# Control System (ECE411) Lectures 13 & 14

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1 Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016

2 Steady-State Error Analysis Remark: For a unity feedback system (H(s) = 1): e(t) = r(t) c(t) E(s) = R(s) C(s) = R(s) R(s)M(s) = E(s) = [1 M(s)]R(s) where M(s) is the closed loop transfer function. Thus, e ss = lim s 0 se(s) = lim s 0 s[1 M(s)]R(s) For a unit step R(s) = 1 s, we get e ss = [1 M(0)] Note: The above results could sometimes be used for cases when H(s) 1 (tracking error). Example Given a unity feedback system shown below with closed loop transfer function M(s) = K (s 2 +2s+2)(s+a),

3 Steady-State Error Analysis-Cont. (a) find K and a such that e ss = 1.5 to unit ramp input, (b) find e ss for unit step input. Part (a): First, we find the open-loop transfer function G(s) from M(s) using, M(s) = G(s) 1 + G(s) = K (s 2 + 2s + 2)(s + a) = G(s) = K (s 2 + 2s + 2)(s + a) K = K s 3 + (a + 2)s 2 + (2a + 2)s + (2a K) Now, in order to avoid a Type 0 system which yields e ss for ramp input, 2a K = 0 = K = 2a. For a unit ramp input: e ss = 1 K v where K v = lim s sg(s) Thus, e ss = 1 K v = 1.5 = K v = 2 3

4 Steady-State Error Analysis-Cont. Using G(s) in K v = lim s sg(s) and K v = 2 3 gives, K K v = lim s s 2 + (a + 2)s + (2a + 2) = K 2a + 2 = 2 3 Solving for K and a using the above equation and K = 2a gives a = 2, and K = 4. Part (b): Using the result from part (a): G(s) = 4 s(s 2 + 4s + 6) which is obviously Type 1 system = e ss = 0 to unit step input.

5 Transient Analysis Transient Response Transient response allows for determining whether or not a system is stable and, if so, how stable it is (i.e. relative stability) as well as the speed of response when a step reference input is applied. A typical time-domain response of a second order system (closed loop) to a unit step input is shown.

6 Transient Analysis-Cont. Key Definitions: 1 Max Overshoot (M p ) M p = cmax css c ss c max : max value of c(t), c ss : steady-state value of c(t) %max overshoot = 100 M p M p determines relative stability: Large M p less stable 2 Delay time (t d ): Time for c(t) to reach 50% of its final value. 3 Rise time (t r ): Time for c(t) to rise from 10% to 90% of its final value. 4 Settling time (t s ): Time for c(t) to decrease and stay within a specified (typically 5%) of c ss. Desirable characteristics: Small M p, small t d, quick t r and fast t s (cannot be accomplished simultaneously).

7 Transient Response of 2 nd -Order Control System Consider a control system with closed-loop transfer function, M(s) = C(s) R(s) = ω 2 n s 2 + 2ξω n s + ωn 2, M(0) = 1 Characteristic Equation : ρ(s) = s 2 + 2ξω n s + ω 2 n = 0 has the following roots, s 1,2 = ξω n ± jω n 1 ξ2 = α ± jω These are depicted in the following figure. cos θ = ξ, tan θ = 1 ξ 2 ξ

8 Transient Response of 2 nd -Order Control System-Cont. Response to unit step input (R(s) = 1 s ) is C(s) = ω 2 n s(s 2 + 2ξω n s + ωn 2 ) }{{} (s+α) 2 +ω 2 Use PFE, time-domain response is found to be c(t) = 1 }{{} c ss + damping {}}{ e αt 1 ξ 2 sin[ωt θ], t 0 } {{ } c tr(t) α = ξω n : Damping Factor- Controls the rate of rise time and decay time i.e. α controls damping and speed of response. Can control oscillations by changing ω. Can control damping by changing ξ.

