# The equivalence between exponential form of Fourier series and trigonometric form of fourier series

The equivalence between exponential form of Fourier series and trigonometric form of fourier series. I do not know How they are equivalent, could anyone explain this for me please?

• How much do you know about $L^2[0,2\pi]$? – user228113 May 19 '17 at 10:04
• I know to the extent that the partial sum of Fourier series is convergent in L^2. – Emptymind May 19 '17 at 10:24
• I'm not sure I understand. The relation $e^{in \theta} = \cos( n \theta) + i sin(n \theta)$ allows us to convert a exponential Fourier serie into a trigonometric serie and vice-versa. Was that your question ? – user171326 May 19 '17 at 10:43

Given that $$\sqrt{-1} = j$$, the exponential fourier series can describe periodic functions, with period $$T_c$$, that take on complex values. It can be denoted as follows

$$f(t) = \sum_{n = - \infty}^{\infty}X_ne^{jn\omega_c t}$$



$$Xn = \frac{\left< f(t), e^{jn\omega_c t} \right>_{PP}}{{\| e^{jn\omega_c t} \|}^2} = \frac{1}{T}\int_{a -T/2}^{a + T/2} f(t)e^{-jn\omega_c t}dt$$

$$X_n = |X_n|\angle \theta_n$$

Here, $$f(t)$$ can take on complex values because both $$X_n$$ and $$e^{jn\omega_c t}$$ take on complex values. If you want to describe a complex function with other functions, those other functions have to be complex too. (Notice how the inner product takes the complex conjugate of the second function in $$X_n$$)



The trigonometric fourier series can only describe periodic functions, with period $$T_c$$, that take on real values (The major difference). It can be denoted as follows

$$f(t) = a_0 + \sum_{n = 1}^{\infty}a_n cos(n\omega_c t) + \sum_{n = 1}^{\infty}b_n cos(n\omega_c t)$$



$$a_v = \frac{a_0}{2} = \frac{\left< f(t), 1 \right>_{PP}}{{\| 1 \|}^2} = \frac{1}{T}\int_{a-T/2}^{a+T/2} f(t)dt$$

$$a_n = \frac{\left< f(t), cos( n\omega_c t) \right>_{PP}}{{\| cos( n\omega_c t) \|}^2} = \frac{2}{T}\int_{a-T/2}^{a+T/2} f(t)cos( n\omega_c t)dt$$

$$b_n = \frac{\left< f(t), sin( n\omega_c t) \right>_{PP}}{{\| sin( n\omega_c t) \|}^2} = \frac{2}{T}\int_{a-T/2}^{a+T/2} f(t)sin( n\omega_c t)dt$$

Here the trigonometric functions, and the constant $$1$$, are restricted to take on only real values, and thus $$f(t)$$ can only take on real values.



### My attempt to show the connection between the two fourier series

For real values of $$f(t)$$ the exponential fourier series can be simplified to

$$f(t) = X_0 + 2\sum_{n = 1}^{\infty}|X_n|cos(n\omega_c t + \theta_n)$$

This should remind you of the compact trigonometric fourier series which can be denoted as follows

$$f(t) = \frac{C_0 cos(\theta_0)}{2} + \sum_{n = 1}^{\infty}C_ncos(n\omega_c t + \theta_n)$$



$$C_n = \sqrt{(a_n)^2 + (b_n)^2}$$

$$\theta_n = \begin{cases} -arctan(\frac{b_n}{a_n}) &; \space a_n \geq 0 \\ \pi -arctan(\frac{b_n}{a_n}) &; \space a_n \lt 0 \end{cases}$$

We can confirm the first term, knowing $$b_0 = 0$$, by looking at the following evaluation

$$\theta_0 = \begin{cases} 0 &; \space a_n \geq 0 \\ \pi &; \space a_n \lt 0 \end{cases}$$

$$\theta_0 = sign(a_n)$$

$$C_0 = \sqrt{ (a_n)^2} = |a_n|$$

$$C_0 cos(\theta_0) = |a_n| sign(a_0) = a_0$$



We can thus make the following observation regarding the different coefficients

$$C_n = 2|X_n| \space , \space \space n \geq 0$$



### Conclusion

From the above definitions, we can derive the following combining definition

$$X_n = \frac{a_n - jb_n}{2}$$