12.7 Steady State Error

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1 Lecture Notes on Control Systems/D. Ghose/ Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there was a steady state error between the output and the reference input. In fact, the same steady state error occurs even in the case of the second order system. Why does this steady state error occur? Note that the plant (which has the transfer function G(s)) is driven by u e, where e is the error signal. If the error goes to zero then the output will also be driven to zero. So there should be a non-zero input to the plant to eep it close to the reference input. But our objective is to mae sure that the steady state output matches the reference input. How can we ensure this? There are basically three possible methods by which we can achieve this objective by driving the steady state error to zero or to very small values. Two of these are almost obvious but are fraught with practical difficulties. The third one is the best and gives rise to the idea behind integral control. Method 1: Use large control gain. Since, for both first and second order systems, the steady state output of the closed-loop system is given by,wehave, 1as So, by using large values of, we can ensure that the steady state output is as close to 1 as we want. In fact, using large also improves system response as we have seen earlier. But there is a serious problem. Loo at the input to the system u(t) e(t) [r(t) y(t)] Initially, y(0) 0 and r(0) 1. So u(0) r(0), that is, the input to the plant is initially times the reference step input. This signal later drops to low values. Now, thin of any common electrical device that operates at the line voltage of 30 V. If 10, then initially the voltage applied to the device will be 300 V, enough to damage it completely! Same problem for electronic components that operate with voltages ranging between 5 V to 1 V. None of them are designed to withstand voltages that are 10 times the specified voltage. Method : Command shaping: Scale the commanded or reference input by. 1 Y (s) τs+1 +1 R(s) τs+1 1 R(s) For which, when R(s) 1/s, and y(t) 1 e t/ τ 1ast

2 Lecture Notes on Control Systems/D. Ghose/ Figure 1.9: Command shaping control This appears to offer the solution we are looing for. However, this method will wor only when we now the open-loop gain exactly, which is unliely since in a practical system, gains are measured inaccurately and they may vary over operating ranges and differing conditions. Suppose, the actual open-loop gain is not but is off by an amount δ. So,theopenloop gain is + δ. Then, the closed-loop gain is, ( ) + δ +1 + δ +1 + δ +1 + δ +1 + δ δ +1 + ( +1)δ ( + δ +1) 1 δ 1+ δ + ( ) 1+ δ ( 1 δ ) δ ( 1 δ ) +1 + (Expanding in Taylor series) δ δ ( 1 δ ) (Ignoring nd order terms in δ) +1 δ δ (Again ignoring nd order terms in δ) δ 1+ ( +1) Thus, a non-zero steady state error still remains unless is nown exactly and does not change during operation. Method 3: Integral control. For y 0 in steady state, G(s) has to be driven by u 0. But, if we ill the steady state error completely then this condition is no longer satisfied. So, instead of u e let us try u e. Then e r y can go to zero while u e 0.

3 Lecture Notes on Control Systems/D. Ghose/ Figure 1.10: Integral control The corresponding bloc diagram is as shown below. The plots alongside show that even when e 0, the area under the e curve which represents e is non-zero. Consider a first-order system. The open loop response is given by, The closed loop response is given by, G(s) 1 τs+1 Y ol (s) G(s)R(s) 1 τs+1 R(s) Y c (s) G(s)K(s) 1+G(s)K(s) τs + s + R(s) 1 s τs s τs+1 τ s + s τ + τ R(s) which is a second order system. Since, ζω n 1 τ is a constant, the settling time ( 4/(ζω n )), is a constant too, irrespective of the value of. What happens when you use the exact expression? Steady state response to a step input is given by, Y c (s) τ s + s τ + τ 1 s

4 Lecture Notes on Control Systems/D. Ghose/ The system is stable for all positive values of. This can be shown by a simple application of the Routh-Hurwitz criterion. s 1 τ s 1 1 τ 0 s 0 τ 0 The final value theorem can be applied to show that So there is no steady state error. Finally, we can see that y c ( ) lim s 0 sy c (s) 1 ω n which implies that the system is τ ζ 1 τ Underdamped if ζ<1 > 1 4τ Overdamped if ζ>1 < 1 4τ So, for a first order system, integral control produces zero steady state error for step inputs but runs the ris of oscillations in the output if the gain value is large. Let us consider a general second order system and examine how integral control affects its performance. Let the open loop system be given by, G ol (s) ω n s +ζω n s + ω n Then, the closed loop system with integral control would be, G c (s) ωn s +ζω n s+ωn s ωn s +ζω ns+ω n s 1+ ωn s(s +ζω n s + ωn)+ω n ωn s 3 +ζω n s + ωns + ωn

