The discrete-time Fourier series representation of a signal x[n] with period N is written as $\mathrm{x}[\mathrm{n}]=\sum _{\mathrm{k}=0}^{\mathrm{N}-1}{\mathrm{a}}_{\mathrm{k}}{\mathrm{e}}^{\mathrm{j}(2\mathrm{k}\mathrm{n}\pi /\mathrm{N})}$. A discrete-time periodic signal with period N = 3, has the non-zero Fourier series coefficients: a_-3 = 2 and a₄ = 1. The signal is

The correct option is:
$1+2{\mathrm{e}}^{\left(\mathrm{j}\frac{2\pi }{6}\mathrm{n}\right)}\cos \left(\frac{2\pi }{6}\mathrm{n}\right)$

Step-by-Step Derivation and Explanation:

1. Periodic Property of Fourier Series Coefficients:
For a discrete-time periodic signal with fundamental period N, the Fourier series coefficients _ak are also periodic with the same period N.
This periodicity is expressed as:
${\mathrm{a}}_{\mathrm{k}}={\mathrm{a}}_{\mathrm{k}+\mathrm{m}\mathrm{N}}$
for any integer m.

Given that the period is N = 3, we can determine the coefficients in the fundamental period (for k = 0, 1, 2) from the given non-zero coefficients _a-3 = 2 and _a4 = 1:
• For k = 0: Since _a0 = _{a-3 + 3} = _a-3, we have:
${\mathrm{a}}_0=2$
• For k = 1: Since _a1 = _{a4 - 3} = _a4, we have:
${\mathrm{a}}_1=1$
• For k = 2: Since there are no other non-zero coefficients specified, the remaining coefficient in the period is:
${\mathrm{a}}_2=0$

2. Reconstructing the Signal:
The synthesis equation for the discrete-time Fourier series is:
$\mathrm{x}[\mathrm{n}]=\sum _{\mathrm{k}=0}^{\mathrm{N}-1}{\mathrm{a}}_{\mathrm{k}}{\mathrm{e}}^{\mathrm{j}(2\mathrm{k}\mathrm{n}\pi /\mathrm{N})}$

Substituting N = 3 and the coefficients _a0 = 2, _a1 = 1, and _a2 = 0, we get:
$\mathrm{x}[\mathrm{n}]=2{\mathrm{e}}^{\mathrm{j}(0)}+1{\mathrm{e}}^{\mathrm{j}(2\pi /3)\mathrm{n}}+0$
Simplifying this:
$\mathrm{x}[\mathrm{n}]=2+{\mathrm{e}}^{\mathrm{j}\frac{2\pi }{3}\mathrm{n}}$

3. Verification of the Correct Option:
Let us simplify the correct option expression using Euler's formula to show it is equivalent to the signal x[n]:
The option is:
$1+2{\mathrm{e}}^{\mathrm{j}\frac{2\pi }{6}\mathrm{n}}\cos \left(\frac{2\pi }{6}\mathrm{n}\right)$

Simplifying the fraction inside the exponential and cosine terms:
$\frac{2\pi }{6}=\frac{\pi }{3}$

Now rewrite the expression:
$1+2{\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}\cos (\frac{\pi }{3}\mathrm{n})$

Using Euler's identity for the cosine term:
$\cos \left(\frac{\pi }{3}\mathrm{n}\right)=\frac{{\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}+{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{3}\mathrm{n}}}{2}$

Substitute this back into the expression:
$1+2{\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}\left(\frac{{\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}+{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{3}\mathrm{n}}}{2}\right)$
Multiply the factors:
$=1+{\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}({\mathrm{e}}^{\mathrm{j}\frac{\pi }{3}\mathrm{n}}+{\mathrm{e}}^{-\mathrm{j}\frac{\pi }{3}\mathrm{n}})$
Distribute the exponential term:
$=1+{\mathrm{e}}^{\mathrm{j}\frac{2\pi }{3}\mathrm{n}}+{\mathrm{e}}^0$
Since any base to the power of 0 is 1:
$=1+{\mathrm{e}}^{\mathrm{j}\frac{2\pi }{3}\mathrm{n}}+1$
$=2+{\mathrm{e}}^{\mathrm{j}\frac{2\pi }{3}\mathrm{n}}$

This matches our reconstructed signal x[n] perfectly, confirming the correctness of the option.