regarding linear systems

Hi, I have one doubt abt openloop systems.

Suppose there is a openloop stable transfer funcion 1/(s*(s+1)). If i give step input to this, the response will be unbounded. How can i say the openloop system is unstable.

adding to this, If i give finite amount of voltage to a DC motor, the position or shaft angle goes to infinity as time increases. can we say the DC motor underloop is unstable for finite amount of input. (The DC motor has an integral term..like the tf mentioned above)

Any comments on this.

Best regards srinivas

Reply to
kickee
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You need to define what "stable" means in a way appropriate to the circumstance. A motor running continuously may (or may not) have a stable speed, but it doesn't have a stable position.

All linear equations are in a sense small-signal approximations. You can make a very linear spring by winding 10 meters of wire on a one-cm mandrel, but there's no sense in calculating the force it exerts when stretched to 15 meters -- that's outside the spring's linear range.

Jerry

Reply to
Jerry Avins

I mean the state variables, x, x(dot) should be bounded for bounded input.

Reply to
kickee

Cool. Now define "x".

Fred

Reply to
Fred Marshall

Umm.. x - shaft angle of the motor. x(dot)-it's velocity.

Reply to
kickee

How about x = angular velocity; x(dot) = angular acceleration? By most definitions, the planets have stable orbits.

Jerry

Reply to
Jerry Avins

"kickee" wrote in news:1138593515.546837.17140 @z14g2000cwz.googlegroups.com:

Why is a step bounded?

Reply to
Scott Seidman

To stop people falling off, of course !! ;-)

Cameron:-)

Reply to
Cameron Dorrough

How about x = angular position; x(dot) = angular velocity; x(double dot) = angular acceleration

In the time honoured tradition of successive differentiation

Reply to
Fulliautomatix

As you wish. But then x isn't bounded.

Jerry

Reply to
Jerry Avins

Indeedly not

This is why you are a control engineer and I am but an egg

Reply to
Fulliautomatix

You system is asymptotically unstable because it has a pole a zero (along the imaginary axis)

This is because the pole at zero is an integrator. It accumulates the input.

Reply to
François Chap

Youve just said it - use the final value theorem

well the longer you simulate it for the greater the distance travelled (rotated) by the shaft will grow and grow

Reply to
Setanta

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