DTFS Basics and Given Information

The question asks for the representation of a discrete-time periodic signal $x[n]$ using its Discrete-Time Fourier Series (DTFS). The general formula for the DTFS representation is given as:

x[n] = \sum_{k = 0}^{N - 1} a_k e^{j \frac{2k n \pi}{N}}

Key information provided:

• The period of the signal is N = 3.
• The non-zero Fourier series coefficients are a_{-3} = 2 and a_4 = 1.

From the period N = 3, we know the fundamental frequency is $\omega_0 = \frac{2\pi}{N} = \frac{2\pi}{3}$.

Periodic Coefficients Mapping

A crucial property of DTFS coefficients is their periodicity with the same period as the signal. This means a_k = a_{k + mN} for any integer m.

We need to find the equivalent coefficients within the fundamental range k = 0, 1, ..., N-1 (which is k = 0, 1, 2 for N=3).

• For a_{-3}: Using the periodicity property, a_{-3} = a_{-3 \pmod{3}}. Since -3 is a multiple of 3, -3 mod 3 = 0. Thus, a_{-3} = a_0. Given a_{-3} = 2, we get a_0 = 2.
• For a_4: Using the periodicity property, a_4 = a_{4 \pmod{3}}. Since 4 = 1*3 + 1, 4 mod 3 = 1. Thus, a_4 = a_1. Given a_4 = 1, we get a_1 = 1.

Since only a_{-3} and a_4 are specified as non-zero, we infer that other coefficients within the fundamental period, like a_2, are zero.

Signal Reconstruction

Now we can reconstruct the signal $x[n]$ using the standard DTFS formula with the determined coefficients and fundamental frequency:

x[n] = \sum_{k=0}^{N-1} a_k e^{j k \omega_0 n}

For N=3, this expands to:

x[n] = a_0 e^{j \cdot 0 \cdot n \frac{2\pi}{3}} + a_1 e^{j \cdot 1 \cdot n \frac{2\pi}{3}} + a_2 e^{j \cdot 2 \cdot n \frac{2\pi}{3}}

Substituting the values a_0 = 2, a_1 = 1, and a_2 = 0:

x[n] = 2 \cdot e^{j \cdot 0} + 1 \cdot e^{j \frac{2\pi}{3} n} + 0

x[n] = 2 + e^{j \frac{2\pi}{3} n}

Analyzing the Options

Let's simplify each option using Euler's formula \cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2} to see which one matches our derived signal x[n] = 2 + e^{j \frac{2\pi}{3} n}.

Note that the options use the frequency term $\frac{2\pi}{6} = \frac{\pi}{3}$.

Option 1: \rm 2 + 2e^{- \left( j \frac{2\pi}{6} n \right) }\cos \left( \frac{2 \pi}{6} n\right)

Simplifies to: 2 + 2e^{- j \frac{\pi}{3} n} \left( \frac{e^{ j \frac{\pi}{3} n} + e^{- j \frac{\pi}{3} n}}{2} \right) = 2 + 1 + e^{- j \frac{2\pi}{3} n} = 3 + e^{- j \frac{2\pi}{3} n}. This does not match.

Option 2: \rm 1 + 2e^{ \left( j \frac{2\pi}{6} n \right) }\cos \left( \frac{2 \pi}{6} n\right)

Simplifies to: 1 + 2e^{ j \frac{\pi}{3} n} \left( \frac{e^{ j \frac{\pi}{3} n} + e^{- j \frac{\pi}{3} n}}{2} \right) = 1 + e^{ j \frac{\pi}{3} n} e^{ j \frac{\pi}{3} n} + e^{ j \frac{\pi}{3} n} e^{- j \frac{\pi}{3} n} = 1 + e^{ j \frac{2\pi}{3} n} + 1 = 2 + e^{ j \frac{2\pi}{3} n}. This matches our derived signal.

Option 3: \rm 1 + 2e^{ \left( j \frac{2\pi}{3} n \right) }\cos \left( \frac{2 \pi}{6} n\right)

Simplifies to: 1 + 2e^{ j \frac{2\pi}{3} n} \left( \frac{e^{ j \frac{\pi}{3} n} + e^{- j \frac{\pi}{3} n}}{2} \right) = 1 + e^{ j \frac{2\pi}{3} n} e^{ j \frac{\pi}{3} n} + e^{ j \frac{2\pi}{3} n} e^{- j \frac{\pi}{3} n} = 1 + e^{ j \pi n} + e^{ j \frac{\pi}{3} n}. The frequency components here are incorrect for the DTFS sum with N=3.

Option 4: \rm 2 + 2e^{ \left( j \frac{2\pi}{6} n \right) }\cos \left( \frac{2 \pi}{6} n\right)

Simplifies to: 2 + 2e^{ j \frac{\pi}{3} n} \left( \frac{e^{ j \frac{\pi}{3} n} + e^{- j \frac{\pi}{3} n}}{2} \right) = 2 + e^{ j \frac{\pi}{3} n} e^{ j \frac{\pi}{3} n} + e^{ j \frac{\pi}{3} n} e^{- j \frac{\pi}{3} n} = 2 + e^{ j \frac{2\pi}{3} n} + 1 = 3 + e^{ j \frac{2\pi}{3} n}. This does not match.

Final Conclusion

Option 2, when simplified using Euler's formula, yields the expression 2 + e^{j \frac{2\pi}{3} n}. This corresponds to a DTFS representation with coefficients a_0 = 2 and a_1 = 1, which are consistent with the given non-zero coefficients a_{-3} = 2 and a_4 = 1 for a signal with period N = 3.