Time Differentiation and Integration Properties of Continuous-Time Fourier Series

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Fourier Series

If $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as,

$$\mathrm{x(t) \:=\: \sum_{n= -\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t} \:\:\dotso\:(1)}$$

Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by,

$$\mathrm{C_{n} \:=\: \frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)\:e^{-jn\omega_{0} t}dt \:\:\dotso\:(2)}$$

Time Differentiation Property of Fourier Series

If $x(t)$ is a periodic function with time period T and with Fourier series coefficient $C_{n}$. If

$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$

Then, the time differentiation property of continuous-time Fourier series states that

$$\mathrm{\frac{dx(t)}{dt}\overset{FS}{\leftrightarrow}jn\omega_{0}C_{n}}$$

Proof

By the definition of continuous time Fourier series, we get,

$$\mathrm{x(t) \:=\: \sum_{n= -\infty}^{\infty}\:C_{n}e^{jn\omega_{0} t} \:\:\dotso\:(3)}$$

By taking time differentiation on both sides of the equation (3), we have,

$$\mathrm{\frac{dx(t)}{dt} \:=\: \sum_{n= -\infty}^{\infty}\:C_{n}\frac{d(e^{jn\omega_{0} t})}{dt}}$$

$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt} \:=\: \sum_{n= -\infty}^{\infty}C_{n}e^{jn\omega_{0} t}(jn\omega_{0})}$$

$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt} \:=\: \sum_{n= -\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0} t} \:\:\dotso\:(4)}$$

$$\mathrm{\because \: \sum_{n= -\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0}t} \:=\: FS^{-1}[jn\omega_{0}C_{n}] \:\:\dotso\: (5)}$$

From equation (4) & (5), we obtain,

$$\mathrm{\frac{dx(t)}{dt}\overset{FT}{\leftrightarrow}jn\omega_{0}C_{n}\:\:(Hence\:Proved)}$$

Time Integration Property of Continuous-Time Fourier Series

If $x(t)$ is a periodic function with time period T and with Fourier series coefficient $C_{n}$. Then, if

$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$

Then, the time integration property of continuous time Fourier series states that

$$\mathrm{\int_{-\infty}^{t}x(\tau)d \tau\overset{FS}{\leftrightarrow}\frac{C_{n}}{jn\omega_{0}};\:\:C_{0}=0}$$

Proof

By the definition of continuous time Fourier series, we get,

$$\mathrm{x(t)\:=\:\sum_{n= -\infty}^{\infty}C_{n}e^{jn\omega_{0}t} \:\:\dotso\:(6)}$$

By taking time integration on both sides of equation (6), we have,

$$\mathrm{\int_{-\infty}^{\infty}x(\tau)d \tau \:=\:\int_{-\infty}^{t}\left [\sum_{n= -\infty}^{\infty} C_{n}\:e^{jn\omega_{0} \tau}\right ]d \tau}$$

Rearranging the integration and summation, we get,

$$\mathrm{\int_{-\infty}^{t}x(\tau)d\tau \:=\: \sum_{n=-\infty}^{\infty}C_{n}\int_{-\infty}^{t}\:e^{jn\omega_{0} \tau}d\tau}$$

$$\mathrm{\Rightarrow\:\int_{-\infty}^{t}x(\tau)d\tau\:=\:\sum_{n=-\infty}^{\infty}C_{n}\left[\frac{e^{jn\omega_{0} \tau}}{jn\omega_{0}}\right]_{-\infty}^{t}}$$

$$\mathrm{\Rightarrow\:\int_{-\infty}^{t}x(\tau)d\tau \:=\: \sum_{n=-\infty}^{\infty}C_{n}\left[\frac{e^{jn\omega_{0} t}}{jn\omega_{0}} \:-\: \frac{e^{-\infty}}{jn\omega_{0} } \right ]}$$

$$\mathrm{\Rightarrow\:\int_{-\infty}^{t}x(\tau)d\tau \:=\: \sum_{n=-\infty}^{\infty}C_{n}\left(\frac{e^{jn\omega_{0} t}}{jn\omega_{0}} \right ) \:=\: \sum_{n=-\infty}^{\infty}\left(\frac{C_{n}}{jn\omega_{0}} \right)e^{jn\omega_{0} t} \:\:\dotso\: (7)}$$

$$\mathrm{\because \: \sum_{n= -\infty}^{\infty}\left(\frac{C_{n}}{jn\omega_{0}} \right)\:e^{jn\omega_{0}t} \:=\: FS^{-1}\left( \frac{C_{n}}{jn\omega_{0}} \right ) \:\:\dotso\:(8)}$$

Hence, from equation (7) & (8), we get,

$$\mathrm{\int_{-\infty}^{t}x(\tau)d\tau\overset{FS}{\leftrightarrow}\frac{C_{n}}{jn\omega_{0}};\:\:C_{0} \:=\: 0 \:\: (Hence,\:Proved)}$$