ELG 3150 Introduction to Control Systems. TA: Fouad Khalil, P.Eng., Ph.D. Student

Transcription

1 ELG 350 Introduction to Control Systems TA: Fouad Khalil, P.Eng., Ph.D. Student

2 My agenda for this tutorial session I will introduce the Laplace Transforms as a useful tool for you to tackle control systems analysis. I will give examples on how to derive the transfer function of different systems represented by their mathematical models. I will talk about solving the system equation to obtain the system response and I will give examples regarding that. 2

3 Laplace Transforms We transform the system (model), which is identified by a differential equation (transfer function), from time domain to frequency domain. d D dt = = s s = σ + jω We normally assume zero initial conditions at t=0. If any of the initial conditions are non-zero, then they must be added. 3

4 How to implement time-to-frequency transformation Example However, in control systems analysis we would rather use tables to implement such transformation 4

5 How can we carry out control system analysis using Laplace Transforms. We convert the system transfer function (differential equation) to the s-domain using Laplace Transform by replacing d/dt or D with s. 2. We convert the input function to the s-domain using the transform tables. 3. We combine algebraically the input and the transfer function to find out an output function. 4. We Use partial fractions to reduce the output function to simpler components. 5. We convert the output equation from the s-domain back to the time-domain to obtain the response using Inverse Laplace Transforms according to the tables. 5

6 Laplace Transforms Properties L: Laplace Transform TIME DOMAIN f t ( ) f s ( ) Kf t ( ) f ( t) + f 2 ( t) f 3 ( t ) + df( t ) dt d 2 f( t ) dt 2 d n f( t ) dt n FREQUENCY DOMAIN KL[ f( t) ] f ( s ) + f 2 ( s ) f 3 ( s ) + sl[ f( t) ] f( 0 ) s 2 L[ f( t) ] sf( 0 ) df( ) dt s n L[ f( t) ] s n f( 0 ) s n 2 df( ) dt d n f( ) dt n t 0 f( t ) dt f( t a)u( t a), a > 0 L f( t [ )] s L[ f( t) ] e a s e a t f( t) f( s a ) f( at), a > 0 tf t ( ) t n f t ( ) f( t ) t --f s a ā - df( s ) ds ( ) ndn f ( s ) ds n s f( u) du 6

7 Laplace Transforms make things easy, for example consider the following convolution process: which can be performed as simple multiplication in frequency domain using Laplace Transforms: 7

8 Laplace Transforms Table TIM E D OM A IN δ( t) A t e at sin ( ω t) c os( ω t) te at unit i mpulse step ramp exponent ial deca y FREQ U EN CY D O MA IN A --- s ---- s 2 t s 3 t n, n > 0 n! s n s + a ω s 2 + ω 2 s s 2 + ω ( s + a ) 2 t 2 e a t ! ( s + a ) 3 8

9 TIME DOMAIN e at sin ( ωt ) e at cos( ωt) e at sin( ωt) e at B cosωt 2 A e α t cos ( βt + θ ) 2t A e αt cos( βt + θ) C ab sin ω ωt ( c a)e at ( c b)e bt b a e at e b t b a FREQUENCY DOMAIN ω ( s + a ) 2 + ω 2 s + a ( s + a ) 2 + ω 2 ω ( s + a ) 2 + ω 2 Bs + C ( s + a) 2 + ω 2 complex conjugate A A s + α βj s + α + βj complex conjugate A A ( s + α βj) ( s + α + βj) 2 s + c ( s + a )( s + b) ( s + a )( s + b) 9

10 Example For this simple mechanical system obtain the transfer function in s-domain (i.e. dynamical model) Kd Ks F = MD 2 x + K d Dx + K s x M x F( t) = x( t) MD 2 + K D + K d s F L F( t ) x( t) F( s) = = Ms 2 + K x( s) d s + K s 0

