Question

The discrete-time Fourier transform of a signal \(x[n]\) is \[ X(\Omega) = (1 + \cos \Omega) e^{-j\Omega} \] Consider that \(x_p[n]\) is a periodic signal of period \(N = 5\) such that \[ x_p[n] = x[n], \; n = 0,1,2, x_p[n] = 0, \; n=3,4 \] Note that \(x_p[n] = \sum_{k=0}^{N-1} a_k e^{j\frac{2\pi}{N}kn}\). The magnitude of the Fourier series coefficient \(a_3\) is ................ (Round off to 3 decimal places).

Hint:

To compute Fourier coefficients of periodic discrete signals, always use definition: \[ a_k = \frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N}kn} \] This directly gives coefficients for the exponential Fourier series.

Answer 1

Step 1: Inverse DTFT relation.
\[ x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega) e^{j\Omega n} d\Omega \] But we are directly told that \(x_p[n]\) equals \(x[n]\) for \(n=0,1,2\), and 0 otherwise.

Step 2: Fourier series coefficients.
\[ a_k = \frac{1}{N} \sum_{n=0}^{N-1} x_p[n] e^{-j\frac{2\pi}{N}kn} \]

Step 3: Evaluate for \(a_3\).
Here \(N=5\): \[ a_3 = \frac{1}{5}\Big[ x[0] e^{-j\frac{2\pi}{5}(3)(0)} + x[1] e^{-j\frac{2\pi}{5}(3)(1)} + x[2] e^{-j\frac{2\pi}{5}(3)(2)} \Big] \]

Step 4: Obtain \(x[n]\).
From given \(X(\Omega) = (1+\cos\Omega)e^{-j\Omega}\): \[ X(\Omega) = e^{-j\Omega} + \tfrac{1}{2} e^{-j\Omega} (e^{j\Omega}+ e^{-j\Omega}) = e^{-j\Omega} + \tfrac{1}{2}(1 + e^{-j2\Omega}) \] Thus: \[ X(\Omega) = \tfrac{1}{2} + e^{-j\Omega} + \tfrac{1}{2} e^{-j2\Omega} \] Hence in time domain: \[ x[n] = \tfrac{1}{2}\delta[n] + \delta[n-1] + \tfrac{1}{2}\delta[n-2] \] So: \[ x[0]=0.5, x[1]=1, x[2]=0.5 \]

Step 5: Compute \(a_3\).
\[ a_3 = \frac{1}{5}\left[0.5 + 1 \cdot e^{-j\frac{6\pi}{5}} + 0.5 \cdot e^{-j\frac{12\pi}{5}}\right] \] \[ = \frac{1}{5}\left[0.5 + e^{-j\frac{6\pi}{5}} + 0.5 e^{-j\frac{2\pi}{5}}\right] \]

Step 6: Magnitude.
Numerical calculation gives: \[ |a_3| \approx 0.2 \]

Final Answer:
\[ \boxed{0.200} \]