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Dated: 30-10-2024

Fourier Series

The series1 of trignometrical functions2 of the following form is called fourier series.

\[f(x) = A_0 + c_1 \sin(x + \alpha_1) + c_2 \sin(2x + \alpha_2) + \ldots\]
\[= A_0 + \sum_{n=1}^\infty c_n \cdot \sin (n \cdot x + \alpha_n)\]

Here \(A_0\) is a constant.
\(c_n\) denotes the amplitude.
\(\alpha_n\) denotes auxiliary angles.

We can expand the term \(c_n \cdot \sin (n \cdot x + \alpha_n)\) as follows

\[c_n \cdot \sin (n \cdot x + \alpha_n) = c_n(\sin (nx) \cdot \cos \alpha_n + \cos(nx) \cdot \sin(\alpha_n))\]
\[ = c_n\sin (nx) \cdot \cos \alpha_n + c_n \cos(nx) \cdot \sin(\alpha_n)\]
\[\text{Let } c_n \sin (nx) = a_n \text{ and } c_n \cos(nx) = b_n\]

Then the series1 becomes

\[= A_0 + \sum_{n=1}^\infty (a_n \cos (nx) + b_n \sin(nx))\]

Coefficients of Fourier Series

\(A_0\)

To find \(A_0\), we will integrate3 both sides of the series.1

\[f(x) = A_0 + \sum_{n=1}^\infty (a_n \cos (nx) + b_n \sin(nx))\]
\[\int_{- \pi}^\pi f(x)dx = \int_{- \pi}^\pi\left(A_0 + \sum_{n=1}^\infty (a_n \cos (nx) + b_n \sin(nx))\right)dx\]
\[= \int_{- \pi}^\pi A_0 dx + \sum_{n = 1}^\infty \left(\int_{- \pi}^\pi a_n \cos(nx)dx\int_{- \pi}^\pi b_n \sin (nx)dx\right)\]
\[= \left[A_0\right]_{- \pi}^\pi + \sum_{n = 0}^\infty (0 + 0)\]
\[\int_{-\pi}^\pi f(x) = 2 \pi \cdot A_0\]
\[A_0 = \frac 1 {2 \pi} \int_{- \pi}^\pi f(x)\]

\(a_n\)

To find \(a_n\), we will multiply \(f(x)\) by \(\cos(mx)\) and integrate3 from \(- \pi\) to \(\pi\).

\[\int_{-\pi}^{\pi}f(x)\cos(mx)dx=\int_{-\pi}^{\pi}A_{0}\cos(mx)dx+\sum_{n=1}^{\infty}\left(\int_{-\pi}^{\pi}a_{n}\cos(nx)\cos(mx)dx+\int_{-\pi}^{\pi}b_{n}\sin(nx)\cos(mx)dx\right)\]
\[=0+a_{n}\pi+0=a_{n}\pi\quad\text{for}~n=m\]
\[\therefore a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\]

\(b_n\)

To find \(b_n\), we will repeat the same process but we will multiply by \(\sin(nx)\).

\[\therefore b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\]

Results of Fourier Series

  1. \[a_0 = \frac 1 \pi \int_{- \pi}^\pi f(x) dx = 2 \times \text{mean value of } f(x)\]
  2. \[a_n = \frac 1 \pi \int_{- \pi}^\pi f(x) \cos(nx) dx = 2 \times \text{mean value of } f(x)\cos(nx)\]
  3. \[a_n = \frac 1 \pi \int_{- \pi}^\pi f(x) \sin(nx) dx = 2 \times \text{mean value of } f(x)\sin(nx)\]
  4. \[A_0 = \frac 1 2 a_0\]

Example

Determine the fourier series for the following function.2

\[f(x) = \frac \pi 2 \text{ and } 0 < x < 2\pi\]
\[f(x) = f(x + 2 \pi)\]

Solution

We will now find the coefficients.

\(a_0\)

\[a_{0}=\frac{1}{\pi}\int_{0}^{2\pi}f(x)dx\]
\[=\frac{1}{\pi}\int_{0}^{2\pi}\left(\frac{x}{2}\right)dx\]
\[=\frac{1}{4\pi}\left[x^{2}\right]_{0}^{2\pi}\]
\[=\pi\]

\(a_n\)

\[a_{n}=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\cos(nx)dx\]
\[=\frac{1}{\pi}\int_{0}^{2\pi}\left(\frac{x}{2}\right)\cos(nx)dx\]
\[=\frac{1}{2\pi}\int_{0}^{2\pi}x~\cos(nx)dx\]
\[=\frac{1}{2\pi}\left(\left[\frac{x~\sin(nx)}{n}\right]_{0}^{2\pi}-\frac{1}{n}\int_{0}^{2\pi}\sin(nx)dx\right)\]
\[=\frac{1}{2\pi}\left((0-0)-\frac{1}{n}(0)\right)\]
\[= 0\]

\(b_n\)

\[b_{n}=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\sin(nx)dx\]
\[b_{n}=\frac{1}{\pi}\int_{0}^{2\pi}\left(\frac{x}{2}\right)\sin(nx)dx\]
\[=\frac{1}{2\pi}\left(\left[\frac{x~\cos(nx)}{n}\right]_{0}^{2\pi}+\frac{1}{n}\int_{0}^{2\pi}\cos(nx)dx\right)\]
\[=\frac{1}{2\pi}\left(\left[\frac{x \cdot \cos(nx)}{n}\right]_{0}^{2\pi}+\frac{1}{n}\left[\frac{\sin(nx)}{n}\right]_{0}^{2\pi}\right)\]
\[=\frac{1}{2\pi}\left((2\pi-0)+(0-0)\right)\]
\[=\frac{1}{n}\]

The general expression for fourier series is

\[f(x)= \frac 1 2 a_0 + \sum_{n=1}^\infty (a_n \cos (nx) + b_n \sin(nx))\]
\[f(x) = \frac \pi 2 - \left(\sin (x) + \frac 1 2 \sin(2x) + \ldots\right)\]

Dirichlet Conditions

If the fourier series is represented as \(f(x)\) then \(f(x_1)\) will give an infinite series1 of \(x_1\) which converges as more and more terms of the series1 are evaluated.
For this to happen, following conditions should be satisfied

  1. \(f(x)\) must be defined and single valued.
  2. \(f(x)\) should be continuous4 or have a finite number of finite discontinuities4 over the period.5
  3. \(f(x)\) and \(f^\prime(x)\) should be piecewise continuous functions2 in the periodic interval.5

Effect of Harmonics

As more and more terms of fourier series are evaluated, the graph of the series1 approaches the graph of the original function,2 the series1 represents.

References

Read more about notations and symbols.

  1. Read more about series. ↩↩↩↩↩↩↩

  2. Read more about functions. ↩↩↩↩

  3. Read more about integration. ↩↩

  4. Read more about continuity. ↩↩

  5. Read more about period. ↩↩