Inverse Discrete-Time Fourier Transform

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The inverse discrete-time Fourier transform (IDTFT) is the process of finding the discrete-time sequence x(n) from its frequency response $\mathrm{X(\omega)}$.

Mathematically, the inverse discrete-time Fourier transform is defined as -

$$\mathrm{x(n) \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: X(\omega) \:e^{j\omega n}\: d\omega\:\:\dotso\: (1)}$$

The solution of the equation (1) for x(n) is useful for analytical purposes, but it is very difficult to evaluate for typical functional forms of $\mathrm{X(\omega)}$. Therefore, an alternate method of determining the values of the discrete-time sequence x(n) follows directly from the definition of the Fourier transform, i.e.,

$$\mathrm{X(\omega) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) e^{-j\omega n} \:=\: \dotso \:+\: x(-3) e^{j3\omega} \:+\: x(-2) e^{j2\omega} \:+\: x(-1) e^{j\omega} \:+\: x(0) \:+\: x(1) e^{-j\omega} \:+\: x(2) e^{-j2\omega} \:+\: x(3) e^{-j3\omega}\:\: \dots\: (2)}$$

Hence, from the equation of $\mathrm{X(\omega)}$, we can say that if $\mathrm{X(\omega)}$ can be expressed as a series of complex exponentials as given in the equation (2), then x(n) is simply the coefficient of $\mathrm{e^{-j\omega n}}$.

Numerical Example 1

Find the inverse discrete-time Fourier transform of $\mathrm{X(\omega)}$.

$$\mathrm{X(\omega) \:=\: e^{-j\omega} \quad \text{for} \quad -\pi \:\leq\: \omega \:\leq\: \pi}$$

Solution

The given Fourier transform is,

$$\mathrm{X(\omega) \:=\: e^{-j\omega} }$$

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

$$\mathrm{F^{-1}[X(\omega)] \:=\: x(n) \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: X(\omega) e^{j\omega n}\: d\omega}$$

$$\mathrm{\therefore\:x(n) \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: e^{-j\omega}\: e^{j\omega n} \:d\omega \:=\: \frac{1}{2\pi} \int_{-\pi}^{\pi}\: e^{j\omega (n-1)}\: d\omega}$$

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

$$\mathrm{\Rightarrow\:x(n) \:=\:\frac{1}{\pi(n\:-\:1)}\left[\frac{e^{j\pi(n-1)}\:-\:e^{-j\pi(n-1)}}{2j}\right] \:=\: \frac{\sin(\pi(n\:-\:1))}{\pi(n\:-\:1)}}$$

$$\mathrm{\therefore\:F^{-1}[X(\omega)] \:=\: x(n) \:=\: \frac{\sin(\pi(n\:-\:1))}{\pi(n\:-\:1)}}$$

Numerical Example 2

Determine the function $\mathrm{x(n)}$, i.e., the inverse discrete-time Fourier transform of $\mathrm{X(\omega)}$.

$$\mathrm{X(\omega) \:=\: 4 \:+\: e^{-j\omega} \:+\: 2e^{-j2\omega} \:+\: 3e^{-j4\omega}}$$

Solution

The given Fourier transform is,

$$\mathrm{X(\omega) \:=\: 4 \:+\: e^{-j\omega} \:+\: 2e^{-j2\omega} \:+\: 3e^{-j4\omega} \:\:\dotso\:(3)}$$

Now, from the definition of the discrete-time Fourier transform, we have,

$$\mathrm{X(\omega) \:=\: \sum_{n=-\infty}^{\infty} \:x(n) e^{-j\omega n} \:=\: \dotso \:+\: x(-3)\: e^{j3\omega} \:+\: x(-2)\: e^{j2\omega} \:+\: x(-1) e^{j\omega} \:+\: x(0) \:+\: x(1) e^{-j\omega} \:+\: x(2) e^{-j2\omega} \:+\: x(3) e^{-j3\omega} \:+\: x(4) e^{-j4\omega}\: \dotso\:(4)}$$

Comparing equations (3) and (4), we get,

$$\mathrm{x(0) \:=\: 4, \quad x(1) \:=\: 1, \quad x(2) \:=\: 2, \quad x(3) \:=\: 0, \quad x(4) \:=\: 4}$$

$$\mathrm{\therefore\:x(n) \:=\: \{4,\: 1,\: 2,\: 0,\: 4\}}$$