Fourier Series of the Full Wave Rectified Sine Wave: sin(x) in [0, 2π]

The function fx sin(x) in the interval [0, 2π] is often referred to as the full wave rectified sine wave. This article will delve into the Fourier series representation of this function. Fourier series are a powerful tool for expressing functions as sums of simpler trigonometric functions, and in the case of the full wave rectified sine wave, they help in understanding the behavior and properties of the function.

What is the Full Wave Rectified Sine Wave?

A full wave rectified sine wave is a periodic function derived by taking the absolute value of the sine function. In other words, it is the result of rectifying a sine wave, meaning converting any negative values to positive. For more detailed information, you can refer to the article on Fourier series of full wave rectified sine wave.

Understanding Fourier Series

Fourier series are used to express periodic functions as sums of sine and cosine functions. They are particularly useful in signal processing, communications engineering, and many other fields where periodic signals are analyzed. The basic idea is that any periodic function can be decomposed into a series of sine and cosine terms, which simplifies the analysis of the function.

Odd Functions and Fourier Series Decomposition

Given that both the function fx sin(x) and the sine function itself are odd functions, we can use the Fourier sine series for each one separately. An odd function has the property that f(-x) -f(x). Therefore, the Fourier series for an odd function will only contain sine terms (bn terms) and no cosine terms (an terms).

Deriving the Fourier Sine Series for sin(x)

The Fourier sine series for a function fx in the interval [0, L] is given by:

[ f(x) sum_{n1}^{infty} b_n sinleft(frac{nπx}{L}right)]

where the coefficients b_n are given by the formula:

[ b_n frac{2}{L} int_{0}^{L} f(x) sinleft(frac{nπx}{L}right) dx]

For our function fx sin(x) in the interval [0, 2π], we can substitute L 2π:

[ b_n frac{2}{2π} int_{0}^{2π} sin(x) sinleft(frac{nπx}{2π}right) dx frac{1}{π} int_{0}^{2π} sin(x) sin(nx) dx]

Using the trigonometric identity for the product of sine functions:

[ sin(x) sin(nx) frac{1}{2} [cos((n-1)x) - cos((n 1)x)]]

We can substitute this into the integral:

[ b_n frac{1}{π} int_{0}^{2π} frac{1}{2} [cos((n-1)x) - cos((n 1)x)] dx]

Integration of cosine terms from 0 to 2π will yield 0 unless the arguments of the cosine terms are 0. Therefore, for all n other than n1:

[ b_n 0]

For n 1, the integral simplifies to:

[ b_1 frac{1}{π} int_{0}^{2π} sin^2(x) dx frac{1}{π} int_{0}^{2π} frac{1 - cos(2x)}{2} dx frac{π}{π} 1]

So, the Fourier sine series for fx sin(x) in the interval [0, 2π] is:

[ f(x) sin(x)]

This means that the Fourier series of fx sin(x) is simply sin(x) itself, confirming that the function is already a sine wave.

Conclusion

In conclusion, the Fourier series of the full wave rectified sine wave fx sin(x) in the interval [0, 2π] is simply sin(x). This is because the function is an odd function and its Fourier series only contains sine terms, and in this specific case, the series simplifies to the function itself.

Keywords

Fourier series, full wave rectified sine wave, sine series

References

You can refer to the Wikipedia article on Fourier series of full wave rectified sine wave for more detailed information.