Question

Hi everyone! Can someone please discuss with me some properties of inverse laplace transforms? Specifically, for a problem like L^-1{s/(s^2+4s+5)...my book shows how to solve it first making it look this way>>>
L^-1{s/((s+2)^2+1) by completing the square then solving...is this "completing the square" method only used when you can't neatly organize the denominator like (s+1)(s-4) just for example? Thanks! :o)

Answer 1

its a process yes

Answer 2

I would really just like to ask a few theory questions so I can understand better :o)

Answer 3

Do you have a few minutes?

Answer 4

i have limited knowledge, but i might be able to help

Answer 5

Okay, well when we are asked to take the inverse laplace transforms of easily recognizable functions like 1/s^5 or 1/s^2+64, we can just kind of easily go backward correct?

Answer 6

yes, if we have a table of laplace transforms we can incert them back and forth

Answer 7

**invert them back and forth

Answer 8

okay great, so next question...

Answer 9

if we see something like L^-1{s/(s^2+4s-5) we can factor the bottom to (s+5)(s-1), in which case we can just simply do some partial fraction decomp and go from there correct?

Answer 10

correct

Answer 11

ok next question...

Answer 12

if we see something like L^-1{s/(s^2+4s+5), then we CAN NOT make the bottom into (something)(something) because it simply won't factor...
is that when we ABSOLUTELY HAVE TO USE completing the square?

Answer 13

itll factor, it just has imaginary roots, if you are comfortable with imaginary roots then by all means attempt it that way

Answer 14

uhm yeah, for my sanity, let's assume it won't factor :o) then would my statement be correct?

Answer 15

L(i e^(it)) = sL(e^(it)) - e^0

1 = sL(e^(it)) - L(i e^(it)) , but constants pull out

1 = sL(e^(it)) - i L(e^(it)) , but constants pull out

1/(s-i) = L(e^(it)) right?

Answer 16

got some duplication when i copied a row lol

Answer 17

I appreciate the effort, but remember the whole sanity statement above? :o~

Answer 18

imaginary roots are indicative of trig; sin and cos

Answer 19

hey, if im not allowed any sanity ... no one is!! :)

Answer 20

lol

Answer 21

okay next question...and don't forget my sanity plea!

Answer 22

I actually would like to rephrase what I said above a little...
if we see something like L^-1{s/(s^2+4s+5) and can't turn the denominator into (something)(something) without using imaginary numbers, is that when we ABSOLUTELY MUST USE the 1st translation theorem (which involves completing the square) ?

Answer 23

yes, completing the square allows us to transform the results into something more recognizable.

Answer 24

but we only use that in order to set ourselves up to use the 1st translation theorem though correct?

Answer 25

sryy, but i dont know what a first translation thrm is ...

Answer 26

okay...well it says this>>>
L{e^at times f(t)}=F(s-a)

That's kind of what I am trying to understand myself...
do we use this property when we can't easily factor the denominator and easily do partial fraction decomp?

Answer 27

L(y') = sL(y) - y(0)

L(y'') = sL(y') - y'(0)

L(y'') = s(sL(y) - y(0)) - y'(0)

L(y'') = s^2 L(y) - sy(0) - y'(0)

cos(at) > -a sin(at) > -a^2 cos(at)

L(-a^2 cos(at)) = s^2 L(cos(at)) - s cos(0) +a sin(0)

L(-a^2 cos(at)) = s^2 L(cos(at)) - s

s = s^2 L(cos(at)) - L(-a^2 cos(at))

s = s^2 L(cos(at)) +a^2 L(cos(at))

s = (s^2+a^2) L(cos(at))

s/(s^2+a^2) = L(cos(at))

which is similar to what you have, so its prolly a version of cos

Answer 28

that was just me remembering stuff, not pertinant in the least prolly

Answer 29

I am starting to think that for my example if we see something like L^-1{s/(s^2+4s+5), then the 1st translation theorem won't work...
I may only work for problems like L{e^5t times t^3}, which now that I look at it more closely, isn't even an example of an inverse laplace transform

Answer 30

does that make sense what I just wrote?

