Appendix ALinear AlgebraIn this appendix we review the notation and concepts from linear algebrathat are used in the text.A.1 Linear spaces and operatorsA.2 Jordan Form117118 APPENDIX A. LINEAR ALGEBRAAppendix BLaplace TransformsThe Laplace transform is an essential part of the language of control. Onlya few elementary properties are needed for basic control applications. Thereis a beautiful theory for Laplace transforms which makes it possible to usemany powerful tools of the theory of functions of a complex variable to getdeep insights into the behavior of systems.B.1 Basic ConceptsThe Laplace transform maps a time function f : R+→ R to a functionF = Lf : C → C of a complex variable. It is defined byF (s) =Z∞0e−stf(t)dt. (B.1)The transform has some properties which makes it very well suited to dealwith linear systems.First we observe that the transform is linear becauseL(af + bg) = aF (s) + bF (s) = aZ∞0e−stf(t)dt + bZ∞0e−stg(t)dt=Z∞0e−st(af(t) + bg(t))dt = aLf + bLg(B.2)Next we show that the Laplace transform of the derivative of a functionis related to the Laplace transform of the function in a very simple way.If F(s) is the Laplace transform of a function f(t) then the transform ofthe derivative df/dt is given by Next we will calculate the transform of the119120 APPENDIX B. LAPLACE TRANSFORMSderivative of a function, i.e. f0(t) =df(t)dt. We haveLdfdt=Z∞0e−stf0(t)dt = e−stf(t)¯¯¯∞0+ sZ∞0e−stf(t)dt = −f(0) + sLfwhere the second equality is obtained by integration by parts. We thusobtain the following important formula for the transform of a derivativeLdfdt= sLf − f(o) = sF (s) − f(0) (B.3)This formula is particularly simple if the initial conditions are zero becauseit follows that differentiation of a function corresponds to multiplication ofthe transform with s.Since differentiation corresponds to multiplication with s we can expectthat integration corresponds to division by s. This is true as can be seen bycalculating the Laplace transform of an integral. We haveLZt0f(τ)dτ =Z∞0³e−stZt0f(τ)dτ´dt= −e−stsZt0e−sτf0(τ)dτ¯¯¯∞0+Z∞0e−sτsf(τ)dτ =1sZ∞0e−sτf(τ)dτhenceLZt0f(τ)dτ =1sLf =1sF (s) (B.4)Integration of a time function thus corresponds to dividing the Laplace trans-form by s. This is consistent with the fact that differentiation of a timefunction corresponds to multiplication of the transform with s.Consider a linear time-invariant system where the initial state is zero.The relation between the input u and the output y of is given by the con-volution integraly(t) =Z∞0g(t −τ)u(τ)dτsee (2.20). We will now consider the Laplace transform of such an expression.We haveY (s) =Z∞0e−sty(t)dt =Z∞0e−stZ∞0g(t −τ)u(τ)dτ dt=Z∞0Zt0e−s(t−τ)e−sτg(t −τ)u(τ)dτ dt=Z∞0e−sτu(τ)dτZ∞0e−stg(t)dt = G(s)U(s)B.2. ADDITIONAL PROPERTIES 121The result can be written as Y (s) = G(s)U (s) where G, U and Y are theLaplace transforms of g, u and y. The system theoretic interpretation isthat the Laplace transform of the output of a linear system is a productof two terms, the Laplace transform of the input U(s) and the Laplacetransform of the impulse response of the system G(s). A mathematicalinterpretation is that the Laplace transform of a convolution is the productof the transforms of the functions that are convoluted. the fact that theformula Y (s) = G(s)U(s) is much simpler than a convolution is one reasonwhy Laplace transforms have become popular in control.B.2 Additional PropertiesFor the sake of completeness we will give a few Laplace transforms and someof their properties.The transform of f1(t) = e−atis given byF1(s) =Z∞0e−(s+a)tdt = −1s + ae−st¯¯¯∞0=1s + aBy setting a = 0 we obtain the Laplace transform for a step. Differentiatingthe above equation we find that the transform of the function f2(t) = te−atisF2(s) =1(s + a)2Repeated differentiation shows that the transform of the function f3(t) =tne−at0n! isF3(s) =1(s + a)n+1Setting a = 0 in f1we find that the transform of the unit step functionf4(t) = 1 isF4(s) =1sSimilarly we find by setting a = 0 in f3that the transform of f5= tn/n! isF5(s) =1sn+1Setting a = ib in f1we find that the transform of f(t) = e−ibt= cos bt −i sin bt isF (s) =1s + ib=s − ibs2+ b2=ss2+ b2− ibs2+ b2122 APPENDIX B. LAPLACE TRANSFORMSSeparating real and imaginary parts we find that the transform of f6(t) =sin bt and f7(t) = cos bt areF6(t) =bs2+ b2, F7(t) =ss2+ b2Proceeding in this way it is possible to build up tables of transforms thatare useful for hand calculations.The behavior of the time function for small arguments is governed by thebehavior of the Laplace transform for large arguments. Or more preciselythat the value of f(t) for small t is thus equal to sF (s) for large s. This isshown as follows.lims→∞sF (s) = lims→∞Z∞0se−stf(t)dt = lims→∞Z∞0e−vf(vs)dv = f(0)This result, which requires that the limit exists, is called is the initial valuetheorem. The converse is also true, we havelims→0sF (s) = lims→0Z∞0se−stf(t)dt = lims→0Z∞0e−vf(vs)dv = f(∞)The value of f (t) for large t is thus equal to sF (s) for small s, the result iscalled the final value theorem. These properties are very useful for qualita-tive assessment of a time functions and Laplace
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