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# CONTROL MECH ENGR SYS EML 4312

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This 18 page Class Notes was uploaded by Paige Cruickshank on Friday September 18, 2015. The Class Notes belongs to EML 4312 at University of Florida taught by Prabir Barooah in Fall. Since its upload, it has received 4 views. For similar materials see /class/206702/eml-4312-university-of-florida in Engineering Mechanical at University of Florida.

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Date Created: 09/18/15

NOTES ON LAPLACE TRANSFORM TRANSFER FUNCTION AND BIBO STABILITY Dr Prabir Barooah EML 4312 Spring 2009 January 23 2009 1 Laplace transform The Laplace transform of a signal is de ned by Ye uw 3 aw 56 C 1 The integral exists has a well de ned nite value only if the signal yt grows with t at a rate slower than the exponential e decays or vice versa The rate of decay depends on the complex number 8 Therefore for a given signal yt for the Laplace transform to exist the value of 8 must be such that the integral above convergesi The region of convergence of a Laplace transform Ys is the region in the complex plane C that s can take values in so that Ys is nite and well de ned Example 1 Let us evaluate the Laplace transform of the complex signal b39 10 eltabjgtr 5 t3 0 0 tlt 0 By applying the de nition of the Laplace transform we get yt zt Xs zte dt eltabj 5 dt 0 0 1 7 11111 a b 7 s ltTH0ltgt 7 5703 if Res gt a unde ned otherwise eltaeReltsgtltbeImltsgtgtjgtw 1 1 57abj Res gt a In general the Laplace transform of a signal ep where p is a complex number is Sip with a ROC given by Res gt RepI Note that the strict inequality is important Hence the Laplace transform of the signal eabj is with a region of convergence The Laplace transform Ys of a signal yt is a complex number whose values depend on the complex argument 8 Therefore Laplace transform can be thought of as a complex function that maps one complex plane the s plane to another complex plane the Ys planeI For a complex number 30 6 ROC the Laplace transform Y80 is a complex number Exercise plot the values of Ys for a few complex numbers 5 Sketch the signals e t1t sketch the signal that is ob tained by delaying it by 1 sec and the signal ob tained by advancing it by 1 sec Find mathematical expressions for the delayed as well as the advanced ver sions 11 Useful properties of the Laplace transform The following properties ofthe Laplace transform will be very useful in analyzing LTl systems speci ed in terms of differential equations To state one of the properties welll need the unit step signal 1t which is de ned as H E0 9 F A 1 t gt 0 1 t 7 2 0 t lt 0 Linearity a1t 593 a 1t b yt7 where zt are signals are ab are complex scalarsi Transform of a derivative signal lf Ys is the Laplace transform of a signal yt then the transform of the signal y39 is WW 8W8 90 This can be shown by integration by parts 9 7 manlttgtdt 7 lam 7 lt7sgte ylttgtdtl 7 7ylt0gt ms 00 0 where we have used the fact that since e yt is integrable for all values of s in the ROC otherwise the Laplace transform Ys would not existl e yt 7gt 0 as t 7gt 00 Transform of a integral signal lf Ys is the Laplace transform of a signal yt then L low 7143 It is left as an exercise for you to show that this true using the previous resulti Time shifting lf Ys is the Laplace transform of yt then the Laplace transform of the shifted signal yt 7 7391t 7 739 is L yt 7 T1t 7 7 e STYsi To prove it yt 7 T1t7 T 3 0 yt 7 aw 7 6 0 yt 7 Te 5quot wdt e7 oo yne quot dn where 77 3 t 7 739 e STYlEs Note that yt 7 7 e Ysl When the time shift 739 is positive yt 7 739 is the signal yt delayed by 739 units of time When 739 is negative yt 7 739 is the signal yt advanced by 739 units of time 1why 5i Convolution The convolution zt yt of two signals zt and yt is de ned as 9W 975 3 ITyt 7 7d7 3 The Laplace transform of the convolution of two signals is given by the product of their Laplace transforms 2t W t W gt Z s X sY s 4 To prove it ltzlttgt t w m m t m 6 0 Otmmt 7 we e s dt AmAtI MOi e SWTdt We will now reverse the order of the integration To do so notice that in the t 7 739 plane t along the horizontal axis and 739 along the vertical axis the area over which the integrand is integrated consists of that part of the first quadrant that lies below the line t 7quot The integration