Question

Let \( X(e^{j\omega}) \) represent the discrete-time Fourier transform of a 4-length sequence \( x[n] \), where \( x[0] = 1 \), \( x[1] = 2 \), \( x[2] = 2 \), and \( x[3] = 4 \). \( X(e^{j\omega}) \) is sampled at \( \omega = \frac{2\pi k}{3} \) to generate a periodic sequence in \( k \) with period 3, where \( k \) represents an integer. Let \( y[n] \) represent another sequence such that its discrete Fourier transform \( Y[k] \) is given as \( Y[k] = X(e^{j\omega}) \) for \( 0 \leq k \leq 2 \). The value of \( y[0] \) is ________ (in integer).

Hint:

To calculate the value of \( y[0] \), sum the sampled values of \( X(e^{j\omega}) \) at the appropriate points based on the periodicity and the given conditions.

Answer 1

Step 1: Find \( X(e^{j\omega}) \) using the inverse discrete Fourier transform (IDFT).
The sequence \( x[n] \) is given by: 
\[ x[0] = 1, \quad x[1] = 2, \quad x[2] = 2, \quad x[3] = 4 \] 
The discrete-time Fourier transform \( X(e^{j\omega}) \) of \( x[n] \) is: 
\[ X(e^{j\omega}) = \sum_{n=0}^{3} x[n] e^{-j\omega n} \] 
Substituting the values of \( x[n] \), we get: 
\[ X(e^{j\omega}) = 1 + 2e^{-j\omega} + 2e^{-j2\omega} + 4e^{-j3\omega} \] 
Step 2: Sample \( X(e^{j\omega}) \) at \( \omega = \frac{2\pi k}{3} \).
Sampling the above \( X(e^{j\omega}) \) at \( \omega = \frac{2\pi k}{3} \) gives: 
\[ X\left(e^{j\frac{2\pi k}{3}}\right) = 1 + 2e^{-j\frac{2\pi k}{3}} + 2e^{-j\frac{4\pi k}{3}} + 4e^{-j2\pi k} \] 
Since \( e^{-j2\pi k} = 1 \), we simplify: 
\[ X\left(e^{j\frac{2\pi k}{3}}\right) = 1 + 2e^{-j\frac{2\pi k}{3}} + 2e^{-j\frac{4\pi k}{3}} + 4 \]