Time Convolution and Frequency Convolution Properties of Discrete-Time Fourier Transform

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Discrete-Time Fourier Transform (DTFT)

The Fourier transform of a discrete-time sequence is known as the discrete-time Fourier transform (DTFT). Mathematically, the DTFT of a discrete-time sequence x(n) is defined as:

$$\mathrm{F[x(n)] \:=\: X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j \omega n}}$$

Time Convolution Property of DTFT

Statement

The time convolution property of DTFT states that the discrete-time Fourier transform of the convolution of two sequences in the time domain is equivalent to the multiplication of their discrete-time Fourier transforms. Therefore, if:

$$\mathrm{x_1(n)\:\overset{FT}\longleftrightarrow\: X_1(\omega) \quad \text{and} \quad x_2(n) \:\overset{FT}\longleftrightarrow\: X_2(\omega)}$$

Then,

$$\mathrm{F[x_1(n)\:\cdot\:x_2(n)]\:=\:X_1(\omega)\:\cdot\: X_2(\omega)}$$

Proof

From the definition of the DTFT, we have:

$$\mathrm{F[x(n)] \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j \omega n}}$$

$$\mathrm{\therefore\:F[x_1(n)\:\cdot\:x_2(n)]\:=\: \sum_{n=-\infty}^{\infty}\: \left[ x_1(n)\:\cdot\:x_2(n) \right] e^{-j \omega n}}$$

But, the convolution of two sequences is defined as:

$$\mathrm{x_1(n)\:\cdot\:x_2(n)\:=\: \sum_{k=-\infty}^{\infty} \:x_1(k)\: x_2(n \:-\: k)}$$

$$\mathrm{\therefore\:F[x_1(n)\:\cdot\:x_2(n)] \:=\: \sum_{n=-\infty}^{\infty}\: \left[ \sum_{k=-\infty}^{\infty}\: x_1(k)\: x_2(n \:-\: k) \right] e^{-j \omega n}}$$

By interchanging the order of summations, we get:

$$\mathrm{F[x_1(n)\:\cdot\:x_2(n)]\:=\: \sum_{k=-\infty}^{\infty} \:x_1(k) \:\sum_{n=-\infty}^{\infty}\: x_2(n \:-\: k) e^{-j \omega n}}$$

Substituting (n - k) = m and n = (m + k) in the second summation, we have:

$$\mathrm{F[x_1(n)\:\cdot\:x_2(n)] \:=\: \sum_{k=-\infty}^{\infty}\: x_1(k)\: \sum_{m=-\infty}^{\infty}\: x_2(m)\: e^{-j \omega (m\:+\:k)}}$$

$$\mathrm{\Rightarrow\:F[x_1(n)\:\cdot\:x_2(n)] \:=\: \sum_{k=-\infty}^{\infty}\:x_1(k)\:e^{-j \omega k}\: \sum_{m=-\infty}^{\infty}\: x_2(m)\: e^{-j \omega m}}$$

$$\mathrm{\therefore\:F[x_1(n)\:\cdot\:x_2(n)] \:=\: X_1(\omega)\: X_2(\omega)}$$

Therefore, the convolution of sequences in the time domain is equal to the product of their spectra in the frequency domain.

Frequency Convolution Property of DTFT

Statement

The frequency convolution property of DTFT states that the discrete-time Fourier transform of the multiplication of two sequences in the time domain is equivalent to the convolution of their spectra in the frequency domain. Therefore, if:

$$\mathrm{x_1(n)\:\overset{FT}\longleftrightarrow\: X_1(\omega) \quad \text{and} \quad x_2(n)\:\overset{FT}\longleftrightarrow\: X_2(\omega)}$$

Then,

$$\mathrm{F[x_1(n) x_2(n)] \:=\: X_1(\omega)\:\cdot\: X_2(\omega)}$$

Proof

From the definition of the DTFT, we have:

$$\mathrm{F[x(n)] \:=\: X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: e^{-j \omega n}}$$

$$\mathrm{F[x_1(n) x_2(n)] \:=\: \sum_{n=-\infty}^{\infty} \:\left[ x_1(n) x_2(n) \right] e^{-j \omega n}}$$

But, from the definition of the inverse DTFT, we have:

$$\mathrm{x_1(n) \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: X_1(\theta)\: e^{j \theta n}\: d\theta}$$

$$\mathrm{F[x_1(n) x_2(n)]\:=\: \sum_{n=-\infty}^{\infty} \:\left[ \frac{1}{2\pi} \int_{-\pi}^{\pi}\: X_1(\theta)\: e^{j \theta n} \:d\theta \right]\: e^{-j \omega n}\: x_2(n)}$$

Now, by interchanging the order of summation and integration, we get:

$$\mathrm{F[x_1(n) x_2(n)] \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi} \:X_1(\theta) \left[ \sum_{n=-\infty}^{\infty}\: x_2(n) \:e^{-j (\omega\:-\:\theta) n} \right]\: d\theta}$$

$$\mathrm{F[x_1(n) x_2(n)] \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: X_1(\theta)\: X_2(\omega \:-\: \theta)\: d\theta}$$

$$\mathrm{F[x_1(n)x_2(n)]\:=\:X_1(\omega)\:\cdot\:X_2(\omega)}$$