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The angular frequency corresponding to a specified filter cutoff frequency. For a first-order RC low-pass filter, the half-power convention gives ${\omega }_c=1/(RC)$ and $f_c=1/(2\pi RC)$.

In a quasi-one-dimensional inviscid channel with depth $H$ and width $b(x)=b_0{e}^{\mu x}$, the linear shallow water equations are ${\eta }_t+(H/b)(bu{)}_x=0$ and $u_t+g{\eta }_x=0$. Combining them gives ${\eta }_{tt}=gH({\eta }_{xx}+\mu {\eta }_x)$.

For a time dependence $e^{i\omega t}$, write the elevation amplitude as $N(x)=e^{-\mu x/2}V(x)$. The spatial equation becomesThere is a travelling spatial phase only when the bracket is positive, so the cutoff frequency is ${\omega }_c=\mu \sqrt{gH}/2$. Above it, an outgoing wave has a phase proportional to $\omega t-kx$ and amplitude proportional to $e^{-\mu x/2}$; below it the spatial response is evanescent. Geometric widening introduces the cutoff even though a constant-width channel has the nondispersive speed $\sqrt{gH}$.

For a linear filter with chosen nonzero reference voltage gain $H_0$, define its effective noise bandwidth byThis is the width of an ideal rectangular linear filter with gain magnitude $|H_0|$ passing the same mean square of input white noise. ConsequentlyUse the full voltage transfer from the open-circuit voltage source to the readout, including any electrical load, when applying this formula. A nominal signal bandwidth stated between two cutoff frequencies does not specify the integral: gradual tails and a nonuniform passband matter. For an ideal rectangular unit-gain passband $[f_1,f_2]$, $B_{\mathrm{n}}=f_2-f_1$. For an unloaded RC low-pass filter with $H_0=1$ and cutoff frequency $f_c=1/(2\pi RC)$,The last equality follows by substituting $x=f/f_c$ and integrating $1/(1+x^2)$ as $\arctan x$. Using the nominal $f_c$ as the noise bandwidth underestimates the variance by a factor $\pi /2$, or the root mean square voltage by $\sqrt{\pi /2}$.

Let $K(f)$ be the complete current-to-voltage frequency response from detector input to readout, in $V/A$. For a real stable linear time-invariant system, the shot contribution to the output voltage variance isIf $K(f)=R_{F}GH(f)$ with feedback electrical resistance $R_F$, a reference voltage gain $G$, and dimensionless normalized frequency response $H$, thenThis is the effective noise bandwidth for unit reference magnitude. The transimpedance amplifier factor $R_F$ converts electric current to voltage; $G$ multiplies voltage. Neither is an additional noise source merely because it amplifies an existing fluctuation. In contrast, a real amplifier adds input voltage noise and input current noise, and its feedback resistor contributes thermal electrical noise. Those terms need their own transfers to the output. In particular, input voltage noise generally has a different transfer from an injected detector electric current. Simply multiplying every input specification by $R_{F}G$ is incorrect.

For a unit-gain exponential low-pass filter, with time constant $\tau >0$,The cutoff frequency $f_c$ is not $B_{\mathrm{n}}$. An independent time-domain derivation sums independent event contributions to giveThe same reasoning for a rectangular averaging gate has $h=1/T$ on $[0,T]$ and zero elsewhere, hence effective noise bandwidth $1/(2T)$ and variance $qI/T$, agreeing with the shot noise counting model. Equal numerical durations $T=\tau$ do not make these two different linear filters equivalent.

An ideal operational amplifier with a grounded noninverting input, input resistor $R_i$, and parallel feedback resistor $R_f$ and capacitor $C$ implements a low-pass filter. The virtual short and Kirchhoff's current law give $V_{\mathrm{out}}/V_{\mathrm{in}}=-Z_f/R_i$, where $Z_f=R_f/(1+i\omega R_{f}C)$. The DC gain magnitude is $R_f/R_i$ and the cutoff frequency is $1/(2\pi R_{f}C)$. The capacitance reduces high-frequency feedback electrical impedance, causing the gain magnitude to fall.

Use complex phasors with time dependence $e^{i\omega t}$ and assume an unloaded output. The capacitor has electrical impedance $Z_C=1/(i\omega C)$. The voltage divider formed with the resistor gives the transfer functionFor input $V_i(t)=V_{i0}\cos \omega t$, the output amplitude and phase shift areso $V_o(t)=|{\tilde{V}}_o|\cos (\omega t-{\tan }^{-1}(\omega RC))$.

