# Is this a correct way of thinking of Fourier transforms

I am working on my understanding of various transforms and I have been thinking about the Fourier transform, what i "does" to the function it is applied to.

The way I see it:

The function $f$ that is transformed is multiplied with $\exp(icx)$ which essentially describes a rotating vector in the complex plane.

• If $f$ is periodic cosine and the period of $f$ does not match the period of $\exp(icx)$ the "terms" in the integral will vary and cancel each other leading to a value of zero for the transform.

• If $f$ is periodic cosine and the period of $f$ matches the period of $\exp(icx)$ the "terms" in the integral will be constant and the value of the transform will be $\infty$.

• If $f$ is periodic but not a cosine it can be decomposed to a sum of cosines and the different cosine terms of this sum will work as above resulting in a spectra for $f$.

If this is somewhat correct I wonder:

2. Is there a similar way of thinking about the Laplace transform?

Please forgive the non mathematical language, I'm neither a math major nor is English my mother tongue.

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I really think that en.wikipedia.org/wiki/Fourier_transform will answer all your questions. – Josh Dec 12 '12 at 10:53
You're not completely wrong but I don't think that's the best way to think at it either. Periodic functions are to be represented with Fourier series, period (pun intended :-) ). Now there clearly is relationship between Fourier series and Fourier integrals, but that's another story. – Giuseppe Negro Jan 28 '13 at 23:43

Suppose you have a function $f\colon \mathbb{R}\to \mathbb{C}$ which is $T$-periodic. Then the following family of sinusoidal functions is compatible with $f$ in the sense that a period of $f$ contains an integer number of periods of each one of them: $$\sin\left( \frac{2\pi n}{T}t\right),\quad \cos\left( \frac{2\pi n}{T}t\right),\quad n=0, 1, 2, \ldots$$ Now the theory of Fourier series tells us that we can decompose $f$ as the superposition of those sinusoidal functions: $$f(t)=\frac{a_0}{2}+\sum_{n=1}^\infty \left( a_n\cos\left( \frac{2\pi n}{T}t\right)+b_n\sin\left( \frac{2\pi n}{T}t\right)\right),$$ which is usually rewritten in complex notation $$\tag{D} f(t)=\sum_{n=-\infty}^\infty c_n e^{i\frac{2\pi n}{T}t}.$$ (Here "D" stands for "discrete"). The map $f\to (c_n)$ is a kind of Fourier transform. It is a coordinate change in the sense that $f(t)$ is a representation of the original information as a function of $t$ (say, time) while $c_n$ is a representation of the same thing as a function of $n$, which is a number indicating frequency.
If $f$ is not periodic, we may regard it as a function having "infinite period", and construct an analogue of the previous decomposition by letting $T\to \infty$ with some care. The result is the following formula: $$\tag{C}f(t)=\int_{-\infty}^\infty \tilde{f}(\nu)e^{2\pi i \nu t}\, d\nu,$$ ("C" stands for "continuum") where $$\tilde{f}(\nu)=\int_{-\infty}^\infty f(t)e^{-2\pi i t\nu}\, dt$$ is the Fourier transform. Exactly as before, the map $f(t)\to \tilde{f}(\nu)$ is a coordinate change. It serves the purpose of switching between a representation of a piece of information as a function of $t$ (time) and a representation of the same information as a function of $\nu$, which is a frequency.