Every time function x(t) can be expressed as a Fourier series, that is, a sum of trigonometric functions of a base frequency and its harmonics:

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

(ID 14342)

The base frequency
u_k of the Fourier series is defined as a function of the time T of the time series x(t):

$ \nu_k = \displaystyle\frac{ k }{ T }$

(ID 14343)

To calculate the coefficient a_k of the series

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

the function x(t) must be integrated weighted by the cosine of the corresponding frequency:

$ a_k = \displaystyle\frac{1}{ T } \displaystyle\int_0^T x(t) \cos(2 \pi \nu_k t ) dt$

(ID 14347)

To calculate the coefficient b_k of the series

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

the function x(t) must be integrated weighted by the sine of the corresponding frequency:

$ b_k = \displaystyle\frac{1}{ T } \displaystyle\int_0^T x(t) \sin(2 \pi \nu_k t ) dt $

(ID 14348)

To estimate the integral

$ a_k = \displaystyle\frac{1}{ T } \displaystyle\int_0^T x(t) \cos(2 \pi \nu_k t ) dt$

you can discretize the function x(t) and replace the integral with a sum:

$ a_k = \displaystyle\frac{1}{ T } \displaystyle\sum_{n=0}^{ N -1} x_n \cos(2 \pi \nu_k n \Delta t )$

(ID 14349)

To estimate the integral

$ b_k = \displaystyle\frac{1}{ T } \displaystyle\int_0^T x(t) \sin(2 \pi \nu_k t ) dt $

you can discretize the function x(t) and replace the integral with a sum:

$ b_k = \displaystyle\frac{1}{ T } \displaystyle\sum_{n=0}^{ N -1} x_n \sin(2 \pi \nu_k n \Delta t )$

(ID 14350)

The coefficients of the Fourier transform

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

can be regrouped as a complex number by defining

$ X_k = a_k - i b_k $

(ID 14352)

Fourier transform

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

you can with Euler\'s relation

$ x(t) = \displaystyle\sum_{k=-\infty}^{\infty}( a_k \cos 2 \pi \nu_k t + b_k \sin 2 \pi \nu_k t )$

the definition

$ X_k = a_k - i b_k $

and the decreticization of time

$ t_n = n \Delta t $

redefine as the discrete transforms on the complex space of the time series x_n equal to

$ x_n = \displaystyle\sum_{k=0}^{N-1} X_k e^{ i 2 \pi \nu_k n \Delta t }$

(ID 14351)

To calculate the coefficient X_k of the series

$ x_n = \displaystyle\sum_{k=0}^{N-1} X_k e^{ i 2 \pi \nu_k n \Delta t }$

the function x(t) must be integrated weighted by the cosine of the corresponding frequency:

$ X_k = \displaystyle\frac{1}{ T } \displaystyle\int_{0}^{ T } x(t) e^{ i 2 \pi \nu_k t } dt$

(ID 14353)

To estimate the integral

$ X_k = \displaystyle\frac{1}{ T } \displaystyle\int_{0}^{ T } x(t) e^{ i 2 \pi \nu_k t } dt$

you can discretize the function x(t) and replace the integral with a sum:

$ X_k = \displaystyle\frac{1}{ T } \displaystyle\sum_{ n =0}^{ N -1} x_n e^{ i 2 \pi \nu_k n \Delta t }$

(ID 14354)

If the complex coefficient is

$ X_k = a_k - i b_k $

Its magnitude is defined as

$ r_k = \sqrt{ a_k ^2 + b_k ^2}$

(ID 14355)

If the complex coefficient is

$ X_k = a_k - i b_k $

the phase can be calculated from

$ \phi_k = \arctan\displaystyle\frac{ b_k }{ a_k }$

(ID 14356)