Signals & Systems – Conjugation and Autocorrelation Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

Fourier Transform

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

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

Conjugation Property of Fourier Transform

Statement − The conjugation property of Fourier transform states that the conjugate of function x(t) in time domain results in conjugation of its Fourier transform in the frequency domain and ω is replaced by (−ω), i.e., if

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

Then, according to conjugation property of Fourier transform,

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

Proof

From the definition of Fourier transform, we have

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

Taking conjugate on both sides, we get

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

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

Now, by replacing (ω) by (−ω), we obtain,

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

$$\mathrm{\therefore F[x^*(t)]=X^*(-\omega)}$$

Or, it can also be represented as,

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

Autocorrelation Property of Fourier Transform

The autocorrelation of a continuous-time function 𝑥(𝑡) is defined as,

$$\mathrm{R(\tau)=\int_{-\infty}^{\infty}x(t)x^*(t-\tau)dt}$$

Statement − The autocorrelation property of Fourier transform states that the Fourier transform of the autocorrelation of a single in time domain is equal to the square of the modulus of its frequency spectrum. Therefore, if

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

Then, by the autocorrelation property of Fourier transform,

$$\mathrm{R(\tau)\overset{FT}{\leftrightarrow}|X(\omega)|^2}$$

Proof

By the definition of autocorrelation, we have,

$$\mathrm{R(\tau)=\int_{-\infty}^{\infty}x(t)x^*(t-\tau)dt}$$

Then, from the definition of Fourier transform, we get,

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

By interchanging the order of integration, we get,

$$\mathrm{F[R(\tau)]=\int_{-\infty}^{\infty}x(t)[\int_{-\infty}^{\infty}x^*(t-\tau)e^{-j\omega t}d\tau]dt}$$

Substituting [(𝑡 − 𝜏) = 𝑢] in the second integration,

$$\mathrm{\tau=(t-u)\:and\:d\tau=du}$$

$$\mathrm{\therefore F[R(\tau)]=\int_{-\infty}^{\infty}x(t)[\int_{-\infty}^{\infty}x^*(u)e^{-j\omega(t-u)}du]dt}$$

$$\mathrm{ \Rightarrow F[R(\tau)]=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\int_{-\infty}^{\infty}x^*(u)e^{j\omega u}du}$$

$$\mathrm{ \Rightarrow F[R(\tau)]=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt[\int_{-\infty}^{\infty}x(u)e^{-j\omega u}du]^*}$$

Therefore

$$\mathrm{F[R(\tau)]=X(\omega)X^*(\omega)=|X(\omega)|^2}$$

Or, it can also be represented as,

$$\mathrm{R(\tau)\overset{FT}{\leftrightarrow}|X(\omega)|^2}$$

raja
Updated on 03-Dec-2021 13:34:42

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