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DTFS of 1 + 3cos(2πn/N + π/3)

Short answer

Using Euler's formula, x[n] = 1 + 1.5 e^{jπ/3} e^{j(2π/N)n} + 1.5 e^{-jπ/3} e^{-j(2π/N)n}. Coefficients: a_0 = 1, a_1 = 1.5 e^{jπ/3}, a_{-1} = 1.5 e^{-jπ/3}, others zero. Magnitudes: 1, 1.5, 1.5; phases: 0°, 60°, -60°. Spectra repeat with period N.

Obtain fourier series coefficients of the signal x[n] = 1 + 3cos(2πn/N + π/3), where N is the fundamental time period of x[n]. Also, plot magnitude and phase spectrum of the coefficients.

With fundamental period NN, ω0=2π/N\omega_0 = 2\pi/N. Expand the cosine using Euler's formula:

x[n]=1+3cos⁡(2πnN+π3)=1+32ejπ/3ej(2π/N)n+32e−jπ/3e−j(2π/N)n\begin{aligned} x[n] &= 1 + 3\cos\left(\frac{2\pi n}{N} + \frac{\pi}{3}\right) \\ &= 1 + \frac{3}{2}e^{j\pi/3}e^{j(2\pi/N)n} + \frac{3}{2}e^{-j\pi/3}e^{-j(2\pi/N)n} \end{aligned}

Fourier series coefficients

Comparing with x[n]=∑k=⟨N⟩akejk(2π/N)nx[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}:

a0=1,a1=32ejπ/3=0.75+j1.299,a−1=32e−jπ/3=0.75−j1.299a_0 = 1, \qquad a_1 = \frac{3}{2}e^{j\pi/3} = 0.75 + j1.299, \qquad a_{-1} = \frac{3}{2}e^{-j\pi/3} = 0.75 - j1.299

All other coefficients in one period are zero, and ak+N=aka_{k+N} = a_k (so aN−1=a−1a_{N-1} = a_{-1}, aN+1=a1a_{N+1} = a_1).

kkaka_k∣ak∣\lvert a_k\rvert∠ak\angle a_k
−1-11.5e−jπ/31.5e^{-j\pi/3}1.5−60∘-60^\circ
0110∘0^\circ
11.5ejπ/31.5e^{j\pi/3}1.5+60∘+60^\circ

Magnitude spectrum

 |a_k|  (one period shown; repeats every N)
   1.5     |     |
     1     |  |  |
       -o--+--+--+--o--> k
       -2 -1  0  1  2

Phase spectrum

 angle a_k (deg)
    60           |
       -o--+--o--+--o--> k
   -60     |
       -2 -1  0  1  2

The magnitude is even and the phase odd in kk (real signal), and both repeat with period NN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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