Fourier transform formulas

$\omega$ in rad/s. The $1/2\pi$ sits only in the inverse, so it reappears in duality, the multiplication property, Parseval's theorem and the formula for $x(0)$.

$f$ in Hz; symmetric, with no factor outside the integrals. Convert with $X(f)=X(\omega ){|}_{\omega =2\pi f}$, and for impulses use $\delta (\omega )=\delta (f)/2\pi$, so $2\pi \delta (\omega )$ becomes $\delta (f)$.

Normalised sinc, with zeros at $x=\pm 1,\pm 2,\dots$. Some texts use $\sin x/x$ under the same name; every sinc result below is also written in explicit $\sin$ form, so nothing depends on the convention.

$X(s)$ is the bilateral Laplace transform. Example: $e^{-at}u(t)$, $a>0$, has ROC $\text{Re}(s)>-a$, which contains the axis. If the ROC boundary is the $j\omega$ axis (poles on the axis, as for $u(t)$ with ROC $\text{Re}(s)>0$), the Fourier transform exists only with impulses and the substitution misses them. If the ROC excludes the axis altogether (as for $e^{2t}u(t)$, ROC $\text{Re}(s)>2$), there is no Fourier transform.

Pulse of width $T$ and height 1. $X(0)=T$, the pulse area. First nulls at $\omega =\pm 2\pi /T$, i.e. $f=\pm 1/T$: a narrower pulse has a wider main lobe.

The dual of the rect–sinc pair. Peak value $W/\pi$ at $t=0$, zeros at $t=k\pi /W$ ($k\ne 0$). This is $h(t)$ of the ideal lowpass filter with cutoff $W$ rad/s.

$e^{-a|t|}=e^{-at}u(t)+e^{at}u(-t)$, so its transform is $\frac{1}{a+j\omega }+\frac{1}{a-j\omega }$. For $a<0$ both signals grow and have no Fourier transform; $a=0$ gives $u(t)$ or $1$, which need impulses.

The impulse in $\mathcal{F}\{u(t)\}$ is the transform of its average value: $\frac{1}{2}\cdot 2\pi \delta (\omega )=\pi \delta (\omega )$. $\mathrm{sgn}(t)$ has zero average, so it has no impulse.

A narrow Gaussian in time (large $a$) gives a wide Gaussian in frequency. In the $f$ form, $e^{-\pi t^2}\leftrightarrow e^{-\pi f^2}$ exactly.

Every Fourier-series coefficient of the impulse train equals $1/T$, which gives impulses of area $2\pi /T$. This pair is the basis of sampling theory.

Real $a\ne 0$. Compressing in time ($|a|>1$) stretches the spectrum and lowers it. Since $x(at-b)=x(a(t-b/a))$, the delay is $b/a$, not $b$.

In the $f$ form there is no $2\pi$: $X(t)\leftrightarrow x(-f)$. For an even $x$, $x(-\omega )=x(\omega )$.

Each shifted copy has half the height. With $\sin {\omega }_{0}t$ instead: $\frac{1}{2j}\left[X(\omega -{\omega }_0)-X(\omega +{\omega }_0)\right]$. If the copies do not overlap, the energy is halved.

Differentiation removes any DC impulse: $\frac{d}{dt}u(t)=\delta (t)$, and $j\omega \left[\pi \delta (\omega )+\frac{1}{j\omega }\right]=1$ because $\omega \delta (\omega )=0$.

The impulse term vanishes only when $X(0)=\int xdt=0$. Dividing by $j\omega$ alone is the classic error.

The $1/2\pi$ belongs to the multiplication property in the $\omega$ form only. The convolution property is why an LTI output is $Y(\omega )=H(\omega )X(\omega )$.

Put $\omega =0$ in the forward integral, or $t=0$ in the inverse. In the $f$ form, $x(0)=\int X(f)df$. If $x$ jumps at $t=0$, the second integral gives the midpoint of the jump.

For energy signals. Generalised form: $\int x(t)y^{\ast }(t)dt=\frac{1}{2\pi }\int X(\omega )Y^{\ast }(\omega )d\omega$.

