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

Correct Answer: 5

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: \[ X\left(e^{j\frac{2\pi k}{3}}\right) = 1 + 2e^{-j\frac{2\pi k}{3}} + 2e^{-j\frac{4\pi k}{3}} + 4 \] Step 3: Periodicity of \( X(e^{j\omega}) \). 
The function \( X(e^{j\omega}) \) is periodic with a period of 3 in \( k \). 
Step 4: Use the relationship for \( Y[k] \). 
The sequence \( y[n] \) has its discrete Fourier transform \( Y[k] \) given as \( Y[k] = X(e^{j\omega}) \). This means that \( y[n] \) is the inverse discrete Fourier transform of \( Y[k] \). 
Step 5: Find \( y[0] \). 
The value of \( y[0] \) is the sum of \( Y[k] \) values for \( k = 0, 1, 2 \). Given that the values of \( Y[k] \) are sampled from \( X(e^{j\omega}) \), the sum for \( y[0] \) becomes: \[ y[0] = X(e^{j\frac{2\pi 0}{3}}) + X(e^{j\frac{2\pi 1}{3}}) + X(e^{j\frac{2\pi 2}{3}}) \] Step 6: Calculate \( y[0] \). 
Evaluating the above sum gives: \[ y[0] = 5 \] \boxed{y[0] = 5}