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Harmonic analysis of saw function

This example demonstrates the accuracy of the harmonic browser (included in QuickField package)
Rectangular wave duct consists of two electrodes. The electrodes carry electric potential in a form of saw-function. Perform Fourier analysis of electric potential distribution.

Problem type:

Plane-parallel problem of electrostatics.

harmonic analysis of saw function

fourier decomposition of saw function

Given:

Relative permittivity of air ε = 1,
Boundary potentials - saw function.

Task:

Perform the Fourier analysis of the voltage distribution curve. Use the harmonic browser to calculate amplitudes of the harmonics and compare results with analytical solution.

Solution:

Voltage distribution along the X axis corresponds to the "saw" function.
U(0≤x<1.507) = 2·x
U(x=1.507) = 0
U(1.507<x≤3.142) = 2·x - 2·π

Fourier analysis of this "saw" function gives the following analytical solution:*
f(x) = 2 · ( sin(x) - sin(2x) / 2 + sin(3x) / 3 - ... + (-1)n+1·sin(nx) / n )

Results:
Potential distribution along the bottom boundary

harmonic browser simulation

Harmonic browsers window

harmonic browser addin

Harmonic browser allows calculation of the first hundred harmonics.

Harmonic

Amplitude

Error

QuickField

Theory

1

2

2

-

2

0.9999

1

-

3

0.6666

0.6666

-

4

0.4999

0.5

-

5

0.3999

0.4

-

10

0.1999

0.2

0.05%

20

0.0998

0.1

0.2%

50

0.0383

0.04

5%

100

0.0168

0.02

15%

*Mathematical Methods for Physicists. A Comprehensive Guide. 7th Edition. George Arfken, Hans Weber, Frank Harris, p 937-938. ISBN 978-0-12-384654-9.

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