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- Laplace Transform
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- Z Transform
- Z-Transforms (ZT)
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- What is Inverse Z Transform?
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- Convolution Property of Z Transform
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- Discrete Fourier Transform
- Discrete-Time Fourier Transform
- Properties of Discrete Time Fourier Transform
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- Time Shifting and Frequency Shifting Properties of Discrete Time Fourier Transform
- Inverse Discrete-Time Fourier Transform
- Time Convolution and Frequency Convolution Properties of Discrete-Time Fourier Transform
- Differentiation in Frequency Domain Property of Discrete Time Fourier Transform
- Parseval’s Power Theorem
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- What is Mean Square Error?
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- Region of Convergence
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- Zero Order Hold and its Transfer Function
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- What is the Frequency Response of Discrete Time Systems?
- Basic Elements to Construct the Block Diagram of Continuous Time Systems
- BIBO Stability Criterion
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- Rayleigh’s Energy Theorem
Linearity and Frequency Shifting Property of Fourier Transform
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,
$$\mathrm{cos\:\omega_{0} t \:=\: \left [\frac{e^{j\omega_{0} t} \:+\: e^{-j\omega_{0} t}}{2} \right ]}$$
$$\mathrm{\therefore\:x(t) \:=\: cos\:\omega_{0} t\:u(t) \:=\: \left [\frac{e^{j\omega_{0} t} \:+\: e^{-j\omega_{0} t}}{2}\:\cdot\:u(t) \right ]}$$
Now, the Fourier transform of $x(t)$ is,
$$\mathrm{F[x(t)] \:=\: F[cos\:\omega_{0} t\:u(t)]\:=\:F\left [\frac{e^{j\omega_{0} t}\:+ \:e^{-j\omega_{0} t}}{2}\: \cdot \: u(t) \right ]}$$
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,
$$\mathrm{F[cos\:\omega_{0} t\:u(t)] \:=\: \left [\frac{\pi}{2}\delta (\omega \:-\: \omega_{0}) \:+\: \frac{\pi}{2}\delta (\omega \:+\: \omega_{0}) \:+\: \frac{j\omega}{\omega_{0}^{2} \:+ \:(j\omega)^{2}}\right ]}$$