Question

Derive the relation between trigonometric Fourier series and exponential Fourier series.​

Answer 1

Step-by-step explanation:

Fourier series is for periodic signals and Fourier transform is for aperiodic signals. Fourier series is used to decompose signals into basic elements. While Fourier transform are used to analyse signal in another domain.

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Answer 2

Tip:

• The trigonometric Fourier series is a periodic function of period $T_0=\frac{2\pi}{\omega_0}$. If the function $g(t)$ is periodic with period $T_0$, then a Fourier series representing $g(t)$ over an interval $T_0$ will also represent $g(t)$ for all $t$.
• The main result from the exponential Fourier series analysis is that an arbitrary periodic signal can approximate by summing individual cosine terms with specified amplitudes and phases. This result serves as much of the conceptual and theoretical framework for the field of signal analysis.

Explanation:

• in the given question we have the terms trigonometric Fourier series and exponential Fourier series.​
• We have to derive the relation between them.
• We will derive this with the help of tip.

Steps:

Step 1 of 2:

Trigonometric Fourier Series:

∴  $x(t)=a_0+\sum_{n=1}^{\infty}(a_nCosn\omega_0t+b_nSinn\omega_0t)~~~~~(t_0<t<t_0+T)$  ...(1)

Exponential Fourier Series:

∴     $f(t)=\sum_{n=-\infty}^{\infty}F_ne^{jn\omega_0t}$     $(t_0<t<t_0+T)$ …(2)

Step 2 of 2:

From (1) and (2), we get:

$x(t)=a_0+\sum_{n=1}^{\infty}(a_nCosn\omega_0t+b_nSinn\omega_0t)$

$x(t)=\sum_{n=-\infty}^{\infty}=−\inftyF_ne^{jn\omega_0t}\\=F_0+F_1e^{j\omega_0t}+F_2e^{j2\omega_0t}+\ldots+F_ne^{jn\omega_0t}+\ldotsF_{−1}e^{−j\omega_0t}+F_{−2e^{−j2\omega_0t}+\ldots+F_{−n}e^{−jn\omega_0t}+\ldots$

$=F_0+F_1(Cos\omega_0t+jSin\omega0t)+F_2(Cos2\omega0t+jSin2\omega_0t)+\ldots+F_n(Cosn\omega_0t+jSinn\omega_0t)+\lsots+F_{−1}(Cos\omega_0t−jSin\omega_0t)+F_{−2}(cos2\omega_0t−jSin2\omega_0t)+\ldots+F_{−n}(cosn\omega_0t−jSinn\omega_0t)+\ldots$∴  $x(t)=F_0+\sum_{n=1}^{\infty}((F_n+F_{-n})Cosn\omega_0t+j(F_n−F_{-n})Sinn\omega_0t)$ ...(3)

Comparing equation (1) and (3), we get

$a_0=F_0\\ a_n=F_n+F_{-n}\\b_n=j(F_n-F_{-n})$

Similarly,

$F_n=\frac{1}{2}(a_n-jb_n)\\ F_{-n}=\frac{1}{2}(a_n+jb_n)$

Final Answer:

This is the relation between Trigonometric Fourier Series and Exponential Fourier Series.