Transcription of Laplace transform with a Heaviside function
1 Laplace transform with a Heaviside functionby Nathan GriggThe formulaTo compute the Laplace transform of a Heaviside function times any other function , useL{uc(t)f(t)}=e csL{f(t+c)}.Think of it as a formula to get rid of the Heaviside function so that you can just computethe Laplace transform off(t+c), which is words: To compute the Laplace transform ofuctimesf, shiftfleft byc, take theLaplace transform , and multiply the result bye cs. Remember that to shift left, youreplacetwitht+ other way to write the formulaYou will sometimes see the formula written asL{uc(t)f(t c)}=e csF(s),whereF(s) isthe Laplace transform off(t). This is a correct formula that says the same thing as thefirst formula, but it is aterribleway to compute the Laplace transform .
2 It is, however, aperfectly fine way to compute theinverseLaplace transform . Rewrite it asL 1{e csF(s)}=uc(t)f(t c).In words: To compute the inverse Laplace transform ofe cstimesF, find the inverseLaplace transform ofF, call itf, then shiftfright bycand multiply byuc. Rememberthat to shift right, you replacetwitht it worksRight now you are probably thinking,Don t prove it to me! I trust you!Mathematiciansbelieve that understanding a proof is crucial to understanding a statement, because that show our brains work. Sometimes we go a little too far and forget that there are other greatways to understand this case, though, you should see the proof. It will help you me!L{uc(t)f(t)}= 0e stuc(t)f(t)dtdefinition of Laplace transform = ce stf(t)dtbecause the integral from 0 tocis 0= 0e s(t+c)f(t+c)dtshift left byc=e cs 0e stf(t+c)dtpull out constante cs=e csL{f(t+c)}definition of Laplace transformHere is the same thing in pictures:uc(t)f(t) 0e stuc(t)f(t)dt 0e s(t+c)f(t+c)dtThe last integral simplifies toe csL{f(t+c)}because at this point we are treatingsasa by looking at these pictures, you see the key idea:Shifting left bycallows usto get rid of the Heaviside algebra requiredIff(t) =t2, thenf(t+c) = (t+c)2.
3 How do you take the Laplace transform of (t+c)2?You have to rearrange things, in this case expanding tot2+ 2ct+ example: Iff(t) =e2tthenf(t+c) =e2(t+c). If you want to take the Laplacetransform, you need to expand a harder one: Iff(t) = sint, thenf(t+c) = sin(t+c). If you want to take theLaplace transform , you need to do some trigonometric magic. Ifcis a multiple of /2 or , you can probably figure it out by drawing some triangles. Otherwise, pull out your trigidentities!1 This is not a product ruleOne common misconception about this Laplace transform formula is that it is a kind ofproduct rule, that the Laplace transform ofuc(t) timesf(t) is the Laplace transform ofuc(t) times the Laplace transform off(t).
4 It is not! There is no product rule for not convinced? Here you go:L{uc(t)}=e cssL{t}=1s2L{uc(t) t}=e cs(1s2+cs)L{t}=1s2L{t2}=2s3L{t t2}=6s4 Once again, there is no product rule. There is only a fancy way to take a step functionfrom inside a Laplace transform operator and bring it might need sin(t+c) = cos(c) sin(t) + sin(c) cos(t), or maybe cos(t+c) = cos(c) cos(t) sin(c) sin(t).
