

We can calculate the overshoot too if you like, but knowing that there is at least some overshoot tells us the response is underdamped.

Laplace Transforms were made to make this stuff easier, might as well use them

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#### DGElder

Joined Apr 3, 2016
351
You have treated the 8 ohm resistor as if it were zero ohms. Why?

#### MrAl

Joined Jun 17, 2014
11,578
You have treated the 8 ohm resistor as if it were zero ohms. Why?
Hi,

I assume you were talking to watersnick because i think it is clear that i included the 8 ohm'er.
It's always good to mention at least who the reply was for if not quote it

#### DGElder

Joined Apr 3, 2016
351
Hi,

I assume you were talking to watersnick because i think it is clear that i included the 8 ohm'er.
It's always good to mention at least who the reply was for if not quote it
Yes, the OP left out the inconvenient 8 ohm resistor leaving a simple critically damped parallel RLC circuit. Adding the 8 ohms back in will dissipate energy more quickly making the circuit overdamped. There are two equations (two loops or two nodes) in two unknowns to solve this circuit, not the one the OP used.

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#### MrAl

Joined Jun 17, 2014
11,578
Yes, the OP left out the inconvenient 8 ohm resistor leaving a simple critically damped parallel RLC circuit. Adding the 8 ohms back in will dissipate energy more quickly making the circuit overdamped. There are two equations (two loops or two nodes) in two unknowns to solve this circuit, not the one the OP used.
Hi,

I had not considered the shorting out of the 8 ohms resistor, i'll have to check that, thanks.

Sorry though i dont understand what you mean about dissipating the energy faster with the 8 ohm resistor as making the circuit overdamped, i think you meant "underdamped" ? That would happen if the energy dissipated more quickly. The response i got for the original circuit came out underdamped. I will go over it again without the 8 ohm (shorting it) and see what happens. That would explain what he did at first.

LATER:
Yes you are right, i got critically damped with the 8 ohm shorted out also. So for some reason he shorted that out, and that was probably the whole problem.