Answer 31

now we are getting to something that is not in my knowledge base. so i dont want to lead you astray with it. if we are talking convolutions then im just not adept at them.

@hartnn took a laplace course i beleive and was pretty well versed with them but i just dont see them around anymore.

Answer 32

that's an awesome list amistre!

Answer 33

I can tell you that my example is not a convolution, it's from a couple sections before those

Answer 34

gimme just a sec to write out an inverse laplace transform using the 1st translation theorem, then it will make sense to you I'm sure, at least enough to guide me...one sec...

Answer 35

y y'
e^at f(t) --> a e^at f(t) + e^at f'(t)

using the definition:

L(y') = sL(y) - y(0)

L(a e^at f(t) + e^at f'(t)) = sL(e^at f(t)) - f(0)

a L(e^at f(t)) + L(e^at f'(t)) = sL(e^at f(t)) - f(0)

L(e^at f'(t)) = sL(e^at f(t))-a L(e^at f(t)) - f(0)

L(e^at f'(t)) = (s-a) L(e^at f(t)) - f(0)

Answer 36

L^-1{s/(s^2+4s+5)
then...
L^-1{s/(s+2)^2 +1)
then...
L^-1{(s + 2 - 2) / (s+2)^2 +1)
then split into...
L^-1{(s + 2 ) / (s+2)^2 +1) - L^-1{( 2 / (s+2)^2 +1)
which then equals...
e^-2tcost - 2e^-2tsint

so this would be an example of doing an inverse laplace transform by using the 1st translation theorem which was L{e^at times f(t)}=F(s-a)

Answer 37

so the whole reason I wrote that out was to find out this...
when we can't easily factor the bottom without using imaginary numbers, MUST WE ALWAYS USE the 1st translation theorem?
I mean, I don't know another way, and I am just trying to corral all possible method to better understand them...does that make sense?

Answer 38

well, im not sure of all the methods yet, im trying to learn them myself.

the FTT does seem to be appropriate for it. unless like i said you become comfortable with imaginary roots :) thats the only other idea i got

Answer 39

okay great...i'm thinking the same thing myself

Answer 40

when i plug s/((s+2)^2+1) into the wolf, it actually spits back imaginaries :)

Answer 41

okay so let me recap then...
1) if doing the inverse laplace transform of easily recognizable function, use your table and go backwards
2) If you can easily factor the bottom, do partial fraction decomp to solve
3)If you CAN'T easily factor the bottom without using imaginary number, use the 1st translation theorem (aka, completing the square and splitting the transform)

how do these hints sound so far?

Answer 42

I know...I plugged it into wolf also, which didn't help me at all

Answer 43

they sound like a good recipe for approaching solutions.

Answer 44

okay...can you gimme a few minutes to think up a couple more questions so I can add to this recipe list?

Answer 45

sure .... youre the expert at the moment ;)

Answer 46

haha...not really...just trying to talk my way through this mess with your help!

Answer 47

thats what i do ... work it out till it makes sense :) or give up and have a donut for my efforts lol

Answer 48

mmm donut! (spoken of course in the voice of Homer Simpson)! :o)

Answer 49

by the way, if you know anyone you could tag to confirm my recipe list, that would be great...gimme a few minutes to keep going

Answer 50

@rational @zepdrix @kainui
Hi guys, if you get a chance, could you skim this post and give a thumbs up or down for my recipe list of solving laplace transforms? Thanks so much! Here is the list again...

okay so let me recap then...
1) if doing the inverse laplace transform of easily recognizable function, use your table and go backwards
2) If you can easily factor the bottom, do partial fraction decomp to solve
3)If you CAN'T easily factor the bottom without using imaginary number, use the 1st translation theorem (aka, completing the square and splitting the transform)

how do these hints sound so far?