above is done by first letting 739 vary from 0 to t and then letting t vary from 0 to 00 To reverse the order of integration that is to integrate first over t and then over 739 while integrating over the same area we must first let t vary from 739 to 00 and then let 739 vary from 0 to t This implies that ltzlttgt t w new Tendth At 17 yt 7 aw dT At 17 0 yt 7 T1t 7 aw dT gamma5W 7 lt0tx739eind739gt Ys 7 X3Ys where we have used the previously proved result that yt 7 T1t 7 7 is vie 12 Useful pairs of signals and their Laplace transforms The impulse or the Dirac delta function 6t is defined as 00 0 ft6tdt ft6t W s 7m 0 for every function It can be shown that this implies f3 6tdt 12 The impulse function is 0 everywhere except at 0 where it is arbitrarily large but in such a manner that the total area under it is 1 The unit step signal has been defined earlieri 2How What does this mean take Laplace transform both sides The unit ramp signal tt is de ned as A t t gt 0 t t 7 6 lt gt 0 K 0 lt gt It is left as an exercise for you to show the following 1 m i 1A 1 217 7 HS 9 g l tt A W0 32wg F sinw0t i 5 I 5 s1nw0t A mi The signals mentioned above are the most important ones in the analysis of LTl systems and in the design of linear control systems 13 Inverse Laplace transform The inverse of a Laplace transform Ys de ned as 04J3900 m lolt5 Yltsgte5 ds lt7 747100 where 00 is any real number so that 00 jw is inside the ROC of Ys for every possible am Looking at the limits of integration it is obvious that the integration is being carried out in the complex plane along a vertical line passing through the real axis at 00 You can think of the inverse transform as something that recovers the signal yt from the Laplace transform Ysr The inverse transform is also linearr That is if Xs and Ys are Laplace transforms of two signals 1t and yt with associated ROCs A and B such that Am B is not empty then the inverse Laplace transform of Xs Ys is given by L 1 Xs Ys 1t Evaluation of the inverse transform involves complex integration but usually the signal can be recovered from Ys by partial fraction expansion and using knowledge of Laplace transform of complex exponential signals 2 Transfer function Consider the following linear differential equation that models the dynamics of a mass m that is acted on by an external force u when the mass has a spring and damper attached to it mit bdct km ut L cl z0z390r s to of Take the Laplace transform of both sides to get m32Xs 7 810 7 bsXs 7 kXs Us lt82 S gtXSUSSb m m m 10 310 which can be rearranged into 1 5711795lt0gtilt0gt YS828U5 52S 9 HA5 Yzc5 10 1 39 1 73 1010 Yltsgt7828rltsgt g s 10gt W 1HcSUS 5103018 11 This shows that the output consists of two terms one due to the input and the other due to the initial conditions We will deal extensively with causal linear time invariant LTl systems with a single input and a single output SISO described by a linear constant coef cient ordinary dif ferential equation of the following form3 M a1y 1 7172 0471719 any b1um bgum l 39 39 39 bm71 bmubmlu 12 where y is the n th derivative of y and aibj7s are parameters that do not change with time Taking Laplace transform of both sides of the equation above we get W 7 HcltsgtUltsgt he s 13gt where the complex function A amsm am713m71 a18 a0 Hcltsgt 1 8 n713n7 518 50 is called the continuous time transfer function from u to y for the system 12 and 03 depends as it does in on the initial conditions 14 Upon taking the inverse Laplacetransform of 13 we get Mt 1Hc8US 5109018 which shows that the output of the system is a superposition of the response due to the input and the response due to initial conditions If all the initial conditions are 0 then 13 reduces to WS Hc8US 15 Since yt can be recovered by taking the inverse Laplace transform of the right hand side it shows that when all initial conditions are zero the transfer function together with the input completely determines the output Poles and zeros The roots of the denominator polynomial of a transfer function HC 8 are called the poles of the system and the roots of its numerator polynomial are called the zeros of the systemi 3Note that the coef cient of the highest derivative of y is 1 This is the standard way of expressing such equations note that the transfer func tion is from Us to Ys not from ut to yt SO continuouatime trans fer function is really a mis nomer 1n