# Steady State Errors. Recall the closed-loop transfer function of the system, is

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2 E( R( C(.. () C( E( G( (2) R( Substituting the (2) into () and rearrange, yields, E( (3) G( By applying the Final Value Theorem, we have: e ss e( t) se( t s0 sr( s 0 G(. (4) Eq. (4) shows that the steady state error depends upon the input R( and the forward transfer function G(. The effects of various types of standard test signals on the steady-state errors are investigated next. Note: Since we are concerned with the difference between the input and output of a feedback control system after steady-state has been reached, our discussion is ited to STBLE systems. - Ess for Step (Positional) input Steady-state error for step function input, r(t) = u(t) R( = s, is the magnitude of the input Then from eq.(4), we get sr +G Let us define: s(/ + G = + G = + G Position error constant: K p = G (dimensionles Thus e ss = +K p P a g e 2

3 2- Ess for Ramp (Velocity) input Steady-state error for ramp function input, r(t) = t R( = s 2, : is the slope of the ramp sr + G s(/s 2 ) + G = s( + G) = s + sg = sg Let us define: Velocity error constant: K v = sg (sec ) Thus, e ss = K v 3- Ess for Parabolic (cceleration) input Steady-state error for parabolic function input, r(t) = 2 t2 R( = s 3, sr + G s(/s 3 ) + G = s 2 ( + G) = s 2 + s 2 G = s 2 G Let us define: The static cceleration error constant: K a = s 2 G (sec 2 ) Thus, e ss = K a P a g e 3

4 Note: Control systems may be classified according to their ability to track polynomial inputs, or in other words, their ability to reach zero steady-state to step-inputs, ramp inputs, parabolic inputs and so on. Kp, Kv, Ka :ability to reduce steady state error. If error constant is infinite, we can achieve zero steady state error (ccurate tracking). System types The open-loop transfer function of a unity feedback system can be written in standard form: where, G( = K (s+z )(s+z 2 ). s n (s+p )(s+p 2 ) The term of s n in the denominator of G( indicates that there are n integrations in the forward path. ccording to the value of n, systems can be classified into different types. Specifically, a system is called Type 0 system if n = 0, Type system if n =, or Type 2 system if n = 2, and so on. Note that this classification is different from that of the system order. Examples: G G G s s s s Ks 2 s s 3 2 s 2 s s s K 2s 3 s K 0.5s Type 0 Type Type 3 Relationships between input, system type, static error constants, and steady-state errors can be summarized in the following table: P a g e 4

5 System Input Step Ramp Parabolic 0 + K p K v 0 0 K a Conclusions: Note:. Ramp and parabolic inputs should not be applied to type 0 system. 2. Parabolic input should not be applied to type system. 3. For type 3, ess = 0 for all inputs. ess for a linear combination of inputs can be superimposed. The method of finding ess fails for sinusoidal inputs. Example : The open-loop T.F of a unity feedback system is given by: G( = 00 s 2 (s+4)(s 2 +5s+25) Find static error constants and the ess of the system when it is subjected to an input of Solution r(t) = 2 + 4t + 2t 2 Taking L.T for input, we get R( = 2 s + 4 s s 3 e ss = + K p + 2 K v + 3 K a K p = G = 00 s 2 (s+4)(s 2 +5s+25) = P a g e 5

6 K v = sg = 00 s(s + 4)(s 2 + 5s + 25) = K a = s 2 G = 00 (s + 4)(s 2 + 5s + 25) = 00 (4)(25) = So; e ss = = 4 H.W : Repeat the pervious example for the system with : G( = 5(s+) s 2 (s+2)(s+5) H.W 2: The open-loop T.F of a unity feedback system is given by: G( = K (s+)(s+2) Find the ess of the system with input is a unit step when: ) K =, 2) K = 0, then discuss the result Example 2: For the system as shown below : R + - G( C G( = 0 s 2 (s+2) Find ess for unit step, unit ramp, and unit parabolic inputs. Solution the ch. eq. will be : s 3 + 2s = 0 the system will be UNSTBLE.and error analysis is meaningless. Example 3: Given the control system shown in the figure below, find the value of K so that there is 0% error in the steady state. R Let the input will be unit step, unit ramp, and unit parabolic. Solution + - K(s + 5) s(s + 6)(s + 7)(s + 8) C P a g e 6

7 Since the system type is, the error stated in the problem must apply to a ramp input; only a ramp yields a finite error for in a type system. Thus, e ss = K v = K v = 0. K v = 0 Therefore, K v = sg 0 = sg = 5K = 5K k = = 672 H.W 3: Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t 2 u(t) by first evaluating the static error constants. Steady State Error for Non-Unity Feedback systems Control system often do not have unity feedback because of the compensation used to improve performance or because of the physical model for the system. general feedback system, is shown in the following figure. To convert a nonunity feedback system to a unity feedback system, form a unity feedback system by adding and subtracting unity feedback paths, as shown in figure. Next combine H( with the negative unity feedback as shown in figure. P a g e 7

8 Finally combine the feedback system consisting of G( and [H( ] as shown in figure. Notice that the final figure shows E(=R(-C( explicitly and we can use all algorithms explained before. Example 4: For the system shown in the figure, find the steady state error for a unit step input. Solution We can calculate the steady-state error anymore : The negative value for steady-state error implies that the output step is larger than the input step. P a g e 8

9 H.W 4: Calculate the error constants and determine ess for a unit step, ramp and parabolic functions response of the following system. R + - s(s + 2) C 5(s + ) (s + 5) Steady-state response to disturbance inputs Feedback control systems are used to compensate for disturbance or unwanted inputs enter a system The following figure shows a feedback control system with a disturbance, D(, injected between the controller and the plant. We now re-derive the expression for steady-state error with the disturbance included. P a g e 9

10 C( = C r ( + C d ( C( = G G 2 + G G 2 R( + Sub. in; E( = R( C( G 2 + G G 2 D( The error E( is given by the following formula; E( = + G G 2 R( G 2 + G G 2 D( To find steady-state value of the error, we apply the final value theorem to equation of E( and obtain; ess = s E( = ess = e r ( ) + e d ( ) s + G (G 2 ( Where; Er: is the steady-state error due to R(, R( Ed: is the steady-state error due to disturbance D(. How to reduce the error due to disturbance? If it is assume for example that D( is a step disturbance, D( = s the last equation, we get sg 2 ( + G (G 2 ( D(, substituting this value into e d ( ) = sg 2 ( + G (G 2 ( D( = G 2 ( + G ( The value G ( is sometimes called dc gain of the system. The last formula shows that the ess due to step disturbances can be reduced by increasing the dc gain of the controller G. P a g e 0

11 Example 5: Consider a system shown in the figure, find the steady state error due to unit step disturbance D(. Solution The system is stable!! Check that e d ( ) = G 2 ( + G ( = = 0.00 H.W 5: Consider a system shown in the figure, find the steady state error due to unit step disturbance D(. H.W 6: Evaluate the steady-state error for the system described by the following equation for unit step and unit ramp inputs. Use the final value theorem. P a g e

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