IN 227 Control Systems Design Lecture 4 Instructor: G R Jayanth Department of Instrumentation and Applied Physics Ph: 22933197 E-mail: jayanth@isu.iisc.ernet.in Laplace transform of signals operated on by some common operators Differentiation st f (t )e dt fe 0 • • st 0 sf (t )e st dt 0 Thus, L( f ) sL( f ) f (0) More generally, L( f ( n) ) s n L( f ) s n1 f (0) s n2 f (0) .. f n1 (0) Integration t 0 e st f ()d e dt f ()d s t st 1 f (t )e st dt s 0 0 0 1 f 1 (0) L ( fd ) L ( f ) Thus , where, f 1 (0) fd s s t • Time delay L( f (t )1(t )) e s f ()e 0 es L( f (t )) s d Laplace transform of functions under a convolution integral t • Convolution integral x(t ) u ()g (t )d 0 • t 0 0 Strictly speaking, g (t ) g (t )1(t ) . Thus, x(t ) u()g (t )1(t )d u ()g (t )1(t )d 0 0 0 0 L( x(t )) x( s) u ()g (t )1(t )d e st dt g (t )1(t )e st dt u ()d • Changing variables and defining t 0 0 0 0 X ( s) g ()1()e s ( ) d u ()d g ()1()e s d e su ()d • • Thus, we get: X (s) G(s)U (s) , where, G(s) and U(s) are the Laplace transforms of the impulse response and the input respectively. G(s), the ratio of the output of a given system X(s) to its input U(s) in the Laplace domain, when all initial conditions are zero, is called the ‘transfer function’ of the system. It is easy to see that if U(s)=1, then X(s)=G(s). In other words, a transfer function is the Laplace Transform of the impulse response of the system. Usefulness of Laplace transforms • Generally, a physical system is a cascade of several subsystems. Thus, from a practical standpoint, the solutions can get laborious to calculate with increasing complexity and quickly become inaccessible to design insights. An electromechanical system like a motor would begin with the electrical subsystem whose output becomes the input for the first mechanical sub-system. The output of the first mechanical subsystem would become the input to the second subsystem and so on. Assuming, that there are n subsystems with outputs x1, x2,..xn, and all such subsystems are considered to be of second order (for simplicity), the cascade of subsystems can be represented by the equations x1 a1 x1 a2 x1 u x2 b1 x2 b2 x2 x1 x1 .. xn 1 xn • • • • x m1 xn 1 m2 xn 1 2 n In order to calculate the output of the ith sub-system, it is necessary to know the following two variables: the impulse response of the ith subsystem and the output of the (i-1)th subsystem. Thus to know the effect of the actual input u on the final output xn, one needs to make 2n calculations. Furthermore, as engineers, we also wish to be able to study the dependence of the final response to changes in parameter of one or more of the subsystems, say, be performing ‘k’ iterations. Thus, the total number of calculations we would need to perform is n+nk=n(k+1). This makes it extremely laborious and the complexity tends to dilute our understanding of the role of a specific parameter on the dynamics for the overall system. Alternate route#1: One can reduce the n differential equations into a single 2nth order differential equation between xn and u, obtain the characteristic equation of the system and study the dependence of its roots on the parameters of the system. Alternate route#2: Use Laplace transforms. Modeling systems using Laplace transforms • If we apply Laplace Transform to the second problem posed at the beginning, we get: • Thus, we see that the input and output relationship can be readily derived for this case to be: (m s m2 ) 1 ( s 1) Xn 2 .. 2 1 U (s) 2 ( s a1s a2 ) ( s b1s b2 ) (s 1s 2 ) • It is now possible to solve the above equation in 2 steps. We also see that Xn and U are now directly related and that the dynamics of each intermediate subsystem is separately represented in the overall expression. Thus, if we develop sufficient intuition about the relationship between transfer functions and time domain responses, we can inspect just one equation (given above) in order to engineer the dynamics of the overall system. Thus as a consequence of reducing differential equations into an algebraic equation, the overall problem of engineering the dynamics of physical systems is greatly simplified. • ( s 2 a1s a2 ) X 1 ( s) U ( s) ( s 2 b1s b2 ) X 2 ( s) ( s 1) X 1 ( s) .. ( s 2 1s 2 ) X n ( s) (m1s m2 ) X n 1 ( s) Laplace Transforms of physical systems • • • The Laplace transform ‘transports’ us from time-domain into a new domain, viz., the s-plane. To obtain a working knowledge of the transformed systems, it is imperative that we establish connections between linear systems in the time-domain and their transformed counterparts. This lecture presents theorems and their consequences that realize this goal and in the process, enhances our familiarity with the splane. Region of convergence Initial value theorem: The initial value theorem states that if F(s) is the Laplace transform of f(t) then, lim sF (s) f (0) . The proof is given on the s right. This theorem connects the Laplace transform of a system for ‘very large’ s to the initial conditions in the time-domain at t=0+. Im(s) ρ Re(s) Proof of initial value theorem L( f ) L( f ) fe st dt sF ( s) f (0) 0 fe st dt sF ( s ) f (0) 0 fe st dt 0 0 0 0 0 fe st dt fe st dt f (0) f (0) fe st dt lim L( f ) f (0) f (0) lim fe st dt f (0) f (0) s s 0 f (0) f (0) lim sF ( s) f (0) s lim sF (s) f (0) s Laplace Transforms of physical systems • • • Consequence of initial value theorem on the nature of physical transfer functions A physical linear time invariant system has the following input-output relationship: x( n) a1x( n1) .. an x b1u ( m) b2u ( m1) .. bm1u m m 1 Thus its transfer function is given by G(s) b1s b2 s .. bm1 • The response of the system to a step input is given by X (s) G(s) • sX ( s) lim G( s) f (0) . Thus we can From initial value theorem, it follows that lim s s G( s) infer the initial response of the system to a step input by evaluating lim s • Now consider the general transfer function of a linear time-invariant system: G(s) s n a1s n 1 .. an 1 s m n lim G ( s ) b1 m n s 0 m n • It can be readily seen that • Thus, if m>n, then the initial response to a step input is infinitely large. Such systems are called non-causal. This cannot happen in physical systems. The outputs for the other two cases are physically reasonable. Thus, in physical systems, the degree of the numerator and denominator polynomials of the transfer function always have to satisfy m≤n If m=n, then such transfer functions are called “proper” transfer functions. If m<n, then the transfer functions are called “strictly proper” transfer functions. • • b1s m b2 s m1 .. bm1 s n a1s n 1 .. an Laplace Transforms of physical systems Proof of final value theorem • • • • Final value theorem: The final value theorem states that if F(s) is the sF (s) f () . The proof is given on Laplace transform of f(t) then, lim s 0 the right. A crucial assumption that is made in evaluating the limit above is that the limit exists. Since the Laplace transform of a time domain function exists only within the ‘region of convergence’, the region of convergence should include s=0 for the final value theorem to hold. In other words, the exponent ρ of the exponential function eρt that bounds | f (t )| should be strictly less than 0, i.e., ρ<0 (Fig. below right). For signals satisfying the above assumption, this theorem connects the Laplace transform of a system for ‘very small’ s to the final conditions in the time-domain. It follows that for a step input to a plant, the final value of the response is given by X () lim sX (s) lim sG(s) 1 lim G(s) s 0 • • s 0 s L( f ) fe st dt sF ( s) f (0) 0 lim L( f ) lim fe st dt lim sF ( s) f (0) s 0 s 0 lim fe dt s 0 st 0 0 s 0 fdt f () f (0) 0 f () lim sF ( s) s 0 Region of convergence Im(s) s 0 The transfer function G(s) is the Laplace transform of the impulse response g(t). The impulse response of a system is bounded if the roots of the characteristic equation of the system are less than zero. Such a system is said to be stable. For stable systems, the exponent ρ<0. Thus, meaningful application of the final value theorem is possible only for stable systems. ρ Re(s) Inverse Laplace transform for physical systems • A general physical system with arbitrary initial conditions and arbitrary forcing is given by x( n ) a1 x( n 1) .. an x b1u ( m) b2u ( m1) .. bm1u (m n) initialconditions : x(0), x(0),.., x n1 (0) • • • • • The Laplace transform X(s) of the response x(t) can be readily derived to be b1s m b2 s m1 .. bm1 1 X ( s) U (s) n (c1 ( s) x(0) c2 ( s) x(0) cn ( s) x(0)) n n 1 n 1 s a1s .. an s a1s .. an where, ci(s) (i=1,..n) are n polynomials of degree n-1 or lesser. The time domain response x(t) can be obtained by taking the inverse