# NCEES 127



## Ricky (Oct 25, 2007)

I see how the Open Loop Transfer Function was arrived at, however, I'm having problems with the Closed Loop Transfer Function. The problem is asking for the closed-loop damping ration. Any last minute assistance on this one would be greatly appreciated.

Thanks,


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## benbo (Oct 25, 2007)

Ricky said:


> I see how the Open Loop Transfer Function was arrived at, however, I'm having problems with the Closed Loop Transfer Function. The problem is asking for the closed-loop damping ration. Any last minute assistance on this one would be greatly appreciated.
> Thanks,


Never fear. If nobody else answers I'll answer when I get home and look at the problem. I know that's the last minute, but I'm sure someone else will answer first. Or you can post some more details here and I can give it a shot. Be sure to post the correct answer if you do, so I can make sure I'm not giving you a bum steer. Good luck.


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## Flyer_PE (Oct 25, 2007)

Ricky said:


> I see how the Open Loop Transfer Function was arrived at, however, I'm having problems with the Closed Loop Transfer Function. The problem is asking for the closed-loop damping ration. Any last minute assistance on this one would be greatly appreciated.
> Thanks,


I'm far from a controls guy but here's my attempt at this one.

If you have the EERM, look in Section 6 of Chapter 63 for the block diagram that applies here.

G1 = 50/(20s2+40s)

G2 = 1

The closed loop transfer function is G1/(1+G1G2)

The rest is just algebra to get the denominator in a form that gives you the damping ratio.

Jim


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## benbo (Oct 25, 2007)

IFR_Pilot said:


> I'm far from a controls guy but here's my attempt at this one.
> If you have the EERM, look in Section 6 of Chapter 63 for the block diagram that applies here.
> 
> G1 = 50/(20s2+40s)
> ...


This logic looks right, although I don't have the problem. Jim is usually right. Assuming these numbers, plug into the formula

Then you would mult top and bottom by 20s^2+40s and you would end up with

50/(20s^2+40s+50)

Then divide top and bottom by 20

2.5/(s^2+2s+2.5)

You need natural frequency, wn = sqrt(2.5) = 1.58

then the s^1 term equals 2(psi)wn

So 2(psi)1.58 = 2

Something like that. I may have made an error or two here.

I'll check it out in the night if you are still confused.


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## Ricky (Oct 25, 2007)

http://engineerboards.com/style_emoticons/...-smiley-048.gif

Got it. Thanks again for your responses. 20 hours to exam time....


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## benbo (Oct 25, 2007)

Glad you got it. I already saw a math mistake I made above - dividing 50/20 I got 1. I corrected it. duh. Anyway, you got the problem and that's the important part.


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