Fourier Transform of Complex and Real Functions

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Transform

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

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

And the inverse Fourier transform is defined as,

$$\mathrm{x\left ( t \right )=\frac{1}{2\pi }\int_{-\infty }^{\infty}X\left ( \omega \right )e^{j\omega t}d\omega}$$

Fourier Transform of Complex Functions

Consider a complex function 𝑥(𝑡) that is represented as −

$$\mathrm{x\left ( t \right )=x_{r}\left ( t \right )+jx_{i}\left ( t \right )}$$

Where, 𝑥𝑟 (𝑡) and 𝑥𝑖 (𝑡) are the real and imaginary parts of the function respectively.

Now, the Fourier transform of function 𝑥(𝑡) is given by,

$$\mathrm{F\left [ x\left ( t \right ) \right ]=X\left ( \omega \right )=\int_{-\infty }^{\infty}x\left ( t \right )e^{-j\omega t}dt=\int_{-\infty}^{\infty}\left [ x_{r}\left ( t \right )+jx_{i}\left ( t \right ) \right ]e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow X\left ( \omega \right )=\int_{-\infty}^{\infty}\left [ x_{r}\left ( t \right )+jx_{i}\left ( t \right ) \right ]\left [ \cos \omega t-j\sin \omega t \right ]dt}$$

$$\mathrm{\Rightarrow X\left ( \omega \right )=\int_{-\infty}^{\infty}\left [ x_{r}\left ( t \right )\cos \omega t+x_{i}\left ( t \right )\sin \omega t \right ]dt+j\int_{-\infty}^{\infty}\left [ x_{i}\left ( t \right )\cos \omega t-x_{r}\left ( t \right )\sin \omega t \right ]dt}$$

Therefore, the Fourier transform of complex function is,

$$\mathrm{X(\omega )=X_{r}\left ( \omega \right )+jX_{i}\left ( \omega \right )}$$

Where,

$$\mathrm{X_{r}(\omega )=\int_{-\infty}^{\infty}\left [ x_{r}\left ( t \right )\cos \omega t+x_{i}\left ( t \right )\sin \omega t \right ]dt}$$

And

$$\mathrm{X_{i}(\omega )=\int_{-\infty}^{\infty}\left [ x_{i}\left ( t \right )\cos \omega t-x_{r}\left ( t \right )\sin \omega t \right ]dt} $$

Inverse Fourier Transform of Complex Functions

From the definition of inverse Fourier transform, we have,

$$\mathrm{x\left ( t \right )=\frac{1}{2\pi }\int_{-\infty }^{\infty}X\left ( \omega \right )e^{j\omega t}d\omega}$$

$$\mathrm{=\frac{1}{2\pi }\int_{-\infty }^{\infty}\left [ X_{r}\left ( \omega \right )+jX_{i}\left ( \omega \right ) \right ]\left [ \cos \omega t+j\sin \omega t \right ]d\omega} $$

$$\mathrm{\Rightarrow x\left ( t \right )=\frac{1}{2\pi }\int_{-\infty }^{\infty}\left [ X_{r}\left ( \omega \right )\cos \omega t-X_{i}\left ( \omega \right )sin \omega t \right ]d\omega+j\frac{1}{2\pi }\int_{-\infty }^{\infty}\left [ X_{r}\left ( \omega \right )\sin \omega t+X_{i}\left ( \omega \right )cos \omega t \right ]d\omega}$$

Therefore,

$$\mathrm{x\left ( t \right )=x_{r}\left ( t \right )+jx_{i}(t)}$$

Where,

$$\mathrm{x_{r}\left ( t \right )=\frac{1}{2\pi }\int_{-\infty }^{\infty}\left [ X_{r}\left ( \omega \right )\cos \omega t-X_{i}\left ( \omega \right )sin \omega t \right ]d\omega}$$

And

$$\mathrm{ x_{i}\left ( t \right )=\frac{1}{2\pi }\int_{-\infty }^{\infty}\left [ X_{r}\left ( \omega \right )\sin \omega t+X_{i}\left ( \omega \right )cos \omega t \right ]d\omega}$$

Fourier Transform of Real Functions

Case I – When 𝑥(𝑡) is a real function,

$$\mathrm{x_{i}\left ( t \right )=0\; \; and\; \; X\left ( -\omega \right )=X^{\ast }\left ( \omega \right )}$$

Hence, the Fourier transform of the real and imaginary parts of the function is,

$$\mathrm{X_{r}\left ( \omega \right )=\int_{-\infty }^{\infty}x\left ( t \right )\cos \omega t\; dt} $$

$$\mathrm{X_{i}\left ( \omega \right )=-\int_{-\infty }^{\infty}x\left ( t \right )\sin \omega t\; dt}$$

$$\mathrm{\therefore X\left ( \omega \right )=\int_{-\infty }^{\infty}x\left ( t \right )\cos \omega t\; dt-j\int_{-\infty }^{\infty}x\left ( t \right )\sin \omega t\; dt}$$

Case II – When 𝑥(𝑡) is real and even,

$$\mathrm{X_{r}\left ( \omega \right )=\int_{-\infty }^{\infty}x\left ( t \right )\cos \omega t\; dt=2\int_{0 }^{\infty}x\left ( t \right )\cos \omega t\; dt} $$

$$\mathrm{X_{i}\left ( \omega \right )=0}$$

$$\mathrm{\therefore X\left ( \omega \right )=2\int_{0 }^{\infty}x\left ( t \right )\cos \omega t\; dt}$$

Case III – When 𝑥(𝑡) is real and odd,

$$\mathrm{X_{r}\left ( \omega \right )=0}$$

$$\mathrm{X_{i}\left ( \omega \right )=jX\left ( \omega \right )=-j\int_{-\infty }^{\infty}x\left ( t \right )\sin \omega t\; dt}$$

$$\mathrm{\Rightarrow X_{i}\left ( \omega \right )=-j2\int_{0 }^{\infty}x\left ( t \right )\sin \omega t\; dt}$$

$$\mathrm{\therefore X\left ( \omega \right )=-j2\int_{0 }^{\infty}x\left ( t \right )\sin \omega t\; dt}$$

If 𝑥𝑒 (𝑡) and 𝑥𝑜 (𝑡) are the even and odd parts of the function 𝑥(𝑡), then for a non-symmetric function, we have,

$$\mathrm{F\left [ x\left ( t \right ) \right ]=X\left ( \omega \right )=X_{r}\left ( \omega \right )+jX_{i}\left ( \omega \right )} $$

$$\mathrm{\Rightarrow X\left ( \omega \right )=\int_{-\infty }^{\infty}x_{e}\left ( t \right )\cos \omega t\; dt-j\int_{-\infty }^{\infty}x_{0}\left ( t \right )\sin \omega t\; dt=X_{e}\left ( \omega \right )+X_{0}\left ( \omega \right )}$$

raja
Published on 15-Dec-2021 11:30:45
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