Laplace transform of X(s). j 1 Inverse Laplace transform: The inverse Laplace transform of a function F(s) is defined to be f (t ) F ( s)e st ds 2j j where ‘σ’ is any constant within the region of convergence. However, since a given time domain signal has a unique Laplace transform, we do not generally compute the integral. Instead we try to break the given transform into parts whose corresponding time domain signals can be easily recognized. This can be readily done for linear systems with inputs, each of whose Laplace transforms can be expressed as the ratio of two polynomials. In this case the two polynomials can be factored and subsequently split into partial fractions. Assuming that the input U(s) can be written as U (s) e1s q e2 s q 1 .. eq 1 s r d1s n 1 .. d r , the overall response, after factoring each of the numerator and denominator polynomials of G and U, can be written as X ( s) G(s)U (s) K • ( s z 1 )( s z 2 )..( s z m q ) ( s p 1 )( s p 2 )..( s p n r ) If we can obtain the inverse transform for this case, we can also obtain the inverse transform for the response to initial conditions, since it involves solving the same kind of problem. This is attempted next. If m+q≤n+r the RHS in the equation above can be expressed as K (s z 1 )(s z 2 )..(s z mq ) A1 A2 .. Anr ( s p 1 )( s p 2 )..( s p n r ) ( s p 1 ) ( s p 2 ) (s p nr ) where, Ai (i=1..n) are constants. Inverse Laplace transform for physical systems • • If the roots of the denominator polynomial do not repeat, it can be shown thatAi [(s n rpi ) X (s)] s p We can easily recognize that for this case, the time domain solution is given by xu (t ) Ai e p t i 1..n r i i i 1 • If a particular root repeats s times, then the solution can be written as where, An+r-s+k are given by Anr s k s A A1 A2 X ( s) .. n r s k k ( s p 1 ) ( s p 2 ) k 1 ( s p s ) 1 d s k [( s ps ) s X ( s)] k 1,..s (k 1)! ds s k nr s s pi t Ae e pst An r s k t k 1 i • The corresponding inverse transform is given by: xu (t ) • Note that in either case, the nature of the response is decided only by the roots of the denominator polynomial. The numerator polynomial only fixes the constants Ai. Thus, if e p t (i 1,..n) can be labeled as the ‘modes’ of the system, then the numerator polynomial only decides the proportion in which the ‘modes’ are combined in the overall response. Henceforth pi shall be called the poles of the system. For simplicity, let us continue with the case of non-repeating roots. The response of the system to the specified initial n conditions can be obtained by the same manner described above to be x0 (t ) Bi e p t and the overall response is: i 1 k 1 i • i x(t ) xu (t ) x0 (t ) i 0 • • • • n Role of initial conditions: Let us now focus just on the response to the homogeneous equation: x0 (t ) Bi e p t i 0 We notice that if any of the pi≤0 then, in general, we have an exponentially growing solution. Although theoretically we can carefully set our initial conditions such that the coefficient Bi of the growing term is zero, in practice, this is not possible. Any small disturbance is likely to change the initial conditions and restore the exponential growth of x0. Thus for such a case, the homogeneous solution x0(t)would soon overwhelm the inhomogeneous solution xu(t). On the other hand, if all pi>0, then x0(t) will die down with time and would soon become insignificantly small. Thus, the only response worth discussing after this initial transient is xu(t). Thus we notice that in either case, the exact initial conditions are irrelevant, albeit for different reasons. Thus, henceforth, we shall focus only on the input-output relationship, i.e., the transfer function G(s) without regard to the initial conditions. i Inverse Laplace transform for physical systems • Significance of the modes on the overall response: The response x(t) of a system to a given input u(t) is given by the t convolution integral x(t ) u () g (t )d . Note that in setting the lower limit to 0 instead of -∞, we assume that the 0 input is applied at t=0. This is okay because we are dealing with time invariant systems and we can always shift coordinates to the fix the origin at that point in time when the input is applied. • n Taking the inverse Laplace transform of the response to an impulse input U(s)=1, we see thatg (t ) Ai e p t . Thus, i n t x(t ) Ai u ()e pi (t ) d i 1 0 t i 1 , i.e., the response x(t) is the linear combination of n “areas under the curve” u ()e p (t ) d . If we i 0 assume that p1<p2<..