5.4 UnitStepFunctionsandPeriodicFunc- tions

1 unit step functions AND periodic FUNCTIONS157 Which implies thaty(t) =t2solves the DE. (One may easily checkthat, indeedy(t) =t2does solve the DE/IVP. ExercisesIn 1-8, solve the ODE/IVP using the Laplace + 4y + 3y= 0, y(0) = 1, y (0) = + 4y + 3y=t2, y(0) = 1, y (0) = 3y + 2y= sint, y(0) = 0, y (0) = 3y + 2y=et, y(0) = 1, y (0) = 2y=t2, y(0) = 1, y (0) = 4y=e2t, y(0) = 0, y (0) = + 3ty y= 6t, y(0) = 0, y (0) = +ty 3y= 2t, y(0) = 0, y (0) = 1 (You will need integrationby parts or use technology) unit step functions and periodic Func-tionsIn this section we will see that we can use the Laplace transform tosolve a new class of problems efficiently. In particular, we will be ableto consider discontinuous forcing functions .)

2 First, we make a unit step FunctionLetu(t) ={0t <01t >0 This function is also called a Heaviside the graphs of (a)u(t), (b)u(t 1), (c)u(t) u(t 1)(d)(sint) [u(t) u(t 1)]158 CHAPTER 5. LAPLACE TRANSFORMSF igure :Plots of (a)-(d) in Exercise unit step functions AND periodic FUNCTIONS159 Figure :Plot ofu(t a) u(t b), which is 1 on(a, b)Solution: Note that the general plot ofu(t a) u(t b),wherea < bis shownin the plot below: We can use unit step functions to write any case-defined, up to thepoints where the discontinuity points of the unit step (t) = 0t <1t21< t <2 5 2< t <3sintt >3in terms of unit step :We may rewrite this function as160 CHAPTER 5. LAPLACE TRANSFORMSf(t) =t2[u(t 1) u(t 2)] 5[u(t 2) u(t 3)] + (sint)u(t 3)Note that this can be further simplified asf(t) =t2u(t 1) (5 +t2)u(t 2) + (sint+ 5)u(t 3) Below, we describe how to express a case defined function using unitstep a Case-Defined FunctionThe functionf(t) = f1(t)t0< t < t1f2(t)t1< t < (t)tn 1< t < tncan be rewritten asf(t) =f1(t)[u(t t0) u(t t1)] +f2(t)[u(t t1) u(t t2)] +.}

3 +fn(t)[u(t tn 1) u(t tn)]orf(t) =n j=1fj(t)[u(t tj 1) u(t tj)]Note that iff(t) = f1(t)t0< t < t1f2(t)t1< t < (t)tn 1< tthen we would expressf(t) asf(t) =f1(t)[u(t t0) u(t t1)] +f2(t)[u(t t1) u(t t2)] +.. unit step functions AND periodic FUNCTIONS161+fn 1(t)[u(t tn 2) u(t tn 1)] +fn(t)u(t tn 1)Laplace Transforms of step FunctionsLaplace Transform ofu(t a)Fora 0,L[u(t a)](s) =e ass, s >0 More generally,Laplace Transform ofu(t a)f(t a) (Pre-Shift Theorem)Fora 0,L[u(t a)f(t a)](s) =e asL[f(t)](s)Proof:By definitionL[u(t a)f(t a)] = 0e stu(t a)f(t a)dtSinceu(t a) = 0 fort < a,andu(t a) = 1 fort > a,this integralbecomes ae stf(t a) aanddw=dt. Then this integral becomes 0e s(w+a)f(w)dwore sa 0e swf(w)dw=e saL[f(w)](s) =e saL[f(t)](s) We will call this Theorem the Pre-Shift Theorem, since it requires usto rewrite the variablettot ain order to use the result as the nextexamples 5.

