Trigonometric Fourier Series (TFS)

$sin nomega_0 t$ and $sin momega_0 t$ are orthogonal over the interval $(t_0, t_0+{2pi over omega_0})$. So $sinomega_0 t,, sin 2omega_0 t$ forms an orthogonal set. This set is not complete without {$cos nomega_0 t$ } 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. {$sin nomega_0 t,,cos nomega_0 t$ } where n=0, 1, 2…

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

$$ x(t) = a_0 cos0omega_0 t+ a_1 cos⁡ 1omega_0 t+ a_2 cos2 ⁡omega_0 t +…+ a_n cos⁡ nomega_0 t + … $$

$$ + b_0 sin⁡ 0omega_0 t + b_1 sin⁡ 1omega_0 t +…+ b_n sin⁡ nomega_0 t + … $$

$$ = a_0 + a_1 cos⁡ 1omega_0 t + a_2 cos 2⁡ omega_0 t +…+ a_n cos⁡ nomega_0 t + …$$

$$ + b_1 sin⁡ 1omega_0 t +…+ b_n sin⁡ nomega_0 t + …$$

$$ therefore x(t) = a_0 + sum_{n=1}^{infty} (a_n cos⁡ nomega_0 t + b_n sin⁡ nomega_0 t ) quad (t_0

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

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

$$a_n = {int_{t_0}^{t_0+T} x(t)· cos⁡ nomega_0 t,dt over int_{t_0}^{t_0+T} cos ^2 nomega_0 t, dt}$$

$$b_n = {int_{t_0}^{t_0+T} x(t)· sin nomega_0 t,dt over int_{t_0}^{t_0+T} sin ^2 nomega_0 t, dt}$$

$$text{Here}, int_{t_0}^{t_0+T} cos ^2 nomega_0 t, dt = int_{t_0}^{t_0+T} sin ^2 nomega_0 t, dt = {Tover 2}$$

$$therefore a_n = {2over T}· int_{t_0}^{t_0+T} x(t)· cos⁡ nomega_0 t,dt$$

$$b_n = {2over T}· int_{t_0}^{t_0+T} x(t)· sin nomega_0 t,dt$$

Exponential Fourier Series (EFS)

Consider a set of complex exponential functions $left{e^{jnomega_0 t}right} (n=0, pm1, pm2…)$ which is orthogonal over the interval $(t_0, t_0+T)$. Where $T={2pi over omega_0}$ . This is a complete set so it is possible to represent any function f(t) as shown below

$ f(t) = F_0 + F_1e^{jomega_0 t} + F_2e^{j 2omega_0 t} + … + F_n e^{j nomega_0 t} + …$

$quad quad ,,F_{-1}e^{-jomega_0 t} + F_{-2}e^{-j 2omega_0 t} +…+ F_{-n}e^{-j nomega_0 t}+…$

$$ therefore f(t) = sum_{n=-infty}^{infty} F_n e^{j nomega_0 t} quad quad (t_0

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

$$ F_n = {int_{t_0}^{t_0+T} f(t) (e^{j nomega_0 t} )^* dt over int_{t_0}^{t_0+T} e^{j nomega_0 t} (e^{j nomega_0 t} )^* dt} $$

$$ quad = {int_{t_0}^{t_0+T} f(t) e^{-j nomega_0 t} dt over int_{t_0}^{t_0+T} e^{-j nomega_0 t} e^{j nomega_0 t} dt} $$

$$ quad quad quad quad quad quad quad quad quad ,, = {int_{t_0}^{t_0+T} f(t) e^{-j nomega_0 t} dt over int_{t_0}^{t_0+T} 1, dt} = {1 over T} int_{t_0}^{t_0+T} f(t) e^{-j nomega_0 t} dt $$

$$ therefore F_n = {1 over T} int_{t_0}^{t_0+T} f(t) e^{-j nomega_0 t} dt $$

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⁡ nomega_0 t + b_n sin⁡ nomega_0 t) … … (1)$

$ x(t) = Sigma_{n=-infty}^{infty} F_n e^{j nomega_0 t}$

$quad ,,, = F_0 + F_1e^{jomega_0 t} + F_2e^{j 2omega_0 t} + … + F_n e^{j nomega_0 t} + … $

$quad quad quad quad F_{-1} e^{-jomega_0 t} + F_{-2}e^{-j 2omega_0 t} + … + F_{-n}e^{-j nomega_0 t} + … $

$ = F_0 + F_1(cos omega_0 t + j sinomega_0 t) + F_2(cos 2omega_0 t + j sin 2omega_0 t) + … + F_n(cos nomega_0 t+j sin nomega_0 t)+ … + F_{-1}(cosomega_0 t-j sinomega_0 t) + F_{-2}(cos 2omega_0 t-j sin 2omega_0 t) + … + F_{-n}(cos nomega_0 t-j sin nomega_0 t) + … $

$ = F_0 + (F_1+ F_{-1}) cosomega_0 t + (F_2+ F_{-2}) cos2omega_0 t +…+ j(F_1 – F_{-1}) sinomega_0 t + j(F_2 – F_{-2}) sin2omega_0 t+… $

$ therefore x(t) = F_0 + Sigma_{n=1}^{infty}( (F_n +F_{-n} ) cos nomega_0 t+j(F_n-F_{-n}) sin nomega_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,

$F_n = frac12 (a_n – jb_n )$

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