Example As a simple example, consider the differential equation: x ϩ x ϭ 0 x(0) ϭ 0 (14.3) x . (0) ϭ 1 Taking the Laplace transform of the equation yields: s 2 X(s) Ϫ sx(0) Ϫ x . (0) ϩ X(s) ϭ 0 (14.4) Algebraic manipulation gives: X(s) ϭ (14.5) Consequently, from the table of Laplace transform pairs: x(t) ϭ sin(t) (14.6) For more examples, see Ogata (1997), Raven (1995), Kuo (1995), and Franklin et al. (1994). Due to the convolution property of Laplace transforms, aconvenient representation of a linear control system is the block diagram illustrated in Figure 14.1.In such a block diagram, each block contains the Laplace transform of the differential equation representing that component of the control system that relates the block’s input to its output. Arrows between blocks indicate that the output from the preceding block is transferred to the input of the subsequent block. The output of the preceding block multiplies the contents of the block to which it is an input. Simple algebra will yield the overall transfer function of a block diagram representation for a system. 1 ᎏ s 2 ϩ 1 Fundamentals of Control Theory 14-3 TABLE 14.1 Laplace Transform Pairs for Basic Functions F(t) F(s) 1 Unit impulse, δ (t) 1 2 Unit step, 1(t) 3 t 4 t n , n ϭ 1, 2, 3, … 5 e Ϫat 6 t n e Ϫat 7 sin ω t 8 cos ω t 9 e Ϫat cos bt 10 e Ϫat sin bt b ᎏᎏ (s + a) 2 + b 2 s + a ᎏᎏ (s + a) 2 + b 2 s ᎏ s 2 + ω 2 ω ᎏ s 2 + ω 2 n! ᎏ (s + a) n+1 1 ᎏ s ϩ a n! ᎏ s nϩ1 1 ᎏ s 2 1 ᎏ s © 2006 by Taylor & Francis Group, LLC Example The transfer function for the system illustrated in Figure 14.2 can be computed by observing that: E(s) ϭ R(s) Ϫ Y(s)S(s) (14.7) and Y(s) ϭ E(s)C(s)A(s)P(s) (14.8) which can be combined to yield ϭ (14.9) A more complete exposition on block diagram algebra can be found in any of the previously cited undergraduate texts. Note that the numerator and denominator of the transfer function will typically be polynomials in s.The denominator is called the characteristic equation for the system. As entry 5 in Table 14.1 shows, if the characteristic polynomial has any roots with a positive real part, then the system will be unstable because it will correspond to an exponentially increasing solution. Given a reference input R(s), determine the response of the system by multiplying the transfer function by the reference input, and perform a partial fraction expansion (i.e., expand): Y(s) ϭ ϭ ϩ ϩ … ϩ (14.10) where each term in the sum on the right-hand side of the equation is similar to one of the entries in Table 14.1. The contribution to the response of each individual term can be determined by referring to a Laplace transform table and can be superimposed to determine the overall solution: y(t) ϭ y 1 (t) ϩ y 2 (t) ϩ … ϩ y n (t) (14.11) where each term in the sum is the inverse Laplace transform of the corresponding term in the partial fraction expansion. C n ᎏ s Ϫ p n C 2 ᎏ s Ϫ p 2 C 1 ᎏ s Ϫ p 1 R(s)C(s)A(s)P(s) ᎏᎏᎏ 1 ϩ C(s)A(s)P(s)S(s) C(s)A(s)P(s) ᎏᎏᎏ 1 ϩ C(s)A(s)P(s)S(s) Y(s) ᎏ R(s) 14-4 MEMS: Introduction and Fundamentals Input Controller Process Output Sensor Actuator − + FIGURE 14.1 Typical block diagram representation of a control system. R(s) E(s) C(s) P(s) Y(s) S(s) A(s) − + FIGURE 14.2 Generic block diagram including transfer functions. © 2006 by Taylor & Francis Group, LLC Example For the block diagram in Figure 14.2 if C(s) ϭ , A(s) ϭ 1, P(s) ϭ ,S(s) ϭ 1 and R(s) ϭ (a unit step input), then: Y(s) ϭ ϭ Ϫ ϭϪ Ϫ ϭϪ Ϫ (14.12) where ω d ϭ ω n ͙ 1Ϫ ෆ ζ 2 ෆ .Referring to Table 14.1 of Laplace transform pairs and assuming that ζ Ͻ 1, y(t) ϭ 1 Ϫ e Ϫ ζω