I Poles & zeros. I First-order systems. I Second-order systems. I E ect of additional poles. I E ect of zeros. I E ect of nonlinearities

Transcription

1 EE C28 / ME C34 Lecture Chater 4 Time Resonse Alexandre Bayen Deartment of Electrical Engineering & Comuter Science University of California Berkeley Lecture abstract Toics covered in this resentation I Poles & zeros I First-order systems I Second-order systems I E ect of additional oles I E ect of zeros I E ect of nonlinearities I Lalace transform solution of state equations / 6 2 / 6 Chater outline 4.2 Poles, zeros, & system resonse 3 / 6 4 / Poles, zeros, & system resonse 4.2 Poles, zeros, & system resonse Definitions, [,. 63] Definitions, [,. 63] Poles of a TF Zeros of a TF I Values of the Lalace transform variable, s, thatcausethetf to become infinite I Values of the Lalace transform variable, s, that cause the TF to become zero I Any roots of the denominator of the TF that are common to the roots of the numerator I Any roots of the numerator of the TF that are common to the roots of the denominator Figure: a. system showing inut & outut, b. ole-zero lot of the system; c. evolution of a system resonse Figure: a. system showing inut & outut, b. ole-zero lot of the system; c. evolution of a system resonse 5 / 6 6 / 6

2 4.2 Poles, zeros, & system resonse System resonse characteristics, [,. 63] 4.2 Poles, zeros, & system resonse System resonse characteristics, [,. 63] I Poles of a TF: Generate the form of the natural resonse I Poles of a inut function: Generate the form of the forced resonse Figure: E ect of a real-axis ole uon transient resonse I Pole on the real axis: Generates an exonential resonse of the form e t, where is the ole location on the real axis. The farther to the left a ole is on the negative real axis, the faster the exonential transient resonse will decay to zero. I Zeros and oles: Generate the amlitudes for both the forced and natural resonses Figure: E ect of a real-axis ole uon transient resonse 7 / 6 8 / 6 Intro, [,. 66] I st -order system without zeros TF I Unit ste inut TF G(s) = C(s) R(s) = a s + a R(s) =s I System resonse in frequency domain C(s) =R(s)G(s) = a s(s + a) I System resonse in time domain c(t) =c f (t)+c n (t) = e at Figure: st -order system; ole-lot 9 / 6 0 / 6 Characteristics, [,. 66] Characteristics, [,. 66] I Time constant, a : The time for e at to decay to 37% of its initial value. Alternatively, the time it takes for the ste resonse to rise to 63% of its final value. Figure: st -order system resonse to a unit ste I Exonential frequency, a: The recirocal of the time constant. The initial rate of change of the exonential at t =0,since the derivative of e at is a when t =0. Since the ole of the TF is at a, thefarther the ole is from the imaginary axis, the faster the transient resonse. Figure: st -order system resonse to a unit ste / 6 2 / 6

3 Characteristics, [,. 66] Characteristics, [,. 66] I Rise time, T r : The time for the waveform to go from 0. to 0.9 of its final value. The di erence in time between c(t) = 0.9 and c(t) = 0.. I 2% Settling time, T s : The time for the resonse to reach, and stay within, 2% (arbitrary) of its final value. The time when c(t) =0.98. T r = 2.2 a T s = 4 a Figure: st -order system resonse to a unit ste Figure: st -order system resonse to a unit ste 3 / 6 4 / 6 General form, [,. 68] I 2 finite oles: Comlex ole air determined by the arameters a and b I No zeros Figure: General 2 nd -order system 5 / 6 6 / 6 Overdamed resonse, [,. 69] Underdamed resonse, [,. 69] I ole at origin from the unit ste inut I System oles: 2 real at, 2 I Natural resonse: Summation of 2 exonentials c(t) =K e t + K 2 e 2t I Time constants:, 2 I ole at origin from the unit ste inut I System oles: 2 comlex at d ± j! d I Natural resonse: Damed sinusoid with an exonential enveloe I Time constant: c(t) =K e dt cos(! d t ) I Frequency (rad/s):! d d 7 / 6 8 / 6

4 Underdamed resonse characteristics, [,. 70] Undamed resonse, [,. 69] I Transient resonse: Exonentially decaying amlitude generated by the real art of the system ole times a sinusoidal waveform generated by the imaginary art of the system ole. I Damed frequency of oscillation,! d : The imaginary art art of the system oles. I Steady state resonse: Generated by the inut ole located at the origin. I Underdamed resonse: Aroaches a steady state value via a transient resonse that is a damed oscillation. Figure: 2 nd -order ste resonse comonents generated by comlex oles I ole at origin from the unit ste inut I System oles: 2 imaginary at ±j! I Natural resonse: Undamed sinusoid I Frequency:! c(t) =A cos (! t ) 9 / 6 20 / 6 Critically damed resonse, [,. 69] Ste resonse daming cases, [,. 72] I ole at origin from the unit ste inut I System oles: 2 multile real I Natural resonse: Summation of an exonential and a roduct of time and an exonential I Time constant: c(t) =K e t + K 2 te t I Note: Fastest resonse without overshoot I Overdamed I Underdamed I Undamed I Critically damed Figure: Ste resonses for 2 nd -order system daming cases 2 / 6 22 / 6 Secification, [,. 73] I Natural frequency,! n I The frequency of oscillation of the system without daming I Daming ratio, = I General TF where Exonential decay frequency Natural frequency (rad/s) G(s) = b s 2 + as + b = = Natural eriod (s) 2 Exonential time constant! 2 n s 2 +2! n s +! 2 n a =2! n, b =! 2 n, = a 2! n,! n = b 23 / 6 24 / 6

