7.0 Laplace Transform Applications In this chapter we will examine many applications of Laplace transforms. While it is possible to go back to the time-domain to determine properties of a system, it is often more convenient to be able to determine these properties in the s-domain directly. 7.1 Characteristic Polynomial, Characteristic Modes, and the Impulse Response Consider a transfer function of the form (()))(NsHDss= where and are polynomials in s with no common factors. is called the characteristic polynomial of the system. The poles of the system are determined from and these give us most of the information we need to completely characterize the system. The time-domain functions that correspond to the poles of the transfer function are called the characteristic modes of the system. To determine the characteristic modes of a system it is often easiest to think of doing a partial fraction expansion and determining the resulting time functions. Finally, the impulse response is a linear combination of characteristic modes. A few examples will help. ()Ns ()Ds ()Ds()Ds Example 7.1.1. Consider the transfer function 12 3 4221() a11(1)(3) 1a3ssaassHsssss+=+ + +++ + +=1s) The characteristic polynomial is 2(1)(3)( sDs s s++=12 3() a () a ()uut tut t=+ + and the characteristic modes are and . The impulse response is a linear combination of these characteristic modes, . (), (), (),tut tut e ut− 3()teut−()ht a34()tta uee−−+ t Example 7.1.2. Consider the transfer function 12 3 4223111(1)((3) 1 ( 1) 3)saa a ass s s s s sHs−=+ + +++=1+++ The characteristic polynomial is 2() ( 1)( 3)Ds ss s=++)t12 3() ()()ttut e tea ut a ut−−=+ + and the characteristic modes are and . The impulse response is a linear combination of these characteristic modes, . (), (), (),ttut e ut te ut−− 3(teu−() a34()ht a ute−+t ©2009 Robert D. Throne 1Example 7.1.3. Consider the transfer function 12222222131122()113 13 1322 22 22Hs a assssss== = +++⎛⎞ ⎛⎞ ⎛⎞⎛⎞ ⎛⎞ ⎛⎞++ ++ ++⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜+⎟⎜⎟⎝⎠ ⎝⎠ ⎝⎠⎝⎠ ⎝⎠ ⎝⎠2 The characteristic polynomial is 2() 1Ds s s=++ and the characteristic modes are 123cos ( )2tteut−⎛⎞⎜⎟⎜⎟⎝⎠ and 12sintut−⎛⎞⎜⎟⎜⎟⎝⎠3( )2te. The impulse response is the linear combination 112212() cos33si((2n)2ttht a t ut a t uee−−⎛⎞ ⎛⎞=+⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠)t. There are a few things to keep in mind when finding the characteristic modes of a system: • There are as many characteristic modes as there are poles in the transfer function • For complex conjugate poles of the form djσω−±the characteristic modes will be of the form and . Note that we could combine these into the form cos( ) ( )tdetσω−teσ−ut utsin( ) (deω) ( )tutωθ+)ttσ−(dcos Example 7.1.4. If a transfer function has poles at 1, 1, 2 3 ,j−−−± and 25 j−±3) ()tut the characteristic modes will be , , and . 22cos(3) (), sttut e−(), (), in(tt teut teut e−− −(2 ) ( )etut5cos(2 ) ( )tetu−t5sint− 7.2 Asymptotic Stability We have previously introduced the concept of Bounded Input Bounded Output, or BIBO, stability. As we have seen, an LTI system is BIBO stable if |()|hdλλ∞−∞<∞∫ Another useful definition of stability, which is used often in control systems, is that of asymptotic stability. A system is defined to be asymptotically stable if all of its characteristic modes go to zero as t , or equivalently, if →∞lim ( ) 0tht→∞=. A system is defines to be asymptotically marginally stable is all of its characteristic modes are bounded as t , or equivalently, if →∞limt→∞| ( ) |ht M≤ for some constant M. If a system is neither asymptotically stable or asymptotically marginally stable, the system is asymptotically unstable. In determining asymptotic stability, the following mathematical truths should be remembered: ©2009 Robert D. Throne 2lim 0 0 0lim ( ) 0 0()cos( )sin( )nattatdtddt e for all n and ae cos t for all au t is boundedt is boundedt is boundedωθωθωθ−→∞−→∞=>>+= >++ Example 7.2.1. Assume a system has poles at -1, 0. and -2. Is the system asymptotically stable? The characteristic modes for this system are and . Both and go to zero as . does not go to zero as , but it is bounded. Hence the system is asymptotically marginally stable. ( ), ( )teut ut− 2()teut−t →∞()teut− 2()teut−t →∞()ut Example 7.2.2. Assume a system has poles at -1, 1, and 23j−±. Is the system asymptotically stable? The characteristic modes for this system are , , , and . All of the modes go to zero as t except for , which goes to infinity. Hence the system is asymptotically unstable. ( )teut−→∞( )teut2cos(3 ) ( )tetu−()teutt t2sin(3 ) ( )tetu− Example 7.2.3. Assume a system has poles at -1, -1, 2 j−±. Is the system asymptotically stable? The characteristic modes for this system are , , , and . All of the characteristic modes of the system go to zero as , so the system is asymptotically stable. ( )teut−( )tte u t−2cos( ) ( )tetu−t →∞t t2sin( ) ( )tetu− From these examples, if should be clear that a system will be asymptotically stable if all of the poles of the system are in the left half plane (all of the poles have negative real parts). This is a very easy test to remember. 7.3 Settling Time and Dominant Poles Once we think about representing the impulse response as a linear combination of characteristic modes, we can define asymptotic stability in terms of the way these modes behave as t . Another benefit of this approach is that we can think of the settling time of a system, i.e., the time the system takes to reach 2% of its final value, in terms of the settling time of each of its characteristic modes. When we talk about the settling time of a system, we assume →∞• the system is asymptotically stable • the poles of the system are distinct (no repeated poles) • the input to the system is a step Let’s assume our system has transfer function ()Hs with corresponding impulse response 11 22() () () ()nnht a t a t a tφφφ=+++" ©2009 Robert D. Throne 3Here the are the coefficients we determine using the partial fraction expansion, and the ia()itφ are the characteristic modes, i.e., 1()iiiaatspφ↔+ Now let’s assume the input to our system is a step of amplitude A and we want to use partial fractions to determine the