Here $\tau =RC=16.0\mu \mathrm{s}$, and the cutoff frequency is $f_c=1/(2\pi RC)=9.947\mathrm{kHz}$. This is an RC low-pass filter: at small frequency the capacitor has large electrical impedance and the output nearly equals the input; at large frequency it increasingly shunts the signal to ground. The amplitude then falls as $1/f$, and the phase shift approaches $-90^{\circ }$.

No numerical input amplitude is supplied, so the output voltages are specified as fractions of $V_{i0}$. An application is suppression of high-frequency electrical noise on a slowly varying signal. A finite load changes the transfer function and must be included if its electrical impedance is comparable with the circuit's output electrical impedance.

For an ideal operational amplifier under negative feedback, the grounded non-inverting terminal makes the inverting terminal a virtual ground, and no electric current enters either input. The feedback resistor and capacitor have parallel electrical impedanceKirchhoff's current law at the inverting node gives ${\tilde{V}}_i/R_1+{\tilde{V}}_o/Z_f=0$, and henceWith the stated components, $R_{2}C=16.0\mu \mathrm{s}$ and $R_2/R_1=5$. Thus this inverting low-pass amplifier has exactly the same ideal cutoff frequency and normalized frequency response as the passive RC low-pass filter, with output multiplied by $-5$:Its amplitude ratios at the three stated frequencies are $4.99975$, $3.52616$, and $0.97560$. One principal choice of phase shift is $180^{\circ }-{\tan }^{-1}(\omega \tau )$: respectively $179.424^{\circ }$, $134.848^{\circ }$, and $101.252^{\circ }$. Adding or subtracting $360^{\circ }$ describes the same output; the extra inversion must be retained.

negative feedback and the ideal operational amplifier's virtual short hold the inverting node at zero potential, with no current entering either input. The parallel feedback resistor and capacitor have electrical impedanceKirchhoff's current law at the inverting node gives $V_{\mathrm{in}}/R_i=-V_{\mathrm{out}}/Z_f$. For the component values read from the circuit,This inverting low-pass amplifier has a DC gain magnitude of 1000 and cutoff frequency $f_c=(2\pi R_{f}C{)}^{-1}=0.15915$ Hz. At the corner, $|G|=707.1$. Above it the gain approaches $1/(2\pi f{R}_{i}C)=159.15/f$, with $f$ in Hz: a slope of $-1$ on logarithmic axes, or $-20$ dB per decade on a Bode plot. At $0.01,1,10,100$ Hz the magnitudes are approximately $998,157.2,15.91,1.592$. These are voltage ratios, not decibels.

An empty perfectly conducting rectangular electromagnetic waveguide hasThe lowest transverse electric mode has indices $(1,0)$ or $(0,1)$. A simply connected hollow metallic guide has no transverse electromagnetic mode. With $a=b=10^{-2}\mathrm{m}$, the two lowest modes are degenerate and their cutoff frequency isPropagation with nonzero axial energy transport requires $f>f_c$; the equality is the lower limiting frequency, with zero axial group velocity. A filling medium would replace $c$ by $1/\sqrt{\mu ϵ}$.

The lowest transverse electric mode has cutoff frequency $c/(2a)\simeq 15.0\mathrm{GHz}$; propagation requires frequency strictly above cutoff.

Full derivation: solution.

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

The vacuum wave equation gives ${\omega }^2=c^2[k_g^2+(\pi /a{)}^2]$. The cutoff frequency is $f_c=c/(2a)=1.499\mathrm{GHz}$. In a lossless empty guide, energy travels at group velocity,The larger phase velocity $\omega /k_g=c^2/v_E$ is not the speed of energy transport.

Use the complex amplitude convention $v(t)=\mathrm{Re}(\hat{V}{e}^{\mathrm{i}\omega t})$, where $\omega =2\pi f$ is the angular frequency and $f$ is the frequency in cycles per second. Substitution into the RC low-pass filter equation replaces the time derivative with multiplication by $\mathrm{i}\omega$, giving its continuous-time frequency responseHere $\varphi$ is the output-minus-input phase difference; a negative value means the output lags. Equivalently, the voltage divider formed from the electrical impedances $R$ and $1/(\mathrm{i}\omega C)$ gives the same ratio. For an input $A\cos (\omega t)$, the sustained output is $A|H|\cos (\omega t+\varphi )$ after the initial exponential decay has died away.

