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



Discrete-Time Fourier Transform

The Fourier transform of a discrete-time sequence is known as the discrete-time Fourier transform (DTFT).

Mathematically, the discrete-time Fourier transform (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 Shifting Property of Discrete-Time Fourier Transform

Statement

The time-shifting property of discrete-time Fourier transform states that if a signal x(n) is shifted by k in time domain, then its DTFT is multiplied by $\mathrm{e^{-j\omega\:k}}$. Therefore, if

$$\mathrm{x(n)\:\overset{FT}\longleftrightarrow\:X(\omega)}$$

Then

$$\mathrm{x(n\:-\:k)\:\overset{FT}\longleftrightarrow\:e^{-j\omega\:k}\:X(\omega)}$$

Where, k is an integer.

Proof

From the definition of discrete-time Fourier transform, we have,

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

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

Substituting (n − k) = m then n = (m + k) in the above summation, we get,

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

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

$$\mathrm{\therefore\: F[x(n\:-\:k)] \:=\: e^{-j \omega k}\: X(\omega) \quad \text{(Time delay)}}$$

Similarly

$$\mathrm{F[x(n\:+\:k)]\:=\:e^{j \omega k} X(\omega); \quad \text{(Time advance)}}$$

Frequency Shifting Property of Discrete-Time Fourier Transform

Statement

The frequency shifting property of DTFT states that the multiplication of a discrete-time sequence by $\mathrm{e^{j \omega_0 k}}$ in time domain corresponds to the shift by $\mathrm{\omega_{0}}$ in the frequency domain. Therefore, if

$$\mathrm{x(n)\:\overset{FT}\longleftrightarrow\:X(\omega)}$$

Then

$$\mathrm{e^{j \omega_0 n}\: x(n)\:\overset{FT}\longleftrightarrow\: X(\omega \:-\: \omega_0)}$$

Proof

From the definition of DTFT, we have,

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

$$\mathrm{\therefore\: F[e^{j \omega_0 n}\: x(n)] \:=\: \sum_{n=-\infty}^{\infty}\: e^{j \omega_0 n}\:x(n) e^{-j \omega n}}$$

$$\mathrm{\Rightarrow\: F[e^{j \omega_0 n} x(n)] \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j (\omega - \omega_0) n}}$$

$$\mathrm{\therefore\: F[e^{j \omega_0 n} x(n)] \:=\: X(\omega \:-\: \omega_0)}$$

The time shifting property and frequency shifting property of DTFT are dual of each other.

Numerical Example 1

Using the time shifting property of DTFT, find the DTFT of sequence

$$\mathrm{x(n)\:=\:u(n\:−\:1)\:+\:u(n\:+\:2)}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n)\:=\:u(n\:−\:1)\:+\:u(n\:+\:2)}$$

Since the DTFT of a unit step sequence defined as −

$$\mathrm{F[u(n)]\:=\:\frac{1}{1\:-\:e^{-j\omega}}}$$

Hence, using the time shifting property of DTFT $\mathrm{\left[\text{i.e., }\:x(n\:-\:k)\:\overset{FT}\longleftrightarrow\:e^{-j\omega k}\:X(\omega) \right]}$, we get,

$$\mathrm{F[u(n\:-\:1)] \:=\: e^{-j \omega} F[u(n)] \:=\: \frac{e^{-j \omega}}{1 \:-\: e^{-j \omega}}}$$

And

$$\mathrm{F[u(n\:+\:2)] \:=\: e^{j 2 \omega} F[u(n)] \:=\: \frac{e^{j 2 \omega}}{1 \:-\: e^{-j \omega}}}$$

$$\mathrm{\therefore\: F[u(n\:-\:1) \:+\: u(n\:+\:2)] \:=\: \frac{e^{-j \omega}}{1 \:-\: e^{-j \omega}} \:+\: \frac{e^{j 2 \omega}}{1 \:-\: e^{-j \omega}}}$$

Numerical Example 2

Using the frequency shifting property of DTFT, find the DTFT of the sequence

$$\mathrm{x(n) \:=\: e^{j 2 \omega}\: u(n)}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n) \:=\: e^{j 2 \omega}\: u(n)}$$

Since the DTFT of a unit step sequence is given by,

$$\mathrm{F[u(n)]\:=\:\frac{1}{1\:-\:e^{-j\omega}}}$$

Now, using the frequency shifting property of DTFT $\mathrm{\left[\text{i.e., }\:e^{j\omega_0\:n}\:x(n)\:\overset{FT}\longleftrightarrow\:X(\omega\:-\:\omega_0) \right]}$, we get,

$$\mathrm{F[e^{j 2 \omega} u(n)] \:=\: \left[\frac{1}{1 \:-\: e^{-j \omega}}\right]_{\omega \:=\: (\omega \:-\: 2)}}$$

$$\mathrm{\therefore\: F[e^{j 2 \omega} u(n)] \:=\: \frac{1}{1 \:-\: e^{-j (\omega \:-\: 2)}}}$$

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