Which Equation to use?

Thread Starter

woolerland

Joined Oct 19, 2010
3
Let me use an example to explain my problem.

http://i43.tinypic.com/akwpaa.png

at t=0, the switch closes.

lets say we use the transfer function approach and we obtain the poles to this function.

the function for current or voltage will be either:

=a(1-e^-p1*t)+b(1-e^-p2*t)

or

= a(e^-p1*t)+b(e^-p2*t)

in a generic case == a(1*e^-t/T)+b(1*e^t/T) where T= period.

i know from experience that this is case 2 as when the switch closes, it becomes a RLC circuit thus the steady state will equal 0.

is there any real stratergy to know which equation to use?
 

t_n_k

Joined Mar 6, 2009
5,455
One probably needs to keep in mind at least a couple of things.

The circuit damping will have significant bearing on the equation type.

The initial and final (boundary) values for the circuit parameters need to be ascertained. This will normally point to the existence or otherwise of any constant terms. Unless a fixed source exists somewhere in the circuit to provide steady state end-of-transient values or there is zero damping with initial conditions provided, then the existence of constant or steady-state terms would be discounted.

In the end I guess one either has to gain a progressively informed "appreciation" for the likely circuit response and the representative equations or otherwise derive the circuit transient behavior from first principles. Either way that usually means doing lots of practice problems.
 

WBahn

Joined Mar 31, 2012
29,976
The equations you offered up are for a first-order system and this system is second order.

[edit: I was assuming that your p1 and p2 are real because of your "generic case" statement. Are you allowing for the poles to be complex?]

Forgetting the specifics of this circuit and just dealing with some of the questions you raised in terms of picking between the two equations (i.e., assuming that one of them really is the correct solution), all you have to do is consider the initial and steady state cases.

Before we go further, however, let's clean up your description some.

You say:

at t=0, the switch closes.
This isn't good enough, though most people will infer the correct interpretation. Instead, you should say something like:

"... at t=0, the switch closes after having been open a long time."

The meaning here is clear (even though still not explicit), namely, the switch as been open long enough for all transients from the prior opening to have died out.

... thus the steady state will equal 0
The steady state what? The voltage or the current? Where? If you are restricting the discussion only to the branch with the RLC series components, then you are correct in this case. But note that the branch with the switch in it does NOT go to zero current in steady state and, were there a resistor in that branch next to the switch, the voltage across the capacitor would also not go to zero in steady state. Were the inductor moved into the branch with the switch, then the current in it would not go to zero in steady state.

In general, alway be wary of making sweeping generalizations.
 
Last edited:

t_n_k

Joined Mar 6, 2009
5,455
Hi woolerland,

Since this an overdamped case the roots will be real and different. Your assumption of a general solution comprising two decaying exponential terms is valid.
 

WBahn

Joined Mar 31, 2012
29,976
Hi woolerland,

Since this an overdamped case the roots will be real and different. Your assumption of a general solution comprising two decaying exponential terms is valid.
I haven't analyzed the circuit and I certainly can't tell that it is overdamped by inspection. That doesn't mean that there isn't a simple means of doing so. Is there?

I don't think I agree with the assertion that his assumption of that particular general solution is valid, because it would appear that he made that assumption without doing so in the context of knowing he was working with an overdampled system (unless the exponents in question are complex, and part of his post implies they are and part of it implies they aren't, so I just don't know).
 

t_n_k

Joined Mar 6, 2009
5,455
Thanks WBahn,

I assumed the OP was aware that this is an overdamped case. I did the quick calculation for a series LCR circuit based on my understanding that a value of R>2√(L/C) ohms would make this an overdamped situation.

In any event we probably should await the OP's return at which time they [hopefully] elaborate their level of understanding - although experience tells me the hoped for "return" isn't a certainty.

Nice to have you around with some useful comments and advice.:)

Regards,

t_n_k
 

WBahn

Joined Mar 31, 2012
29,976
Thanks WBahn,

I assumed the OP was aware that this is an overdamped case. I did the quick calculation for a series LCR circuit based on my understanding that a value of R>2√(L/C) ohms would make this an overdamped situation.

In any event we probably should await the OP's return at which time they [hopefully] elaborate their level of understanding - although experience tells me the hoped for "return" isn't a certainty.

Nice to have you around with some useful comments and advice.:)

Regards,

t_n_k
It's been so long since I've worked with such circuits that I had completely forgotten crossover parameters such as these. Well, not completely, since the first thing I would have done after analyzing it would have been to solve for the conditions for which it were critically damped.

You are right that, in this case, the rule for critical damping in a series LRC circuit applies since, with the switch closed, the rest of the circuit is isolated and the response is independent. Now, put any thing in that same branch as the switch and all bets are off.

Thanks for the compliment; likewise, I enjoy reading your responses, too.

Now if I could just get someone to answer the question I asked in my very first post -- and which was my sole reason for joining! -- I would be ecstatic!
 

Thread Starter

woolerland

Joined Oct 19, 2010
3
sorry for the way my original post was posted, it was somewhat unclarified.

the generic case was presuming the roots were real.

anyway my knowledge was wrong. the standard way of laying solutions out is Ae(t*p1) + Be(t*p2) + (steady state solution) +Ne(t*pn)

where n is the order of the system.

i think: In this case i know the steady state was 0 as the transfer function has a product of P which if we set to 0 to find the steady state will equal 0 but i have to test this theory first.
 

WBahn

Joined Mar 31, 2012
29,976
I still don't think you have ever said what it is you are solving for? A transfer function relates one signal to another, but you haven't specified either. It isn't enough to just talk about "the transfer function" as though there were only one transfer function in a circuit.
 

WBahn

Joined Mar 31, 2012
29,976
As danced around earlier, my problem is centered on saying that the equations are for the general case, and not a special case of the system being overdamped. Nor did he give any indication that he had a basis for claiming that the system if overdamped.
 

Thread Starter

woolerland

Joined Oct 19, 2010
3
ok, i'm going to put this topic on hold.

it seems i've been jumping the gun on my studies. The online notes are incomplete and vital theory is missing. i had a very enlightening lecture this morning which pretty much solved all my problems. I'm going to have a crack at some problems and see how it goes.

Thanks anyway.

p.s. sorry for my clarity been rubbish, my only excuse is that this is my first time i've had the need to actually ask for help online. My original point was to know which equation would you use to display the voltage/current of the resistor in the RLC loop but this was also solved in this mornings lecture.
 

WBahn

Joined Mar 31, 2012
29,976
p.s. sorry for my clarity been rubbish, my only excuse is that this is my first time i've had the need to actually ask for help online.
Don't worry about that, it is not unusual at all. Up to this point, you have probably interacted with people that were very familiar with the problem you were working on, such as your instructor or classmates. As a result, they have the necessary context in order to understand very poor descriptions from you (and, of course, it works the other way, too). This is not completely a bad thing, since it allows for pretty efficient communication. But, as a result, you have inadvertently trained yourself to assume that the person you are describing something to will be able to understand what is being asked based on limited information and you are now finding out this isn't the case. Most of us have been there (and occassionally visit the place and have to relearn to be more explicit). On line is particularly difficult because you are almost completely limited to the written word and perhaps a diagram (which is invaluable IF you mark it up pretty good). It takes a while to learn how to be clear and specific and the only way to learn is by not being clear and specific and having to correct things.
 
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