The band formula is for real $x(t)$. The factor 2 counts the matching negative-frequency band, which carries the same energy because $|X|$ is even.

The phase of $H$ has no effect on output energy.

${\omega }_0=2\pi /T_0$. In the $f$ form, $X(f)={\sum }_k{c}_k\delta (f-k{f}_0)$ with $f_0=1/T_0$.

$X_1(\omega )$ is the transform of one period of $x(t)$ (taken as zero outside that period). The coefficients are samples of $X_1(\omega )/T_0$.

Putting $s=j\omega$ is valid when the system is BIBO stable ($\int |h|dt<\infty$), so the ROC contains the $j\omega$ axis. For a stable system described by $\sum a_k{y}^{(k)}=\sum b_k{x}^{(k)}$, $H(j\omega )=\frac{\sum b_k(j\omega {)}^k}{\sum a_k(j\omega {)}^k}$.

Real $h(t)$, stable system, input applied for all time. The frequency is unchanged; only amplitude and phase change. Superpose for several tones.

Here $K>0$. The conditions are needed only across the band the signal occupies. Adding a multiple of $2\pi$ to the phase changes nothing, and a phase of $-\omega t_d\pm \pi$ gives $y(t)=-Kx(t-t_d)$, an inverted but undistorted copy. Any other constant phase offset changes the waveform (for a narrowband bandpass signal it shifts only the carrier phase; see below).

$\theta$ in radians and $\omega$ in rad/s give seconds. Use the unwrapped phase for ${\tau }_p$. For $\theta =-\omega t_d$, ${\tau }_p={\tau }_g=t_d$ at every frequency; conversely, ${\tau }_p={\tau }_g$ at every frequency only when $\theta =-\omega t_d$ for some constant $t_d$. In terms of $f$: ${\tau }_p=-\frac{\theta }{2\pi f}$ and ${\tau }_g=-\frac{1}{2\pi }\frac{d\theta }{df}$.

The envelope (the information) is delayed by the group delay, also called envelope delay. The carrier is delayed by the phase delay. If ${\tau }_g$ varies across the band, the envelope is dispersed.

A length-$N$ FIR filter with symmetric coefficients has $H(e^{j\Omega })=e^{-j\Omega (N-1)/2}A(\Omega )$ with $A$ real, so its phase is linear (apart from jumps of $\pi$ where $A$ changes sign). Antisymmetric coefficients, $h[n]=-h[N-1-n]$, give $H(e^{j\Omega })=j{e}^{-j\Omega (N-1)/2}A(\Omega )$: the same group delay $(N-1)/2$ with a constant $\pi /2$ phase offset. Example: $h[n]=\{1,2,1\}$ for $n=0,1,2$ gives $e^{-j\Omega }(2+2\cos \Omega )$, a delay of 1 sample.

At the 3-dB point the phase is $-45^{\circ }$. The 10–90% step rise time is $t_r=RC\ln 9\approx 2.2RC$, so $t_r{f}_{3\text{dB}}\approx 0.35$.

Gain 1 at every $\omega \ne 0$, phase $-90^{\circ }$ for $\omega >0$. Its $h(t)=1/(\pi t)$ is non-causal, so the ideal Hilbert transformer is also unrealisable.

Any interval of length $2\pi$ may replace $-\pi$ to $\pi$. The DTFT is the $z$-transform on the unit circle, $X(e^{j\Omega })=X(z){|}_{z=e^{j\Omega }}$, valid when the ROC of $X(z)$ contains $|z|=1$.

Put $\Omega =0$ or $\Omega =\pi$ in the forward sum, or $n=0$ in the inverse integral. More generally $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }X(e^{j\Omega })e^{j\Omega k}d\Omega =x[k]$.

The frequency integral spans one period, $2\pi$ wide, never $-\infty$ to $\infty$.

Real $h[n]$, stable system ($\sum |h[n]|<\infty$), input applied for all $n$. ${\Omega }_0$ and ${\Omega }_0+2\pi$ give the same sequence and the same response.