5 Lecture Notes on Control Systems/D. Ghose/ This is a third order system which has 3 real poles or one real pole and a pair of complex conjugate poles. Let us chec the stability of the system. s 3 1 ω n s ζω n ω n s 1 ζω n ω n ζ 0 s ω n 0 From which we can see that the system will be stable if 0 <<ζω n. So, a second order system with integral control is stable only when this condition on is met. Let us apply a step input and chec the steady state response by applying the final value theorem, assuming that the value of is so chosen as to maintain stability. Then, So, Y (s) which ills the steady state error. ωn s 3 +ζω n s + ωns + ωn 1 s y( ) lim s 0 sy (s) Summary of Results In this section we summarize the results obtained in the previous sections. consider, We will Class of systems: First order and Second order Class of control: Proportional and Integral Class of inputs: Step

6 Lecture Notes on Control Systems/D. Ghose/ Step input P-Control K(s) Integral Control K(s) s First Order System Open loop: Closed loop: Closed loop: G(s) 1 G τs+1 c (s) 1 τ G s+1 c (s) /τ s +(1/τ)s+(/τ) (Second order system) Gain 1 Gain Gain 1 Time constant τ Time constant τ T r.τ Tr.τ T r π ω n π τ T s 3.9τ Ts 3.9τ T s 4 ζω n 8τ Steady state error 0 Steady state error 1 Steady state error 0 Second Order System Open loop: Closed loop: Closed loop: ωn G(s) s +ζω ns+ωn ω G c (s) n s +ζω ns+ωn() G c (s) ωn() s +ζω ns+ωn() ω n s 3 +ζω ns +ω ns+ω n y( ) 1 y( ) y( ) 1 Stable Stable <ζω n (stable) ζω n (unstable) Problem Set 6 1. Let G(s) 1 be a first order system. Assume some value for τ and plot the sτ+1 open loop response to a step and a ramp input.. For the same system consider a feedbac with just P-control. For three different values of gain K plot the time response for a unit step and a unit ramp input. 3. Now consider a integral control. Select two values of K, one maing the system

7 Lecture Notes on Control Systems/D. Ghose/01 11 underdamped and the other overdamped and plot the time responses to step and ramp inputs. 4. Consider a second order system G(s) s +ζω ns+ω and assume reasonable values n for ζ and ω n. Do the same as above. Only in PI-control, select K such that one system is stable and the other unstable. 5. Compare the rise time, settling time, steady state error, etc. of the above time responses and comment on them. ω n Note: Use MATLAB. Select parameter values that are different from other students! 1.9 Disturbance Signals One of the properties of a feedbac system is that of disturbance rejection. Disturbance signals can appear at the system input, the system output, or at the sensor. Figure 1.11: A feedbac system with many input signals In order to find the effect of these disturbance signals we need to find the closed-loop transfer functions from each noise term to the output Y. In the following, we will drop the argument s (the laplace variable) from the equations for clarity. Y D o + Y D o + G(U + D i ) D o + G(KE + D i ) D o + G[K(R R )+D i ]

8 Lecture Notes on Control Systems/D. Ghose/ D o + G[K{R H(Y + N)} + D i ] D o + G[KR KHY KHN + D i ] D o + GKR GKHY GKHN + GD i So, where, Y 1 D KG o + R G Do D o + G R R + G N N + G Di D i G Do G R G N G Di 1 KG KGH G KGH N + G D i are the closed-loop transfer functions from the corresponding disturbance inputs to the output, when all other inputs are considered to be zero. So, a design objective would be to select K(s) andh(s) so that KG Y R R should have a good response Y Do,Y N,Y Di should all be small However, there has to be a trade-off since these four transfer functions are not independent and there are only two parameters K(s) andh(s). For example, note that, G Do + G N 1 + KGH 1 So, if the effect of the sensor noise N on the output is small, then the effect of the output disturbance D o has to be large, provided that the level of both disturbance signals are the same.