# Signals and Systems â€“ Fourier Transform of Periodic Signals

The Fourier series can be used to analyse only the periodic signals, while the Fourier transform can be used to analyse both periodic as well as non-periodic functions. Therefore, the Fourier transform can be used as a universal mathematical tool in the analysis of both periodic and aperiodic signals over the entire interval. The Fourier transform of periodic signals can be found using the concept of impulse function.

Now, consider a periodic signal $\mathit{x\left(t\right )}$ with period $\mathit{T}$. Then, the expression of $\mathit{x\left(t\right )}$ in terms of exponential Fourier series is given by,

$$\mathrm{\mathit{x\left(t\right)=\sum_{n=-\infty }^{\infty } C_{n}\:e^{jn\omega _{\mathrm{0}}t}}}$$

Where $\mathit{C_{n}}$ be the Fourier coefficients and are given by,

$$\mathrm{\mathit{C_{n}=\frac{\mathrm{1}}{T}\int_{-T/\mathrm{2}}^{T/\mathrm{2}}x\left(t\right)\:e^{-jn\omega _{\mathrm{0}}t}\:dt}}$$

Therefore, the Fourier transform of signal $\mathit{x\left(t\right )}$ is,

$$\mathrm{\mathit{F\left [ x\left(t\right) \right ]=X\left(\omega\right)=F\left [ \sum_{n=-\infty }^{\infty }\:C_{n}\:e^{jn\omega _{\mathrm{0}}t} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left(\omega\right)=\sum_{n=-\infty }^{\infty }\:C_{n}\:F\left [ e^{jn\omega _{\mathrm{0}}t} \right ]}}$$

By using frequency shifting property $[\mathit{i.e.,e^{j\omega_{\mathrm{0}}t}\:x\left(t\right)\overset{FT}{\leftrightarrow}X\left(\omega -\omega_{\mathrm{0}}\right)}]$ of Fourier transform, we get,

$$\mathrm{\mathit{F\left [ \mathrm{1}\:.e^{jn\omega _{\mathrm{0}}t} \right ]=\mathrm{2}\pi\delta\left(\omega -n\omega _{\mathrm{0}}\right)}}$$

Therefore, the Fourier transform of a periodic function is,

$$\mathrm{\mathit{X\left(\omega\right)\mathrm{=}\mathrm{2}\pi\:\sum_{n=-\infty }^{\infty }\:C_{n}\:\delta\left(\omega -n\omega _{\mathrm{0}}\right)}}$$

Hence,

• The Fourier transform of a periodic function consists of a series of equally spaced impulses and these impulse are located at the harmonic frequencies of the signal.

• The area or strength of each impulse is given by,

$$\mathrm{\mathit{\mathrm{Area\: of\: impulse} = \mathrm{2}\pi C_{n}}}$$