This preview shows page 153 - 157 out of 227 pages.
Example 3.EvaluateZ2π01sinθ+ 2dθ.Answer.We putz=ejθand getZ2π01sinθ+ 2dθ=ZC1(z-1z)12j+ 2dzjzwhereCis the unit circle. NowZCdz2jz+ (z2-1)12=ZC2dzz2-1 + 4jz.The roots of the bottom arez=-4j±√-16 + 42= (-2±√3)j,soz2+ 4jz-1 = (z+ (2 +√3)j)(z+ (2-√3)j).C153
Observe that (-2 +√3)jis insideC, but (-2-√3)jis not, since(|(-2 +√3)j|= 2-√3<1,|(-2-√3)j|= 2 +√3>1.We also note that the pole at (-2 +√3)jissimple. ThenZC2z2+ 4jz-1dz= 2πj22√3j=2π√3.As a final comment, we recall thatf(x) is even ifff(x) =f(-x). Supposef(x) is even, thenZ∞0f(x)dx=12Z∞-∞f(x)dx.We can thus evaluateR∞0f(x)dxby evaluatingR∞-∞f(x)dxand dividing by 2. Asan example considerR∞011+x2dx.Heref(x) =11+x2forf(x) =f(-x) andfiseven. We thus haveR∞011+x2dx=12·πfrom Example 1. Note also thatf(x) isoddiff(x) =-f(-x), soZ∞-∞f(x)dx=limR→∞ZR-Rf(x)dxbutZR-Rf(x)dx=Z0-Rf(x)dx+ZR0f(x)dx=-ZR0f(x)dx+ZR0f(x)dx= 0,andR∞-∞f(x)dx= 0! Unfortunately, we can’t doR∞0f(x)dxiff(x) isodd.154
III.hInverting the Laplace Transform and ApplicationsWe now come one of the main reasons for all the work we have done: invertingthe Laplace Transform. The basic idea—constructing suitable paths in the complexplane—is similar to the previous section.You may recall that giveny=f(t)(usuallytis actually time),L(f)(s)—the Laplace Transform off—is given byL(f)(s) =Z∞0e-stf(t)dt.In elementary courses,sis thought of as being a real number, but in practicalproblems this isnotthe case and you should think ofsas being complex. Indeed,s=jwwithw= frequency is of particular significance.So we wish to think asL(f)(s) being given for any complexs.The formulas you hopefully recall holdunchanged forscomplex. In particular,L(eat)(s) = 1/(s-a) if Re (s)>Re (a).Now letf(s) be the transform of some functionh(t). We knowf(s) and wishto findh(t), i.e., we wish to invert the transform. In practice,f(s) will have somepoles at some pointss1, . . ., smof the complex plane. As is often the case in practicewe start by assumings1, . . ., small lie in the left half plane. (We shall deal withthe other case later.) We construct the pathC=C1+C2withC2a semicircle ofradiusRas shown.C2CR1s1s2sms3sj R- j RThink ofRas being large. Ifsis any point inside the semicircle (as shown), then155
sincefis analytic inside and onCwe havef(s) =12πjZCf(z)z-sdz.Think ofsas being temporarily fixed, and assume|f(z)| →0 as|z| → ∞. Thisis the case for most “practical” functions but not for all. The latter case is morecomplicated and we do not deal with it in this course. Look atC2: ifzis onC2then|f(z)| →0 as|z|(i.e.,R) approaches∞, and 1/|z-s|also goes to zero as|z| → ∞. One can formally show that as a consequenceRC2f(z)/(z-s)dz→0 asR→ ∞, and a detailed proof is given in theoretical courses. So, sincef(s) neverchanges asR→ ∞, we getf(s) =12πjZCf(z)z-sdz=12πjZC1f(z)z-sdz+ZC2f(z)z-sdz-→12πjZ