Answer 51

hey @amistre...I am question whether or not my example of using the 1st translation theorem, actually is an example...I have a weird feeling it may not have been an example after all and the reason I am thinking this way is because I have a different example that I 100% know uses the FTT...here is that example...

L{e^-2t times cos4t}
so you recognize that L{e^-2t} via the tables would become 1/(s+2)
then...
L{cos4t} via the tables would become s/(s^2 + 4^2) or s/(s^2 + 16)
now...
the normal laplace transform theorem says L{f(t)}=F(s)
and the 1st translation theorem states L{e^at times f(t)}=F(s-a)
so...
where ever we find an "s", we replace it with "s-a"...and from e^-2t, we know that a=-2 so we can say s--->s+2
so I guess the general idea is this...
we only need the e^-2t in order to find the "a",
once we have the "a", we use the laplace transform of the "other function" cos4t, which was s/(s^2 + 16)...so then we replace s for s-a which is s-(-2) = s+2 so...
the answer is then s+2/( (s+2)^2 + 16)...
so to sum it up, we only use the poor little e^-2t to find a, we then find the laplace transform of the other function, replace s with s-a, plop in the -2, and that is the answer and we don't worry about the e^-2t anymore because we already got what we needed from it
does this make sense and is it correct? lol whew! :o)

Answer 52

https://web.stanford.edu/~boyd/ee102/laplace-table.pdf

Answer 53

its a consequence of e^(at) f(t) = F(s-a)

Answer 54

Now that I think about it, the following DOES NOT APPEAR TO BE USING F.T.T. because we never replace any "s" with an "s-a">>>
L^-1{s/(s^2+4s+5)
then...
L^-1{s/(s+2)^2 +1)
then...
L^-1{(s + 2 - 2) / (s+2)^2 +1)
then split into...
L^-1{(s + 2 ) / (s+2)^2 +1) - L^-1{( 2 / (s+2)^2 +1)
which then equals...
e^-2tcost - 2e^-2tsint

so can you confirm that this DOES NOT USE 1st translation theorem?

Answer 55

\[\frac{s}{(s+a)^2+1^2}\]
\[\frac{s+a-a}{(s+a)^2+1^2}\]
\[\frac{s+a}{(s+a)^2+1^2}-a\frac{1}{(s+a)^2+1^2}\]

Answer 56

so is this using FTT or not using FTT?

Answer 57

i'm confused

Answer 58

e^(at) f(t) or e^(-at) f(t)

not sure if theres a difference

F(s-a) or F(s--a) seems to be irrelevant

Answer 59

so is your answer "I'm not sure"

Answer 60

pretty much, im not sure the mechanics behind it all.

i cant see why its not applicable, since a is just a number, can be positive or negative.

Answer 61

well can you skim the logic in this problem for me and tell me if it is correct?
>>>
L{e^-2t times cos4t}
so you recognize that L{e^-2t} via the tables would become 1/(s+2)
then...
L{cos4t} via the tables would become s/(s^2 + 4^2) or s/(s^2 + 16)
now...
the normal laplace transform theorem says L{f(t)}=F(s)
and the 1st translation theorem states L{e^at times f(t)}=F(s-a)
so...
where ever we find an "s", we replace it with "s-a"...and from e^-2t, we know that a=-2 so we can say s--->s+2
so I guess the general idea is this...
we only need the e^-2t in order to find the "a",
once we have the "a", we use the laplace transform of the "other function" cos4t, which was s/(s^2 + 16)...so then we replace s for s-a which is s-(-2) = s+2 so...
the answer is then s+2/( (s+2)^2 + 16)...
so to sum it up, we only use the poor little e^-2t to find a, we then find the laplace transform of the other function, replace s with s-a, plop in the -2, and that is the answer and we don't worry about the e^-2t anymore because we already got what we needed from it

is this logic sound?

