System type when H(s) is not 1.

HI everybody!

I have a question and I can't find an answer.

It is easy to find the system type and the stationary error when the feedback H(s)=1. Just observe how many poles the G block has in the origin and you'll get the type number, and the stationary error is calculated using the following expression: 1 lim s. R(s) . ---------------- s->0 1 + G

where R(s) is the signal input.

But, when H(s) doesn't equal to 1, that expression is not valid.... so, how may I calculate it?

Please, could you help me out?

Regards... Alejandro

But if I am given a system which

Reply to
Alejandro
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Is this for a homework assignment?

I assume you mean H(s) is your feedback gain.

If H(s) depends on s but goes to 1 at DC (s = 0), then the relationship still holds (try it - do the math).

If H(s) goes to some other constant (call it C) at DC then either your error will be 1 - 1/C, or you wanted the overall gain to be 1/C in the first place. By the way, if you have a type I or II system and an "intereseting" disturbance input before H(s) you can regulate the system so the output from the feedback is servoed to zero. This is how rate integrating gyros are used to measure rate, and how force-balance accelerometers measure force. They turn the usual "improve a poor plant with good sensors" paradigm into "improve a poor sensor with a good plant".

If H(s) has a zero at s = 0, then you'll have a nice velocity regulator. This is what gyro stabilized systems are. You can extend this to a double zero, and you'll be regulating acceleration, etc.

If H(s) has a pole at s = 0 then your system output will always settle to 0, assuming no disturbance inputs between your output and the input to H(s).

Reply to
Tim Wescott

First of all, thanks for answering.

Is this for a homework assignment?

  • No, I´m trying to pass a control systems exam and I don't have this topic clear enough.

I assume you mean H(s) is your feedback gain. If H(s) depends on s but goes to 1 at DC (s = 0), then the relationship still holds (try it - do the math).

  • Yes, I did it, and it is ok.

If H(s) goes to some other constant (call it C) at DC then either your error will be 1 - 1/C, or you wanted the overall gain to be 1/C in the first place. By the way, if you have a type I or II system and an "intereseting" disturbance input before H(s) you can regulate the system so the output from the feedback is servoed to zero. This is how rate integrating gyros are used to measure rate, and how force-balance accelerometers measure force. They turn the usual "improve a poor plant with good sensors" paradigm into "improve a poor sensor with a good plant".

If H(s) has a zero at s = 0, then you'll have a nice velocity regulator. This is what gyro stabilized systems are. You can extend this to a double zero, and you'll be regulating acceleration, etc.

If H(s) has a pole at s = 0 then your system output will always settle to 0, assuming no disturbance inputs between your output and the input to H(s).

  • You said the stationary error will be 1 - 1/C, but, is that the error to an impulse input, to a ramp input, or to a parabolic input?? And I still don't know how to get the system type when H(s) is not 1.

Anyway, I tried the expression you gave me, and right now I'm really puzzled, because I solved a problem using two methods. First, in order to get the stationary error to an impulse input, I did E(s)=R(s)-M(s).R(s) and I got the error. Then,the other method, I used 1-1/C to get the error, and the results are not the same, here's what I did:

Be the system:

---- R(s) +| | ------ -------| |------------| G(s) |--------------- C(s) | | ------ | ---- | - | | | | | ------ | --------------| H(s) |------------ ------

where G(s)=1/(s+1) H(s)=1/(s+2)

R(s)=1/s (an impulse)

First method:

(s+2) M(s)= --------------- (s+1)(s+2)+1

c(inf)= lim s . R(s) . M(s) = 2/3

s->0

r(inf) = lim s. R(s) = 1 s->0

e(inf) = r(inf)-c(inf) = 1/3

This error is ok, because I simulated it with simmulink, and I could see in the scope the difference between the impulse and the system response when t->inf, and it's 1/3. Second method:

C= H(0)= 1/2

e(inf) = 1-1/C = 1-2= -1

As you may see, the errors don´t match, what am I doing wrong?

Again, thanks for helping me.

Alejandro

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Reply to
Alejandro

What you are doing wrong is listening to me when I am oversimplifying my answer. Let me try to clarify things by throwing more mud in the water:

For a stable system who's type and input match (i.e. type 1 with a step, type 2 with a ramp, etc.), the controller will just keep pushing harder until it's input is zero. If you define the error as the difference between the output and the input, then my (slightly too) clever 1/C relationship holds. If you define the error as the output of the summation block then it will go to zero.

How you define the "type" of a system when there is an integrator or a differentiator in the feedback is really about how your particular community views control loops. As an example, when I work on gyro stabilized loops I can either look at it as a position loop or a velocity loop. If it's a position loop then it has a differentiator in the feedback, in which case the _overall_ loop type is one less than the number of integrators in the controller. From the perspective of the summation output, the loop type is exactly the number of integrators. With a velocity loop it's easier: the feedback looks like a low-pass filter and the loop type is just the number of integrators in the controller.

So when you talk about loop type you need to be careful, and if it's at all ambiguous you need to define what you mean that day. In your case this means that whatever your instructor is expecting as an answer is the right answer, so you need to look at your notes and your book and distill the correct response from that.

Reply to
Tim Wescott

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