Fourier series formulas

Holds for real or complex $x(t)$. $c_0$ is the average (DC) value. Plotting $|c_k|$ and $\angle c_k$ against $k{\omega }_0$ gives the two-sided line spectrum.

For real $x(t)$, with $a_0$ the DC value (convention above).

Here $k=1,2,3,\dots$; the factor is $2/T_0$ for $a_k$ and $b_k$ but $1/T_0$ for $a_0$.

For real $x(t)$. This is the one-sided spectrum: $C_k$ is the physical peak amplitude of the $k$-th harmonic, and ${\theta }_k$ its phase.

For real $x(t)$ and $k\ge 1$. ${\theta }_k$ is the angle of the phasor $a_k-j{b}_k$, so $\cos {\theta }_k=a_k/C_k$ and $\sin {\theta }_k=-b_k/C_k$. The shortcut ${\theta }_k=-{\tan }^{-1}(b_k/a_k)$ is right only when $a_k>0$; when $a_k<0$, add or subtract $\pi$.

When the Dirichlet conditions hold, the series equals $x(t)$ wherever $x$ is continuous and the midpoint of the jump at a discontinuity $t_0$. The partial sums are taken symmetrically, $k=-N$ to $N$. A square wave switching between $0$ and $5$ has series value $2.5$ at every switching instant.

For an even signal the DC value is also $a_0=\frac{2}{T_0}{\int }_0^{T_0/2}x(t)dt$. Under half-wave symmetry every even-$k$ coefficient is zero, so these formulas are used for odd $k$ only.

A shift leaves every $|c_k|$ and the power unchanged. A shift of half a period gives $c_k{e}^{-jk\pi }=(-1{)}^k{c}_k$.

Differentiating removes the DC and boosts high harmonics; integrating smooths. The integration rule needs zero DC in $x(t)$. Any other running integral (textbooks often write the lower limit as $-\infty$) differs from this one only by a constant, so the DC of the result must be found from the waveform.

The trigonometric forms need real $x(t)$. A DC level $a_0$ carries power $a_0^2$ and a sinusoid $A\cos (k{\omega }_{0}t+\varphi )$ carries $A^2/2$. The RMS value is $\sqrt{P}$.

RMS of all harmonics above the fundamental divided by the RMS of the fundamental (DC excluded). For a symmetric square wave, $\mathrm{THD}=\sqrt{{\pi }^2/8-1}\approx 0.483$.

$x_p(t)$ is one period of $x(t)$ (zero outside that period) and $X_p(j\omega )={\int }_{-\infty }^{\infty }{x}_p(t)e^{-j\omega t}dt$ is its Fourier transform in $\omega$ (rad/s). The series coefficients are samples of the single pulse's transform at the harmonics, scaled by $1/T_0$. The transform of the periodic signal itself is a train of impulses of area $2\pi c_k$ at $\omega =k{\omega }_0$; in the Hz convention it is $X(f)={\sum }_k{c}_k\delta (f-k{f}_0)$, with $c_k=\frac{1}{T_0}{X}_p(k{f}_0)$. Example: a rectangular pulse of height $A$ and width $\tau$ has $X_p(j\omega )=A\tau \mathrm{sinc}\left(\frac{\omega \tau }{2\pi }\right)$, which gives the pulse-train coefficients below.

$x=+A$ on $(0,T_0/2)$ and $-A$ on $(-T_0/2,0)$: odd and half-wave symmetric. Power $A^2$, of which the fundamental carries $8/{\pi }^2\approx 0.81$. Shifted to be even ($+A$ for $|t|<T_0/4$): $\frac{4A}{\pi }\left(\cos {\omega }_{0}t-\frac{1}{3}\cos 3{\omega }_{0}t+\frac{1}{5}\cos 5{\omega }_{0}t-\cdots \right)$.

