regarding linear systems

Jan 27, 2006 12 Replies

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kickee 20 years ago

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

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Jerry Avins 20 years ago

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

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kickee 20 years ago

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

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Fred Marshall 20 years ago

Cool. Now define "x".

Fred

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kickee 20 years ago

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

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Jerry Avins 20 years ago

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

Jerry

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Scott Seidman 20 years ago

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

Why is a step bounded?

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Cameron Dorrough 20 years ago

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

Cameron:-)

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Fulliautomatix 20 years ago

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

In the time honoured tradition of successive differentiation

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Jerry Avins 20 years ago

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

Jerry

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Fulliautomatix 20 years ago

Indeedly not

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

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François Chap 20 years ago

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.

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Setanta 20 years ago

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

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