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Steady-state errors

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0.0. 4.4 1

Steady-state errors

• Let us consider the following feedback system:

- - -

6 G(s) (1)

r(t) e(t) c(t)

If the input signal r(t) is a unitary step, unitary ramp or unitary parabola:

r(t) = R0u(t), r(t) = R0t, r(t) = R0 2 t2

the corresponding steady-state errors are given by the following formulas:

ep = R0

1 + Kp , ev = R0

Kv , ea = R0 Ka where Kp, Kv and Ka:

Kp = lim

s→0G(s), Kv = lim

s→0sG(s), Ka = lim

s→0s2G(s) are the position, velocity and acceleration constants.

• Using the Laplace transform method:

e = lim

s→0s E(s) = lim

s→0s

 1

1 + G(s) R(s)



• Qualitative time behaviors of the three output responses:

diagram

(2)

4.4. STEADY-STATE ERRORS 4.4 2

• Step input signal:

er = R0 1 + Kp

If the system is of type 0 the position constant coincides with the static gain (Kp= K) and the steady-state error is finite. If the system is of type 1 or 2 the position constant is infinite (Kp= ∞) and the steady-state error is zero.

• Ramp input signal:

er = R0 Kv

If the system is of type 0, the velocity constant is zero (Kv= 0) and then the steady-state error is infinite. If the system is of type 1, the velocity constant is finite (Kv = K) and the steady-state error is finite (R0/Kv).

If the system is of type 2, the velocity constant is infinite (Kv = ∞) and and the steady-state error is zero.

• Parabola input signal:

er = R0 Ka

If the system is of type 0 or type 1, the acceleration constant is zero (Ka = 0) and the steady-state error is infinite. If the system is of type 2, the acceleration constant is finite (Ka= K) and and the steady-state error is finite (R0/Ka).

• Internal model Principle: referring to system (1), an input signal r(t) can be neutralized with a zero steady-state error if all the poles of its Laplace transform function R(s) are also poles of the G(s) function.

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