Height $A$, width $\tau$, centred on $t=0$ (even, so $c_k$ is real); the second form is for $k\ne 0$. With duty cycle $d=\tau /T_0$: DC $=Ad$, power $=A^{2}d$, and $c_k=0$ whenever $kd$ is a nonzero integer. At $d=1/2$ the even harmonics vanish; at $d=1/3$ every third harmonic vanishes.

$x(t)=A\left(1-4|t|/T_0\right)$ for $|t|\le T_0/2$: $+A$ at $t=0$, $-A$ at $t=\pm T_0/2$. Even and half-wave symmetric; power $A^2/3$. The unipolar version between $0$ and $A$ is $\frac{A}{2}+\frac{4A}{{\pi }^2}{\sum }_{k\text{ odd}}\frac{1}{k^2}\cos k{\omega }_{0}t$.

$x(t)=2At/T_0$ on $(-T_0/2,T_0/2)$, rising from $-A$ to $+A$ and jumping back. Odd but not half-wave symmetric, so every harmonic is present; power $A^2/3$. The ramp $x(t)=At/T_0$ on $(0,T_0)$ is $\frac{A}{2}-\frac{A}{\pi }{\sum }_{k=1}^{\infty }\frac{1}{k}\sin k{\omega }_{0}t$.

$c_k=1/T_0$ for every $k$: a flat line spectrum that never decays. Impulses of area $I$ at $t=n{T}_0$ give $c_k=I/T_0$. The average power is infinite.

${\omega }_1$ is the frequency of the sine before rectification. The period halves to $\pi /{\omega }_1$, so the fundamental is ${\omega }_0=2{\omega }_1$ and only even multiples of ${\omega }_1$ appear. DC $=2A/\pi$; RMS $=A/\sqrt{2}$ (power $A^2/2$, the same as the unrectified sine).

The period is unchanged. DC $=A/\pi$, fundamental amplitude $A/2$, then even harmonics only. RMS $=A/2$ (power $A^2/4$). It equals half of (the sine plus its full-wave rectified version).

For a piecewise-smooth periodic $x(t)$. "Decays as" describes the envelope of $|c_k|$; individual coefficients can still be zero, as in a pulse train. Each differentiation multiplies $c_k$ by $jk{\omega }_0$. A jump in $x$ itself ($m=0$) gives $1/k$; a kink ($m=1$) gives $1/k^2$; impulses in $x$ give no decay. The smoother the waveform, the faster its spectrum falls and the fewer terms a good approximation needs.

Steady state of a BIBO-stable LTI system, after any switch-on transient has died out. $H(j\omega )$ is needed only at the harmonic frequencies $k{\omega }_0$.

For real $h(t)$, so that $H(-j\omega )=H^{\ast }(j\omega )$. DC passes with gain $H(0)$. Output power by Parseval: $P_y={\sum }_k|H(jk{\omega }_0){|}^2|c_k{|}^2$.

$⟨N⟩$ means any $N$ consecutive integers, e.g. $0$ to $N-1$. The coefficients are periodic, $a_{k+N}=a_k$. For real $x[n]$, $a_{-k}=a_{N-k}=a_k^{\ast }$. $a_0$ is the average over one period.

The $N$-point DFT of one period is $N$ times the DTFS coefficients. Time shift: $x[n-n_0]\leftrightarrow a_k{e}^{-jk{\Omega }_0{n}_0}$. Through a stable LTI system: $a_k\to a_{k}H(e^{jk{\Omega }_0})$, where $H(e^{j\Omega })$ is the DTFT of $h[n]$.

The pulse is $x[n]=1$ for $|n|\le N_1$ and $0$ for the rest of the period, repeated every $N$ samples ($2N_1+1<N$). It is even, so $a_k$ is real. The ratio-of-sines form is for $k\ne 0,\pm N,\pm 2N,\dots$; at those $k$, $a_k=(2N_1+1)/N$, the average. The ratio of sines is the discrete counterpart of the sinc envelope of a continuous pulse train.