5 Resonse as a function of, [,. 75] I Poles s,2 =! n ±! n 2 Table: 2 nd -order resonse as a function of daming ratio 25 / 6 26 / 6 Ste resonse, [,. 77] Ste resonse, [,. 77] Transfer function C(s) =! 2 n s(s 2 +2! n s +! 2 n)...artial fraction exansion... = (s +! n )+! s + 2 n 2 (s +! n ) 2 +! n( 2 2 ) Time domain via inverse Lalace transform where c(t) = e!nt cos (! n 2 )t +...trigonometry & exonential relations... = 2 e!nt cos(! n 2 ) = tan ( ) 2 sin (! 2 n 2 )t 27 / 6 28 / 6 Resonses for values, [,. 78] Resonse secifications, [,. 78] Resonse versus lotted along a time axis normalized to! n I Lower roduce a more oscillatory resonse I! n does not a ect the nature of the resonse other than scaling it in time Figure: 2 nd -order underdamed resonses for daming ratio values I Rise time, T r : Time required for the waveform to go from 0. of the final value to 0.9 of the final value I Peak time, T : Time required to reach the first, or maximum, eak Figure: 2 nd -order underdamed resonse secifications 29 / 6 30 / 6

6 Resonse secifications, [,. 78] I I Evaluation of T, [,. 79] T is found by di erentiating c(t) and finding the zero crossing after t = 0, which is simlified by alying a derivative in the frequency domain and assuming zero initial conditions. Overshoot, %OS: The amount that the waveform overshoots the steady state, or final, value at the eak time, exressed as a ercentage of the steady state value Settling time, Ts : Time required for the transient s damed oscillations to reach and stay within ±2% of the steady state value L[c (t)] = sc(s) =!n2 s2 + 2!n s +!n2...comleting the squares in the denominator...setting the derivative to zero T = Figure: 2nd -order underdamed resonse secifications 3 / 6!n 2 32 / 6 Evaluation of %OS, [,. 80] Evaluation of Ts, [,. 79] %OS is found by evaluating Find the time for which c(t) reaches and stays within ±2% of the steady state value, cfinal, i.e., the time it takes for the amlitude of the decaying sinusoid to reach 0.02 cmax cfinal %OS = 00 cfinal e where cmax = c(t ), cfinal = 2 Settling time 00 given %OS Figure: %OS vs. Aroximated by ln( %OS 00 ) Ts = 33 / 6 2 = ) 4!n 34 / 6 Location of oles, [,. 82] A recise analytical relationshi between Tr and cannot be found. However, using a comuter, Tr can be found. Designate!n t as the normalized time variable 2. Select a value for 3. Solve for the values of!n t that yield c(t) = 0.9 and c(t) = The normalized rise time!n Tr is the di erence between those two values of!n t for that value of Evaluation of Tr, [,. 8] ln(0.02!n Ts = =q 2 + ln2 ( %OS 00 ) This equation is a conservative estimate, since we are assuming that cos(!n 2 t )=...substitution %OS = e!n t Figure: Normalized Tr vs. for a 2nd -order underdamed resonse I Natural frequency,!n : Radial distance from the origin to the ole I Daming ratio, : Ratio of the magnitude of the real art of the system oles over the natural frequency cos( ) = 35 / 6!n =!n Figure: Pole lot for an underdamed 2nd -order system 36 / 6

7 Location of oles, [,. 82] Location of oles, [,. 83] I Damed frequency of oscillation,! d : Imaginary art of the system oles! d =! n 2 I Exonential daming frequency, d: Magnitude of the real art of the system oles I T / horizontal lines T =! = n 2! d I T s / vertical lines T s = 4! n = 4 d I %OS / radial lines I Poles s,2 = d =! n d ± j! d Figure: Pole lot for an underdamed 2 nd -order system %OS = e = cos( ) 2 00 Figure: Lines of constant T, T s,and %OS. Note: T s2 <T s, T 2 <T, %OS < %OS / 6 38 / 6 Underdamed systems, [,. 84] I T / horizontal lines T =! = n 2! d I T s / vertical lines T s = 4! n = 4 d I %OS / radial lines %OS = e = cos( ) 2 00 Figure: Ste resonses of 2 nd -order systems as oles move: a. with constant real art, b. with constant imaginary art, c. with constant 39 / 6 40 / 6 E ect on the 2 nd -order system, [,. 87] E ect on the 2 nd -order system, [,. 87] I Dominant oles: The two comlex oles that are used to aroximate a system with more than two oles as a second-order system I Conditions: Three ole system with comlex oles and a third ole on the real axis s,2 =! n ± j! n 2, s 3 = r I Ste resonse of the system in the frequency domain C(s) = A s + B(s +! n)+c! d (s +! n ) 2 +! 2 d I Ste resonse of the system in the time domain + D s + r c(t) =Au(t)+e!nt (B cos(! d t)+c sin(! d t)) + De rt 4 / 6 42 / 6

