When is the Fourier transform implied?

by friend
Tags: fourier, implied, transform
 P: 982 I understand that if you have a system that is linear and time invariant, that you can perform a Fourier transform on it. But that doesn't mean you need to Fourier transform it. Or does it? Is a linear, time invariant system equivalent to or in some way implies a Fourier transform? Or is the Fourier transform just an optional procedure for arbitrary purposes? What information is derived from the Fourier transform that might reproduce the original signal? And if you can reproduce the original signal, does that mean that the Fourier transform is in some way equivalent to the original signal? Or are there systems that are closed under Fourier transformation, whose functions are transforms of themselves?
 Sci Advisor P: 6,106 Are you asking a mathematics or physics question? Mathematically (for well behaved functions) a function and its transform are duel, that is take the transform of the transform (with a sign switch in the exponent) and you will get back the original function. Physically, if the function is supposed to be a time domain signal, the transform is a representation in the frequency domain.
HW Helper
P: 1,391
 Quote by friend I understand that if you have a system that is linear and time invariant, that you can perform a Fourier transform on it. But that doesn't mean you need to Fourier transform it. Or does it? Is a linear, time invariant system equivalent to or in some way implies a Fourier transform? Or is the Fourier transform just an optional procedure for arbitrary purposes? What information is derived from the Fourier transform that might reproduce the original signal? And if you can reproduce the original signal, does that mean that the Fourier transform is in some way equivalent to the original signal? Or are there systems that are closed under Fourier transformation, whose functions are transforms of themselves?
Applying the Fourier transform to a linear, time invariant differential equation turns the equation from a differential equation in the time-domain variable $y(t)$ into an algebraic equation for the frequency domain variable $\hat{y}(\omega)$. It is sometimes quicker to solve things this way. For example, when you have a forcing function, taking the fourier transform can allow you shortcut the bothersome Green's function technique, at least when the time can be any real value. No information is lost during the fourier transform (just... reorganized, in some sense), so by solving the equation in the frequency-domain you can get the solution in time-domain by inverse fourier transforming. Again, sometimes this is simpler than solving the original differential equation. It may also admit some approximation techniques that are more difficult in the time domain.

As for your question as to whether or not there are any functions that are their own fourier transform, the answer is yes. If we take our Fourier transform convention to be

$$\hat{y}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t),$$

(which is the unitary version of the fourier transform), where $\hat{y}(\omega)$ is the fourier transform of $y(t)$, then $y(t) = \exp(-t^2/2)$ and $y(t) = \mbox{sech}(\sqrt{\pi/2}t)$ are two examples of functions which solve the integral equation

$$y(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t).$$

(Note the absence of the hat on the left hand side to indicate that it is the same function y(t) that occurs in the integrand).

Similarly, the functions $e^{-t^2/2}H_n(t)$, where the $H_n(t)$ are Hermite polynomials, solve the associated eigenvalue integral equation problem,

$$\lambda y(\omega) = \int_{-\infty}^\infty dt \frac{e^{-i\omega t}}{\sqrt{2\pi}} y(t),$$

where the eigenvalues are $\lambda_n = (-i)^n$. Note that when n is a multiple of 4, y(t) is again its own fourier transform.

P: 982
When is the Fourier transform implied?

 Quote by Mute As for your question as to whether or not there are any functions that are their own Fourier transform, the answer is yes. If we take our Fourier transform convention to be $$\hat{y}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t),$$ (which is the unitary version of the Fourier transform), where $\hat{y}(\omega)$ is the Fourier transform of $y(t)$, then $y(t) = \exp(-t^2/2)$ and $y(t) = \mbox{sech}(\sqrt{\pi/2}t)$ are two examples of functions which solve the integral equation $$y(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dt e^{-i\omega t} y(t).$$ (Note the absence of the hat on the left hand side to indicate that it is the same function y(t) that occurs in the integrand).
Since you seem to have a genuine interest in these matters, let me tell you what I'm up to.

I've stumbled across a novel approach to derive Feynman's path integral from the Dirac delta function. The Dirac delta function has the following properties:

1)
$$\int_{ - \infty }^{ + \infty } {{\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}){\rm{d}}{{\rm{x}}_1}} = 1$$
and also,

$$\int_{ - \infty }^{ + \infty } {{\rm{f(}}{{\rm{x}}_1}{\rm{)\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}){\rm{d}}{{\rm{x}}_1}} = {\rm{f(}}{{\rm{x}}_0})$$
where if we let ${\rm{f(}}{{\rm{x}}_1}) = {\rm{\delta (x - }}{{\rm{x}}_1})$, we get,

2)
$$\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})$$
This is on the wikepedia.org site for the Dirac delta function about half way down under section 4.3 for Translation. I've also seen this in Lowell S. Brown's book, Quantum Field Theory, page 30.Yet the above can be iterated to get,

$$\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_2}){\rm{\delta (}}{{\rm{x}}_2}{\rm{ - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} d{x_2} = } \int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_2}){\rm{\delta (}}{{\rm{x}}_2}{\rm{ - }}{{\rm{x}}_0})d{x_2}} = {\rm{\delta (x - }}{{\rm{x}}_0})$$
and if iterated an infinite number of times one gets,

3)
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - }}{{\rm{x}}_n}){\rm{\delta (}}{{\rm{x}}_n}{\rm{ - }}{{\rm{x}}_{n - 1}}) \cdot \cdot \cdot {\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1} = {\rm{\delta (x - }}{{\rm{x}}_0})$$
This is explicitly written out in Prof. Hagen Kleinert's book, Path Integrals i Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, page 91.

