I Laplace transform. I Transfer function. I Conversion between systems in time, frequencydomain, and transfer


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1 EE C128 / ME C134 Feedback Control Systems Lecture Chapter 2 Modeling in the Frequency Domain Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley Lecture abstract Topics covered in this presentation I Laplace transform I Transfer function I Conversion between systems in time, frequencydomain, and transfer function representations I Electrical, translational, and rotationalmechanical systems in time, frequencydomain, and transfer function representations I Nonlinearities I Linearization of nonlinear systems in time, frequencydomain, and transfer function representations September 10, 2013 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Chapter outline Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 History interlude The Laplace transform definitions, [1, p. 35] PierreSimon Laplace I I French mathematician and astronomer I Pioneered the Laplace transform I AKA French Newton I...all the e ects of nature are only mathematical results of a small number of immutable laws. I What we know is little, and what we are ignorant of is immense. Laplace transform Inverse Laplace transform where L[f(t)] = F (s) = L 1 [F (s)] = 1 2 j Z 1 0 = f(t)u(t) Z +j1 j1 s = + j! f(t)e st dt F (s)e st ds Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
2 Laplace transform table, [1, p. 36] f(t) (t) 1 u(t) tu(t) t n u(t) e at u(t) sin(!t)u(t) cos(!t)u(t) e at sin(!t)u(t) e at cos(!t)u(t) F (s) 1 s 1 s 2 n! s n+1 1 s+a! s 2 +! 2 s s 2 +! 2! (s+a) 2 +! 2 s+a (s+a) 2 +! 2 Some basic algebraic operations, such as multiplication by exponential functions or shifts have simple counterparts in the Laplace domain Theorem (Frequency shift) Theorem (Time shift) L[e at f(t)] = F (s + a) L[f(t T )] = e st F (s) Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Theorem (Linearity) Theorem (Di erentiation) Theorem (Scaling) L[c 1 f 1 (t)+c 2 f 2 (t)] = c 1 F 1 (s)+c 2 F 2 (s) Examples L d n f dt n = s n F (s) nx k=1 s n k dk 1 f dt k 1 (0 ) L[f(at)] = 1 a F s a L df dt = sf (s) f(0 ) Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Theorem (Integration) h R i L t 0 f( )d = F (s) s Theorem (Final value) L[f(1)] = lim s!0 sf (s) To yield correct finite results, all roots of the denominator of F (s) must have negative real parts, and no more than one can be at the origin. Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
3 Partial fraction expansion, [1, p. 37] To find the inverse Laplace transform of a complicated function, we can convert the function to a sum of simpler terms for which we know the Laplace transform of each term Theorem (Initial value) L[f(0+)] = lim s!1 sf (s) To be valid, f(t) must be continuous or have a step discontinuity at t =0, i.e., no impulses or their derivatives at t =0. Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 F (s) = N(s) D(s) How F (s) can be expanded is governed by the relative order between N(s) and D(s) 1. O(N(s)) < O(D(s)) 2. O(N(s)) O(D(s)) and the type of roots of D(s) 1. Real and distinct 2. Real and repeated 3. Complex or imaginary Bayen (EECS, UCB) Feedback Control Systems September 10, / The TF The transfer function, [1, p. 44] 2.3 The TF General nth order, linear, timeinvariant di erential equation a n d n c(t) dt n + a n 1 d n 1 c(t) dt n a 0 c(t) = b m d m r(t) dt m + b m 1 1 r(t) dt m b 0 r(t) d m Under the assumption that all initial conditions are zero the transfer function (TF) from input, c(t), to output, r(t), i.e., the ratio of the output transform, C(s), divided by the input transform,r(s) is given by G(s) = C(s) R(s) = b ms m + b m 1 s m b 0 a n s n + a n 1 s n a 0 Also, the output transform, C(s) can be written as C(s) =R(s)G(s) Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Electrical network TFs, [1, p. 47] Table: Voltagecurrent, voltagecharge, and impedance relationships for capacitors, resistors, and inductors Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
4 Electrical network TFs, [1, p. 48] Electrical network TFs, [1, p. 59] Example (Resistorinductorcapacitor (RLC) system) the capacitor voltage, V C (s), to the input voltage, V (s) Figure: RLC system Example (Inverting operational amplifier system) the output voltage, V o (s), to the input voltage V i (s) Figure: Inverting operational amplifier system Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / Translational mechanical system TFs 2.5 Translational mechanical system TFs Translational mechanical system TFs, [1, p. 61] Table: Forcevelocity, forcedisplacement, and impedance translational relationships for springs, viscous dampers, and mass Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / Translational mechanical system TFs Translational mechanical system TFs, [1, p. 63] 2.6 Rotational mechanical system TFs Example (Translational inertiaspringdamper system) the position, X(s), to the input force, F (s) Figure: Physical system; block diagram Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
5 2.6 Rotational mechanical system TFs Rotational mechanical system TFs, [1, p. 69] 2.6 Rotational mechanical system TFs Rotational mechanical system TFs, [1, p. 63] Table: Torqueangular velocity, torqueangular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia Example (Rotational inertiaspringdamper system) the position, 2 (s), tothe input torque, T (s) Figure: Physical system; schematic; block diagram Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 2 Modeling in the frequency domain Nonlinearities, [1, p. 88] 2 Modeling in the frequency domain Common physical nonlinearities found in nonlinear (NL) systems Figure: Some physical nonlinearities Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Linearization, [1, p. 89] Motivation I Must linearize a NL system into a LTI DE before we can find a TF Linearization procedure 1. Recognize the NL component and write the NL DE 2. Linearize the NL DE into an LTI DE 3. Laplace transform of LTI DE assuming zero initial conditions 4. Separate input and output variables 5. Form the TF Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
6 Linearization, [1, p. 89] Linearization, [1, p. 89] 1 st order linearization I Output, f(x) I Input, x I Operating at point A, [x 0,f(x 0 )] I Small changes in the input can be related to changes in the output about the point by way of the slope of the curve, m a, at point A [f(x) f(x 0 )] m a (x x 0 ) f(x) m a x f(x) f(x 0 )+m a x Figure: Linearization about point A General linearization via Taylor series expansion f(x) =f(x 0 )+ df dx (x x 0 ) x=x 0 + d2 f 1! dx 2 (x x 0 ) 2 x=x ! For small excursions of x from x 0, we can neglect higherorder terms. The resulting approximation yields a straightline relationship between the change in f(x) and the excursion away from x 0. Neglecting higherorder terms yields f(x) f(x 0 ) df dx x=x 0 (x x 0 ) or 5 f = f x m x=x 0 which is a linear relationship between f(x) and x for small excursions away from x 0. Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Linearization, [1, p. 92] Bibliography Example (NL electrical system) the inductor voltage, V L (s), to the input voltage, V (s). The NL resistor voltagecurrent relationship is defined by i r =2e 0.1vr,wherei r and v r are the resistor current and voltage, respectively. Also the input voltage, v, is a smallsignal source. Figure: NL electrical system Norman S. Nise. Control Systems Engineering, Bayen (EECS, UCB) Feedback Control Systems September 10, / 34 Bayen (EECS, UCB) Feedback Control Systems September 10, / 34
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