8 E ect on the 2 nd -order system, [,. 88] 3 cases for the real ole, r I r is not much greater than! n I r! n I Assuming exonential decay is negligible after 5 time constants I The real ole is 5 farther to the left than the dominant oles I r = Figure: Comonent resonses of a 3-ole system: a. ole lot, b. comonent resonses: non-dominant ole is near dominant 2 nd -order air, far from the air, and at 43 / 6 44 / 6 E ect on the 2 nd -order system, [,. 9] E ect on the 2 nd -order system, [,. 9] Assume a grou of oles and a zero far from the oles....artial-fraction exansion... I E ects on the system resonse I Residue, or amlitude I Not the nature, e.g., exonential, damed sinusoid, etc. I Greater as the zero aroaches the dominant oles I Conditions: Real axis zero added to a two-ole system Figure: E ect of adding a zero to a 2-ole system 45 / 6 s + a T (s) = (s + b)(s + c) = A s + b + B s + c ( b + a)/( b + c) ( c + a)/( c + b) = + s + b s + c If the zero is far from the oles, then a b and a c, and n o T (s) a /( b+c) = s+b + a (s + b)(s + c) /( c+b) s+c Zero looks like a simle gain factor and does not change the relative amlitudes of the comonents of the resonse. 46 / 6 E ect on the 2 nd -order system, [,. 9] E ect on the 2 nd -order system, [,. 9] Another view... I Resonse of the system, C(s) I System TF, T (s) I Add a zero to the system TF, yielding, (s + a)t (s) I Lalace transform of the resonse of the system (s + a)c(s) = sc(s)+ ac(s) I Resonse of the system consists of 2 arts I The derivative of the original resonse I A scaled version of the original resonse 3 cases for a I a is very large I Resonse! ac(s), a scaled version of the original resonse I a is not very large I Resonse has additional derivative comonent roducing more overshoot I a is negative right-half lane zero I Resonse has additional derivative comonent with an oosite sign from the scaled resonse term 47 / 6 48 / 6

9 Non-minimum-hase system, [,. 92] Non-minimum-hase system: System that is causal and stable whose inverses are causal and unstable. I Characteristics: If the derivative term, sc(s), is larger than the scaled resonse, ac(s), the resonse will initially follow the derivative in the oosite direction from the Figure: Ste resonse of a non-minimum-hase system scaled resonse. 49 / 6 50 / 6 Saturation, [,. 96] Dead zone, [,. 97] Figure: a. e ect of amlifier saturation on load angular velocity resonse, b. Simulink block diagram Figure: a. e ect of dead zone on load angular dislacement resonse, b. Simulink block diagram 5 / 6 52 / 6 Backlash, [,. 98] Figure: a. e ect of backlash on load angular dislacement resonse, b. Simulink block diagram 53 / 6 54 / 6

10 Lalace transform solution of state equations, [,. 99] State equation Outut equation ẋ = Ax + Bu y = Cx + Du Lalace transform of the state equation zx(s)...combining all the X(s) terms...solving for X(s) (si x(0) = AX(s) + BU(s) A)X(s) = x(0) + BU(s) X(s) =(si A) x(0) + (si A) BU(s) adj(si A) = x(0) + BU(s) det(si A) 55 / 6 Lalace transform solution of state equations, [,. 99] State equation ẋ = Ax + Bu Outut equation y = Cx + Du Lalace transform of the state equation Y (s) =CX(s)+DU(s) 56 / 6 Eigenvalues & TF oles, [,. 200] Eigenvalues of the system matrix A are found by evaluating det(si A) =0 The eigenvalues are equal to the oles of the system TF....for simlicity, let the outut, Y (s), and the inut, U(s), be scalar quantities, and further, to conform to the definition of a TF, let x(0) = 0 Y (s) h i U(s) = C adj(si A) det(si A) B + D = C adj(si A)B + D det(si A) det(si A) The roots of the denominator are the oles of the system. 57 / 6 58 / 6 Time domain solution of state equations, [,. 203] Linear time-invariant (LTI) system I Time domain solution Z t e A(t x(t) =e At x(0) + 0 I Zero-inut resonse e At x(0) I Zero-state resonse / convolution integral ) Bu( )d Time domain solution of state equations, [,. 203] Linear time-invariant (LTI) system I Frequency domain unforced resonse L[x(t)] = L[e At x(0)] = (si A) x(0) I Lalace transform of the state-transition matrix e At x(0) I Characteristic equation I State-transition matrix Z t e A(t 0 ) Bu( )d e At det(si A) =0 I??? equation L [(si h i A) ]=L adj (si A) det (si A) = (t) 59 / 6 60 / 6

11 Bibliograhy Norman S. Nise. Control Systems Engineering, / 6