Trigonometric Fourier Series

PERIODIC FUNCTION:

A function f(t) is said to be periodic if,

                                                           f(t+T)=f(t); for all values of t

where T is some positive real number. This T is interval between two successive repetitions and is called the period of f(t).

A sine wave having a period of $T = \frac{{2\pi }}{\omega }$ is common example of periodic function.

DIRICHLET CONDITION:

For a given function f(t),

1. f(t) is periodic having a period of T.
2. f(t) is single valued everywhere.
3. In case it is discontinuous, f(t) has a finite number of discontinuities in any one period.
4. f(t) has a finite number of maxima and minima in any one period.
5. The integral, $T = \frac{{2\pi }}{\omega }$ exists and is finite.

The function f(t) may represent either a voltage or current waveform.

TRIGONOMETRIC FOURIER SERIES:

According to Fourier theorem, a function f(t) which satisfy the Dirichlet condition, may be represented in trigonometric form by the infinite series.

$$\begin{array}{l} f(t) = {a_0} + {a_1}\cos {\omega _0}t + {a_2}\cos 2{\omega _0}t + .... + {a_n}\cos n{\omega _0}t + {b_1}\sin {\omega _0}t + {b_2}\sin 2{\omega _0}t + .... + {b_n}\sin n{\omega _0}t\\  = {a_0} + \sum\limits_{n = 1}^\infty  {\left( {{a_n}\cos n{\omega _0}t + {b_n}\sin n{\omega _0}t} \right)} \end{array}$$

since ${\omega _0} = \frac{{2\pi }}{T}$ above equation can be written as

$f(t) = {a_0} + \sum\limits_{n = 1}^\infty  {\left( {{a_n}\cos \frac{{2\pi n}}{T}t + {b_n}\sin \frac{{2\pi n}}{T}t} \right)}$

where, ${\omega _0}$ is the fundamental angular frequency, T is the period and ${a_0}$,${a_n}$ and ${b_n}$ are constant which are depend on n and f(t).

FOURIER ANALYSIS:

Evaluation of fourier constant called fourier analysis.

1. Value of ${a_0}$:

$f(t) = {a_0} + \sum\limits_{n = 1}^\infty  {\left( {{a_n}\cos n\omega t + {b_n}\sin n\omega t} \right)}$

Integrate over a period t=0 to t=T.

$\int\limits_0^T {f(t)} dt = \int\limits_0^T {{a_0}} dt + \sum\limits_{n = 1}^\infty  {\int\limits_0^T {({a_n}\cos n\omega t + {b_n}\sin n\omega t} } )dt = {a_0}T$

Hence, ${a_0} = \frac{1}{T}\int\limits_0^T {f(t)} dt$

                             = mean value of f(t) between the limits 0 to T i.e. over one cycle or period.

2. Value of ${a_n}$:

$f(t) = {a_0} + \sum\limits_{n = 1}^\infty  {\left( {{a_n}\cos n\omega t + {b_n}\sin n\omega t} \right)}$

multiply both sides of the fourier series by $\cos k\omega t$ and integrate between limits t=0 to t=T.

$\int\limits_0^T {f(t)\cos k\omega t} dt = \int\limits_0^T {{a_0}\cos k\omega t} dt + \sum\limits_{n = 1}^\infty  {\int\limits_0^T {({a_n}\cos n\omega t\cos k\omega t + {b_n}\sin n\omega t\cos k\omega t} } )dt$

=${a_k}\frac{T}{2}$

Therefore,

${a_k} = \frac{2}{T}\int\limits_0^T {f(t)\cos k\omega t} dt$

3. Value of ${b_n}$:

$f(t) = {a_0} + \sum\limits_{n = 1}^\infty  {\left( {{a_n}\cos n\omega t + {b_n}\sin n\omega t} \right)}$

multiply both sides of the fourier series by $\sin k\omega t$ and integrate between limits t=0 to t=T.

$\int\limits_0^T {f(t)\sin k\omega t} dt = \int\limits_0^T {{a_0}\sin k\omega t} dt + \sum\limits_{n = 1}^\infty  {\int\limits_0^T {({a_n}\cos n\omega t\sin k\omega t + {b_n}\sin n\omega t\sin k\omega t} } )dt$

=${b_k}\frac{T}{2}$

Therefore,

${b_k} = \frac{2}{T}\int\limits_0^T {f(t)\sin k\omega t} dt$