9 Transient Response of 2 nd -Order Control System-Cont. τ = 1/α: Time Constant Large α = small τ = signal decays quickly. ξ: Damping Ratio (ratio between actual damping factor and the damping factor for critically damped (ξ = 1 = s 1,2 = ω n ). ω n : Natural Undamped Frequency (ξ = 0 = s 1,2 = ±jω n i.e. purely oscillatory with frequency ω n ) ω = ω n 1 ξ2 : Conditional Frequency

10 Transient Response-Different Damping Cases (a) Underdamped: 0 < ξ < 1, s 1,2 = ξω n ± jω n 1 ξ 2 Characteristics: Small rise time (t r ), large overshoot (M p ). (b) Critically Damped: ξ = 1, s 1,2 = ω n (repeated real roots) Characteristics: No overshoot, slow/large rise time.

11 Transient Response-Different Damping Cases (c) Overdamped: ξ > 1 s 1,2 = ξω n ± ω n ξ2 1 (two real distinct roots) Characteristics: No overshoot, very large rise time. (d) Undamped (Oscillatory): ξ = 0, s 1,2 = ±jω n

12 Transient Response-Different Damping Cases (e) Negatively damped (unstable): ξ < 0, s 1,2 = ξω n ± jω n 1 ξ 2

13 Transient Response: Performance Measures 1. Peak Time (t max ) To find the peak time (time at which the step response reaches its maximum), we take the derivative of the step response and set it to zero. dc(t) dt = 0 dc(t) dt = ξe ξωnt 1 ξ 2 sin [ωt θ] + e ξωnt ω cos[ωt θ] 1 ξ 2 Using ω = ω n 1 ξ2 and trig identities, we can simplify the above equation as: dc(t) dt = ωn e ξωnt sin(ωt), t 0, 1 ξ 2 Now, dc(t) dt = 0 = sin (ωt) = 0 or when t (i.e. final value) The first condition gives the extrema points (maxima and minima) of c(t), i.e. ωt = nπ = t = nπ ω n 1 ξ 2 The first maximum (Max overshoot) of c(t) happens for n = 1. Thus, π t max = ω n 1 ξ 2

14 Transient Response-Performance Measures-Cont. Note: Although Max and Min of c(t) occur at periodic interval, the response is NOT periodic due to damping (unless ξ = 0). 2. Max Overshoot (M p ) To find M p, we substitute t max in expression for c(t). This yields, c max = c(t) t=tmax = 1 + e πξ 1 ξ 2 Thus, using the fact that c ss = 1, we get M p = c max 1 = e Or in percentage, πξ 1 ξ 2 πξ 1 ξ 2 %Max Overshoot = 100e As can be seen, Max Overshoot is solely a function of ξ. Hence, Larger ξ = smaller M p (ξ < 1). But this would increase the delay time and rise time as seen next. For t d, t r, and t s only approximate equations can be obtained. These are given next.

15 Transient Response-Performance Measures-Cont. 3. Delay Time (t d ) For t d, we set c(t) = 0.5 and solve for t d, t d ξ ω n, 0 < ξ < 1 t d 1+0.6ξ+0.15ξ2 ω n : wider range of ξ and more accurate. 4. Rise Time (t r ) t r ξ ω n, 0 < ξ < 1 t r = 1+1.1ξ+1.4ξ2 ω n, wider range of ξ and more accurate. 5. Settling Time(t s ) t s 4/ξω n As can be seen, Small ξ = smaller t d and t r but larger t s. Optimum Range for ξ: 0.5 ξ 0.8

16 for 2 nd Order Systems 1. Gain Controller Consider servo control system below: Steady State Error Analysis: Loop transfer function: G(s) = = Type 1 = e ss = 0 for r(t) = u s (t) K s(js+b) For a unit ramp input e ss = 1 K v and K v = lim s 0 sg(s)h(s) = K B Thus, e ss to unit ramp = e ss = B K which implies Small e ss Requires Large Gain K.