n t cos( ω d t) ϩ sin( ω d t) (14.13) 14.2.2 Control System Analysis and Design Control system analysis and design consider primarily stability and performance. The stability of a system with the closed-loop transfer function (note that in such a case a controller has already been specified): T(s) ϭ (14.14) is determined by the roots of the denominator, or characteristic equation. It is possible to determine whether the system is stable without actually computing the roots of the characteristic equation. A nec- essary condition for stability is that each of the coefficients a i appearing in the characteristic equation be positive. Because this is a necessary condition, if any of the a i are negative, then the system is unstable, but the converse is not necessarily true. Even if all the a i are positive, the system may still be unstable. Routh (1975) devised a method to check necessary and sufficient conditions for stability. The method is to construct the Routh array, defined as follows: Row n s n : 1 a 2 a 4 … Row nϪ1 s nϪ1 : a 1 a 3 a 5 … Row nϪ2 s nϪ2 : b 1 b 2 b 3 … Row nϪ3 s nϪ3 : c 1 c 2 c 3 … Ӈ Ӈ Ӈ Ӈ Ӈ Ӈ Row 2 s 2 : * * Row 1 s 1 : * Row 0 s 0 : * b 0 s m ϩ b 1 s mϪ1 ϩ … ϩ b m ᎏᎏᎏ s n ϩ a 1 s nϪ1 ϩ … ϩ a n ζ ᎏ ͙1 ෆ Ϫ ෆ ζ ෆ 2 ෆ ω d ᎏᎏ (s ϩ ζω n ) 2 ϩ ω d ζω n ᎏ ω d s ϩ ζω n ᎏᎏ (s ϩ ζω n ) 2 ϩ ω d 1 ᎏ s ζω n ᎏᎏ (s ϩ ζω n ) 2 ϩ ω d s ϩ ζω n ᎏᎏ (s ϩ ζω n ) 2 ϩ ω d 1 ᎏ s s ϩ 2 ζω n ᎏᎏ s 2 ϩ 2 ζω n s ϩ ω 2 n 1 ᎏ s ω 2 n ᎏᎏ s(s 2 ϩ 2 ζω n s ϩ ω 2 n ) 1 ᎏ s ω 2 n ᎏ s ϩ 2 ζω n 1 ᎏ s Fundamentals of Control Theory 14-5 © 2006 by Taylor & Francis Group, LLC in which the a i are from the denominator of Equation (14.14). b i and c i are defined as: The basic result is that the number of poles in the right-half plane (i.e., unstable solutions) is equal to the number of sign changes among the elements in the first column of the Routh array. If they are all positive, the system is stable. When a zero is encountered, it should be replaced with a small positive con- stant ε which will then be propagated to lower rows in the array. The result can be obtained by taking the limit as ε → 0. Example Construct the Routh array and determine the stability of the system described by the transfer function: ϭ (14.15) The Routh array is s 4 : 1 9 8 s 3 : 4 10 0 s 2 : ϭ 26 ϭ 32 0 (14.16) s 1 : ϭ 33 0 0 s 0 : ϭ 40.6 0 0 The system is stable because there are no sign changes in the elements in the first column of the array. One aspect of performance concerns the steady-state error exhibited by the system. For example, from the time-domain solution of the previous example, as t → ∞ , y(t) → 1. However, the final value theorem can be used to determine this without actually solving for the time-domain solution. Example Determine the steady-state value for the time-domain function y(t) if its Laplace transform is given by Y(s) ϭ ω 2 n /s(s 2 ϩ 2 ζω n s ϩ ω 2 n ). Because all the solutions of s 2 ϩ 2 ζω n s ϩ ω 2 n ϭ 0 have a negative real part, all the poles of sY(s) lie in the left half of the complex plane. Therefore, the final value theorem can be applied to yield: lim t→∞ y(t) ϭ lim s→0 sY(s) ϭ lim s→0 s ϭ 1 (14.17) which is identical to the limit of the time-domain solution as t → ∞ . ω 2 n ᎏᎏ s(s 2 ϩ 2 ζω n s ϩ ω 2 n ) Ϫ(0 Ϫ 1056) ᎏᎏ 26 Ϫ(128 Ϫ 260) ᎏᎏ 4 Ϫ(0 Ϫ 32) ᎏᎏ 1 Ϫ(10 Ϫ 36) ᎏᎏ 1 1 ᎏᎏᎏ s 4 ϩ 4s 3 ϩ 9s 2 ϩ 10s ϩ 8 Y(s) ᎏ R(s) 14-6 MEMS: Introduction and Fundamentals b 1 ϭ Ϫ det ΄ ΅ a 1 a 2 a 3 1 a 1 b 2 ϭ Ϫ det ΄ ΅ a 1 a 4 a 5 1 a 1 b 3 ϭ Ϫ det ΄ ΅ a 1 a 6 a 7 1 a 1 c 1 ϭ Ϫ det ΄ ΅ b 1 a 3 b 2 a 1 b 1 c 2 ϭ Ϫ det ΄ ΅ b 1 a 5 b 3 a 1 b 1 c 3 ϭ Ϫ det ΄ ΅ b 1 a 7 b 4 a 1 b 1 © 2006 by Taylor & Francis Group, LLC 14.2.2.1 Proportional–Integral–Derivative (PID) Control Perhaps the most common control implementation is so-called proportional–integral–derivative (PID) control, where the commanded control input (the output of the “controller” box in Figures 14.1 and 14.2) is equal to the sum of three terms: one term proportional to the error signal (the input to the “controller” box in Figures 14.1 and 14.2), the next term proportional to the derivative of the error signal, and the third term proportional to the time integral of the error signal. From Figure 14.2, C(s) ϭ K P ϩ (K I /s) ϩ K d s,whereK P is the proportional gain, K I is the integral gain, and K d is the derivative gain. A simple analy- sis of a second-order system shows that increasing K P and K I generally increases the speed of the response at the cost of reducing stability. Increasing K d generally increases damping and stability of the response. With K I ϭ 0, there may be a nonzero steady-state error, but when K I is nonzero, the effect of the integral control effort is to typically eliminate steady-state error. Example — PID Control of a Robot Arm Consider a robot arm illustrated in Figure 14.3. Linearizing the equations of motion about θ ϭ 0 (the configuration in Figure 14.3) gives: I θ ϩ mg θ ϭ u (14.18) where I is the moment of inertia of the arm, m is the mass of the arm, θ is the angle of the arm, and u is atorque applied to the arm. For PID control, u ϭ K p ( θ desired Ϫ θ actual ) ϩ K d ( θ . desired Ϫ θ . actual ) ϩ K I ͵ t 0 ( θ desired Ϫ θ actual )dt (14.19) If I ϭ 1 and m ϭ 1/g, the block diagram representation for the system is illustrated in Figure 14.4.Thus, the closed-loop transfer function is T(s) ϭ (14.20) Figure 14.5 illustrates the step response of the system for proportional control (K P ϭ 1, K I ϭ 0, K d ϭ 0), PD control (K P ϭ 1, K I ϭ 0, K d ϭ 1), and PID control (K P ϭ 1, K I ϭ 1, K d ϭ 1). Note that for proportional and PD controls, there is a final steady-state error that is eliminated with PI control. (Also note that both of these facts could be verified analytically using the final value theorem.) Finally, note that K d s 2 ϩ K p s ϩ K I ᎏᎏᎏ s 3 ϩ K d s 2 ϩ (K p ϩ I)s ϩ K I Fundamentals of Control Theory 14-7 , FIGURE 14.3 Robot arm model. © 2006 by Taylor & Francis Group, LLC the system response for pure proportional control is oscillatory, whereas with derivative control the response is much more damped. The subjects contained in the subsequent sections consider controller synthesis issues. For PID con- trollers, tuning methods exist. Refer to the undergraduate texts cited previously or to the papers by Ziegler and Nichols (1942, 1943). 14.2.2.2 The Root Locus Design Method As mentioned previously in the discussion of PID control, various rules of thumb can be determined to relate system performance to changes in gains, however, a systematic approach is more desirable. Because pole locations determine the characteristics of the response of the system (recall the partial fraction expansion), one natural design technique is to plot how pole locations change as a system parameter or control gain is varied [Evans, 1948, 1950]. Because the real part of the pole corresponds to exponential solutions, if all the poles are in the left-half plane, the poles closest to the j ω -axis will dominate the system response. If we focus a second-order system of the form: H(s) ϭ (14.21) ω 2 n ᎏᎏ s 2 ϩ 2 ζω n s ϩ ω 2 n 14-8 MEMS: Introduction and Fundamentals K p + K d s + K I s 1 s 2 +1 desired