Answer 62

\[L(e^{-at})=\frac{1}{s+a} =F(s-(-a))\]

Answer 63

oh I like what you just typed there...makes sense
so it almost does seem that the problem is using the FTT then

Answer 64

\[L(cos(nt))=\frac{s}{s^2+n^2}=F(s)\]
\[L(e^{at}cos(nt))=\frac{s-a}{(s-a)^2+n^2}=F(s-a)\]
\[L(e^{-at}cos(nt))=\frac{s-(-a)}{(s-(-a))^2+n^2}=F(s-(-a))\]

Answer 65

so sum up in words the significance of those identities if you would please

Answer 66

its the FTT

Answer 67

lol...that was nicely summed! :o)

Answer 68

I am so sick of talking about the FTT BLEH! :O/

Answer 69

heheh :)

Answer 70

thank you for hanging in there with me! Are you at least learning anything from this also?

Answer 71

im seeing relationships, but i would have to take it to paper to understand the mechanics behind it, under it?

Answer 72

so I'm not bothering you too badly then?

Answer 73

lol, its 1am, hard to be bothered by learning :)

Answer 74

2am, now the computer is caught up with the wall clock lol

Answer 75

time is relative!

Answer 76

remember this problem>>> L^-1{s/(s^2+4s+5)
just trying to use my logic on that other problem, all I would need would be to find out what "a" is and then replace it with "s-a" but you can't find your way to the end of the problem like that...it appears you have to do the tricks like +2-2...
then only after you split it up, can you recognize what "a" actually is>>>
L^-1{s/(s^2+4s+5)
then...
L^-1{s/(s+2)^2 +1)
then...
L^-1{(s + 2 - 2) / (s+2)^2 +1)
then split into...
L^-1{(s + 2 ) / (s+2)^2 +1) - L^-1{( 2 / (s+2)^2 +1)
which then equals...
e^-2tcost - 2e^-2tsint

so long story short, it may appear to be using the FTT, but the mechanics to solving the problem is different...would you agree?

Answer 77

basically, each problem has its own nuances is what I am saying, even though they are based on the same FTT

Answer 78

splitting it up is irrelevant, the algebra lets us split it up. the results are still FTT regardless

Answer 79

yeah, that's kinda what I was trying to say! :o)

Answer 80

splitting it allows us to use the linearlity property of the transform. thats all

Answer 81

yes...I agree!

Answer 82

hey amistre...can you tell me why this would be repeated linear factors?
(s+1)/ s^2(s+2)^3

Answer 83

s*s*(s+2)*(s+2)*(s+2) are all linear factors

Answer 84

i get the s+2 is repeated, but the s^2 confuses me

Answer 85

oh okay!...simply put, both the s+2 and the s are repeated! :o)

Answer 86

yeah, they are repeated linear factors

Answer 87

(s^2+3) is not linear ...

Answer 88

that was another example of when to use the partial fraction decomp to solve inverse laplace transforms...so we went through distinct earlier, 1/(s-1)(s+2)(s+4) and now repeated s+2/ s^2(s+2)^3...there is one more type it looks like..."irreducible"!
3s-2/ s^3(s^2+4)

Answer 89

thats still splitable, but the top of the non linear factor has to be 1degree less then it

Answer 90

unless you want to split it into complex linears

Answer 91

so are you saying you can't split it because degree 3 is more than the degree of the numerator?

Answer 92

s^3 can be topped as Ax^2 + Bx + C since all multiples of it will simplify to that after all.

Answer 93

\[\frac{1}{s^3(s^2+a)}=\frac{A_0+A_1s+A_2s^2}{s^3}+\frac{B_0+B_1s}{s^2+a}\]

Answer 94

I am confused why you said "thats still splitable, but the top of the non linear factor has to be 1degree less then it"
where you just adding info? or was there some other reason you brought that up?

Answer 95

just making sure that the split up is accurate

the lag is getting unbearable on this post

Answer 96

I'm not getting a lag...is your screen freezing up?

Answer 97

thanks @amistre64 for all your insight! :o)
here's a medal for you!