# NCEES 128



## jdd18vm (Feb 19, 2008)

Can someone explain how to get the net difference for the Poles (4) and Zeros (2) in the Transfer Function. I'm missing the 2nd zero and 4th pole. (I guess with 1 and 3 I'd get 2 and the same -40dB/decade...but

100s (s+1)

(s+2)(s+3)(s+4)^2

thanks Benbo for the other solution.

John


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## benbo (Feb 19, 2008)

jdd18vm said:


> Can someone explain how to get the net difference for the Poles (4) and Zeros (2) in the Transfer Function. I'm missing the 2nd zero and 4th pole. (I guess with 1 and 3 I'd get 2 and the same -40dB/decade...but
> 100s (s+1)
> 
> (s+2)(s+3)(s+4)^2
> ...


You have a zero at the origin (because of the s term), and -1 because of the s+1 term.

You have poles at -2 and -3 because of the s+3 and s+2, and you have a double pole at -4 (because it is squared you have to count it twice).

Correct me anyone if I am incorrect.


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## jdd18vm (Feb 19, 2008)

benbo said:


> You have a zero at the origin (because of the s term), and -1 because of the s+1 term.
> You have poles at -2 and -3 because of the s+3 and s+2, and you have a double pole at -4 (because it is squared you have to count it twice).
> 
> Correct me anyone if I am incorrect.


Thanks, that makes sense

so regardless of the numerical value it (ie 100) in the 100s, it's at the origin?


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## benbo (Feb 19, 2008)

jdd18vm said:


> Thanks, that makes sense
> so regardless of the numerical value it (ie 100) in the 100s, it's at the origin?


Yes -

THese plots are basically on the complex plane - real and imaginary axis.

all real poles are on the x (or real) axis. The thing that tells where on the real axis the pole (or zero) is you find from the (s-a) - it is the a. So s is techncally s-0, meaning a real pole or zero at the origin.

Complex conjugate poles are located off the real axis. But I doubt you will have to worry about those in the AM.

As far as the 100 - it is a little more complicated, it is a scale factor that basically tells where the plot starts. For the AM test you probably don't have to worry too much about that.


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## jdd18vm (Feb 19, 2008)

benbo said:


> Yes - THese plots are basically on the complex plane - real and imaginary axis.
> 
> all real poles are on the x (or real) axis. The thing that tells where on the real axis the pole (or zero) is you find from the (s-a) - it is the a. So s is techncally s-0, meaning a real pole or zero at the origin.
> 
> ...



Right, actually knew most of that , but determining the final equations (basic partial fraction resolution) still struggling with

thanks again


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## benbo (Feb 20, 2008)

I figured you probably knew that - but actually I added a little confusion by bringing in the axes. I conflated two related but different types of plots. THe one with the real and imaginary axis is the pole zero plot, or root locus. That is the one that has the zero at the origin.

The one this problem is talking about (I think) is a Bode magnitude plot.

But the poles and zeros are the same, like I said.

You probably got that, but I wanted to make sure I didn't add to the confusion.


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## Dark Knight (Feb 22, 2008)

I am cleaning my files and found this.

Maybe this is not the best review document for Controls but gave me a fair understanding of what to look for and, as a matter of fact, helped me in one question when I was able to eliminate one of the choices just knowing the kind of system it was and the response.

It is rustic and my writing is terrible so my apologies in advance.


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## jdd18vm (Feb 22, 2008)

BringItOn said:


> I am cleaning my files and found this.
> 
> 
> 
> ...


Thanks BIO you actually emailed this to me way back its in my Binder. I admit I spent nearly no time on controls last go around. That showed. There is a lot I still don't get but certainly am at least getting SOME things out of it this time.

Thanks for your input Benbo


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## ndekens (Apr 8, 2008)

So I understand the poles and stuff but how did they determine that each pole/zero contributes +-20Db?


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## Flyer_PE (Apr 8, 2008)

^^I can't remember the derivation of it. However, for the purposes of the exam, I treated it as a given that each pole or zero affected the slope of the curve by plus or minus 20db/decade. The location of the pole or zero determines at what frequencies the slope of the line changes.

There are others on the board who are better at explaining frequency response. Hopefully, I haven't hacked this up too bad.

Jim


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