characetristic equation

[niga]

Hi there,

let there be a transfer function G(s)= s+2/Ms^2+bs+K

the denominator is the characteristic equation and is equated to zero because ithe roots of this eqn describes the character of time response of the system.

Can someone tell me

Why?Why the denominator?

This may sound stupid but i am new to control systems.

Please do help me.

Thanx a lot!

 

[niga]

oops :lol: sorry abt the spelling its

Characteristic Equation

 

[checkmate]

The roots of the characteristic equations are also called the poles of the system. These poles are perhaps the most important parameter in Control Theory.
Assume that the input is an impulse function, U(s)=1. Then Y(s)=G(s).
If you take the inverse laplace transform of G(s), you will find that the time response will comprise of a summation of exponentials exp(pt), where p are the poles of the system.
You can easily see that a positive p will cause y(t) to go to infinity as t increases.
A highly negative p will cause y(t) to go to zero rapidly as t increases.
Because the time response is exponential in nature, the effect of a highly negative pole dies out quickly as compared to that of a mildly negative pole.
As such, we usually only take note of the 2 least negative poles, and the reduced characteristic equation is now (s-p0)(s-p1)=s^2-2Sw+w^2.
S(Pronounced zeta) is called the damping factor. w(Pronounced omega subscript n) is called the natural frequency. All the response parameters like rise time, overshoot and settling time may be computed from these 2 values.

 

[niga]

and hence the denominator is teh characteristic equation...yeah?

Hey checkmate,thanx for ur help.I think thats quite clear to me.

 

[Agent 009]

Gr8 Explanation! I didn't never delve that much into characteristic equations :? ... Good job, checkmate, a 'real' checkmate, may I add :wink: ...

 

[checkmate]

Actually, this is just basic control theory, and I did make mistakes. We usually note the 2 least negative poles instead of the 2 most negative poles.

 

[niga]

Agent009 is right it was a very good explanation.
thanx again,checkmate.

 

[Agent 009]

Actually, this is just basic control theory, and I did make mistakes. We usually note the 2 least negative poles instead of the 2 most negative poles.

Euh, why? To have a stable system?

 

[checkmate]

The more negative poles are "faster" poles and die out rapidly, leaving the "slower" less negative poles that more closely approximates the real response.

To ensure a stable system, we need to ensure that there are no positive poles in the closed loop transfer function. A pole with zero real part will lead to a marginally stable oscillatory response.

 

[Agent 009]

Yeah, this I know... The thing is, I dunno why :? :?: :?:

 

[niga]

duh..?yeah..why is that?
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Oh!Oh! But I am still an Electronics Newbie.

 

[instruite]

The meaning of unstable system is that the output of the system goes into oscillations beyond control or simply output is just beyond control of the system
and When you say you have a controls system its output should be controllable and which is possible only if the system is stable

 

[niga]

yup,thats ok instruite but....

To ensure a stable system, we need to ensure that there are no positive poles in the closed loop transfer function. A pole with zero real part will lead to a marginally stable oscillatory response.

why is that?
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Oh!Oh!But I am still an Electronics Newbie.

 

[crust]

Consider a pole in the negative plane on the real axis at point -r. If you go through the math, you will find that such a pole when excited dampens out at a rate given by e^-(t/r), so as time increase, the response goes to 0. So now you can see that if you move the pole to the positive plane, that equation just goes to infinity and the system just oscillates with increasing amplitude. If they are exactly on the imaginary axis, then they are totally undamped and will just ring forever.

 

[checkmate]

I've edited my explaination above to correct some mistakes.
Each pole p will contribute an exponential component of Ae^(pt).

A positive pole will mean this exponential explodes to infinity as time increases, hence an unstable system. It doesnt matter if the CLTF has 100 negative poles, but a single positive pole, no matter how small it is, will lead to an unstable output.

A zero pole will lead to DC output at amplitude A.

Purely imaginary conjugate pole pairs will lead to oscillatory outputs. You can do the math easily.

Negative poles will attenuate to zero.

Highly negative poles will attenuate to zero exponentially faster than mildly negative poles.

 

[niga]

okay,i think its pretty much clear and will get clearer as i moe on.
thankyou checkmate.
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Oh!Oh!But Iam still an Electronics Newbie!!!

 

[Agent 009]

Yeah, I remember now :!: ... The stability comes from the zero-at-infinite time (steady state) stuff...

 

[samcheetah]

i remember a nice joke our teacher used to tell us about poles.

once people were going in an aeroplane and the flight attendant announced that the city of warsaw is at your right so all the "Poles" (polish people, not the control systems poles) move to the right side of the plane. so all the "Poles" moved to the right half of the plane, the plane became unstable and it crashed.

so thats why when "Poles" move to the right, the system becomes unstable :lol:

 

[niga]

Hey Crust, thanx for ur response.I understand it now.
Sam, thats a good one on "poles" :lol:

 

[Agent 009]

i remember a nice joke our teacher used to tell us about poles.

once people were going in an aeroplane and the flight attendant announced that the city of warsaw is at your right so all the "Poles" (polish people, not the control systems poles) move to the right side of the plane. so all the "Poles" moved to the right half of the plane, the plane became unstable and it crashed.

so thats why when "Poles" move to the right, the system becomes unstable :lol:

LOL, Hahaha... I'll remember that, 10x! Verry funny ...