Complex Form of the Fourier series

The Fourier series find important applications in geophysical signal processing and signal analysis. However in previous posts I have already described the basics of the Fourier series. Through this post I tried as much as possible, to give you an insight about the complex or exponential form of this Mathematical model.

If we recall the power series of ex that:

ex = 1+x+x2/2!+x3/3!+x4/4!+.....

if we put jθ at x, then ejθ = 1+jθ+(jθ)2/2!+(jθ)3/3!+(jθ)4/4!+.....

ejθ = 1+jθ+j2θ2/2!+j3θ3/3!+j4θ4/4!+.....

But j2= -1

Then, ejθ = 1+jθ - θ2/2! - jθ3/3!+θ4/4!+.....

If we group the like terms, ejθ = (1- θ2/2! +θ4/4!+.....) + j(θ - θ3/3!+...).........(1)

But  from equation 1, (1- θ2/2! +θ4/4!+.....) = cosθ  and j(θ - θ3/3!+...) = jsinθ
Then, ejθ = cosθ + jsinθ.........(2) 

Also, e−jθ = cosθ − jsinθ..........(3)

if we take equation 2 plus equation 3,
ejθ + e−jθ = 2cosθ
By making cosθ the subject, cos θ = (ejθ + e−jθ)/2..........(4)
Also if we take equation 2 minus equation 3, it will give,
ejθ − e−jθ = 2jsinθ
Also by making sinθ the subject, sinθ = (e jθ − e−jθ)/2j..........(5)
Recall  the Fourier series f (x) over any range L, is given as,
f (x) = a0 + Σn=1 to ∞ + ancos(2πnx/L) + bnsin(2πnx/L)...........(6)

Then it can be written as,
f (x) = a0 + Σn=1 to ∞ [an(ej2πnx/L+ e−j2πnx/L/2) + bn(ej2πnx/L− e−j2πnx/L/2)]......(7)
If we Multiply by - j both top and bottom of the bn term in equation 7 above
and  recalling that  j2 = −1 then:
f (x) = a0 + Σn=1 to ∞[an(ej2πnx/L + e−j2πnx/L/2) − jbn(ej2πnx/L− e−j2πnx/L)/2]
Then, f (x) = a0 + Σn=1 to ∞(an − jbn/2)ej2πnx/L + (an + jbn/2)e−j2πnx/L .............(8)
The Fourier coefficients a0, an and bn may be replaced by complex coefficients k0, kn and k−n such that 

k0 = a0 

kn = (an − jbn)/2
k−n =(an + jbn)/2
where k−n represents the complex conjugate of kn
Thus, equation (8) can be written as:
f (x) = k0 + Σn=1 to ∞knej2πnx/L+ Σn=1 to ∞k−ne−j2πnx/L.........(9)

It simplifies to f (x) = Σn = 0 to ∞ knej2πnx/L+ Σn=1 to ∞k−ne−j2πnx/L............(10)
If we change the limits n = 1 to n = ∞ to n = −1 to n = −∞, Since n has been made negative, the exponential term becomes ej2πnx/L and k−n becomes kn. Thus,
f(x) = Σn=0 to ∞knej2πnx/L + Σn= -1 to - ∞ knej2πnx/L..............(11)
We can write equation 8 above as here below,

f(x) = Σn= - ∞ to ∞ (kne j2πnx/L).............(12)
Then equation (12) is the Complex (exponential) form of the Fourier series

Where for range of - L/2 to L/2,  kn =1/L∫f(x) e−j2πnx/L dx.

Thank you for your attention!

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