Question

Another Laplace transform:
\[L(h(t))=L(3tu_2(t)+4(1-u_2(t))\]

Answer 1

So \(u_2(t)\) is the heavyside function for c=2. I need to figure out how to write F(t-c)

Answer 2

@hartnn , @Hero

Answer 3

@zzr0ck3r

Answer 4

so if I write 3(t-2+2) would that be right?
which then equals 3(t-2)+6..?

Answer 5

I hate applied math:( i.e. I suck at applied math.

It has been almost 3 years since I took my one and only dif eq class. Sorry:(

Answer 6

otherwise I would jump on this question

Answer 7

:( darn

Answer 8

thanks anyways

Answer 9

People here love applied math so someone will be able to help; i'm sure.

Answer 10

I'm hoping hartnn stops by again

Answer 11

not quite familiar with \(u_2 (t)\)

Answer 12

ohh \(u(t-2)\)
?

Answer 13

it's the heaviside function so that \[u_2(t)=\left\{\frac{1~~~t<2}{0~~~t\ge 2} \right\}\]

Answer 14

but so the Laplace transform is only for f(t-c)

Answer 15

so I need to change 3t into something equivalent

Answer 16

or if you have another way to integrate \[g(x)=\left\{ \frac{3t~~~0\le t<2}{4~~~t\ge2}\right\} \]

Answer 17

not integrate... laplace transform... ohh dear...

Answer 18

sorry, i got away for few minutes
we can use time shifting property here
\(\Large f(t-a)u(t-a) <-> e^{-as}F(s)\)

Answer 19

but I guess the issue for me is how can I apply the shifting property here. I know I need to shift it twice

Answer 20

\(tu(t-2) <-> e^{(-2s)} L[t+2]\)

Answer 21

t = (t+2) -2

Answer 22

so does that mean that here I would need the laplace transform of 3(t+2)?

Answer 23

or 3(t+2-2)?

Answer 24

e^{-2s} L [3(t+2)]

Answer 25

ok, but I guess why don't I have to subtract 2?

Answer 26

i have adjusted 't' to be in the form of F(t-2)

t = t-2 ... but i can't just subtract 2, i need to add 2 to compensate for the -2
so,
(t+2) -2
and my f(t) became t+2

Answer 27

*f(t-2)
notF

Answer 28

I'm sorry I still don't understand

Answer 29

we have f(t)=3t

Answer 30

so then if a move it... I'd have f(t-2)=3(t-2) yes?

Answer 31

\(3t u(t-2)\)
need to bring this in the form of
f(t-2)u(t-2)
thaths our requirement.
got this?

based on the requirement i will form f(t)

Answer 32

so i need f(t-2) to be 3t

Answer 33

yea, your u(t-2) is the same as my \(u_2(t)\) just different notation

Answer 34

to get f(t) from f(t-2)
i substitute t =t+2

Answer 35

f(t+2 -2) = 3 (t+2)

Answer 36

ohhhh ok, I was looking at it backwards

Answer 37

f(t) = 3(t+2)

Answer 38

ok I follow, so now I can find that. How do I shift a constant, do I have to?

Answer 39

so now bring
\(4(u_2(t))\)
in the form of
\(g(t-2)u_2(t)\)

and tell me what will be g(t)

Answer 40

but it is \(4(1-u_2(t))\)

Answer 41

4- that
Laplace T of 4 is trivial

Answer 42

but can we do that?

Answer 43

L[4] =4/s

Answer 44

since \(1-u_2(t)\) defines a different function, I guess I am just wondering if we are allowed to distribute

Answer 45

but I guess it will become the same, ok point won. So now, for \(4u_2(t)\) I need g(t-2)=4 still right?

Answer 46

if it were 4 u_2 (1-t)
then we cannot distribute 4 and say
u_2 (4-4t)
this will be wrong

but 4(1-u_2 (t)) can definitely be 4-4u_2(t)

Answer 47

yes
g(t-2) =4
so g(t) = ... ?

Answer 48

g(t)=6...?

Answer 49

why 6 ?
you add 2 to 't'

Answer 50

I thought it was 4 before but you said it wasn't

Answer 51

i said it wasn't ???? sorry :P

g(t-2) =4
it doesn't depend on 't'
so g(t) is also 4
that why its called a constant function :)

Answer 52

ohh ok

Answer 53

I must have misunderstood, my bad

Answer 54

ok then I got it from here :) Thank you again

Answer 55

or you can look it like this way
\(g(t-2)=4t^0 \\ g(t) = 4 (t+2)^0= 4\)

Answer 56

and welcome ^_^