# 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

$$\mathrm{C_{-n}=C_{n}^{*}\:\:\:[for\:\:real\:x(t)]}$$