Fourier Transform of the Sine and Cosine Functions

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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.