Dictionary:Fourier series

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Representation of a periodic function by the sum of sinusoidal components whose frequencies are integral multiples of a fundamental frequency. See Fourier transform.

Origins

The name Fourier refers to French mathematician and physicist Jean Baptiste Fourier (21 March 1768 – 16 May 1830). Fourier considered the possibility of using trigonometric series as a solution to problems of heat transfer.

What is Fourier Series?  

For a function $f(t)$ that is absolutely integrable, which is to say that the following integral exists

$\int _{a}^{b}|f(t)|\;dt<\infty .$ This follows by a result known as the Riemann-Lebesque lemma.

We may write $f(t)$ as the convergent trigonometric series expansion

$f(t)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[a_{n}\cos \left({\frac {n\pi t}{T}}\right)+b_{n}\sin \left({\frac {n\pi t}{T}}\right)\right].$ Here, the coefficients are given by the expressions

• $a_{0}={\frac {1}{T}}\int _{-T}^{T}f(t)\;dt$ • $a_{n}={\frac {1}{T}}\int _{-T}^{T}f(t)\cos \left({\frac {n\pi t}{T}}\right)\;dt$ • $b_{n}={\frac {1}{T}}\int _{-T}^{T}f(t)\sin \left({\frac {n\pi t}{T}}\right)\;dt.$ Here, T is a positive quantity representing a natural length scale in the problem.

If we consider functions to be members of a vector space, which is to say a set which is closed under addition and multiplication by a scalar, the notion of Fourier series is analogous to a representation of a vector in terms of its components in a basis, with the Fourier series being the representation of $f(t)$ in terms of its components. The sine and cosine functions play the role of the unit or basis vectors in this representation, and integration plays the role of the "inner" or "dot" product.

Thus, the Fourier coefficients are the projection of $f(t)$ in the direction of the respective $\cos(n\pi t/T)$ and $\sin(n\pi t/T)$ . Several important properties come into play. These are periodicity, evenness and oddness, and orthogonality. A number of trigonometric identities are used in the process of proving the Fourier series theorem.

Deriving the expressions for the Fourier coefficients

In general we are solving for the coefficients $a_{n}$ and $b_{n},$ in the expression

$f(t)=\sum _{n=0}^{\infty }\left[a_{n}\cos \left({\frac {n\pi t}{T}}\right)+b_{n}\sin \left({\frac {n\pi t}{T}}\right)\right]$ On the interval $[-T,T].$ This series representation may be thought of consisting of the sum of two series. These are the cosine series and the sine series, representing the even and odd parts of $f(t).$ The Fourier Cosine series

We consider the case of $f(t)$ purely even and seek to find the coefficients $a_{n}$ in the expression

$f(t)=\sum _{n=0}^{\infty }a_{n}\cos \left({\frac {n\pi t}{T}}\right)$ We do this by multiplying both sides of the expression by $\cos(m\pi t/T)$ and integrating over the range of $[-T,T],$ $\int _{-T}^{T}f(t)\cos \left({\frac {m\pi t}{T}}\right)\;dt=\sum _{n=0}^{\infty }a_{n}\int _{-T}^{T}\cos \left({\frac {m\pi t}{T}}\right)\cos \left({\frac {n\pi t}{T}}\right)\;dt.$ We may solve for the coefficients by applying the property of orthogonality of the cosines. We know that the result of the integration on the right is nonzero if and only if $m=n$ . Hence the expression reduces to

$\int _{-T}^{T}f(t)\;dt=a_{n}\int _{-T}^{T}\cos \left({\frac {n\pi t}{T}}\right)\;dt.$ There are only two cases to consider. These are the cases when $n=0$ and when $n\neq 0.$ Case 1: $m=n=0$ The first is the zero frequency case. Here the expression for $a_{0}$ reduces to

$\int _{-T}^{T}f(t)\;dt=a_{0}2T$ yielding the zero frequency result

$a_{0}={\frac {1}{2T}}\int _{-T}^{T}f(t)\;dt$ which shows that the zero frequency component $a_{0}$ of $f(t)$ is the average of $f(t)$ over the interval $[-T,T].$ Case 2: $m=n\neq 0$ For all other cases, $n$ is nonzero, yielding

$\int _{-T}^{T}f(t)\cos \left({\frac {n\pi t}{T}}\right)\;dt=a_{n}\int _{-T}^{T}\cos ^{2}\left({\frac {n\pi t}{T}}\right)\;dt.$ Applying the double angle formula for cosine yields

$\int _{-T}^{T}f(t)\cos \left({\frac {n\pi t}{T}}\right)\;dt=a_{n}T,$ Yielding the expression for $a_{n}$ for $n\neq 0$ $a_{n}={\frac {1}{T}}\int _{-T}^{T}f(t)\cos \left({\frac {n\pi t}{T}}\right)\;dt.$ The Fourier Sine series

We consider the case of $f(t)$ purely odd and seek to find the coefficients $b_{n}$ in the expression

$f(t)=\sum _{0}^{\infty }b_{n}\sin \left({\frac {n\pi t}{T}}\right)$ We do this by multiplying both sides of the expression by $\sin(m\pi t/T)$ and integrating over the range of $[-T,T],$ $\int _{-T}^{T}f(t)\sin \left({\frac {m\pi t}{T}}\right)\;dt=\sum _{n=0}^{\infty }b_{n}\int _{-T}^{T}\cos \left({\frac {m\pi t}{T}}\right)\sin \left({\frac {n\pi t}{T}}\right)\;dt.$ We may solve for the coefficients by applying the property of orthogonality of the sines. We know that the result of the integration on the right is nonzero if and only if $m=n$ . Hence the expression reduces to