4 LAPLACE TRANSFORMSE xample [u(t 7)t2]Solution:We need to rewritet2in terms oft ((t 7) + 7)2= (t 7)2+ 14(t 7) + :L[u(t 7)t2] =L[u(t 7)(t 7)2] + 14L[u(t 7)(t 7)] + 49L[u(t 7)]=e 7s(2s3+14s2+49s) Example [u(t 4) sin 2t]Solution:We need to rewrite sintin terms oft 4 using a trigono-metric identity. Sosin 2t= sin(2[t 4] + 8) = sin 2(t 4) cos 8 + cos 2(t 4) sin 8 Substituting:L[u(t 4) sin 2t] =L[sin[2(t 4)] cos 8 + cos[2(t 4)] sin 8]=e 4s(cos 8(2s2+ 4)+ sin 8(ss2+ 4)) Inverse Laplace Transforms involvinge as(Backward Pre-Shift Theorem)Fora 0,L 1[e asF(s)] =u(t a)f(t a),whereF(s) =L[f(t)](s). unit step functions AND periodic FUNCTIONS163 Example 1[e 4s1s4]Solution:We know forF(s) =6s4thatf(t) =t3. SoL 1[e 4s1s4]=16L 1[e 4s6s4]=16u(t 4)(t 4)3 We now solve a differential equation arising from a spring mass systemwith discontinuous +y= 10[u(t ) u(t 2 )], y(0) = 0, y (0) = 1and plot its graph from0 t 3.

5 Explain the behavior if this were aspring-mass system and find amplitude of the steady :Taking the Laplace transform of both sides and writingL[y(t)] asY(s), we obtain:s2Y(s) 1 +Y(s) = 10[e s e 2 s]1ssoY(s) = 10[e s e 2 s]1s(s2+ 1)+1s2+ partial fractionsY(s) = 10[e s e 2 s](As+Bs+Cs2+ 1)+1s2+ see thatA= 1, B= 1, C= 0 SoY(s) = 10[e s e 2 s](1s ss2+ 1)+1s2+ 5. LAPLACE TRANSFORMSF igure :Plot of solution with discontinuous forcingThereforey(t) = 10u(t )[1 cos(t )] 10u(t 2 )[1 cos(t 2 )] + sin(t).Since fort >2 ,bothu(t ) andu(t 2 ) equal 1, this will reducetoy(t) = 10 cos(t ) + 10 cos(t 2 ) + sintwhich can be rewritten using trigonometric identities as: 10[costcos( ) + sintsin ] + 10 cost+ sint= 20 cost+ sintso the amplitude is 202+ 1 = 401 plot this example illustrates the effect forcing a particular solution of aspring mass system with a force of 10 Nfromt= tot= 2 that aroundt= the displacement increases to about 20, itis at this time that the forcing is stopped and the spring mass systemcontinues to oscillate at this new amplitude.

6 unit step functions AND periodic periodic FunctionsDefinition function isperiodicif for someT >0, f(t+T) =f(t)for smallest such positive value ofTis called theperiodoff(t).One way to define a periodic function is simply to specify its valueson [0, T] and then extend it. We define thewindowed version of afunctionf(t) to befT(t) ={f(t) 0< t < T0elseorfT(t) =f(t) [u(t) u(t T)]Then we can write:f(t) = k= fT(t kT) = k= f(t kT) [u(t kT) u(t (k+ 1)T)],but note that this function is not actually defined at the values oft= 0, , 2T, ..,since the unit step functions are not defined that if we only only care aboutf(t) whent >0,thenf(t) = k=0fT(t kT) = k=0f(t kT) [u(t kT) u(t (k+ 1)T)].Extending a Piece of a Function to aT- periodic FunctionLetf(t) be a function defined for allt.}

