Exponential Fourier series of A sin ω₀n

The signal is read as the continuous-time sinusoid $x(t) = A\sin\omega_0 t$ (the question writes $x[n]$, but asks for the continuous-time series). The derivation of the general coefficient formula comes first, then the formula is applied.

Exponential Fourier series and its coefficients

A periodic signal with period $T$ and $\omega_0 = 2\pi/T$ can be written as a sum of harmonically related complex exponentials:

$$x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}$$

To find $a_n$, multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period:

$$\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty} a_k \int_0^T e^{j(k-n)\omega_0 t}dt$$

Orthogonality of the exponentials gives

$$\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}$$

so only the $k = n$ term remains, and

$$a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$$

Applying it to x(t) = A sin ω₀t

Using Euler's identity:

$$x(t) = A\,\frac{e^{j\omega_0 t} - e^{-j\omega_0 t}}{2j} = \frac{A}{2j}e^{j\omega_0 t} - \frac{A}{2j}e^{-j\omega_0 t}$$

Comparing with $\sum a_k e^{jk\omega_0 t}$:

$$a_1 = \frac{A}{2j} = -j\frac{A}{2}, \qquad a_{-1} = -\frac{A}{2j} = j\frac{A}{2}, \qquad a_k = 0 \ (k \ne \pm 1)$$

The integral formula gives the same result. For example,

$$a_1 = \frac{1}{T}\int_0^T A\,\frac{e^{j\omega_0 t}-e^{-j\omega_0 t}}{2j}e^{-j\omega_0 t}dt = \frac{A}{2j}\cdot\frac{1}{T}\left[T - 0\right] = \frac{A}{2j}$$

Spectrum

• Magnitude: $|a_1| = |a_{-1}| = A/2$
• Phase: $\angle a_1 = -\pi/2$, $\angle a_{-1} = +\pi/2$

 |a_k|
   A/2  |     |
       -+--o--+--> k
       -1  0  1

 angle(a_k)
 +pi/2  |
       -+--o--+--> k
 -pi/2        |
       -1  0  1

The spectrum has only two lines, at $\pm\omega_0$. The DT signal $A\sin\omega_0 n$ with $\omega_0 = 2\pi/N$ gives the same coefficients $a_{\pm1} = \pm A/(2j)$, repeated every $N$.