What is the inverse discrete-time Fourier transform of the frequency domain representation shown in the figure?

F4 Madhuri Engineering 10.11.2022 D1

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  1. \(\rm x[n]=\frac{\pi}{2}\sin (\Omega_1n)\)
  2. \(\rm x[n]=\frac{3}{2\pi}\sin (\Omega_1n)\)
  3. \(\rm x[n]=\frac{1}{\pi}\sin (\Omega_1n)\)
  4. \(\rm x[n]=\frac{1}{2\pi}\sin (\Omega_1n)\)

Answer (Detailed Solution Below)

Option 4 : \(\rm x[n]=\frac{1}{2\pi}\sin (\Omega_1n)\)
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Detailed Solution

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Concept:

The inverse discrete-time Fourier transform is given by:

\(x[n]={1\over N}{\sum_{-π}^{π}}x(\Omega)e^{-j\Omega n}\)

where, N = Time period of one complete cycle

Calculation:

F4 Madhuri Engineering 10.11.2022 D1

In the given figure, -π to +π makes one complete cycle. So, the time period is 2π.

\(x[n]={1\over 2\pi}({-j\over 2}e^{j\Omega_1 n}+{j\over 2}e^{-j\Omega_1 n})\)

\(x[n]={1\over 2\pi}\times {-j\over 2}(e^{j\Omega_1 n}-e^{-j\Omega_1 n})\)

Multiplying the numerator and denominator by 2j, we get:

\(x[n]={2j\over 2\pi}\times {-j\over 2}({e^{j\Omega_1 n}-e^{-j\Omega_1 n}\over 2j})\)

\(\rm x[n]=\frac{1}{2\pi}\sin (\Omega_1n)\)

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