# Fourier Transform of the Sine and Cosine Functions

Signals and SystemsElectronics & ElectricalDigital Electronics

## Fourier Transform

The Fourier transform of a continuous-time function $x(t)$ can be defined as,

$$\mathrm{x(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t }dt}$$

## Fourier Transform of Sine Function

Let

$$\mathrm{x(t)=sin\:\omega_{0} t}$$

From Euler’s rule, we have,

$$\mathrm{x(t)=sin\:\omega_{0} t=\left[\frac{ e^{j\omega_{0} t}- e^{-j\omega_{0} t}}{2j} \right]}$$

Then, from the definition of Fourier transform, we have,

$$\mathrm{F[sin\:\omega_{0} t]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}sin\:\omega_{0}\: t\: e^{-j\omega t}dt}$$

$$\mathrm{ \Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left[ \frac{e^{j\omega_{0} t}-e^{-j\omega_{0} t}}{2j}\right] e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2j}\left[ \int_{−\infty}^{\infty}e^{j\omega_{0} t}e^{-j\omega t} dt-\int_{−\infty}^{\infty} e^{-j\omega_{0} t}e^{-j\omega t} dt\right]}$$

$$\mathrm{=\frac{1}{2j}\{F[e^{j\omega_{0} t}] -F[e^{-j\omega_{0} t}]\}}$$

Since, the Fourier transform of complex exponential function is given by,

$$\mathrm{F[e^{j\omega_{0} t}]=2\pi\delta(\omega-\omega_{0})\:\:and\:\:F[e^{-j\omega_{0} t}]=2\pi\delta(\omega+\omega_{0})}$$

$$\mathrm{ \therefore\:X(\omega)=\frac{1}{2j}[2\pi\delta(\omega-\omega_{0})-2\pi\delta(\omega+\omega_{0})]}$$

$$\mathrm{\Rightarrow\:X(\omega)=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$

Therefore, the Fourier transform of the sine wave is,

$$\mathrm{F[sin\:\omega_{0}\:t]=-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$

Or, it can also be represented as,

$$\mathrm{sin\:\omega_{0}\:t\overset{FT}{\leftrightarrow}-j\pi[\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})]}$$

The graphical representation of the sine function with its magnitude and phase spectra is shown in Figure-1.

## Fourier Transform of Cosine Function

Given

$$\mathrm{x(t)=cos\:\omega_{0}t}$$

From Euler’s rule, we have,

$$\mathrm{cos\:\omega_{0}t=\left[\frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2}\right]}$$

Then, from the definition of Fourier transform, we have,

$$\mathrm{F[cos\:\omega_{0} t]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}cos\:\omega_{0} t e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left[\frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2} \right]e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}\left[ \int_{−\infty}^{\infty}e^{j\omega_{0} t}e^{-j\omega t} dt+ \int_{−\infty}^{\infty}e^{-j\omega_{0} t}e^{-j\omega t} dt \right]}$$

$$\mathrm{=\frac{1}{2}\{F[e^{j\omega_{0} t}]+ F[e^{-j\omega_{0} t}]\}}$$

$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}[2\pi\delta(\omega-\omega_{0})+2\pi\delta(\omega+\omega_{0})]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\pi[\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0})]}$$

Therefore, the Fourier transform of cosine wave function is,

$$\mathrm{F[cos\:\omega_{0} t]=\pi[\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0})]}$$

Or, it can also be represented as,

$$\mathrm{cos\:\omega_{0} t\overset{FT}{\leftrightarrow}\pi[\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0})]}$$

The graphical representation of the cosine wave signal with its magnitude and phase spectra is shown in Figure-2.

Published on 09-Dec-2021 06:57:00