actual + − FIGURE 14.4 Robot arm block diagram. Amplitude Step response 0 5 10 15 20 25 30 35 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Proportional control PD control PID control Time (sec) FIGURE 14.5 PID control response. © 2006 by Taylor & Francis Group, LLC the poles of the system are as illustrated in Figure 14.6. The terms ω n , ω d , and ζ are the natural frequency, the damped natural frequency, and the damping ratio, respectively. Multiplying H(s) by 1/s (unit step), and performing a partial fraction expansion give: Y(s) ϭ Ϫ Ϫ (14.22) so the time response for the system is y(t) ϭ 1 Ϫ e Ϫ ζω n t cos ω d t ϩ sin ω d t (14.23) where ω d ϭ ω n ͙ 1Ϫ ෆ ζ 2 ෆ and 0 Յ ζ Ͻ 1. Figure 14.7 illustrates plots of the response for various values of ζ . Referring to the previous equation and Figure 14.7, if the damping ratio is increased, the oscillatory nature of the response is increasingly damped. ζ ᎏ ͙ 1 ෆ Ϫ ෆ ෆ 2 ෆ ζω n ᎏᎏᎏ (s ϩ ζω n ) 2 ϩ ω 2 n (1 Ϫ ζ 2 ) s ϩ ζω n ᎏᎏᎏ (s ϩ ζω n ) 2 ϩ ω 2 n (1 Ϫ ζ 2 ) 1 ᎏ s Fundamentals of Control Theory 14-9 Im(s) Re(s) x x n n d FIGURE 14.6 Complex conjugate poles, natural frequency, damped natural frequency, and damping ratio. Amplitude Step response 0 5 10 15 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time (sec) FIGURE 14.7 Step response for various damping factors. © 2006 by Taylor & Francis Group, LLC Because the natural frequency and damping are directly related to the location of the poles, one effec- tive approach to designing controllers is picking control gains based upon desired pole locations. Aroot locus plot is a plot of pole locations as a system parameter or controller gain is varied. Once the root locus has been plotted, pick the location on the root locus with the desired pole locations to give the desired system response. There is a systematic procedure to plot the root locus by hand (refer to the cited under- graduate texts), and computer packages such as Matlab (using the rlocus() and rlocfind() func- tions) make it even easier. Figure 14.9 illustrates a root locus plot for the previously noted robot arm with the block diagram as the single gain K is varied from 0 to ∞ as illustrated in Figure 14.8. Note that for the usual root locus plot, only one gain can be varied at a time. In the previous example, the ratio of the proportional, integral, and derivative gains was fixed, and a multiplicative scaling factor was varied in the root locus plot. Because the roots of the characteristic equation start at each pole when K ϭ 0 and approach each 0 of the characteristic equation as K → ∞ , the desired K can be determined from the root locus plot by find- ing the part of the locus that most closely matches the desired natural frequency ω n and damping ratio ζ (recall Figure 14.7). Typically, control system performance is specified in terms of time-domain conditions, such as rise time, maximum overshoot, peak time, and settling time, all of which are illustrated in Figure 14.10. Rough estimates of the relationship between the time-domain specifications and the natural frequency and damping ratio are given in Table 14.2 [Franklin et al., 1994]. 