Report

F (s) e 0 st f ( t ) dt L f ( t ) Professor Walter W. Olson Department of Mechanical, Industrial and Manufacturing Engineering University of Toledo Laplace Transforms Outline of Today’s Lecture Review Phase Phase Computations Full Bode Plot System Identification Using Bode Plots for System Identification Laplace Transform Inverse Laplace Transform Properties of the Laplace Transform Final Value Theorem Phase s i u ( t ) e i t cos i i sin i y ( t ) G ( i ) e i t Me i t w here M G ( i ) is the m agnitude or gain of G i a nd tan 1 Im G i is the phase angle or argum ent of G i R e G i For a sinusoidal input, phase represents the lag of the system or, alternatively, the processing time of the system to produce an output from the input Phase is measured as an angle A cycle of the input is consider to take 2p radians or 360 degrees Phase is the angular distance it takes for the output to represent the input Thus it is normal that as the frequency increases that the phase also increase In the case where the phase exceeds 180 degrees, the output appears to “lead” the input. This is particularly evident in the range of 270 to 360 degrees. Phase G ( s ) tan 1 Im G i R e G i As with magnitude there are 4 factors to consider which can be added together for the total phase angle. We will consider, in turn, K s n (s a) ( s 2 n s n ) 2 2 The sign will be positive if the factor is in the numerator and negative if the factor is in the denominator Phase Computations K is the angle of a real num ber; therefor e it is alw ays 0 (or 180 if the num ber w ere negative) s ( i n ) n n n If n is an even num ber, then i w ill either be + 1 or -1, both real num ber s for n an odd num ber, i w ill be -1: s 180 n n the angle of com plete turns (n ) 4 2 for n an even num ber, i w ill be + 1: s 0 the angle of com plete turn s (n ) n n 4 2 If n is an o dd num ber, then i w ill either be i or - i , b oth pure im aginary num bers n for n 1 an odd num ber, i w ill be i : s 90 n n the angle of com plete turns (n ) 4 2 for n 1 an ev en num ber, i w ill be - i: s 270 the angle of com plete turns (n ) n n 2 E xam ples: s 90 s 1 1 90 s 2 -180 s 3 s = -270 s 5 90 360 450 4 Phase Computations ( s a ) ( i a ) arctan a T his now involves the variable, , w hich is plotted on the sem ilog sca le W here a , arctan 0 a W here 0 a , arctan 90 a W here 10 a , arctan a line from 0 to 90 10 a a ( s 2 n s n ) ( n 2 2 2 2 2 n 2 2 n i 2 ) arctan n T his now also involves the va riable, 0 a W here n , arctan W here n , arctan W here 0 180 a 10 a , arctan a line from 0 to 180 w ith a correction for 10 a a Matlab Command bode(sys) System Identification It is not unusual for a field engineer to be shown a piece of equipment and then asked if he can put a control system on it or replace the control system for which there are no parts. The task of determining how an unknown structure responds is called “System Identification”. To identify a system, there are many tools are your disposal First and foremost, what should the system structure look like? Motors are often first order transfer functions (G ( s ) K ) which you then sa attempt to identify the constants Perform step tests and see what the response looks like Perform tests with sinusoidal outputs and use the Bode plot to identify the system Apply statistical/time series methods such as ARMAX and RELS Using Bode Plots for System Identification The overall order of the system will be the high frequency phase divided by 90 degrees The exponent of the “s” term will be the slope on the magnitude plot at the lowest frequency divided by 20 Alternatively, the exponent of “s” is the lowest frequency phase divided by 90 degrees. The system gain constant (Kt) in dB will be the height value at the extension of the “s” term line on the magnitude plot to where it crosses1 rps Starting from the left (the lowest frequency) on the magnitude plot, determine the structural components using the change in slopes in increments of 20 degrees either up or down Then by using the intersection of the lines at those places match to the test curve, determine the break frequencies Write the transfer function in the form s2 s2 s s s 2 z1s 2 2 s 1 1 ... 1 1 1 ... 2 2 nz 1 nz 2 b1 b2 bm nz 1 nz 2 G (s) K t 2 2 s s 2 p1s 2 p 2 s s s p s s 1 1 ... 1 1 1 ... 