# Feedback Controller Step Response

Discussion in 'General Electronics Chat' started by MarioMan, Apr 23, 2010.

1. ### MarioMan Thread Starter New Member

Apr 22, 2010
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I'm having an issue conceptualizing the step response of a feedback controller. The controller I have is a voltage follower.

Calculating the transfer function is not terribly dificult.
Vo = A * (Vp - Vo)
Vo = A*Vp + A*Vo
Vo*(1 + A) = A * Vp
Vo = ( A / (1 + A) ) * Vp

So, for large values of A, the output voltage should be the input voltage.

The problem I have is this: upon closer inspection, the system doesn't appear to be BIBO stable, but I know it should be. Here's my thought process. Assume that the entire system is at a state of equilibrium.
Vo = 0 , Vp = 0, and A = 100
Now, apply a unit step function to the input. At some time "x" the input will jump from a zero to a one.
Vp = u(t - x)

If I assume that the signal propogates down the system at the speed of light (just as a convenience), then the input to the amplifier immediately after this step from zero to one will be one. Since Vo is zero at the time of the step function, it makes sense that 1- 0 = 1.
So, 1 gets multiplied by 100. Now Vo is 100. If the way that I'm thinking about this problem is correct (which I'm sure it is not), then the system will oscillate to plus and minus infinity rather quickly.

Any thoughts?

~Mario

2. ### rjenkins AAC Fanatic!

Nov 6, 2005
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A feedback system can oscillate easily if it is poorly designed or wrongly adjusted.

Some of the critical parameters are the phaze shift through the system and the frequency response of the various stages.

In practice, amplifiers have some form of frequency compensation, either added or designed in and feedback systems such as servo controllers usually have a 'response' or stability control which adjusts the amplifier characteristics to match the mechanics being controlled.

http://www.compumotor.com/catalog/cataloga/A36-A38.pdf

That shows the worst case as a decaying oscillation. In practice, adjusting the response further away from the correct setting willl eventually cause full self-sustaining oscillation.

Last edited: Apr 23, 2010
3. ### MarioMan Thread Starter New Member

Apr 22, 2010
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I am assuming an ideal system here. I modelled this feedback controller after an op-amp buffer, which I know to be a stable system.

Also, please humor the assumption that the supply voltages are extremely high. High enough to allow the oscillation I described.

4. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
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If you have an 'ideal' amp there is no delay so no phase shift.

If you add delay in the overall feedback loop, you are simply creating a phase shift oscillator. Any high frequency opamp would oscillate if wired as a buffer with the feedback path routed through a delay line such as a piece of cable.

The supply voltage does not really matter, as long as it is enough for the amp to work properly.

5. ### MarioMan Thread Starter New Member

Apr 22, 2010
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Are you saying that the buffer won't work as drawn? When you say high frequency op-amp, are you refering to the bandwidth (i.e., the location of the amplifier's poles)?

6. ### mik3 Senior Member

Feb 4, 2008
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Yes, he refers to the bandwidth of the op amp.

In the way you describe your idea in post #1, it is like there is a delay in the response of the system. This is the same as a phase shift. If the phase shift is greater than 180 degrees then the system will oscillate. The phase shift depends on the characteristics of the system and the input signal frequency. If the input signal's frequency is such that at a phase shift of 180 degrees the system gain is greater than 1, then it will oscillate. If you assume an ideal system then there is no delay (phase shift) and thus no oscillations.

7. ### russ_hensel Distinguished Member

Jan 11, 2009
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your are applying a step function, which has infinite frequency components. all op amps, even ideal ones have a high frequency roll off at some point ( kind of part of the definition of op amp ) This is required for stability.

And you do not have a feedback controller, you simply have a unity gain amplifier.

8. ### Ghar Active Member

Mar 8, 2010
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Your problem is you are selectively assuming instantaneous responses.

Basically, if you want to assume everything is instantly fast and everything is perfect then the system must satisfy the equation you wrote.

Vo = A/(1+A) Vp

The only fixed node is the input Vp, the error signal and the output depends on the system.

This system has no poles, it's just a pure gain of A/(1+A)

9. ### MarioMan Thread Starter New Member

Apr 22, 2010
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I am aware that I have a unity gain amplifier. This is a feedback controller, as its output is fed back to its input, negatively. I am also aware that the transfer function is A/(1+A).

I am assuming instantenous response because something about the math doesn't add up if you make that assumption. Also, I wanted to study the "ideal op amp" model as a feedback control system. (Again, I am aware that real-world analog systems have bandwidth issues.)

I guess the real question I am trying to ask is: Are feedback systems that lack poles BIBO stable?

If the system is not dynamic (no energy storage, anywhere), then a step input will instantaneously result in the output reading 100. If you follow that logic, then it seems that the subtraction of the output and the input will be negative 99, which would not give a transfer function of A/(1 + A). If there are no poles, and no time delay, it is impossible for the input to the amplifier to be 1 and -99 at the same time.

Dec 5, 2009
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Apr 22, 2010
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