$\int _{-T}^{T}f(t)\;dt=b_{n}\int _{-T}^{T}\sin ^{2}\left({\frac {n\pi t}{T}}\right)\;dt.$ There are only two cases to consider. These are the cases when $n=0$ and when $n\neq 0.$ Case 1: $m=n=0$ The first is the zero frequency case. Here the expression for $b_{0}$ reduces to

$\int _{-T}^{T}f(t)\;dt=0$ because $\sin 0=0$ and because $f(t)$ is odd, this yields the zero frequency result

$b_{0}=0.$ Case 2: $m=n\neq 0$ For all other cases, $n$ is nonzero, yielding

$\int _{-T}^{T}f(t)\sin \left({\frac {n\pi t}{T}}\right)\;dt=b_{n}\int _{-T}^{T}\sin ^{2}\left({\frac {n\pi t}{T}}\right)\;dt.$ Applying the double angle formula for sine yields

$\int _{-T}^{T}f(t)\sin \left({\frac {n\pi t}{T}}\right)\;dt=b_{n}T,$ Yielding the expression for $b_{n}$ for $n\neq 0.$ $b_{n}={\frac {1}{T}}\int _{-T}^{T}f(t)\sin \left({\frac {n\pi t}{T}}\right)\;dt.$ The complex Fourier Series

If we consider repalcing the cosine and sine in the expression

$f(t)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[a_{n}\cos \left({\frac {n\pi t}{T}}\right)+b_{n}\sin \left({\frac {n\pi t}{T}}\right)\right].$ with their respective complex exponential representations, we obtain

$f(t)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[a_{n}\left({\frac e^{i{\frac {n\pi t}{T}}}+e^{-i{\frac {n\pi t}{T}}}}{2}}\right)+b_{n}\left({\frac e^{i{\frac {n\pi t}{T}}}-e^{-i{\frac {n\pi t}{T}}}}{2i}}\right)\right].$ Rearranging the terms we have

$f(t)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[\left({\frac {a_{n}-ib_{n}}{2}}\right)\displaystyle e^{i{\frac {n\pi t}{T}}}+\left({\frac {a_{n}+ib_{n}}{2}}\right)\displaystyle e^{-i{\frac {n\pi t}{T}}}\right].$ If we define $c_{0}=a_{0}/2,$ $c_{n}=(a_{n}-ib)/2,$ and $c_{-n}=(a_{n}+ib)/2,$ then we may write the complex form of the Fourier Series

$f(t)=\sum _{n=-\infty }^{\infty }c_{n}\displaystyle e^{i{\frac {n\pi t}{T}}}$ We may find the coefficients by multiplying both sides by $\exp(-im\pi t/T)$ and integrating over $[-T,T].$ $\int _{-T}^{T}f(t)\displaystyle e^{-i{\frac {m\pi t}{T}}}\;dt=\sum _{n=-\infty }^{\infty }c_{n}\int _{-T}^{T}\displaystyle e^{i{\frac {n\pi t}{T}}}\displaystyle e^{-i{\frac {m\pi t}{T}}}\;dt.$ By the orthogonality of the sines and cosines, only the $m=n$ case is nonzero, yielding

$\int _{-T}^{T}f(t)\displaystyle e^{-i{\frac {n\pi t}{T}}}\;dt=c_{n}\int _{-T}^{T}\;dt$ which yields the result

$c_{n}={\frac {1}{2T}}\int _{-T}^{T}f(t)\displaystyle e^{-i{\frac {n\pi t}{T}}}\;dt.$ Heuristic derivation of the Fourier transform

Beginning with the expression for the complex Fourier coefficients

$c_{n}={\frac {1}{2T}}\int _{-T}^{T}f(t)\displaystyle e^{-i{\frac {n\pi t}{T}}}\;dt$ we set $1/T=\Delta \omega /\pi$ and set $n\Delta \omega =\omega _{n}$ $c_{n}={\frac {\Delta \omega }{2\pi }}\int _{-\pi /\Delta \omega }^{\pi /\Delta \omega }f(t)\displaystyle e^{-i\omega _{n}t}\;dt.$ If we substitute this expression for $c_{n}$ in the expression for the complex Fourier series of $f(t)$ $f(t)=\sum _{n=-\infty }^{\infty }\left({\frac {\Delta \omega }{2\pi }}\int _{-\pi /\Delta \omega }^{\pi /\Delta \omega }f(t)\displaystyle e^{-i\omega _{n}t}\;dt\right)\displaystyle e^{i\omega _{n}t}.$ Rearranging the terms and pass to the continuous limit, we have

$f(t)=\lim _{\Delta \omega \rightarrow 0}\sum _{n=-\infty }^{\infty }\left({\frac {1}{2\pi }}\int _{-\pi /\Delta \omega }^{\pi /\Delta \omega }f(t)\displaystyle e^{-i\omega _{n}t}\;dt\right)\displaystyle e^{i\omega _{n}t}\Delta \omega .$ which suggests the form of a Riemann sum, suggesting the following identity

$f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }f(t)e^{-i\omega t}\;dt\right)e^{i\omega t}\;d\omega ,$ This suggests the following Fourier transform pairs

${\hat {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{\pm i\omega t}\;dt$ and

$f(t)=\int _{-\infty }^{\infty }{\hat {f}}(\omega )e^{\mp i\omega t}\;d\omega .$ The $\pm$ and $\mp$ follow because the exponent sign conventions may vary between authors, but that the exponent sign is on the forward transform is always opposite that of the inverse transform.