Conjugate symmetry of Fourier coefficients

For a real signal $x(t)$, the Fourier series coefficients satisfy conjugate symmetry:

$c_x[-k]=c_x^{\ast }[k].$

The same property holds for the Fourier transform of a real signal: $X(-f)=X^{\ast }(f)$.

Derivation

For real $x$, take the complex conjugate of the analysis equation:

$c_x^{\ast }[k]=\overline{\frac{1}{T}{\int }_{T}x(t)e^{-j2\pi kt/T}dt}=\frac{1}{T}{\int }_{T}x(t)e^{+j2\pi kt/T}dt=\frac{1}{T}{\int }_{T}x(t)e^{-j2\pi (-k)t/T}dt=c_x[-k].$

(We pulled the conjugate inside the integral; $x$ is real so it’s unchanged; the conjugate of $e^{-j\theta }$ is $e^{+j\theta }=e^{-j(-\theta )}$, so we end up with the $-k$ coefficient.)

Two consequences

Magnitude is even, phase is odd. Writing $c_x[k]=|c_x[k]|e^{j{\varphi }_k}$:

$|c_x[-k]|=|c_x[k]|,{\varphi }_{-k}=-{\varphi }_k.$

The magnitudes of $c_x[k]$ and $c_x[-k]$ are equal; the phases are equal in magnitude and opposite in sign.

Negative coefficients are determined by positive. Only compute $c_x[k]$ for $k\ge 0$; the rest follow by conjugation. This cuts the work in half.

Worked verification

For the sawtooth example $x(t)=3-t$ on $0<t<4$, we found $c_x[k]=-2j/(\pi k)$ for $k\ne 0$, purely imaginary with magnitude $2/(\pi |k|)$ (even in $k$) and phase $\pm \pi /2$ (odd in $k$, since $-j$ has phase $-\pi /2$ and $+j$ has phase $+\pi /2$, and the sign flips when $k$ does). ✓

Implications for the trigonometric form

Conjugate symmetry is what makes the trigonometric Fourier series consist of real cosine and sine coefficients. The pair $c_x[k]+c_x^{\ast }[k]=2\mathrm{Re}{c}_x[k]$ is real, hence $a_x[k]$ is real. Similarly $j(c_x[k]-c_x^{\ast }[k])=-2\mathrm{Im}{c}_x[k]$ is real, hence $b_x[k]$ is real.

If $x$ were complex-valued, conjugate symmetry would fail and the trigonometric form wouldn’t exist as a real-coefficient representation.

Special symmetries

• Real and even $x$ → $c_x[k]$ is real and even in $k$. The trigonometric form has only cosines (all $b_x[k]=0$).
• Real and odd $x$ → $c_x[k]$ is purely imaginary and odd in $k$. The trigonometric form has only sines (all $a_x[k]=0$).
• Real but neither even nor odd → $c_x[k]$ has both real and imaginary parts; the trigonometric form has both cosines and sines.

Recognizing parity in $x(t)$ before computing coefficients can simplify integrals significantly. See Even and odd signals.