Define the cutoff frequency by a voltage amplitude ratio $1/\sqrt{2}$ relative to the low-frequency value:On a Bode plot, the voltage gain in decibels isAt the cutoff frequency it is $-3.0103\mathrm{dB}$, and far above that frequency it approaches a decrease of $20\mathrm{dB}$ per factor of ten in $f$. There is a smooth transition rather than an abrupt wall. The squared voltage ratio is one half at the corner; converting that statement to a power ratio requires the same real electrical resistance for the compared voltages. It does not mean that an unloaded capacitor dissipates half the source power.

For $R=2.2\mathrm{k}\Omega$ and $C=47\mathrm{nF}$, the cutoff frequency is $f_c\approx 1.5392\mathrm{kHz}$. A $1\mathrm{V}$ peak input at $0.1f_c$, $f_c$, and $10f_c$ produces peak output amplitudes of approximately $0.9950\mathrm{V}$, $0.7071\mathrm{V}$, and $0.09950\mathrm{V}$, with phase differences $-5.711^{\circ }$, $-45^{\circ }$, and $-84.289^{\circ }$. Thus a wanted oscillation at $0.1f_c$ is almost preserved while an unwanted one at $10f_c$ is reduced by about a factor of ten in voltage. The superposition principle lets these two responses be added when both oscillations occur in the same signal.

The University of Colorado Boulder's Lab 3: Filters, §§7.1–7.4, specifies measuring the resistor and capacitor, then driving the assembled RC low-pass filter with a function generator. A two-channel oscilloscope reads actual input and output using high electrical impedance. Sweep the frequency, record the voltage amplitude ratio, and locate $1/\sqrt{2}$. A control compares a cable with a $10\times$ oscilloscope probe on a resistive voltage divider, isolating measurement capacitance before using the probe for the filter. A slow square wave then lets discharge timing estimate the time constant independently. Agreement supports the RC low-pass frequency response; source output impedance, probe loading, and instrument signal bandwidth limit it. These are prescribed measurements, with no measured dataset supplied here.^[ref]

The two-channel setup can also test the predicted phase difference directly: for positive frequency $f$, measure the output's time lag $\Delta t$ between corresponding features of the two sinusoids. For this single-stage model the phase lies between $-\pi /2$ and $0$, so choose the lag within a quarter cycle and calculate $\varphi =-2\pi f\Delta t$. At the cutoff frequency the predicted lag is one eighth of a cycle. This tests the timing part of the frequency response as well as the voltage ratio.

The step response describes a sudden persistent input change. Suppose the input becomes a constant $V_s$ at $t=0$, and the initial capacitor voltage is $v_0$. Finite electric current leaves that voltage continuous. Subtract $V_s$ from the RC low-pass filter equation: the remaining error satisfies $\tau de/dt+e=0$. Separating variables givesFor an initially uncharged capacitor and a $0$ to $3\mathrm{V}$ step in the constructed example,At one time constant, $t=103.4\mu \mathrm{s}$, the voltage is $1.8964\mathrm{V}$, about $63.2\%$ of its final value. The voltage error then continues its exponential decay. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.

Define settling time relative to the total voltage change: require $|v_{\mathrm{out}}-V_s|\le ϵ|v_0-V_s|$ thereafter, with $0<ϵ<1$. The exponential decay is monotone, soFor $1\%$ settling, $t_{0.01}=4.6052\tau \approx 476.17\mu \mathrm{s}$; the $10\%$ to $90\%$ rise takes $\tau \ln 9\approx 227.19\mu \mathrm{s}$. Reducing the cutoff frequency by doubling $C$ doubles both times. This reciprocal relation is the useful connection between signal bandwidth and response speed: stronger suppression of rapid noise comes with slower tracking of wanted changes.

A guided electromagnetic wave whose electric field has no component along the propagation axis. Its magnetic field can have a longitudinal component. In an empty perfectly conducting rectangular electromagnetic waveguide, transverse wave numbers determine a nonzero cutoff frequency. This differs from a transverse electromagnetic mode, in which both fields are transverse.