7 The periodic extension off(t) viafT(t) is the function with periodTgiven by f(t) = k=0f(t kT) [u(t kT) u(t (k+ 1)T)].Note that this function is actually undefined for:t= 0, T,2T, canbe rewritten as: f(t) =f(t) + k=1[f(t kT) f(t (k 1)T)]u(t kT).166 CHAPTER 5. LAPLACE TRANSFORMSF igure :Plot of periodic function generated byf(t) =ton(0,2)If we only care about this function on a finite interval, we do not needall the terms in this infinite thatf(t) =tand we want to createfT(t)forT= 2and extend it to a periodic function f(t). Plot the graph of f(t)on[0,10]and express f(t)in terms of unite step functions on[0,10].Solution:Effectively, we are takingf(t) =ton the interval (0,2)repeating it, so its graph on [0,10] is in Figure that fort >0, f(t) = k=0(t 2k) [u(t 2k) u(t 2(k+ 1))].

8 Note that this is (after expanding) f(t) =t 2u(t 2) 2u(t 4) 2u(t 6) .. unit step functions AND periodic FUNCTIONS167=t 2 k=1u(t 2k) Example +y= f(t), y(0) = 0, y (0) = 0where f(t)is as in Example :Since f(t) =t 2 k=1u(t 2k)we take the Laplace transform of both sides to obtain:(s2+ 1)Y(s) =1s2 2 k=1e 2kssY(s) =1s2(s2+ 1) 2 k=1e 2kss(s2+ 1)Y(s) =(1s2 1s2+ 1) 2(1s ss2+ 1) k=1e 2kssoy(t) =t sin(t) 2 k=1(1 cos(t 2k))u(t 2k).A plot of the solution fort= 0 tot= 44 is shown. The following is also helpful for a periodic function with windowedversionfT(t).168 CHAPTER 5. LAPLACE TRANSFORMSF igure :Solution of IVP in Example Transform of periodic FunctionsFor a periodic function f(t) with associated windowed versionfT(t) we haveL[ f(t)] =11 e T sFT(s) =FT(s) k=0e kT s,Proof:Since fort >0 we havefT(t) = f(t) [u(t) u(t T)]Since fisT periodc we havefT(t) =f(t)u(t) f(t T)u(t T).

9 Taking the Laplace transform of both sides yields:FT(s) =L[ f(t)] e sTL[ f(t)].Therefore,L[ f(t)] =11 e sTFT(s). unit step functions AND periodic FUNCTIONS169 Note that we have a the form of the sum of an infinite geometric se-quence, namely:11 e sT= 1 +e sT+e 2sT+..SoL[ f(t)] =FT(s) k=0e kT s. ExercisesIn 1-5, write the function in terms of unit step functions and takethe Laplace (t) ={1t <1ett > (t) ={sint t < cost t > (t) ={sin(2t)t <2 0t >2 (t) = 1 0< t <22 2< t <46t > (t) = t20< t <28 t22< t <5e 3tt >56. Solvey + 2y + 4y=u(t 2) u(t 3), y(0) = 0, y (0) = Solvey + 2y + 4y=t2u(t 2) t2u(t 3), y(0) = 0, y (0) = Solvey + 2y + 4y=et[u(t 2) u(t 3)], y(0) = 0, y (0) = Graph the functionf(t) = 1 u(t 1) +u(t 2) u(t 3) +.}}}

10 170 CHAPTER 5. LAPLACE TRANSFORMS10. Solvey +2y +4y=f(t), y(0) = 0, y (0) = 0,wheref(t) is givenin the previous Graph the functionf(t) =t (2t 2)u(t 1) + (2t 4)u(t 2) (2t 6)u(t 3) +..12. Solvey +2y +4y=f(t), y(0) = 0, y (0) = 0,wheref(t) is givenin the previous Considerf(t) =e2tmade into a periodic function f(t) by takingfT(t) whereT= 1.(a) Plot f(t) for 0< t <4.(b) FindL[ f(t)](c)y + 2y + 3y= f(t), y(0) = 0, y (0) = 0,14. Use the differentiation theorem to verify thatL[t u(t a)] =e as1s215. Use appropriate theorems to computeL[tsintetu(t a)]