Modulation Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Transform

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

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

Modulation Property of Fourier Transform

Statement – The modulation property of continuous-time Fourier transform states that if a continuous-time function $x(t)$ is multiplied by $cos \:\omega_{0} t$, then its frequency spectrum gets translated up and down in frequency by $\omega_{0}$. Therefore, if

$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$

Then, according to the modulation property of CTFT,

$$\mathrm{x(t)\:cos\:\omega_{0}t\overset{FT}{\leftrightarrow}\frac{1}{2}[X(\omega-\omega_{0})+X(\omega+\omega_{0})]}$$

Proof

Using Euler’s formula, we get,

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

Therefore,

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

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

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

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

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

$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\frac{1}{2}F[x(t)e^{j\omega_{0} t}]+\frac{1}{2}F[x(t)e^{-j\omega_{0} t}]}$$

$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\frac{1}{2}X(\omega -\omega_{0})+\frac{1}{2}X(\omega + \omega_{0})}$$

Therefore, the Fourier transform is,

$$\mathrm{F[x(t)\:cos\:\omega_{0} t]=\frac{1}{2}[X(\omega -\omega_{0})+X(\omega + \omega_{0})]}$$

Or, it can also be represented as,

$$\mathrm{x(t)\:cos\:\omega_{0} t\overset{FT}{\leftrightarrow}\frac{1}{2}[X(\omega -\omega_{0})+X(\omega + \omega_{0})]}$$

Similarly, when the signal x(t) is multiplied by $sin \:\omega_{0}\:t$, then according to the modulation property of CTFT, the Fourier transform of the signal is given by,

$$\mathrm{x(t)\:sin\:\omega_{0} t\overset{FT}{\leftrightarrow}\frac{1}{2j}[X(\omega -\omega_{0})- X(\omega + \omega_{0})]}$$

Numerical Example

Using modulation property of Fourier transform, find the Fourier transform of $[sin\:\omega_{0}\:t]$.

Solution 

Given,

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

Let $x(t)$ is multiplied by a function $x_{1}(t)$ as,

$$\mathrm{x(t)=x_{1}(t)\cdot sin\:\omega_{0}\:t}$$

Where,

$$\mathrm{x_{1}(t)=1}$$

Also, the Fourier transform of a constant amplitude is given by,

$$\mathrm{F[x_{1}(t)]=F[1]=2\pi\delta(\omega)}$$

Now, using modulation property, we get,

$$\mathrm{F[x(t)]=F[x_{1}(t)\:sin\:\omega_{0}\:t]=\frac{1}{2j}[X(\omega -\omega_{0})- X_{1}(\omega + \omega_{0})]}$$

$$\mathrm{\Rightarrow\:F[(1)\cdot sin\:\omega_{0}\:t]=\frac{1}{2j}[2\pi\delta(\omega - \omega_{0})-2\pi\delta(\omega + \omega_{0})]}$$

$$\mathrm{\Rightarrow\:F[sin\:\omega_{0}t]=\frac{1}{j}[\pi\delta(\omega - \omega_{0})-\pi\delta(\omega + \omega_{0})]}$$

Therefore, the Fourier transform of the given function is,

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

raja
Updated on 02-Dec-2021 12:14:11

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