# NCEES #127



## Kuku (Aug 5, 2008)

Can someone walk me through this problem with step by step instructions. My references aren't helping out at all.

"For the parameters shown in the figure, the closed loop-damping ration is most nearly...."

25 --- 2 / 0.5s+1 --- 1/40s

I seem to be pretty weak in the area of transfer functions and feedback theory.

Thanks for the help.


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## benbo (Aug 5, 2008)

A few of us solved a lot of these problems. There is a post around here that recaps.

I hope you can understand what follows here-

Open loop (call it G(s)) just multiply the boxes together = 50/(20s^2+40s)

Closed loop = G(s)/[1+G(S)]

= [50/(20s^2+40s)]/{ 1+ [50/(20s^2+40s)]}

Multiply top and bottom by (20s^2+40s)

You get =

50/[20s^2+40s+50]

To get the char equation you have to get rid of the 20 in front of s^2 so divide top and bottom by 20

2.5/[s^2+2s+2.5]

Look at denominator =

By definition

w = sqrt(2.5)

2psiw= 2

Psi = .633


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## Kuku (Aug 6, 2008)

Thanks benbo.

I'm finally getting my memory fresh on these types of problems. I know how to represent a resistor, capacitor, and inductor (R, sL, and 1/sC), but what about a voltage source that varies with time? Would it be V/s?


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## benbo (Aug 6, 2008)

Kuku said:


> Thanks benbo.
> I'm finally getting my memory fresh on these types of problems. I know how to represent a resistor, capacitor, and inductor (R, sL, and 1/sC), but what about a voltage source that varies with time? Would it be V/s?


I'm not sure I understand your question. For the test at least, just worry about a DC source and an AC source with a constant frequency omega(w).

For the DC source it doesn't really make much sense to use Laplace. At least not to me. I would analyze that in a different way, doing a transient type analysis where the cap charges up, etc. Basically a cap is open to DC, and starts out with 0 volts and charges up to Vin. The current goes the opposite direction. An inductor is the opposite - start with max voltage and 0 cxurrent and eventually becomes a short circuit. Anyway, that's probably confusing. If you don't understand that, read the sections on transient analysis.

For an AC source (which is really what they are asking about in Laplace problems) it would just be V (just the amplitude) - or technically V(s) I guess. The "s" is basically the frequency. Don't worry about representing it in your analysis until the very end. THe frequency is always going to be the same - only the Amplitude and phase angle change based on the analysis, and you don't worry about that until the end. I would probably use superposition if I have multiple voltage sources. But you're probably not going to see that in the AM. THe main thing is to try to get the problem so you have one output voltage and one input voltage with a network in between - Vo(s)/Vi(s) = T(s)

And if you have sources with different frequencies, well that would be a mess.

If I made a mistake, or somebody can explain better, please do so. I have a feeling I messed up something simple here.


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## Kuku (Aug 6, 2008)

If you got it, check NCEES question 114. You basically have a circuit and have to show the circuit in terms of I(s). It contains a source that is dependent on time u(t). I have no problem solving the majority of that problem, its just they have the u(t) being divided by s and I'm just trying to understand where it came from. My initial answer was A.

Thanks for the help.


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## benbo (Aug 6, 2008)

Kuku said:


> If you got it, check NCEES question 114. You basically have a circuit and have to show the circuit in terms of I(s). It contains a source that is dependent on time u(t). I have no problem solving the majority of that problem, its just they have the u(t) being divided by s and I'm just trying to understand where it came from. My initial answer was A.
> Thanks for the help.


That probably has something to do with the algebra for solving the problem. I don't think that's the first step. Let me look at it. Maybe somebody else will get it before I do.

edit:

Maybe I'm wrong. Let me look at this some more, my memory is a little hazy.


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## Flyer_PE (Aug 6, 2008)

If I'm not mistaken, the term u(t) represents a step function. This translates into the S domain as 1/s.

The result is that a voltage source shown as 50 u(t) = 50 * 1/s


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## benbo (Aug 6, 2008)

Flyer_PE said:


> If I'm not mistaken, the term u(t) represents a step function. This translates into the S domain as 1/s.
> The result is that a voltage source shown as 50 u(t) = 50 * 1/s


You're right. As usual. I forgot about u(t) until I looked in my transform table.

I was sloppy in my explanation above, because it turns out there are functions that are not quite DC and not quite AC - eg the step function.

So it pays to check the table just to make sure, and just use it to transform everything in the circuit. Just in case.


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