Question

laplace transform of heaviside function using shifting theorem

Answer 1

the answer i obtained is \[[\frac{15}{s}-\frac{17}{s ^{2}}-\frac{8}{s ^{3}}]e ^{-2s}+\frac{2}{s ^{3}}\]is it correct?

Answer 2

because i obtained the answer different from the answer given...
dotz\[[\frac{17}{s }+\frac{17}{s ^{2}}+\frac{8}{s ^{3}}]-e ^{-2s}+\frac{2}{s ^{3}}\]and my tutor is well known of giving wrong answer LOL

Answer 3

\[\int\limits_{0}^{\infty} e^{-st} H(t) dt = \int\limits_{0}^{\infty} e^{-st}dt = 1/(-s) e^{-st} ]_0^{\infty} = -1/s (0 - 1) = 1/s\]
how about this

Answer 4

the question requires us to solve using shifting methods.
if using integrals, it should be integrated from 0-2 for t^2 and from 2 to infinity for
(1-t-3t^2)

Answer 5

what is a shift method?

Answer 6

erm first shifting says that\[L[e ^{at}f(t)]=F(s-a)\]where the function is shifted by a units.
the 2nd shifting theorem says that\[L[f(t-a)H(t-a)]=e ^{-as}F(s)\]

Answer 7

so we are just shifting functions left and right on the graph it appears

Answer 8

im not sure I understand the mechanics behind how that helps tho

Answer 9

yes =). this tutorial i have so many answers to the question different from the 1 given =_=

Answer 10

\[\int t^2e^{-st}dt=t^2\frac{e^{-st}}{-s} +\frac{2}{s} \int t e^{-st}dt\]
\[\int t^2e^{-st}dt=t^2\frac{e^{-st}}{-s} +\frac{2}{s} (t\frac{e^{-st}}{-s}+\frac{1}{s}\int e^{-st}dt)\]
\[\int t^2e^{-st}dt=t^2\frac{e^{-st}}{-s} +\frac{2}{s} (t\frac{e^{-st}}{-s}+\frac{1}{s}(\frac{e^{-st}}{-s}))\]

\[\int t^2e^{-st}dt=\frac{t^2e^{-st}}{-s} +\frac{2te^{-st}}{-s^2}+\frac{2e^{-st}}{-s^3}; [0,2]\]

\[\frac{4e^{-2s}}{-s} +\frac{4e^{-2s}}{-s^2}+\frac{2e^{-2s}}{-s^3}-\frac{2}{-s^3}\]

\[\frac{4e^{-2s}}{-s} +\frac{4e^{-2s}}{-s^2}+\frac{2e^{-2s-2}}{-s^3}\]

Answer 11

that -2 got misplaced in the codeing :) its not in the exponent

Answer 12

this is definantly something i got to get better acquainted with

Answer 13

are we shifting the infinity one or the finite one?

Answer 14

erm from what i'd learn, if heaviside function is given, t^2 will be integrated from 0 to 2 while (1-t-3t^2) will be integrated from 2 to infinity

Answer 15

right; but what advantage do we get from a shift?

Answer 16

im so blur in dis topic lol

Answer 17

erm shifted for what im not so sure, but the heaviside functionis mainly used to represent any system what switches on and off, like those involved in electrical or mechanical

Answer 18

which*

Answer 19

so the system will be turned on after the value a, so the function is shifted to a? lol (this statement need confirmation)

Answer 20

the integration defines the area under a curve; so shifting the curve should help us determine the values for the finite and infinite portions

the area of t^2 from 0 to 2 would amount to

t^2 from 0 to infinity - (t+2)^2 from 0 to infinity is what im thinking

Answer 21

but im prolly just babbling like an idiot :)

Answer 22

if we shift the other one such that t = t+2 we can define it from 0 to inf as well

Answer 23

oo

Answer 24

lol im so doomed

Answer 25

oh it aint all that bad :)

an answer is doable, but the method asked isnt my strong point

Answer 26

kay thx. time to sleep =)

Answer 27

http://www.wolframalpha.com/input/?i=integrate+%28t%5E2+e%5E%28-st%29%29+dt+from+0+to+2

this matches what i get on paper for t^2 from 0 to 2

Answer 28

when i do my shifting idea .. still got no clue if its the same thing as Laplaces.

L{t^2} - L{(t+2)^2}

2 2 4 4
--- - ( ---- + ---- + --- )
s^3 s^3 s^2 s

i get the same overall structure except for the e^-2s

Answer 29

if i assume im right :)

the next shift should amount to the same idea as, but all we need is the part from 2 to inf and not really a subtraction of parts:

\[L\{1-(t+2)-3(t+2)^2\}\to\ \frac{1}{s}-\frac{1}{s^2}-\frac{2}{s}-3L\{(t+2)^2\}\]

Answer 30

\[\frac{2}{s^3}-\frac{2e^{-2t}}{s^3}-\frac{4e^{-2t}}{s^2}-\frac{4e^{-2t}}{s}-\frac{1}{s}-\frac{6e^{-2t}}{s^3}-\frac{12e^{-2t}}{s^2}-\frac{12e^{-2t}}{s}\]

\[\frac{2-8e^{-2t}}{s^3}-\frac{16e^{-2t}}{s^2}-\frac{16e^{-2t}+1}{s}\]

maybe .... but i got me doubts

Answer 31

holy == a negative sign just missed out from my sight.
this is my working for 2nd shift theorem.
f(t) can be rewritten as\[f(t)=f(t)[H(t-0)-H(t-2)]+f(t)H(t-2)\]substituting f(t),\[f(t)=t^2[H(t-0)-H(t-2)]+(1-t-3t^2)H(t-2)\]expanding,\[f(t)=t^2H(t-0)-t^2H(t-2)+H(t-2)-tH(t-2)-3t^2H(t-2)\]\[f(t)=H(t-2)-tH(t-2)+t^2H(t-0)-4t^2H(t-2)\]let them be A-B+C-D

for A, L[H(t-2)] is just \[\frac{e ^{-2s}}{s}\]
for B, \[t H(t-2)\]\[= (t-2 +2) H(t-2)\]\[=(t-2)H(t-2)+2H(t-2)\]taking the laplace of B gives\[\frac{e ^{-2s}}{s^2}+\frac{2e^{-2s}}{s}\]

for C, L[t^2 H(t-0)] is\[=\frac{2e ^{-0s}}{s^3}=\frac{2}{s^3}\]

for D, \[t^2 H(t-2)=(t-2+2)^2 H(t-2)\]\[=[(t-2)^2 +2(2)(t-2)+4]H(t-2)\]\[=(t-2)^2H(t-2)+4(t-2)H(t-2)+4H(t-2)\]therefore \[L[4t^2H(t-2)]\]\[=4L[(t-2)^2H(t-2)+4(t-2)H(t-2)+4H(t-2)]\]\[=4L[(t-2)^2H(t-2)] +16L[(t-2)H(t-2)]+16L[H(t-2)]\]\[=\frac{8e ^{-2s}}{s^3}+\frac{16e ^{-2s}}{s^2}+\frac{16e ^{-2s}}{s}\]combining everything, L[f(t)]=L[A]-L[B]+L[C]-L[D]
i missed out a negative sieng when expanding laplace of D == sorry for all the fuss