Linearity and Frequency Shifting Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Transform

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

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

Linearity Property of Fourier Transform

Statement − The linearity property of Fourier transform states that the Fourier transform of a weighted sum of two signals is equal to the weighted sum of their individual Fourier transforms.

Therefore, if

$$\mathrm{x_{1}(t)\overset{FT}{\leftrightarrow}X_{1}(\omega)\:\:and\:\:x_{2}\overset{FT}{\leftrightarrow}X_{2}(\omega)}$$

Then, according to the linearity property of Fourier transform,

$$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$

Where, a and b are constants.

Proof

From the definition of Fourier transform, we have,

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

$$\mathrm{\Rightarrow\:X(\omega)=F[ax_{1}(t)+bx_{2}(t)]=\int_{−\infty}^{\infty}[ax_{1}(t)+bx_{2}(t)]e^{-j \omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}ax_{1}(t)e^{-j \omega t} dt+\int_{−\infty}^{\infty}bx_{2}(t)e^{-j \omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=a\int_{−\infty}^{\infty}x_{1}(t)e^{-j \omega t} dt+b\int_{−\infty}^{\infty}x_{2}(t)e^{-j \omega t}dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=aX_{1}(\omega)+bX_{2}(\omega)}$$

$$\mathrm{\therefore\:F[ax_{1}(t)+bx_{2}(t)]=aX_{1}(\omega)+bX_{2}(\omega)}$$

Or, it can also be written as,

$$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$

Frequency Shifting Property of Fourier Transform

Statement – Frequency shifting property of Fourier transform states that the multiplication of a time domain signal $x(t)$ by an exponential $(e^{j \omega_{0} t })$ causes the frequency spectrum to be shifted by $\omega_{0}$. Therefore, if

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

Then, according to the frequency shifting property,

$$\mathrm{e^{j \omega_{0} t }\:x(t)\overset{FT}{\leftrightarrow}X(\omega - \omega_{0})}$$

Proof 

From the definition of Fourier transform, we have,

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

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

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

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

Or, it can also be represented as,

$$\mathrm{e^{-j \omega_{0} t}x(t)\overset{FT}{\leftrightarrow}X(\omega-\omega_{0})}$$

Similarly,

$$\mathrm{e^{-j \omega_{0} t}x(t)\overset{FT}{\leftrightarrow}X(\omega + \omega_{0})}$$

Numerical Example

Using linearity and frequency shifting properties of Fourier transform, find the Fourier transform of $[cos\:\omega_{0} t\:u(t)]$.

Solution 

Given,

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

Using Euler’s formula, we can write,

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

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

Now, the Fourier transform of $x(t)$ is,

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

Using linearity property $[i.e., ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)]$, we get,

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

Now, using frequency shifting property $[i.e.,e^{j\omega_{0} t }x(t)\overset{FT}{\leftrightarrow}X(\omega - \omega_{0})]$ of Fourier transform, we get,

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

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

$$\mathrm{\Rightarrow\:F[cos\:\omega_{0} t\:u(t)]=\frac{1}{2}\left [\pi\delta (\omega-\omega_{0}) + \pi\delta (\omega + \omega_{0}) + \frac{2j\omega}{\omega_{0}^{2}-\omega^{2}}\right ]}$$

Therefore, the Fourier transform of the given signal is,

$$\mathrm{F[cos\:\omega_{0} t\:u(t)]=\left [\frac{\pi}{2}\delta (\omega-\omega_{0}) + \frac{\pi}{2}\delta (\omega + \omega_{0}) + \frac{j\omega}{\omega_{0}^{2}+(j\omega)^{2}}\right ]}$$

raja
Published on 02-Dec-2021 12:09:14
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