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I was reading about steady error calculation for system with H(s) having Nth order Zero at s=0.

In book it is written that steady error calculation is meaningless when closed loop transfer function(CLTF) M(s) is unstable(And it turns out same is the case here due to pole at origin in CLTF M(s)).

But in this example despite having pole at s=0. He is finding the value of steady state error (ess) for unit step input i.e,(R(s)=1/s). But i am not able to understand the justification he gave in the statement saying "The velocity control system is stable, Although M(s) has a pole at s=0, because objective is to control velocity with step input"

Also i don't get the meaning of the statement in last line that "if H(s) were constant for this type 2 system, the close loop system would be unstable. So,the derivative control in the feedback path also has a stabilizing effect".

I hope you all get the context so i am also telling the name of the book it is "Automatic Control Systems" By Farid Golnaraghi and Benjamin C. Kuo. And this problem is in the Chapter 5: Time domain analysis of control systems.

In Gist below paragraph is main source of problem: I think i am not understanding the term "Velocity control system" and how in this kind of system step input is used to control system output that contains ramp in steady state(figure is also attached for this statement)

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Sorry for such a long post but if you please clear my concept with reference to steady state error chapter i would be very thankful to you.

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The control system tries to make the error signal \$r(t) - b(t)\$ zero. Now note that the feedback element \$H(s)\$ has a zero at the origin. This means that, if the output of the plant is a position signal, then \$\frac{10}{s+5}\$ smooths it with a low pass filter and the \$s\$ then differentiates it. So \$b(t)\$ has units of velocity. This means that the input \$r(t)\$ also has the units of velocity.

Since the command is in units of velocity and the output has units of position, to find the (steady-state) velocity-error, we need to look at \$r(t) - \frac{dy(t)}{dt}\$. The differentiation of the output introduces a zero which cancels the pole at origin. So if we looked only at the velocities in the system, we will not see a pole at the origin and so the system will appear stable.

The position signal still grows unbounded for a step input; but we just happen to define our error as velocity error which is bounded.

Alternate explanation

redrawn system

Since the plant has two poles at origin and the feedback has one zero at origin, we can redraw the system so that the velocity state is separated out from the position state. Now, looking only at the velocity control system, if we redo the analysis, it can be seen to be stable with finite steady state error.

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  • \$\begingroup\$ I Got it now. Very simple, nice and clear explanation. Yes he is looking at velocity error so due to differentiation of output that pole at origin gets cancelled and it works like a charm.Thank you so much. \$\endgroup\$
    – MR LUN
    May 6, 2021 at 15:20

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