Fourier Series Types

Trigonometric Fourier Series (TFS)

$\sin n\omega_0 t$ and $\sin m\omega_0 t$ are orthogonal over the interval $(t_0, t_0+{2\pi \over \omega_0})$. So $\sin\omega_0 t,\, \sin 2\omega_0 t$ forms an orthogonal set. This set is not complete without {$\cos n\omega_0 t$ } because this cosine set is also orthogonal to sine set. So to complete this set we must include both cosine and sine terms. Now the complete orthogonal set contains all cosine and sine terms i.e. {$\sin n\omega_0 t,\,\cos n\omega_0 t$ } where n=0, 1, 2...

$\therefore$ Any function x(t) in the interval $(t_0, t_0+{2\pi \over \omega_0})$ can be represented as

$$x(t) = a_0 \cos0\omega_0 t+ a_1 \cos⁡ 1\omega_0 t+ a_2 \cos2 ⁡\omega_0 t +...+ a_n \cos⁡ n\omega_0 t + ...$$

$$+ b_0 \sin⁡ 0\omega_0 t + b_1 \sin⁡ 1\omega_0 t +...+ b_n \sin⁡ n\omega_0 t + ...$$

$$= a_0 + a_1 \cos⁡ 1\omega_0 t + a_2 \cos 2⁡ \omega_0 t +...+ a_n \cos⁡ n\omega_0 t + ...$$

$$+ b_1 \sin⁡ 1\omega_0 t +...+ b_n \sin⁡ n\omega_0 t + ...$$

$$\therefore x(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos⁡ n\omega_0 t + b_n \sin⁡ n\omega_0 t ) \quad (t_0< t < t_0+T)$$

The above equation represents trigonometric Fourier series representation of x(t).

$$\text{Where} \,a_0 = {\int_{t_0}^{t_0+T} x(t)·1 dt \over \int_{t_0}^{t_0+T} 1^2 dt} = {1 \over T}· \int_{t_0}^{t_0+T} x(t)dt$$

$$a_n = {\int_{t_0}^{t_0+T} x(t)· \cos⁡ n\omega_0 t\,dt \over \int_{t_0}^{t_0+T} \cos ^2 n\omega_0 t\, dt}$$

$$b_n = {\int_{t_0}^{t_0+T} x(t)· \sin n\omega_0 t\,dt \over \int_{t_0}^{t_0+T} \sin ^2 n\omega_0 t\, dt}$$

$$\text{Here}\, \int_{t_0}^{t_0+T} \cos ^2 n\omega_0 t\, dt = \int_{t_0}^{t_0+T} \sin ^2 n\omega_0 t\, dt = {T\over 2}$$

$$\therefore a_n = {2\over T}· \int_{t_0}^{t_0+T} x(t)· \cos⁡ n\omega_0 t\,dt$$

$$b_n = {2\over T}· \int_{t_0}^{t_0+T} x(t)· \sin n\omega_0 t\,dt$$

Exponential Fourier Series (EFS)

Consider a set of complex exponential functions $\left\{e^{jn\omega_0 t}\right\} (n=0, \pm1, \pm2...)$ which is orthogonal over the interval $(t_0, t_0+T)$. Where $T={2\pi \over \omega_0}$ . This is a complete set so it is possible to represent any function f(t) as shown below

$f(t) = F_0 + F_1e^{j\omega_0 t} + F_2e^{j 2\omega_0 t} + ... + F_n e^{j n\omega_0 t} + ...$

$\quad \quad \,\,F_{-1}e^{-j\omega_0 t} + F_{-2}e^{-j 2\omega_0 t} +...+ F_{-n}e^{-j n\omega_0 t}+...$

$$\therefore f(t) = \sum_{n=-\infty}^{\infty} F_n e^{j n\omega_0 t} \quad \quad (t_0< t < t_0+T) ....... (1)$$

Equation 1 represents exponential Fourier series representation of a signal f(t) over the interval (t0, t0+T). The Fourier coefficient is given as

$$F_n = {\int_{t_0}^{t_0+T} f(t) (e^{j n\omega_0 t} )^* dt \over \int_{t_0}^{t_0+T} e^{j n\omega_0 t} (e^{j n\omega_0 t} )^* dt}$$

$$\quad = {\int_{t_0}^{t_0+T} f(t) e^{-j n\omega_0 t} dt \over \int_{t_0}^{t_0+T} e^{-j n\omega_0 t} e^{j n\omega_0 t} dt}$$

$$\quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, = {\int_{t_0}^{t_0+T} f(t) e^{-j n\omega_0 t} dt \over \int_{t_0}^{t_0+T} 1\, dt} = {1 \over T} \int_{t_0}^{t_0+T} f(t) e^{-j n\omega_0 t} dt$$

$$\therefore F_n = {1 \over T} \int_{t_0}^{t_0+T} f(t) e^{-j n\omega_0 t} dt$$

Relation Between Trigonometric and Exponential Fourier Series

Consider a periodic signal x(t), the TFS & EFS representations are given below respectively

$x(t) = a_0 + \Sigma_{n=1}^{\infty}(a_n \cos⁡ n\omega_0 t + b_n \sin⁡ n\omega_0 t) ... ... (1)$

$x(t) = \Sigma_{n=-\infty}^{\infty} F_n e^{j n\omega_0 t}$

$\quad \,\,\, = F_0 + F_1e^{j\omega_0 t} + F_2e^{j 2\omega_0 t} + ... + F_n e^{j n\omega_0 t} + ...$

$\quad \quad \quad \quad F_{-1} e^{-j\omega_0 t} + F_{-2}e^{-j 2\omega_0 t} + ... + F_{-n}e^{-j n\omega_0 t} + ...$

$= F_0 + F_1(\cos \omega_0 t + j \sin\omega_0 t) + F_2(cos 2\omega_0 t + j \sin 2\omega_0 t) + ... + F_n(\cos n\omega_0 t+j \sin n\omega_0 t)+ ... + F_{-1}(\cos\omega_0 t-j \sin\omega_0 t) + F_{-2}(\cos 2\omega_0 t-j \sin 2\omega_0 t) + ... + F_{-n}(\cos n\omega_0 t-j \sin n\omega_0 t) + ...$

$= F_0 + (F_1+ F_{-1}) \cos\omega_0 t + (F_2+ F_{-2}) \cos2\omega_0 t +...+ j(F_1 - F_{-1}) \sin\omega_0 t + j(F_2 - F_{-2}) \sin2\omega_0 t+...$

$\therefore x(t) = F_0 + \Sigma_{n=1}^{\infty}( (F_n +F_{-n} ) \cos n\omega_0 t+j(F_n-F_{-n}) \sin n\omega_0 t) ... ... (2)$

Compare equation 1 and 2.

$a_0= F_0$

$a_n=F_n+F_{-n}$

$b_n = j(F_n-F_{-n} )$

Similarly,

$F_n = \frac12 (a_n - jb_n )$

$F_{-n} = \frac12 (a_n + jb_n )$