Question

The discrete time Fourier series representation of a signal \( x[n] \) with period \( N \) is written as \[ x[n] = \sum_{k=0}^{N-1} a_k e^{j(2\pi kn / N)} \] A discrete time periodic signal with period \( N = 3 \), has the non-zero Fourier series coefficients: \( a_{-3} = 2 \) and \( a_4 = 1 \). The signal is

• A. ( 2 + 2e^{j(2\pi / 6)n} \cos\left( \frac{2\pi}{6} n \right) \)
• B. ( 1 + 2e^{j(2\pi / 6)n} \cos\left( \frac{2\pi}{6} n \right) \)
• C. ( 1 + 2e^{j(2\pi / 3)n} \cos\left( \frac{2\pi}{6} n \right) \)
• D. ( 2 + 2e^{j(2\pi / 6)n} \cos\left( \frac{2\pi}{6} n \right) \)

Hint:

Fourier series representation expresses a periodic signal as a sum of sinusoids with complex exponentials. The coefficients determine the amplitude and phase of each frequency component.

Answer 1

The Correct Option is B

Step 1: Fourier series representation.
The Fourier series representation for a periodic signal \( x[n] \) with period \( N \) involves the coefficients \( a_k \), where each coefficient represents a specific frequency component of the signal. Given that \( N = 3 \), we are working with a periodic signal of period 3, and the given non-zero Fourier series coefficients are \( a_{-3} = 2 \) and \( a_4 = 1 \). Step 2: Constructing the signal.
Using the formula for Fourier series representation: \[ x[n] = a_0 + a_1 e^{j(2\pi n / 3)} + a_2 e^{j(4\pi n / 3)} + a_3 e^{j(6\pi n / 3)} \] Considering the coefficients \( a_{-3} \) and \( a_4 \), we can write the expression for the signal \( x[n] \) in terms of cosine and exponential terms, leading to the correct form as \( 1 + 2 e^{j(2\pi / 6)n} \cos\left( \frac{2\pi}{6}n \right) \), which matches option (B). Step 3: Conclusion.
The correct answer is (B) because it matches the given Fourier series coefficients and the derived signal.