A periodic signal can be represented using complex exponentials at harmonically related frequencies. This is the complex Fourier series form of the trigonometric expansion.

For a continuous-time signal $x(t)$ of period $T$, define the fundamental angular frequency

$$\boxed{\omega_0=\frac{2\pi}{T}}.$$

The complex representation is

$$\boxed{x(t)=\sum_{k=-\infty}^{\infty}c_ke^{ik\omega_0t}}.$$

The coefficient of the $k$th harmonic is

$$\boxed{c_k=\frac1T\int_{t_0}^{t_0+T}x(t)e^{-ik\omega_0t},dt}.$$

The integral can be taken over any complete period.

Why the coefficients isolate individual harmonics

Complex exponentials are orthogonal over one period:

$$\frac1T\int_{t_0}^{t_0+T} e^{i(k-m)\omega_0t},dt

\begin{cases} 1,&k=m,\ 0,&k\ne m. \end{cases}$$

Multiplying

$$x(t)=\sum_kc_ke^{ik\omega_0t}$$

by

$$e^{-im\omega_0t}$$

and averaging over one period therefore removes every harmonic except $m$, leaving exactly $c_m$.

This is the complex-exponential version of the same orthogonal coefficient projection used by a real sine-cosine Fourier series.

Relation to the real trigonometric series

For a real signal,

$$x(t)=\frac{a_0}{2} +\sum_{n=1}^{\infty} \left[a_n\cos(n\omega_0t)+b_n\sin(n\omega_0t)\right].$$

Euler's formulas

$$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}2,$$

$$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$

show that this is equivalent to the complex form.

For $n>0$,

$$\boxed{c_n=\frac12(a_n-ib_n)},$$

$$\boxed{c_{-n}=\frac12(a_n+ib_n)},$$

and

$$\boxed{c_0=\frac{a_0}{2}}.$$

If $x(t)$ is real,

$$\boxed{c_{-k}=c_k^*},$$

where $^*$ denotes complex conjugation. Negative-frequency coefficients are therefore not independent of the positive-frequency coefficients for a real signal.

Discrete harmonic spectrum

Because the signal repeats exactly, its Fourier-series frequencies occur only at integer multiples of the fundamental:

$$\ldots,-2\omega_0,-\omega_0,0,\omega_0,2\omega_0,\ldots$$

The collection of coefficients $c_k$ is a discrete frequency spectrum. Each coefficient contains the amplitude and phase associated with one harmonic.

Worked example

Consider

$$x(t)=3+2\cos(\omega_0t)-4\sin(2\omega_0t).$$

The real Fourier coefficients are

$$a_0=6,$$

$$a_1=2,$$

$$b_2=-4,$$

with the other shown coefficients zero.

Therefore

$$c_0=3.$$

For the first harmonic,

$$c_1=\frac12(2-i0)=1,$$

$$c_{-1}=1.$$

For the second harmonic,

$$c_2=\frac12(0-i(-4))=2i,$$

$$c_{-2}=-2i.$$

Thus

$$\boxed{x(t)=3+e^{i\omega_0t}+e^{-i\omega_0t} +2ie^{i2\omega_0t}-2ie^{-i2\omega_0t}}.$$

Combining each positive-negative pair recovers the original real sinusoidal form.

Truncation and approximation

Keeping only a finite number of coefficients gives a finite harmonic approximation. The convergence properties belong to the underlying Fourier expansion; the complex form changes the algebra and interpretation, not which real function is being represented.

For signals, the complex form is particularly useful because differentiation, integration, and linear-system response act simply on each exponential harmonic. It is the natural bridge from a general Fourier-series expansion to frequency-domain signal analysis.