# 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)
Best regards srinivas
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kickee wrote:

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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Engineering is the art of making what you want from things you can get.
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I mean the state variables, x, x(dot) should be bounded for bounded input.
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Cool. Now define "x".
Fred
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Umm.. x - shaft angle of the motor. x(dot)-it's velocity.
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kickee wrote:

How about x = angular velocity; x(dot) = angular acceleration? By most definitions, the planets have stable orbits.
Jerry
--
Engineering is the art of making what you want from things you can get.
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Jerry Avins wrote:

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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Fulliautomatix wrote:

As you wish. But then x isn't bounded.
Jerry
--
Engineering is the art of making what you want from things you can get.
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Jerry Avins wrote:

Indeedly not
This is why you are a control engineer and I am but an egg
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Why is a step bounded?
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Scott
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To stop people falling off, of course !! ;-)
Cameron:-)
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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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Il n'y a que les imb�ciles qui ne changent pas d'avis. Quoique...

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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