[SOLVED] relationship fourier - laplace

[wasserkasten]

Hello,

I often hear "the Fourier transfer function is equal to the Laplace transfer function with the Laplace variable s replaced by j*w: H(j*w) = H(s)".
Can anyone explain how this applies to cos(w0*t)?

The Fourier transform of cos(w0*t) is "1/2 * ( delta(f-f0) + delta(f+f0) )" (where w0=2*pi*f0 and w=2*pi*f) - see https://www.astro.umd.edu/~lgm/ASTR410/ft_ref2.pdf
But its Laplace transfer function is "s / ( s^2 + w0^2 )".

So the beginning of the transformation could be:
s / ( s^2 + w0^2 )
-> j*w / ( (j*w)^2 + w0^2 ) =
= j*w / ( - w^2 + w0^2 )

-> where do the delta(...) come from (leading to the corresponding Fourier transform)?

Thank you in advance.

 

[jeffrey samuel]

s domain in laplace transform is represented as

s=σ+jw

in fourier transform we represent a signal in jw terms

so the relationship is

Ft= Lt where σ->0

 

[wasserkasten]

ok, but this is what i already said in the first posting. the actual question was:
how to use this fact to find the relationship between the "fourier transform of cosinus" and the "laplace transform of cosinus"?

 

[jeffrey samuel]

substitute instead of s and then put sigma = 0

this will bring your answer out

one more thing what is Cosinus cos(s) or something like that

 

[wasserkasten]

hmm, the question is still open: starting with the laplace transform of cos "s/(s^2+w0^2)" and substituting "s=j*w" i'm stuck in the last step mentionned in the first posting.
how to proceed there (explicite steps) to get to the fourier transform of cos?

 

[jeffrey samuel]

s= sigma +jw

my pal and not sigma*jw

- - - Updated - - -

this becomes more easy now i guess

 

[wasserkasten]

thanks for your quick answer.
i will give it a try:
laplace transform of cos(w0*t) =
s / ( s^2 + w0^2 )
-> substitution of s = "sigma + jw" with sigma=0 leads to
jw / ( -w^2 + w0^2 )
= ?

i don't see the way how to get to the fourier transform of cos "1/2 * (delta(f-f0) + delta(f+f0))" after the last step.
so how would you proceed after the last step above?

 

[jeffrey samuel]

https://www.projectrhea.org/rhea/index.php/HW10_Daniel_Barjum:_Example_illustration_relationship_between_Fourier_and_Laplace_transform_ECE301Fall2008mboutin

this will be more exact my pal I am bit confused here and need to work on it

 

[klystron]

The Fourier transform is the Laplace Transform with the real component equal to zero. In other words you get the Fourier transform if you "stay" on the j omega axis.

 

[wasserkasten]

yes this was clear since the beginning of this thread (see first posting).
i think i've found the answer concerning the laplace-fourier-relationship of the cos transform:
http://en.wikipedia.org/wiki/Laplace_transform#Inverse_Laplace_transform (section 4.3.2)
and
**broken link removed** (slide 7)

there it is stated that the fourier transform of cos can not be simply found by substituting s by jw in the laplace transform of cos.

 

[zorro]

Hi wasserkasten

I didn't follow your links, but anyway let me point out this important fact:

FT) The Fourier transform is an integral that extends from -Inf to +Inf in time. The function cos(w0*t) you are considering "lives" in the whole real t axis too.

LT) Instead, te (usual) Laplace transform (that you are considering) is unilateral, i.e., the integral in t goes from 0 to +Inf.

Yo say that LT end FT are equivalent just replacing s by jw is right only for functions that are 0 for all t<0.
If you calculate the FT for the cos() that starts at t=0 (i.e. is 0 for t<0) you shuld get a coincidence.

I avoided to mention the problem of convergence of the integrals, but I think that what I said answers your question.
Regards

Z

 

[phongphanp]

Fourier transform have 1D and 2D calculation, so the denominator, pi or 2*pi is Difference from Laplace transform and result in , I donn't know Advance Laplace transforms will equal with Fouier transform or not .

Someone had try take a*b as input of two type of Transform , then the output will be easy to see. Someone try other method that control direction , It's very hard .

- - - Updated - - -

Fourier transform have 1D and 2D calculation, so the denominator, pi or 2*pi is Difference from Laplace transform and result in , I donn't know Advance Laplace transforms will equal with Fouier transform or not .

Someone had try take a*b as input of two type of Transform , then the output will be easy to see. Someone try other method that control direction , It's very hard .