2 2 np 1 np 2 a1 a2 aq np 1 np 2 Laplace Transform Traditionally, Feedback Control Theory was initiated by using the Laplace Transform of the differential equations to develop the Transfer Function The was one caveat: the initial conditions were assumed to be zero. For most systems a simple coordinate change could effect this If not, then a more complicated form using the derivative property of Laplace transforms had to be used which could lead to intractable forms While we derived the transfer function, G(s), using the convolution equation and the state space relationships, the transfer function so derived is a Laplace Transform under zero initial conditions Laplace Transform CAUTION: Some Mathematics is necessary! The Laplace transform is defined as For an analytic function f ( t ) (i.e., integrable everyw here and e veryw here less than e F (s) e st 0 s0 t for finite s 0 ) f ( t ) dt L f ( t ) F s is the Laplace transform of f ( t ) s is a com plex num ber T he Inverse L aplace transform is defined as f (t ) 1 2p i a i a i e F ( t ) ds L st 1 F (t ) Fortunately, we rarely have to use these integrals as there are other methods Laplace Transforms Tables are available for determining the Laplace transform of most common functions This table which continues on the next slide is from Modern Control Engineering by K. Ogata 4th ed., 2002 Laplace Transforms Laplace Transform F (s) e st 0 f ( t ) dt L f ( t ) Note that the index on the integral is 0: it is assumed that no dynamics are considered prior to t=0 f (t ) 0 t0 The Laplace is a linear transform: L ( af ( t )) L af t bg t e st 0 af ( t ) dt a e st 0 e st 0 e 0 st f ( t ) dt aL f t af t bg t dt af t dt e st 0 aL f t bL g t bg t dt Some Common Laplace Transforms The Laplace Transform of the Impulse Function t0 0 1 (t ) 0 t< 0 t 0 L (t ) 1 The Laplace Transform of the Step Function 0 t< 0 1( t ) 1 t 0 L 1 t 1 Unit Ramp: 0 t< 0 f (t ) t t 0 1 s 2nd power of t: 0 f (t ) t 2 2 t< 0 t0 1 L ( f ( t )) s 3 The Place Transform of the nth power of t: 0 f (t ) t n n! s The Laplace Transform of a L ( f ( t )) The Laplace Transform of the 2 t< 0 t0 1 L ( f ( t )) s n 1 Some Common Laplace Transforms Laplace Trans Form of the exponentials: 0 f ( t ) at e L ( f ( t )) trigonometric functions: t< 0 t0 1 sa 1 s a 2 0 t< 0 f ( t ) n at t0 t e L ( f ( t )) n 1 s 2 2 0 t< 0 f (t ) cos t t 0 L ( f ( t )) s s 2 2 0 t< 0 f ( t ) at e sin t t 0 L ( f ( t )) n! s a 0 t< 0 f (t ) sin t t 0 L ( f ( t )) 0 t< 0 f ( t ) at t0 te L ( f ( t )) Laplace Transforms of s a 2 2 Examples Find the Laplace Transform of f ( t ) t 4 t 17 2 L f t L t 4 t 17 L t 2 L t 2 L 4 t L 17 2 s 3 4 s 2 2 L 4 t L 17 17 s Find the Laplace Transform of f ( t ) 4 sin(3t 2 ) N otice that this one has a phase (2) and there are no transform s in the tables for this f ( t ) 4 sin(3 t 2 ) 4 sin 3 t cos( 2 ) sin( 2 ) cos 3 t L f ( t ) L 4 s in 3 t cos( 2 ) sin( 2 ) cos 3 t 4 cos 2 L sin 3 t 4 sin 2 L cos 3 t L f ( t ) 4 cos 2 3 s 9 2 4 sin( 2 ) and then evaluate the constants L f (t ) 3.6372 s 4.9938 s 9 2 s s 9 2 Lumped Parameter Model of an Armature Controlled DC Motor Find the transfer function for this system with voltage as the input and angular position as the output using Laplace Transforms R a ia V b V a V oltage L oop: d B ack V oltage: Vb K b dt M otor T orque: T K i a 2 d d b T R otations N S L : J dt dt K d i b a R a dt 2 K ia b d dt J J Va Ra d dt b KK b d K Va dt JR a dt JR a J d 2 2 b K Va s KK b s s (s) JRa JR a J K T . F .( s ) O utput ( s ) Input ( s ) (s) Va s (s) Va s K JR a s bR a K K b s 2 JR a b KK b 2 s s J JR a Important Inverse Transforms 1 at 1 L e sa 1 1 at 1 e L s s a a 1 1 1 bt at e e L s a s b b a 1 2 n L 2 2 s 2 n n 1 n 1 e 2 i n t sin n t 1 2 0< < 1 Examples Find the Inverse Laplace Transform of L 1 F (s) 10 s s 2 10 1 1 1 t F ( s ) L 10 L 10 1 e 2 s s s s 1 Find the Inverse Laplace Transform of F (s) 300 s 36 s 4 300 1 1 L F (s) L 4 300 L 2 2 2 s s 36 s 36 s 1 1 3 300 6 L F (s) L 2 2 6 t sin 6 t 2 216 216 s s 6 1 L 1 300 1 F ( s ) 8.3333 t 1.3889 sin 6 t 2 Properties of the Laplace Transform Laplace Transforms have several very import properties which are useful in Controls d L f ( t ) sF ( s ) f (0 ) dt d2 d 2 L f ( t ) s F ( s ) sf (0 ) f (0 ) dt dt L t 0 f ( t ) dt F (s) s Now, you should see the advantage of having zero initial conditions Final Value Theorem If f(t) and its derivative satisfy the conditions for Laplace Transforms, then lim f ( t ) lim sF ( s ) t s 0 This theorem is very useful in determining the steady state gain of a stable system transfer function Do not apply this to an unstable system as the wrong conclusions will be reached! Example What is the steady state gain (DC Gain) of the system G (s) 3s 2 s s 2 s 10 2 lim g ( t ) lim sG ( s ) lim t s 0 s 0 3s s 2 s s 2 s 10 2 0.6 Summary Laplace Transform Inverse Laplace Transform Properties of the Laplace Transform Final Value Theorem Next Class: Using the Laplace Transform