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Multiplication or Modulation Property 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,
$$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt\:\:… (2)}$$
Modulation or Multiplication Property
Let $x_{1}(t)$ and $x_{2}(t)$ two periodic signals with time period $T$ and with Fourier series coefficient $C_{n}$ and $D_{n}$. If
$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$
$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$
Then, the modulation or multiplication property of continuous time Fourier series states, that
$$\mathrm{x_{1}(t)\cdot x_{2}(t)\overset{FS}{\leftrightarrow}\sum_{k=−\infty}^{\infty}C_{k}\:D_{n-k}}$$
Proof
From the definition of continuous time Fourier series, we get,
$$\mathrm{FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[x_{1}(t)\cdot x_{2}(t)]e^{-jn\omega_{0} t}dt}$$
$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{jk\omega_{0} t}\right )e^{-jn\omega_{0} t}dt}$$
$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{-j(n-k)\omega_{0} t}\right )e^{-jn\omega_{0} t}dt\:\:… (3)}$$
By rearranging the order of integration and summation in equation (3), we obtain,
$$\mathrm{FS[x_{1}(t)\cdot x_{2}(t)]=\sum_{k=−\infty}^{\infty} C_{k}\left ( \frac{1}{T} \int_{t_{0}}^{t_{0}+T} x_{1}(t)e^{-j(n-k)\omega_{0} t}\:dt\right )=\sum_{k=−\infty}^{\infty}C_{k}D_{n-k}}$$
Where,
$$\mathrm{D_{n-k}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)e^{-j(n-k)\omega_{0} t}\:dt}$$
Therefore,
$$\mathrm{x_{1}(t)\cdot x_{2}(t)\overset{FS}{\leftrightarrow}\sum_{k=−\infty}^{\infty}C_{k}D_{n-k}\:\:\:(Hence,\:Proved)}$$
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