Question

f(t) ={ (0, t<4) , ( (t^2 -8t + 20), t=>4) } find the laplace transform

Answer 1

the unit step function must be constructed from \[0, t<4\] and \[t^2 - 8t + 20, t \ge4\]

Answer 2

it would start at 0 +u(t-4)(t^2 -8t + 20) right?

Answer 3

Yeah f(t) can be written as
\[f(t)= (t^2-8t+20) U(t-4)\]

Answer 4

You need to find the LT???

Answer 5

right... does the laplace transform just distribute over into each term?

Answer 6

Yeah it can be distributed but you know that LT is defined as
\[f(s)=\int_{0}^{\infty} f(t) e^{-st} dt\]
so we'll have to use a LT propertu

Answer 7

*property

Answer 8

You know that property?

Answer 9

err no..

Answer 10

Ok
\[\large LT( f(t) U(t-a))= e^{-as} F(s)\]
Have you studied this?
Do you want me to derive this?

Answer 11

ooh yes i know that

Answer 12

Ok so we'll use it
First we'll find the LT of
\[t^2-8t+20\]
Can you find that?

Answer 13

2/(s^3) + -8/(s^2) + 20/s

Answer 14

from the table

Answer 15

Great:)
Now just we need to multiply it by
\[\large e^{-4s}\]
and you have your LT
\[\huge e^{-4s} ( \frac{2}{s^3}-\frac{8}{s^2}+\frac{20
}{s})\]

Answer 16

but the answer contains terms (1+4s)8 for the middle term, and (1+4s+8s^2) on the first in the answer book

Answer 17

It can also be written as
\[\huge e^{-4s} \frac{2-8s+20s^2}{s^3}\]
But it's also not not matching!!!
Could you post the complete answer

Answer 18

sure one sec

Answer 19

Oh no I made a mistake

Answer 20

Don't post the answer, we'll get it

Answer 21

Actually the property is
\[\huge LT( f(t-a) u(t-a))=e^{-as} F(s)\]

Answer 22

hmm

Answer 23

sidetrack: this would work with sin(pit) right? (my next question)

Answer 24

We have
\[f(t)= t^2-8t-20\]
so
\[f(t-4)=t^2-8t+16-8t+32-20=t^2-16t+28\]
so f(t) can be written as
\[f(t)= f(t-4)+8t-48\]
Do you understand this?

Answer 25

oh you plugged in t-4 into all the t's

Answer 26

yeah

Answer 27

You understand the last step, where I write f(t) in terms of f(t-4)??

Answer 28

on second thought no...

Answer 29

OH
WAIT

Answer 30

you added and subtracted 48 to make it equal to our original function

Answer 31

added 8t rather

Answer 32

Yeah but made a mistake

Answer 33

f(t)=t^2-8t+20
so
\[f(t-4)=t^2+16-8t-8t+32+20=t^2-16t+68\]
so
\[f(t)=f(t-4)+8t-48\]

Answer 34

righto

Answer 35

Great:D
So
I can further change this as
\[\huge f(t)=f(t-4)+8(t-4)-16\]
You got this???

Answer 36

hmmm

Answer 37

8(t-4) -16 = 8t -48 ?

Answer 38

Yeah !!!!
Okay so
\[LT(f(t)U(t-4))=LT(f(t-4)+8(t-4)-16)U(t-4))\]
\[= e^{-4s}(LT(f(t))+8 LT(t)-16 LT(1)) \]

Answer 39

what is the point of doing that?

Answer 40

Because we want to use
\[LT[f(t-4) U(t-4)]= e^{-4s} F(s)\]

Answer 41

ah so t-4 will go into each t ok

Answer 42

Yeah!!!

Answer 43

Now you find LT[f(t)] f(t)= t^2-8t+20
and substitute in above expression

Answer 44

so L(f(t-4)) + L(8(t-4)) so on so forth?

Answer 45

Nope
we have
\[e^{-4s} ( LT (f(t)+8LT(t)-16 LT(1))\]

Answer 46

We have used the property which I mentioned earlier

Answer 47

Did you understand?

Answer 48

kinda im at this point \[e ^{-4s} (L(ft))\]

Answer 49

f(t) being f(t-4) +8(t-4) -16

Answer 50

ugh im lost

Answer 51

No here for LT of f(t-4) use LT of f(t)
for LT of (t-4) use LT( t)
and -16 is just -16/s

Answer 52

just curious if this works:

since 0 to 4 is zero we can ignore that part

\[ \int_{4}^{inf} e^{-st}(t^2 -8t + 20) dt=...\]


e^-st
t^2 -8t +20 e^-st/-s
-(2t -8) e^-st/s^2
2 e^-st/-s^3

at inf we go zero sooo, if i push my negative thru

t^2 -8t +20 e^-st/s
2t -8 e^-st/s^2
2 e^-st/s^3


16 -32 +20 e^-4s/s
8 -8 e^-4s/s^2
2 e^-4s/s^3

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

But then I aint got a clue as to what your doing with that U stuff

Answer 53

@amistre64 This would work always
I'm trying to create the form
\[f(t-4) U(t-4)\]
where U is the unit step function
so that I could use the property and write Laplace transform as
\[e^{-4s}F(s)\]

Answer 54

:) yeah, I got no idea how to do it like that

Answer 55

I can write f(t) as
\[f(t)U(t-4)= (f(t-4)+8(t-4)-16) U(t-4)\]
Taking LT now
\[e^{-4s} (F(s)+ 8/s^2-16/s)\]
\[F(s)=LT(f(t)= 2/s^3-8/s^2+20/s\]
so we get
\[e^{-4s}(2/s^3+4/s)\]
I'm making some mistake, I'll check again

Answer 56

oh wait you also got that!!!

Answer 57

i did :) but i make no gaurentees for its correctness ;)

Answer 58

Oh can you bump this!!!

Answer 59

it says bump in 10min

Answer 60

But it's come to the top. Thanks:)

Answer 61

oh, then that means it can bump again in 10 mins :)

Answer 62

Yeah:)

Answer 63

Here's two ways