14-10 MEMS: Introduction and Fundamentals 1+ s + 1 s 1 s 2 +1 desired actual + − K FIGURE 14.8 Robot arm block diagram. −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1.5 −1 −0.5 0 0.5 1 1.5 Real axis Imaginary axis FIGURE 14.9 Root locus for robot arm PID controller. © 2006 by Taylor & Francis Group, LLC Example Returning to the robot arm example, assume the desired system performance has a system rise time less than 1.4 sec, a maximum overshoot less than 30%, and a 1% settling time less than 10 sec. From the first row in Table 14.2, the natural frequency must be greater than 1.29. From the third and fourth rows, the damping ratio should be greater than approximately 0.4. Figure 14.11 illustrates the root locus plot, the pole locations and corresponding gain, and K (rlocfind() is the Matlab command for retrieving the gain value for a particular location on the root locus). These results provide a damping ratio of approximately .45 and a natural frequency of approximately 1.38. Figure 14.12 illustrates the step response of the system to a unit step input verifying these system parameters. 14.2.2.3 Frequency Response Design Methods An alternative approachtocontroller design and analysis is the so-called frequency response method. Frequencyresponse controller design techniques have two main advantages. They provide good controller Fundamentals of Control Theory 14-11 TABLE 14.2 Time-Domain Specifications as a Function of Natural Frequency, Damped Natural Frequency, and Damping Ratio Rise time: t r Х Peak time: t p Х Overshoot: M p ϭ e Ϫ πζ /͙1Ϫ ෆ ζ 2 ෆ Settling time (1%): t s ϭ Note: Results are from Franklin et al. (1994). 4.6 ᎏ ζω n π ᎏ ω d 1.8 ᎏ ω n Time ( sec ) Amplitude 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Maximum overshoot Rise time Peak time Settling time Steady-state error FIGURE 14.10 Time domain control specifications. © 2006 by Taylor & Francis Group, LLC design even with uncertainty with respect to high-frequency plant characteristics, and using experimental data for controller design purposes is straightforward. The two main tools are Bode and Nyquist plots (see [Bode, 1945] and [Nyquist, 1932] for first-source references), and stability analyses are considered first. A Bode plot is a plot of two curves. The first curve is the logarithm of the magnitude of the response of the open-loop transfer function with respect to unit sinusoidal inputs of frequency ω . The second 14-12 MEMS: Introduction and Fundamentals −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1.5 −1 −0.5 0 0.5 1 1.5 Real axis Imaginary axis K = 0 K large K = 2.5 Natural frequency = 1.38 Damping ratio = 0.44 FIGURE 14.11 Selecting pole locations for a desired system response. Time (sec) Amplitude (radians) 0 1.4 2.8 4.2 5.6 7 0 0.2 0.4 0.6 0.8 1 1.2 1.4 FIGURE 14.12 Robot arm step response. © 2006 by Taylor & Francis Group, LLC [...]... system is equal to the dimension of its state space Partial feedback linearization is also possible where the relative degree is less than the dimension of the state space However for such systems, an analysis of the © 2006 by Taylor & Francis Group, LLC 1 4-2 8 MEMS: Introduction and Fundamentals stability of the zero dynamics is necessary In particular, if the relative degree γ Ͻ n, then the change of coordinates... 14.18 plots the Bode plot for a lead compensator for various values of the parameter A Because the lead compensator shifts the phase plot up, by an appropriate choice of the parameter A, the crossover point where the magnitude plot crosses through the value of 0 dB can be shifted to the right, increasing the gain margin © 2006 by Taylor & Francis Group, LLC 1 4-1 6 MEMS: Introduction and Fundamentals. .. is based on the well-known result from complex variable theory called the principle of the argument Consider the (factored) transfer function: G(s) ϭ ͟ (s ϩ zi) i ͟(s ϩ pj) j (14.27) By complex variable theory, ∠G(s) ϭ Σiθi Ϫ Σiϕj, where θi are the angles between s and the zeros zi, and φj