HARMONIC ANALYSIS T MANIKANDAN AP MATHEMATICS MSEC HARMONIC

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HARMONIC ANALYSIS T. MANIKANDAN AP / MATHEMATICS MSEC

HARMONIC ANALYSIS T. MANIKANDAN AP / MATHEMATICS MSEC

HARMONIC ANALYSIS If f(x) is defined analytically in (0, 2π) we know how to

HARMONIC ANALYSIS If f(x) is defined analytically in (0, 2π) we know how to find the Fourier coefficients a 0, an and bn. But in practical Engineering problems, the function f(x) to be expanded in Fourier series is not defined by analytical expression. Instead, we know a few values of the function or its graph. In such cases, the Fourier co-efficients a 0, and bn in Fourier series are calculated by means of approximate integration.

HARMONIC ANALYSIS The process of finding the Fourier series for a function given by

HARMONIC ANALYSIS The process of finding the Fourier series for a function given by numerical value is known as harmonic analysis.

The term a 1 cosx+b 1 sinx is called the fundamental or first harmonic,

The term a 1 cosx+b 1 sinx is called the fundamental or first harmonic, the term a 2 cos 2 x+b 2 sin 2 x is called the second harmonic and so on.

Example: 1 Find the first harmonic of Fourier series of y=f(x) from the given

Example: 1 Find the first harmonic of Fourier series of y=f(x) from the given data x (in degree) y = f(x) 0 60 120 180 240 300 360 1. 98 1. 3 1. 05 1. 3 -0. 88 -0. 25 1. 98 Solution: Since the function is periodic with period 2π, We exclude the last point x = 360. The Fourier series for the first harmonic is

x y Cos x Sin x y cos x 0 1. 98 1 0

x y Cos x Sin x y cos x 0 1. 98 1 0 0 60 1. 3 1. 98 0. 5 0. 866 1. 1258 120 1. 05 0. 65 -0. 5 0. 866 0. 9093 180 1. 3 -0. 525 -1 0 -1. 3 0 240 -0. 88 -0. 5 -0. 866 0. 44 0. 762 300 -0. 25 4. 5 0. 5 -0. 886 -0. 125 1. 12 Total y sin x 0. 2165 3. 013 f(x)=0. 75 + 0. 373 cos x + 1. 004 sin x

Example: 2 Find the first two harmonics of Fourier series of y=f(x) from the

Example: 2 Find the first two harmonics of Fourier series of y=f(x) from the given data x 0 y = f(x) 1 2π 1. 4 1. 9 1. 7 1. 5 1. 2 1 Solution: Since the function is periodic with period 2π, We exclude the last point x = 2π. The Fourier series for the first two harmonic is

x 0 y Cos x 1 1 1 0 0 0. 5 -0. 5

x 0 y Cos x 1 1 1 0 0 0. 5 -0. 5 0. 866 0. 7 -0. 7 1. 2124 -0. 5 0. 866 -0. 95 1. 6454 -1 1 0 0 -1. 7 0 0 -0. 5 -0. 866 -0. 75 -1. 299 -0. 5 -0. 866 0. 6 -1. 0392 1. 4 1. 9 1. 7 1. 5 1. 2 Total 8. 7 0. 5 Cos 2 x Sin 2 x y cos x -1. 1 y cos 2 x y sin x -0. 3 y sin 2 x 0. 5196 -0. 1732

f(x)=1. 45 – 0. 37 cos x – 0. 1 cos 2 x +

f(x)=1. 45 – 0. 37 cos x – 0. 1 cos 2 x + 0. 173 sin x – 0. 577 sin 2 x