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**Chapter 7 Steady State Error <<<4.1>>>**

###Continuous Time Signals###

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**Test waveforms for evaluating steady-state errors of position control systems**

Steady-state error is the difference between the input and the output for a prescribed test input as t →∞

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**Test inputs for steady-state error analysis and design vary with target type**

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**Steady-state error: a. step input; b. ramp input**

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**Closed-loop control system error: a. general representation; b**

Closed-loop control system error: a. general representation; b. representation for unity feedback systems

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**System with: a. finite steady-state error for a step input; b**

System with: a. finite steady-state error for a step input; b. zero steady-state error for step input For pure gain K as in (a), Css= K ess or ess =1/K Css there will always be error But if the pure gain is replaced by an integrator as in (b) the steady state error will be zero.

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**Feedback control system for Example 7.1**

Problem Find the steady-state error for the system in(a) if T(s) = 5/(s2+7s+10) and the input is a unit step. Solution: Using E(s) = R(s)[1-T(s)] We can apply the final value theorem To yield e(∞) =1/2

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**Steady-state error in terms of G(s)**

E(s) =R(s) – C(s), C(s) =E(s)G(s) so E(s) = R(s)/1+G(s) Using final value theorem

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**Steady-state error in terms of G(s)**

For step input, the steady state error will be zero if there is at least one pure integration in the forward path. For ramp input, the steady state error will be zero if there is at least 2 pure integration in the forward path. For parabolic input, the steady state error will be zero if there is at least 3 pure integration in the forward path.

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**Steady state error for systems with no integrations**

Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function. Solution First verify the closed loop system is stable. For input 5u(t) For input 5tu(t) For input 5t2u(t)

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**Steady state error for systems with one integration**

Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function. Solution First verify the closed loop system is stable. For input 5u(t) For input 5tu(t) For input 5t2u(t)

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**Static error constants**

Position constant step input Velocity constant ramp input Acceleration constant parabolic input

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**Steady state error via static error constants**

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

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**Steady state error via static error constants**

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

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**Steady state error via static error constants**

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs. Solution First verify the closed loop system is stable. Thus for step input, For ramp input For parabolic input

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**Feedback control system for defining system type**

System type is the value of n in the denominator, or the number of pure integrations in the forward path.

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**Relationships between input, system type, static error constants, and steady-state errors**

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**Feedback control system for Example 7.6**

Problem Find the value of K so that there is 10% error in the steady state.. Solution: Since the system is Type 1, the error must apply to a ramp input, thus and and K = 672 Applying Routh-Hurwitz we see that the system is stable at this gain.

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**Feedback control system showing disturbance**

Applying final value theorem we obtain

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**Previous system rearranged to show disturbance as input and error as output, with R(s) = 0**

Assume step disturbance D(s) = 1/s We get This shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of G1(s) or decreasing the dc gain of G2(s)

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**Feedback control system for Example 7.7**

Problem Find the steady state error due to a step disturbance for the shown system.. Solution: The system is stable, we find

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**G(s) = G1(s)G2(s) H(s) = H1(s)/G1(s)**

Forming an equivalent unity feedback system from a general non-unity feedback system G(s) = G1(s)G2(s) H(s) = H1(s)/G1(s)

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**Non-unity feedback control system for Example 7.8**

Problem Find the system type, the appropriate error constant, and the steady state error for a unit step input.. Solution: The system is stable, we convert the system into an equivalent unity feedback system The system is Type 0, and Kp= -5/4. The steady-state error is

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**Steady-State error for systems in State Space**

The Laplace transform of the error is E(s) = R(s) –Y(s), and since Y(s) = R(s)T(s). Then E(s) = R(s)[1-T(s)] Using T(s) =Y(s)/U(s) = C(sI-A)-1B + D We have E(s) = R(s)[1-C(sI-A)-1B] Applying final value theorem, we have

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**Steady-State error for systems in State Space**

Example: Evaluate the ss error for the system described by Solution: Substitute A,B, and C in For unit step, R(s)=1/s and e(∞)=4/5, for unit ramp, R(s) = 1/s2 and e(∞)= ∞

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