are the angles between s and the poles pj Thus, a plot of G(s) as s follows a closed contour (in the clockwise... space, called the sliding manifold, upon which the system should indefinitely evolve Because the sliding manifold has a lower dimension than the full state space for the system, a lower order model describes the evolution of the system on the sliding manifold If a stabilizing controller is designed for the sliding manifold, the problem reduces to designing a controller to drive the system to the sliding... etc.) In particular, if the low-frequency asymptote of the magnitude plot has a slope © 2006 by Taylor & Francis Group, LLC 1 4-1 4 MEMS: Introduction and Fundamentals −5 −10 −20 −25 −30 0 −20 To: Y(1) Phase (deg); Magnitude (dB) −15 −40 −60 −80 −100 −1 10 100 10 1 Frequency (rad/sec) FIGURE 14.14 Bode plot for example problem of zero and if the value of this asymptote is denoted by K, then the steady-state... gi(x)ui iϭ1 © 2006 by Taylor & Francis Group, LLC (14.59) Fundamentals of Control Theory 1 4-2 5 where x is a 1 ϫ n vector, the f(x) and gi(x) are smooth vector fields, and the ui are scalar control inputs Note that this is not the most general form for nonlinear systems, as the ui are assumed to enter the equations in an affine manner (i.e., they simply multiply the gi(x) vector fields) For some aerodynamic... a row vector comprised of each of the gains ki Then, the state-space description of the system becomes: x ϭ (A Ϫ BK)x (14.34) x(t) ϭ e(AϪBK)tx(0) (14.35) so that the solution of this equation is © 2006 by Taylor & Francis Group, LLC Fundamentals of Control Theory 1 4-1 9 mg l u x M FIGURE 14.20 Cart and pendulum system where e(AϪBK)t is the matrix exponential of the matrix A Ϫ BK defined by: (A Ϫ... plot of G(s) as s follows a closed contour (in the clockwise direction) in the complex plane will encircle the origin in the clockwise direction the same number of times that there are zeros of G(s) within the contour minus the number of times that there are © 2006 by Taylor & Francis Group, LLC Fundamentals of Control Theory 1 4-1 5 0.35 0.3 Amplitude 0.25 0.2 0.15 0.1 0.05 0 0 0.5 1 1.5 2 2.5 3 3.5 4... within the contour Therefore, an easy check for stability is to plot the open loop G(s) on a contour that encircles the entire left-half complex plane Assuming that G(s) has no right-half plane poles (poles of G(s) itself, in contrast to poles of the closed-loop transfer function), an encirclement of Ϫ1 by the plot will indicate a right-half plane zero of 1 ϩ G(s), which is an unstable right-half plane... Figure 14.14 illustrates the Bode plot for the open-loop transfer function P(s) ϭ (1/2)/(s ϩ 1) The lowfrequency asymptote is approximately at Ϫ6, so 20 log K ϭ Ϫ6 ⇒ K ≈ 0.5012 ⇒ yss ≈ 0.6661, where yss ϭ lim p(t) Figure 14.15 illustrates the unity feedback closed-loop step response of the system, verifying that the t→∞ steady-state value for y(t) is the same as computed from the Bode plot A Nyquist . in the clockwise direction the same number of times that there are zeros of G(s)within the contour minus the number of times that there are 1 ᎏ K 1 ᎏ 1 ϩ K 1 4-1 4 MEMS: Introduction and Fundamentals Fre q uenc y. equations of motion about θ ϭ 0 (the configuration in Figure 14.3) gives: I θ ϩ mg θ ϭ u (14.18) where I is the moment of inertia of the arm, m is the mass of the arm, θ is the angle of the arm, and. closest to the j ω -axis will dominate the system response. If we focus a second-order system of the form: H(s) ϭ (14.21) ω 2 n ᎏᎏ s 2 ϩ 2 ζω n s ϩ ω 2 n 1 4-8 MEMS: Introduction and Fundamentals K p