Question

The discrete time Fourier transform of the signal, \( x(n) = 0.5^{(n-1)}u(n-1) \) is ________.

• A. \( \frac{e^{-j\omega}}{1 - 0.5e^{-j\omega}} \)
• B. \( e^{-j\omega}(1 - 0.5e^{j\omega}) \)
• C. \( \frac{0.5e^{-j\omega}}{1 - 0.5e^{-j\omega}} \)
• D. \( \frac{0.5e^{j\omega}}{1 - 0.5e^{j\omega}} \)

Hint:

Use DTFT of \( a^n u(n) \) and apply time-shifting property: a shift by \( n_0 \) introduces a factor of \( e^{-j\omega n_0} \).

Answer 1

The Correct Option is A

The given signal is: \[ x(n) = 0.5^{n-1} \cdot u(n-1) \] Let us make substitution: \( m = n - 1 \Rightarrow n = m + 1 \). Then: \[ x(n) = 0.5^{n-1}u(n-1) = 0.5^m u(m) \Rightarrow x(m+1) = 0.5^m u(m) \] The DTFT of \( x(n) = a^n u(n) \) is \( \frac{1}{1 - ae^{-j\omega}} \). So for \( 0.5^m u(m) \), we get: \[ X(\omega) = \frac{1}{1 - 0.5e^{-j\omega}} \] Due to time-shifting by 1, DTFT becomes: \[ X(\omega) = e^{-j\omega} \cdot \frac{1}{1 - 0.5e^{-j\omega}} = \frac{e^{-j\omega}}{1 - 0.5e^{-j\omega}} \]