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Why are Fourier series considered to be the subset of Fourier transform? It should have been other way around. Because a non periodic pulse is a subset of periodic pulse with period infinite. So Fourier transform (non periodic signals) is a subset of Fourier series (periodic signals).

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The set of periodic functions is surely a subset of the set of all functions (most of which are aperiodic). So the Fourier series – which allow us to describe periodic functions differently – are a "subset" (in this sentence, the word "subset" is stranger than before) of the Fourier transform – which is a tool to encode an arbitrary function.

Alternatively, we may see the "subset" relationship in the frequency representation, too. Fourier series may be written as a special case of the Fourier transform in which only the frequencies that are multiply of the basic angular frequency $2\pi/T$ where $T$ is the periodicity are allowed. The Fourier transform of a periodic function is a very special kind of a function, a combination of delta-functions $$\tilde f(\omega) = \sum_{n\in{\mathbb Z}} c_n \delta(\omega-n\omega_0) $$ and functions that are a combination of delta-functions like that (determining Fourier series) are a subset of all functions, including distributions (which may be identified with the Fourier transform of the original function).

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I'm not quite sure I understand what you're saying about periodic versus non-periodic pulses, but you can think of a Fourier transform as a projection of a function onto a basis of functions. The same way you can project a vector onto the x, y, or z axes, you can project a function $f(x)$ onto the sine basis or the cosine basis. The Fourier transform projects functions onto the plane wave basis - basically a collection of sines and cosines. A Fourier series is also a projection, but it's not continuous - you sum over $\sin(nx) + \cos(nx)$, whereas in a Fourier transform you preform an integral - a continuous sum.

$$ \int_{-\infty}^{\infty} \;dx\; e^{ikx} f(x) = \int_{-\infty}^{\infty} \;dx\; (\cos(kx) + i\sin(kx)) f(x) \quad\quad x\in\mathbb{R}$$

I think it makes sense that a Fourier series is a subset of a Fourier transform, as the transform takes into account all the same sines and cosines that the series does, but adds in all the noninteger sines and cosines as well.

Hope that helps.

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What my understanding was- Fourier series is used for analyzing periodic signals and Fourier transform for non-periodic signals. Now, Consider f(t)=f(T+t). In this if the period T-> ∞ then we get aperiodic version of periodic signal f(t). So I thought Fourier transform is a subset of Fourier series. The mistake I did was I differentiated this two based on the input signals not based on the output frequency spectrum or the explanation u have given above. I think now you might have understood what I thought previously. Anyway Thanks for yourreply. –  Inquisitive Dec 15 '12 at 14:40
    
@Inquisitive Thanks for the comment, now I understand it too –  wrongusername Mar 3 at 20:49
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I would say that neither is more general, they're different things.

Fourier transform can be defined in a very abstract way, for arbitrary locally compact abelian groups, and sometimes even more general objects. In case of locally compact abelian groups, it turns a function defined on a group into a function defined on its dual, and the formula is quite simple: $\hat f(\xi)=\langle f\mid \xi\rangle=\int_G f \cdot \overline{\xi} \, d\mu$ with $\mu$ the Haar measure.

Fourier series are the instance of Fourier transform where the group considered is the circle group $S^1={\mathbb T}$ (or its finite power), and it so happens that its dual is the group of functions of the form $e^{ikx}$, isomorphic to the group $\bf Z$ of integers (and the Haar measure is just the counting measure); in that way, Fourier transforms are far more general, but that's probably not what you meant to ask about.

What you know as Fourier transform (and what is most commonly referred to when Fourier transforms are invoked) is probably the Fourier transform of a function on the additive group of real numbers $\bf R$ or ${\bf R}^n$, which happen to be isomorphic to their duals, consisting of the functions of the form $e^{i\xi x}$.

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You can zero-extend a period of a periodic function to $\mathbb{R}$ and calculate the Fourier transform of the extension. This is done within pointwise function definition $f:\mathbb{R} \rightarrow \mathbb{C}$ and doesn't need the use of distributions. The zero-extension method has also the property that Fourier coefficients of the original periodic function can be read directly from the values of the transform of the extension. With suitable assumptions on the function also the inverse transform gives correct values within the interior of the period. That's why Fourier series is considered as a subset of Fourier transform.

The other way round it is not as easy. That's because Fourier series handles periodic functions only. A mapping that maps an infinite period to finite nonzero period violates the algebraical structure of the domain and values of the transform of the original aperiodic function cannot be read from the coefficients of the series anymore in a straightforward way. The countable domain $\mathbb{Z}$ of the coefficients isn't the problem because it can be extended to $\mathbb{R}$ by $f(\omega) = \sum_{k=-\infty}^\infty c_k \textrm{sinc}(\omega-k)$, where $\textrm{sinc}(x) = \bigg\{ \begin{eqnarray} & \frac{\sin(\pi x)}{\pi x} & , \ x\neq 0 \\ & 1 & , \ x = 0 \end{eqnarray}$. Now the obvious way to calculate the values of the transform of the original function is to form the Fourier series of the mapped version of the original function. Then it can be inverse mapped to obtain the original function. Values of the transform of the original function can be now obtained by just calculating all the values of the transform of the original function. The procedure consisted of inverse transformation, inverse mapping and transformation, that is not easy to do. However, that kind of procedure exists.

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