I just learned about Fourier series, and this is how I interpreted them: The complex exponentials form a basis for all periodic functions, and the Fourier series essentially decompose the function into a a sum of the basis functions. The Fourier coefficient formula then essentially "projects" the function onto the complex exponential to find how much it "has in common with that exponential". I interpreted the Fourier coefficient formula to be a continuous version of the dot product.

Using this interpretation, shouldn't the formula be as follows (where $T_0$ is the fundamental period) $$ a_k = \int_ 0 ^ {T_0}x(t)e^{\frac{2\pi j kt}{T_0}}dt. $$

To me it seems the formula above finds the projection of the function $x(t)$ onto a basis function. The integral is essentially an extension of the dot product (the dot product is a finite sum of the product of the components, the integral above is an infinite sum of the product of the components).

However, the formula above is obviously incorrect. The correct formula is $$a_k = \frac{1}{T_0}\int_ 0 ^ {T_0}x(t)e^{\frac{-2\pi j kt}{T_0}}dt.$$

1) Can someone explain why there is a negative sign in the power? The basis have positive powers (i.e. the exponents in the Fourier sum don't have negative powers) so shouldn't the exponential of the function "dotted with" also have a positive power?

2) Why is it divided by $T_0$?

  • $\begingroup$ To explain the negative power: note that the "dot product" of anything with itself should be a non-negative number. With that in mind, how should we compute $$ \langle (1,i),(1,i)\rangle?$$ $\endgroup$ Commented Jan 19, 2015 at 0:47
  • $\begingroup$ As for $T_0$: we want to make sure that out "basis functions" have length $1$, for ease of computation. Dividing the "dot product" by a constant changes lengths while maintaining orthogonality. $\endgroup$ Commented Jan 19, 2015 at 0:51
  • $\begingroup$ @Omnomnomnom For the dot product of the imaginary numbers, could we make it equal to the product of the norms? So 2? But then the usual formula of summing the products of the components wouldn't work... $\endgroup$
    – adfasf
    Commented Jan 19, 2015 at 0:55
  • $\begingroup$ How about take the conjugate of all the elements in the second vector? $\endgroup$ Commented Jan 19, 2015 at 0:56
  • $\begingroup$ @Omnomnomnom That makes sense. We would then get the norm of the vector. But then how do we know the rule we came up with - multiplying by the conjugate- extends to cases where both functions being "dotted" are not the same? $\endgroup$
    – adfasf
    Commented Jan 19, 2015 at 0:59

1 Answer 1


The definite integral over a fundamental period annihilates all of the complex exponentials except for the constant term. Thus, to determine "how much of $e^{2\pi in t/T}$ is in $f(t)$" you have to do something to $f$ to turn the "$a_\omega e^{2\pi in t/T}$ term" of interest into a constant term; this is achieved by multiplying by the reciprocal of the complex power, namely $e^{-2\pi in t/T}$. Of course integrating a constant over an interval will yield the constant times the interval's length, so for the original constant we need to normalize by the length of the interval.

Note that inner products defined on complex vector spaces incorporate conjugate-linearity in the second argument, so the inner product works like $\langle \vec{a},\vec{b}\rangle=a_1\overline{b_1}+\cdots+a_n\overline{b_n}$, and the negative power of the complex exponential is in keeping with this practice. If we don't have the conjugate part of this definition, then the bilinear form would no longer be quite as useful for linear algebra purposes - for instance you could not use it to "project" and "extract" coefficients of a vector's representation with respect to a given basis, as is being done in this problem.

  • $\begingroup$ Your answer was perfect, thank you! $\endgroup$
    – adfasf
    Commented Jan 19, 2015 at 1:05

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