Then if we use the gaussian form of the Dirac delta function,

4)
$${\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{1/2}}}}{e^{ - {{({x_1} - {x_0})}^2}/{\Delta ^2}}}$$
with

5)
$${\Delta ^2} = \frac{{2i\hbar }}{m}({t_1} - {t_0})$$
the dirac deltas become
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im{{({x_1} - {x_0})}^2}}}{{2\hbar ({t_1} - {t_0})}}} \right]$$
which can be manipulated to
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})}^2}({t_1} - {t_0})} \right] = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {\left[ {\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{\Delta {x_{1,0}}}}{{\Delta {t_{1,0}}}})}^2}\Delta {t_{1,0}}} \right]$$
or,

6)
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}$$
When this is substituted for each of the dirac deltas in 3) above we get

7)
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta {t_{,n}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{,n}})}^2}\Delta {t_{,n}}}}{{(\frac{m}{{2\pi i\hbar \Delta {t_{n,n - 1}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{n,n - 1}})}^2}\Delta {t_{n,n - 1}}}} \cdot \cdot \cdot {{(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}$$
with the appropriate limits implied.

Then since the exponents add up, 7) above becomes

8)
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta t}})}^{n/2}}{e^{\,\,{\textstyle{-i \over \hbar }}\int_0^t {\frac{m}{2}{{(\dot x)}^2}dt} }}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}$$
Which is Feynman's path integral for a free particle.

This is very curious. I have speculative theories as to why the dirac delta would be fundamental to deriving physics. But the math I've presented here seems pretty straight forward.

However, the use of the gaussian form of the dirac delta seems arbitrary in 4), along with the use of the complex number for the standard deviation in 5). So I am looking for some more fundamental reasons why the complex gaussian would be preferred in this situation. And I wonder if there is not something in the algebra of 2) that dictates the use of the complex gaussian for each of the deltas.

So when I look at 2)

2)

$$\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})$$

and consider that

9)
$${e^{\frac{{-im}}{{2\hbar }}{{(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})}^2}({t_1} - {t_0})}} = {e^{\frac{{-im}}{{2\hbar }}(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})({x_1} - {x_0})}} = {e^{-ip({x_1} - {x_0})/2\hbar }}$$
It seems that 2) can be written as

10)
$$\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = \int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){{(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})}^{1/2}}{e^{ - ip({x_1} - {x_0})/2\hbar }}d{x_1}}$$
which seems suspiciously close to the Fourier transform of the dirac delta. Although I,m not sure whether the factor in front of the exponential or the $2\hbar$ in the exponent can be made to fits in the format of the Fourier transform. This brings up questions as to whether 2) is even a Linear, Translationally Invariant (LTI) system, and if so, does that mean the Fourier transform is automatically implied. And since 3) is equivalent to 2), it would seem that I'm looking for an equation for the deltas such that any number of transforms are equal to one transform. I'm not sure what that means. Any insight anyone might have into these questions would help greatly. Thank you.
 P: 982 Or in other words, What useful properties of a function are gauranteed if the function has a Fourier transform? For example, does a Fourier transform of a function gaurantee that the function is analytical, or smooth, or continuous, or is a distribution, or finite, or what? Does it gaurantee the existence of a dual space for which an inner product is the underlying metric, so that the Fourier transform only exists if the function is in a space that has a metric, and visa versa?
 P: 982 Is the Fourier Transform unique in turning a convolution integral into a multiplication of tranforms? Or are their other transformations that can do this? Thanks.
P: 6,106
 Quote by friend Is the Fourier Transform unique in turning a convolution integral into a multiplication of tranforms? Or are their other transformations that can do this? Thanks.
Laplace transform does something similar.
P: 982
 Quote by friend $$\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})$$
If the above integral exists for the dirac delta function, then must also the Fourier transform of the dirac deltas? For it seems that the above equation can be seen as a convoltuion integral which can be solved with the Fourier transforms method which is gauranteed to have a solution from the above equation. All three delta functions do exist. And the Fourier transform of all three deltas also exist. So does that gaurantee the existence of the Fourier transform? I'm told that the Fourier transform is an automorphism which transforms the deltas into themselves. Does that gaurantee the existence of the FT? What if you add the requirement that the convolution exist? Thanks.
 P: 982 Can the convolution exist without the Fourier Transform?
 P: 1 which is the better way to analyze a signal....time domain or frequency domain???
P: 474
 Quote by friend Then if we use the gaussian form of the Dirac delta function, 4) $${\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{1/2}}}}{e^{ - {{({x_1} - {x_0})}^2}/{\Delta ^2}}}$$ with 5) $${\Delta ^2} = \frac{{2i\hbar }}{m}({t_1} - {t_0})$$ the dirac deltas become $$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im{{({x_1} - {x_0})}^2}}}{{2\hbar ({t_1} - {t_0})}}} \right]$$
Is this definition (in particular the use of imaginary variance) from another source like a textbook or something? If not, I would examine it very carefully to see if the definition still holds. I would expect the Gaussian distribution to have very different properties when the variance is complex, and I fail to see why it should still approach the delta function as the variance tends to zero.

In fact, when I look at the behaviour of the function, it seems to me that the limit you've used (as t1 goes to t0) is undefined. The exponential term is a complex exponential, and so its magnitude will always be one. As t1 goes to t0, its phase will change, and the complex exponential will just rotate around the origin of the complex plane as t1 goes to t0, making the limit undefined.

Sorry if this is a standard definition that I just haven't seen before, but this raised a major red flag for me.

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