MATLAB tfnum den creates a continuoustime transfer function with nu merator and denominator speci ed by mun den Ebr ample tf1 1 2 3 pro 1 duces S 2S3 1n MATLAB zpkzpk creates a continuoustime transfer function with Ze ros poles and gain speci ed by z p and k The MATLAB command impulsesys can be used to plot the impulse re sponse for the system spec i ed in sys Can you nd a bounded in put suCh that if ht is not absolutely integrable the response to this input will be unbounde 7 21 Impulse response and transfer function So far we have been talking about Laplace transform of signals Now we have a Laplace transform HC s that we obtained by manipulating a differential equation A question might arise at this point what is the signal whose Laplace transform is Hcsi The answer to this question is the impulse response The impulse response usually denoted by ht of a LTl system is its output when it is driven by an impulse input 6t with all initial conditions set to 0 It is impossible to determine the impulse response experimentally though it is a useful concept for analysis To verify the above claim let us do the following thought experiment on a system whose transfer function from the input u to the output y is Let the system now be driven by an impulse input 6t In that case W8 H8 5t H87 since the Laplace transform of 6t is l The impulse response ht is the inverse Laplace transform of Ys so that we get Mt 1H8 gt 198 10 Lesson The transfer function of a system from the input to the output is therefore the Laplace transform of its impulse response For an arbitrary input ut since the output yt satis es Ys HsUs when initial conditions are zero from the convolution property of Laplace transforms we get W Mt u Thus the response to an arbitrary input can be determined from the knowledge of the impulse response of the system when initial conditions are 0 r 3 Stability A system is called BIBO stable if the output is bounded for every bounded input It has two equivalent characterizations for LTl systems speci ed by transfer functions in terms of the impulse response and in terms of its poles l A LTl system is BlBO stable if and only if its impulse response ht is absolutely integrable irei BlBO stable ltgt lhtldt lt 00 0 2 A LTl system is BlBO stable if and only if the real part of each pole of its transfer function Hcs is negative If a system is BlBO stable it can be shown that ea 3 r1 mars 16gt is a signal such that 6t A 0 as t A 00 Therefore for a BlBO stable system the output at steady state as t A 00 is given by yss 1HcSUS 17 Thus the transfer function provides information on steady state behavior of the system 88 response of LTI BIBO stable systems to sinusoidal inputs zero LC U105 ejwot gt 9175 Cjw0 jw0t E175 U205 jw0t 3205 G jwo jw0t 6203 Where 61t gt O and 62t gt 0 as t gt 00 When 75 ASi w0t U1t 162057 W6 haVe 3 y1t 9205 A gt y 58t 2 Cjwo 0t G Jw0 7w0t it can be shown that G jw0 Gjw0 ysst A 2339 A 2 GUwO Gjw0 08th jGjw0 Gjw0 Sinwot mywow 2 j2j1mGjw0 cos wot 2jR Gjw0 Si wot A 04 sin wot cos wot Where aw0 2 124mm moo ImGw0t A 39 39w 39w gt ySSt GJw0ej 0t CJw069 0t A 39 o Con GUM COS wot Gjw0 Cjw081nw0t 25 2jImGjw0 cos wot 2jReGjw0 sin wot A 04 sin wot 6 COS wot Where aw0 ReGw0t mo 2 Imam 58 Z A 2 2 a gtyt xOz 3 W W Gjw0A sirlth COS COS wotSin Cb sin wot cos wot oz ReGjw0 gt ysst amwaejwot mew GUwo Gm coswot Ham Guwo sin wot 2 j2jlmGjw0 cos wot 2jReGjw0 sinwot A oz sin wot 3 COS wot Where ozw0 R6Gw0t Memo 2 ImGw0t 04 gt ysst AOz2 52 a2 2 sm wot 042 Z cos wot Gjw0A sin wot cos gb COS wotsin qb Gjw0Asinw0t gbw0 1mGjw0 Where arctan RdGUwo LGQwO GUWOM Z Si lt d0t sinwt gt sinwt in 55 G s s215s8 sin3t mum 1 16 18 20 0 1 30 t second Gain 02 phase 90 sinwt gt sinw 03 b 1 t in 55 12 155 8 7125 sin 006 0404 8 002 gt o 410 27rt 1anmummw ttttttt d w 3 rads mag 13 dB Ie02105 phase 105 w 2pi rads mag 30 dB e 00316 phase 163 Bode diagram Magnitude dB L 1 Phase degree I a In c 100 10 Frequency rads Gjw is Gs restricted to s jw Magnitude dB of Gs Gjw is Gs restricted to s jw Phase deg of Gs Bode plot of 1st order system Magnitude dB Phase degree 10 2O 30 4o 10 45 9o 1o Bode diagram 10 10 Frequency rads Bode plot of second order system real poles 10 600 GO is was 1 G s M 8 2s 30 E 5 V gtgt G tf6001 32 60 39 xx Transfer function g gt m squot2 32 s 60 39 gtgt figurebodeG quot Frequency mussel

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