11 Example For this electrical circuit obtain the transfer function in s-domain

12 Have more examples on how to obtain the transfer function in the s-domain for the given systems: 2

13 Now back to our simple mechanical system to obtain its output response to a step input of magnitude 000 N. Given, Therefore, Assume, x( s ) F( s) F( s) x( s ) K d = = Ms 2 + K d s + K s A = --- s x( s) A = F( s ) = F( s) Ms K d s + K s s 3000 Ns m K s = 2000 N m ---- x ( s ) = M = 000kg A = 000N ( s 2 + 3s + 2)s 3

14 Types of inputs (driving force) 4

15 F( s) = L[ δ ( t)] = 5

16 Solving for system response using Inverse Laplace Transforms Use the following flow chart to determine which way to go to do Inverse Laplace Tarnsforms 6

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18 Performing partial fraction simplification x s ( ) = = = ( s 2 + 3s + 2)s ( s + ) ( s + 2)s A --- s B C s + + s A = lim s = ( s + ) ( s + 2)s s 0 B = lim ( s + ) = ( s + ) ( s + 2)s s C = lim ( s + 2) = ( s + ) ( s + 2)s s x s ( ) = s 2 = ( + 3s + 2 )s s s s + 2 8

19 Now we proceed with the Inverse Laplace Transforms to obtain the system time response x( t) = L [ x( s) ] = L s s s + 2 x( t) = L L L s s + s + 2 x( t) = [ 0.5] + [( )e ] + [( 0.5)e ] x( t) = 0.5 e t + 0.5e 2t 9

20 Try to think about the case where the driving force is an impulse input. So what will be the impulse response? F( s) = L[ δ ( t)] = 20

21 What s about the partial fractions simplification for the repeated roots. Example x( s) = = s 2 ( s + ) --- A s 2 + B --- C s + s C = lim ( s + ) s 2 = s ( s + ) A s = lim s 2 = lim = s 0 ( s + ) s 0 s + B d ---- s d = lim ds s 2 = lim = lim s + s 0 ( s + ) s 0 ds s + [ ( ) 2 ] = s 0 2

22 Have another example F s ( ) = s 2 ( s+ ) s 2 ( s + ) 3 = A --- s 2 + B C --- s D E ( s+ ) 3 ( s + ) 2 ( s + ) 22

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24 s 2 ( s + ) 3 = A ---- s 2 + B C --- s D E ( s + ) 3 ( s + ) 2 ( s + ) A B C D E = lim s s 0 s 2 ( s + ) 3 = lim s 0 ( s + ) 3 = 5 d d lim s s 0 ds s 2 ( s + ) ( 3) = = lim s 0 ds ( s + ) 3 = lim s 0 ( s + ) 4 = = lim ( s + ) s s 2 ( s + ) 3 = lim ---- s s 2 = d lim s +! ds s 2 ( s + ) 3 ( ) d 5 = = lim ---- s s! ds s 2 = lim ( 5) s! s 3 = d lim ( s + ) s 2! ds s 2 ( s + ) d 2 5 = = lim --- s 2! ds s 2 = lim s 2! s 4 = s 2 ( s + ) 3 = s s ( s + ) ( s + ) 2 + ( s + ) 24

25 Special case 25

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27 Initial and Final Value Theorems x( t ) = lim [ sx( s) ] s 0 Final value theorem x( t ) s = lim = lim = = s 0 ( s 2 + 3s+ 2)s s 0 s 2 + 3s+ 2 ( 0) 2 + 3( 0) x( t 0) = lim [ sx( s) ] s Initial value theorem x( t 0) ( s) = lim = = --- = 0 s ( s 2 + 3s+ 2)s (( ) 2 + 3( ) + 2) 27

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32 Conclusion Laplace Transforms are simple, useful and powerful tools to solve for system response and hence analyse the control system BUT we need to prove that!!!! 32

33 as differential equation 33

34 So equation solution (system response) = homogenous response + particular response 34

35 While solving for system response using Laplace Transforms ( s + 0.5) X ( s) = 2 2 X ( s) = s X ( t) = L [ X ( s)] = 2e 0.5t 35

36 Reference Electronic book Dynamic System Modeling and Control by Hugh Jack at ks/model/pdf/model2_6.pdf 36

37 Your Questions 37