ESE 318-02, Fall 2014Lecture 3, Sept. 2, 2014Dirac, Periodic, Derivatives of transformsVisualizing time shifts and shut-offs. Zill 4.3.49-54 (in different order). The original function for all problems is pictured here.Write the functions and Laplace transforms for the following functions graphed below. (Ignore the f(t) in each graph; it is not the same as the original.)Original shifted by b.     sFebtbtfbs UOriginal, shut off from 0 to a.      atfeattfasLUOriginal, shut off after a. Subtract off the latter part.ESE 318-02, Fall 2014            atfesFattfttfasLUUOriginal, shut off after b.          btfesFbttftfbsLUOriginal, shut off from 0 to a and then from b to infinity.             btfeatfebttfattfbsasLLUUOriginal shifted by a, then shut off after b. Need to subtract off shifted function.            batfesFebtatfatatfbsasLUUESE 318-02, Fall 2014Zill 4.R.32. Express f(t) in terms of unit step functions and find L{ f(t)} and L {etf(t)}.Here is how you can do this systematically. First, look at each piece of this function and write what that piece of the function is as if it were the whole function. 221103212112211002tttPartPartPartPartPartPartttttttfThe parts, where “turned on”, have to add up to those functions.                                           21221221222222222111112111112112122110022211021212:2211222122:2221303212221222211ssstsssssessesessFtfeeseessesessFtttttttttttttttfCheckshiftssimpleallttttttttttttfTogetherttPartPartPartPartPartParttttParttParttPartParttPartLUUUUUESE 318-02, Fall 2014Solving a differential equation involving a time-shifted function. Zill 4.3.66.                41411114141110101100042222ssesssYsesssesYssesYYstttftttfyytfyyssssUUAt this point we have a couple of PFEs to do.     ttssssssssFEDDFEDFsEsDDssFEssDssttssssssssssCBAACBAsCsBsAAssCBssAsss2cos12cos44114141414101400144412sin2cos422144114141141114101444141414122412414122222141412224141241412222We have the two pieces, but how do we put them together into one solution? Any e-as part is a time-shifted piece, so replace “t” with “t-a” and multiply by the unit step delayed by a. That is really the key to such problems; understand that part. In this case, only one piece is time-shifted, and a = 1.       11012cos12sin2cos2sin2cos112cos12sin2cos4121414121414141214141tttttttttttty UWhat happens in this case at t = 1?In this case, the second part kicks in, but at t = 1, 1 - cos2(t-1) = 1- cos0 = 0. So there is no sudden step at the transition. That is not always the case.ESE 318-02, Fall 2014The Dirac Delta Function. Let’s first consider the unit impulse function of height 1/(2a) that is centered at time t0 and has a duration of 2a so that it starts at t0-a and ends at t0+a. attattataatttta0000002100See that it integrates to 1. We can describe it in terms of the Unit Step function and find its Laplace transform.            aseeeaseasettattaattattasasstsatsataa22221210000000LUUNow take the limit as duration, a, goes to zero. The result is the Dirac Delta Function. Then we find its Laplace transform.            122''2lim2limlimlimlim00000000000000tesseRulesHopitalLsseseeaseeettttttttttststasasastasasstaaaaaaaLLLLThe Dirac Delta Function is not a normal function. It is zero everywhere except at t = t0, where it is neither finite nor continuous, but it integrates to one. It also does not satisfy the usual “piecewise continuous of exponential order” criteria. But it works, and we use it. You can read a bit more about it in Zill, page 232.Why use it? One reason is that it can be used to model “impulsive” phenomena. Considera mass-spring-damper system where you tap the mass with a hammer.ESE 318-02, Fall 2014Zill 4.5.6                    44220sin2cossincoscos4sin2sincos44sin22sincos1110010422422422tttttttttttttttttttyseessYeeYsYsyyttyyssssUUUUPeriodic functions. A periodic function of period T means f(t+T) = f(t). Take its Laplace transform.                           TstsTsTTstsTssTTsTstTstTststdttfeetftfedttfetftfedfeedTfedttfeTtdttfedttfedttfetf00000011LLLLLZill 4.4.50 The period is 2a (1 from 0 to a, then 0 from a to 2a). See Figure 4.4.7 of Zill.        