<pn , we see that e p (t ) “persists” much longer than the other modes, and thus, its convolution with 1 u(t) possesses a large value. Therefore, if the coefficients of all the modes are comparable, then the first mode contributes the most to the overall response x(t) when the time-scale of the input is larger than 1/p1. This is especially so, by design, in case of feedback systems. • • • • If the poles pi are separated from each other by sufficient amounts, then for relatively slow varying signals, only the pole with the smallest value contributes significantly. Thus, such poles are called slow poles. Likewise, the largest poles, in general, contribute the least to the response to slow varying inputs and are called fast poles. Thus, for a slow varying input, the overall system looks as though it contains only one pole, viz., p1. This observation helps us to simplify systems that might otherwise be described by a transfer function of high degree. Note however that such simplifications should be done with care. If the input is sufficiently fast and we are interested in the initial response to the input (i.e., around t=0), even the fast poles contribute significantly around t=0 and the system does not resemble the simplified version for such cases. With that caveat spelt out, the concepts of slow and fast poles can really simplify how a system looks for the input. Engineers can choose components that ensure sufficient separation between the poles of the overall system and, we, as engineers, should seek to simplify systems where possible. If all the poles are real, the simplest that the overall system can be reduced to is a first order system. If some poles are complex, then the simplest that the overall system can be reduced to is a second order system. Thus, we pay special attention to the response of these two kinds of systems next. First order systems s / z0 1 s / 0 1 • A first order system is one that has the following transfer function: G(s) K • K is the DC gain of the system (easily verified using the final value theorem) and τ0 is called the time constant • Why τ0 is called the time constant becomes evident when we look at the response of a 1 simplified first order system given by G(s) K to a step input. The response is s / 0 1 t / given by x(t ) K[1 e ] Thus, τ0 represents the ‘characteristic time’ in which the system responds to inputs. z0 is called the ‘zero’ of the system. It is said to be a ‘minimum phase’ zero is z0 >0. Else it said to be a ‘non-minimum phase’ zero. Why it is called ‘minimum’ or ‘non minimum phase’ becomes evident when we look at the frequency response of the system. However, from initial value theorem, we see that systems with non-minimum phase zero start off moving in the opposite direction as that of the input at t=0, while those with minimum phase zeros start off in the same direction. 0 • • • Second order systems b1s 2 b2 s b3 s 2 a1s a2 • A second order system is one that., in general, has the following transfer function:G(s) • • The poles of the system can be complex or real depending on the coefficients a1 and a2. Case#1: Poles are real: The transfer function can be expressed as the sum of two first order transfer functions and the response of the overall system is the sum of the responses of these two sub-systems. • Example: G(s) s c 1 . Expanding the right hand side into partial fractions we get, ( s 1)( s 1) 1 1 1 , where, 1 c . Thus, we see that G(s) can be expressed as the sum of G( s) c 1 1 ( s 1) ( s 1) two modes, whose Laplace transforms are 1/(s+1) and 1/(s/β+1). It can be seen that by appropriately choosing the zero ‘c’, we can select one mode or the other: if c→1, α→∞, and thus the second mode is selected. However, if c→β, α→0, and thus the first mode is selected. • • Case 2: Poles are complex: The system behavior can be better understood by changing the variables and rewrite s 2 a1s a2 s 2 2n s 2n Here, ξ is termed the ‘damping ratio’ and ωn is termed as the natural frequency. For this case the step 2 response is in general oscillatory and is given by x(t ) 1 e t [cos d t sin d t ] where d n (1 ) 2 n (1 ) Reference • K Ogata, “Modern Control Engineering”, Pearson (2003)