Even and odd signals

A signal $g(t)$ is even if $g(t)=g(-t)$ for every $t$ (symmetric about the $y$-axis), and odd if $g(t)=-g(-t)$ for every $t$ (antisymmetric about the origin). The canonical even signal is $\cos (\omega t)$; the canonical odd signal is $\sin (\omega t)$.

Not every signal is even or odd. The function $g(t)=t+t^2$ is neither — it’s a sum of an odd part $t$ and an even part $t^2$, but as a whole has no symmetry.

Every signal decomposes

Every signal can be written uniquely as a sum of an even part and an odd part:

$g_e(t)=\frac{g(t)+g(-t)}{2},g_o(t)=\frac{g(t)-g(-t)}{2},$

with $g(t)=g_e(t)+g_o(t)$. You can verify by direct substitution that $g_e$ is even, $g_o$ is odd, and their sum is $g$. So whenever you have a signal whose symmetry you don’t know, you can split it and reason about the parts separately.

Why we care

Symmetry can collapse half the work in any integral over a symmetric interval:

• ${\int }_{-a}^a(\text{even function})dt=2{\int }_0^a$.
• ${\int }_{-a}^a(\text{odd function})dt=0$.

Combined with the product rules — even $\times$ even = even, odd $\times$ odd = even, even $\times$ odd = odd — you can often tell at a glance which integrals are zero and which need computing.

A worked example

Compute

$I={\int }_{-10}^{10}4\mathrm{rect}(t/8)e^{j2\pi t/16}dt.$

The rect factor is $1$ for $|t/8|<1/2$, i.e. $-4<t<4$. Outside it’s zero. So

$I={\int }_{-4}^{4}4e^{j\pi t/8}dt=4{\int }_{-4}^4\cos (\pi t/8)dt+4j{\int }_{-4}^4\sin (\pi t/8)dt.$

Cosine is even, sine is odd, the interval is symmetric. So the sine integral is zero by oddness, and the cosine integral is twice the integral from $0$ to $4$:

$I=8{\int }_0^4\cos (\pi t/8)dt=8\cdot \frac{8}{\pi }\sin (\pi /2)=\frac{64}{\pi }\approx \mathrm{20.37.}$

Spotting the parity cut the work in half and saved us from worrying about the imaginary part. This pattern recurs throughout Fourier series and Fourier transform derivations.

Symmetry in the frequency domain

For a real signal $x(t)$, the Fourier transform $X(f)$ satisfies $X(-f)=X^{\ast }(f)$ — conjugate symmetry. As a consequence, $|X(f)|$ is even in $f$ and the phase $\angle X(f)$ is odd in $f$. If $x$ is also an even real signal, $X$ is purely real (and even). If $x$ is an odd real signal, $X$ is purely imaginary (and odd). See Conjugate symmetry of Fourier coefficients for the discrete-frequency analog.