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Linearity and Conjugation 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,
$$\mathrm{x(t)=\displaystyle\sum\limits_{n=−\infty}^\infty\:C_{n}\:e^{jn\omega_{0}t}\:\:\:… (1)}$$
Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by,
$$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}X(t)e^{-jn\omega_{0}t}\:dt\:\:… (2)}$$
Linearity Property of Continuous Time Fourier Series
Consider two periodic signals $x_{1}(t)$ and $x_{2}(t)$ which are periodic with time period T and with Fourier series coefficients $C_{n}$ and $D_{n}$ respectively. If
$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$
$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$
Then, the linearity property of continuous-time Fourier series states that
$$\mathrm{Ax_{1}(t)+Bx_{2}(t)\overset{FS}{\leftrightarrow}AC_{n}+BD_{n}}$$
Proof
By the definition of Fourier series of a periodic function, we get,
$$\mathrm{FS[Ax_{1}(t)+Bx_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[Ax_{1}(t)+Bx_{2}(t)]e^{-jn\omega_{0}t}\:dt}$$
$$\mathrm{\Rightarrow\:FS[Ax_{1}(t)+Bx_{2}(t)]=A\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{1}(t)\:e^{-jn\omega_{0}t}\:dt\right )+B\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{2}(t)\:e^{-jn\omega_{0}t}\:dt \right )\:\:… (3)}$$
On comparing equations (2) & (3), we have,
$$\mathrm{FS[Ax_{1}(t)+Bx_{2}(t)]=AC_{n}+BD_{n}}$$
$$\mathrm{\therefore\:Ax_{1}(t)+Bx_{2}(t)\overset{FS}{\leftrightarrow}AC_{n}+BD_{n}\:\:(Hence,\:\:Proved)}$$
Conjugation Property of Continuous Time Fourier Series
Let a periodic function $x(t)$ which is periodic with time period $T$ and with Fourier series coefficients $C_{n}$. If
$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$
Then, the conjugation property of continuous time Fourier series states that
$$\mathrm{x^{*}(t)\overset{FS}{\leftrightarrow}C_{-n}^{*}\:\:[for\:complex\:x(t)]}$$
Proof
By the definition of Fourier series, we get,
$$\mathrm{FS[x^{*}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x^{*}(t)\:e^{-jn\omega_{0}t}\:dt}$$
$$\mathrm{\Rightarrow\:FS[x^{*}(t)]=\left (\frac{1}{T} \int_{t_{0}}^{t_{0}+T}x(t)\:e^{jn\omega_{0}t}\:dt\right )^{*}}$$
$$\mathrm{\Rightarrow\:FS[x^{*}(t)]=\left (\frac{1}{T} \int_{t_{0}}^{t_{0}+T}x(t)\:e^{-j(-n)\omega_{0}t}\:dt\right )^{*}=(C_{-n})^{*}}$$
Therefore,
$$\mathrm{x^{*}(t)\overset{FS}{\leftrightarrow}C_{-n}^{*}\:\:[for\:complex\:x(t)]}$$
Conjugate Symmetry Property of Continuous Time Fourier Series
As the conjugation property of Fourier series for a function $x(t)$ states that, if
$$\mathrm{x(t)\overset{FS}{\leftrightarrow}C_{n}}$$
Then,
$$\mathrm{x^{*}(t)\overset{FS}{\leftrightarrow}C_{-n}^{*}\:\:[for\:complex\:x(t)]}$$
The conjugate symmetry property of the Fourier series states that
$$\mathrm{C_{-n}=C_{n}^{*}\:\:\:[for\:\:real\:x(t)]}$$
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