17 for 2 nd Order Systems-Cont. Transient Analysis: The closed-loop transfer function M(s) = G(s) 1+G(s) = K s(sj+b) 1+ K s(js+b) Comparing with standard case, ω 2 n s 2 +2ξω n+ω 2 n = K/J s 2 +Bs/J+K/J M(s) = Since B and J cannot be tweaked (motor parameters), Large K = Reduced ξ = large M p = i.e. less stable. = ω n = K/J and 2ω n ξ = B/J = ξ = B 2 KJ Thus, a simple gain controller (K) won t produce desirable steady-state and transient behavior as a compromise between small steady state error and good relative stability and fast response cannot be achieved.

18 for 2 nd Order Systems-Cont. 2. Proportional-Derivative Controller (PD) Controller transfer function G c (s) = K P + K D s, where K P is the proportional constant, and K D is the derivative constant. Steady State Error Analysis: Loop transfer function: G(s) = G c (s)g p (s) = K P +K D s s(js+b) For a unit ramp input e ss = 1 K v = Still Type 1 = e ss = 0 for r(t) = u s (t) and K v = lim s 0 sg(s)h(s) = K P B = e ss = B K P i.e. it is possible to make e ss to a unit ramp as small as possible by increasing proportional Gain K P.

19 for 2 nd Order Systems-Cont. Transient Analysis: The closed-loop transfer function M(s) = (B + K D)s s 2 + J Gc(s)Gp(s) 1+G c(s)g p(s) = (K P +K D s)/j } {{ } 2ξωn Again, comparing with standard case, M(s) = ω 2 n s 2 +2ξω n+ω 2 n 2ω n ξ = (B+K D) J = ω n = = ξ = B+K D 2 K P J K P J + K P J }{{} and Thus, we can choose: (a) Large K P for small e ss to unit ramp, and (b) Appropriate K D to have 0.5 < ξ < 0.8. However, PD controller adds a zero at s = K P /K D which could have an impact in changing the shape of the response to unit step. Additionally, PD controller is susceptible to noise and difficult to realize. ω 2 n

20 for 2 nd Order Systems-Cont. 3. Tachometer Control ( 1 Using s(js+b) = control as shown. 1 Js+B ) ( 1 s ), we design a rate feedback (tachometer) Alternatively, the above block diagram can be reduced to the typically used tachometer control system.

21 for 2 nd Order Systems-Cont. Steady State Error Analysis: Loop transfer function: G(s)H(s) = K(1+Kts) s(js+b) = Still Type 1 = e ss = 0 for r(t) = u s (t) For a unit ramp input e ss = 1 K v and K v = lim s 0 sg(s)h(s) = K B = e ss = B K i.e. it is possible to make e ss to a unit ramp as small as possible by increasing Gain K. Transient Analysis: The closed-loop transfer function M(s) = K J s 2 + (B+KK t ) J s+ K J Comparing with standard case, M(s) = ω 2 n s 2 +2ξω n+ω 2 n 2ω n ξ = (B+KKt) J = ω n = = ξ = B+KKt 2 KJ K J and

22 for 2 nd Order Systems-Cont. Thus, we can choose: (a) Large Gain K for small e ss to unit ramp, and (b) Appropriate K t to have 0.5 < ξ < 0.8. Note: Tachometer control doesn t have the same issues of the PD controller. Hence, used widely for servo control. Example: For the tach control system below, find K and K t such that max overshoot, M p, to unit step is 0.2 and peak time is 1 second. Then, using these values of K and K t, find t r and t s. M p = e ξπ/ 1 ξ 2 = 0.2 = ξ = t max = π ω n 1 ξ 2 = 1 = ω n = 3.53

23 for 2 nd Order Systems-Cont. But, K = ω 2 n = K = 12.5 Also, ξ = 1+KKt 2 K = = K t = t r = ξ ω n = 0.55sec t s = 4 ξω n = 2.48sec

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