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Time Differentiation and Integration Properties of Continuous-Time Fourier Series
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,
$$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t}… (1)}$$
Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by,
$$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt… (2)}$$
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,
$$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}… (3)}$$
By taking time differentiation on both sides of the equation (3), we have,
$$\mathrm{\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}C_{n}\frac{d(e^{jn\omega_{0} t})}{dt}}$$
$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t}(jn\omega_{0})}$$
$$\mathrm{\Rightarrow\:\frac{dx(t)}{dt}=\sum_{n=−\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0} t}… (4)}$$
$$\mathrm{∵\: \sum_{n=−\infty}^{\infty}(jn\omega_{0}C_{n})e^{jn\omega_{0}t}=FS^{-1}[jn\omega_{0}C_{n}]… (5)}$$
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
$$\mathrm{\int_{−\infty}^{t}x(τ)dτ\overset{FS}{\leftrightarrow}\frac{C_{n}}{jn\omega_{0}};\:\:C_{0}=0}$$
Proof
By the definition of continuous time Fourier series, we get,
$$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0}t}… (6)}$$
By taking time integration on both sides of equation (6), we have,
$$\mathrm{\int_{−\infty}^{\infty}x(τ)dτ=\int_{−\infty}^{t}\left [\sum_{n=−\infty}^{\infty} C_{n}\:e^{jn\omega_{0} τ}\right ]dτ}$$
Rearranging the integration and summation, we get,
$$\mathrm{\int_{−\infty}^{t}x(τ)dτ=\sum_{n=−\infty}^{\infty}C_{n}\int_{−\infty}^{t}e^{jn\omega_{0} τ}dτ}$$
$$\mathrm{\Rightarrow\:\int_{−\infty}^{t}x(τ)dτ=\sum_{n=−\infty}^{\infty}C_{n}\left [ \frac{e^{jn\omega_{0} τ}}{jn\omega_{0}}\right ]_{−\infty}^{t}}$$
$$\mathrm{\Rightarrow\:\int_{−\infty}^{t}x(τ)dτ=\sum_{n=−\infty}^{\infty}C_{n}\left [ \frac{e^{jn\omega_{0} t}}{jn\omega_{0}}-\frac{e^{−\infty}}{jn\omega_{0} } \right ]}$$
$$\mathrm{\Rightarrow\:\int_{−\infty}^{t}x(τ)dτ=\sum_{n=−\infty}^{\infty}C_{n}\left (\frac{e^{jn\omega_{0} t}}{jn\omega_{0}} \right )=\sum_{n=−\infty}^{\infty}\left ( \frac{C_{n}}{jn\omega_{0}} \right )e^{jn\omega_{0} t}… (7)}$$
$$\mathrm{∵\sum_{n=−\infty}^{\infty}\left ( \frac{C_{n}}{jn\omega_{0} } \right )e^{jn\omega_{0} t}=FS^{-1}\left ( \frac{C_{n}}{jn\omega_{0} } \right )… (8)}$$
Hence, from equation (7) & (8), we get,
$$\mathrm{\int_{−\infty}^{t}x(τ)dτ\overset{FS}{\leftrightarrow}\frac{C_{n}}{jn\omega_{0}};\:\:C_{0}=0\:\:(Hence,\:Proved)}$$
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