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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 $f_c=1/(2\pi RC)$.

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 $\sqrt{\pi /2}$.

A low-pass filter preferentially transmits low frequencies and attenuates higher frequencies. It can smooth rapid noise in a slowly varying signal, but it also suppresses wanted variations outside the useful signal bandwidth. An RC low-pass filter gives a concrete continuous-time example; a moving average implements related smoothing on a discrete-time signal.

The useful question is how much change the measurement can tolerate. Preserving the low frequencies of a temperature record while reducing a fast unwanted oscillation can improve readability. Preserving the sharp onset of a pulse requires higher frequencies, so excessive smoothing changes its timing and shape. The RC low-pass frequency response and RC low-pass step response make this signal bandwidth versus settling time tradeoff quantitative.

The output is measured across the capacitor. The voltage divider gives the RC low-pass filter transfer functionThereforeThe amplitude ratio tends to $1$ and the phase shift to zero as $\omega \to 0$. For $\omega \to \infty$, the amplitude ratio is asymptotic to $1/(\omega RC)$ and the phase shift tends to $-\pi /2$.

The half-power amplitude condition for the RC low-pass filter gives $1+(\omega RC{)}^2=2$. Since angular frequency is positive,At this cutoff angular frequency, the output amplitude is smaller by $1/\sqrt{2}$ and the phase shift is $-\pi /4$.

The RC low-pass filter satisfies $RC{\dot{V}}_{\mathrm{out}}+V_{\mathrm{out}}=V$ for $t>0$. With zero initial capacitor voltage, the first-order linear differential equation givesThe output is continuous at switching and approaches $V$ with time constant $RC$.

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.

The passive RC low-pass filter is simple, inexpensive, requires no power supply, and has no operational amplifier saturation or finite bandwidth limitation. It offers no voltage gain; its output impedance is finite, and both source electrical resistance and loading can change the observed frequency response. Its resistor also produces thermal noise.

The active inverting low-pass amplifier provides a low-frequency voltage gain of magnitude five and, within the usable bandwidth, a small output impedance, so it can drive a load with much less change in frequency response. The input electrical resistance is approximately $R_1=1\mathrm{k}\Omega$ and may itself load the source. It needs a power supply and adds electrical noise and offset; finite operational amplifier gain, bandwidth, slew rate, available output swing, and output electric current limit the ideal formula. Its inversion must also suit the application. The best choice depends on required voltage gain, source and load electrical impedances, and the signal's frequency and amplitude.

Let $v_C$ denote the capacitor voltage with the same polarity as the applied voltage. By Kirchhoff's voltage law, Ohm's law and $I=Cd{v}_C/dt$,Solving this first-order differential equation gives the RC low-pass step responseThe capacitor voltage is continuous at switching because an instantaneous finite jump would require an impulsive electric current.

For the square-wave response of an RC low-pass filter, take equal positive and negative half-periods. Let $A=10\mathrm{V}$, $T=1/f$, $h=T/2$ and $\alpha =e^{-h/\tau }$. After the transient, symmetry makes the capacitor voltage alternate between $-v_p$ and $+v_p$. Over the positive half-period,Requiring $v_C(h)=v_p$ givesThe next half-period is the sign-reversed copy. These exact expressions explain the requested sketches:

Use an unloaded RC low-pass filter: a series resistor between the source and the output node, with a capacitor from that node to ground, and measure the output across the capacitor. The components are ideal in this first-order model; “ideal” here does not mean a brick-wall frequency response.

The time constant is $\tau =RC$, henceThe capacitive electrical impedance gives the transfer function $H(i\omega )=1/(1+i\omega RC)$, soOn the magnitude Bode plot, $20{\log }_{10}|H|=-10{\log }_{10}[1+(f/f_c{)}^2]$. The gain approaches $0\mathrm{dB}$ for $f\ll f_c$, is $-3.01\mathrm{dB}$ at $f_c$, and falls with slope $-20\mathrm{dB}$ per decade for $f\gg f_c$. These results assume negligible source resistance beyond the specified $R$ and a sufficiently large load impedance.

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 $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

An RC low-pass filter places a resistor $R>0$ between input and output, with a capacitor $C>0$ from output to their common return. The ideal model prescribes the input electric potential difference and leaves the output unloaded. It is a first-order low-pass filter with unit eventual gain.^[ref]

For electrical resistance $R$ and capacitance $C$, Ohm's law gives the electric current through the resistor as $(v_{\mathrm{in}}-v_{\mathrm{out}})/R$. The capacitor relation gives $Cd{v}_{\mathrm{out}}/dt$. Equating them by Kirchhoff's current law yields the ordinary differential equationThe time constant $\tau$ has units of time. A rapid change in electric potential difference requires a large electric current into the capacitor; the finite electrical resistance limits that rate. A slowly maintained voltage instead lets the capacitor approach the input voltage. The same equation therefore predicts both smoothing of sustained oscillations and a rounded response to a sudden change.

For the original constructed example throughout this article, takeThese stipulated ideal values specify an analytical model, rather than measurements of particular components.

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.

The reset dwell time below threshold in an RC pulse train is the continuous time for which the filtered electric potential difference lies below an electronic comparator's threshold between successive accepted pulses. A positive threshold-reset margin for a periodic RC pulse train ensures a crossing in the ideal model, but does not ensure a long enough low interval for the intended electronics.

Consider the periodic RC low-pass filter response to a high level $V_0>0$ lasting $T$, followed by zero input for a quiet time $D$. Let $\tau >0$ be the time constant, $V_b$ the periodic baseline at the start of the high interval and $V_p$ the peak at its end. Assume a nonhysteretic ideal electronic comparator and $V_b<V_{\mathrm{th}}<V_p$. Solving the two exponential segments gives the rising crossing time $t_{\uparrow }$, measured from the high interval's start, and the falling crossing delay $s_{\downarrow }$, measured from its end:The below-threshold interval runs from the falling crossing to the next rising crossing. ThereforeIncluding $t_{\uparrow }$ matters: the low interval usually extends into the next input high interval. The natural exponential decay underlying these segments is directly testable with the switched-RC apparatus in the MIT laboratory.^[ref]

For an original exact example, take $V_0=1\mathrm{V}$, $\tau =1\mu \mathrm{s}$ and $T=D=\tau \log 2$. The periodic solution in missing comparator reset in an RC pulse train gives $V_b=1/3\mathrm{V}$ and $V_p=2/3\mathrm{V}$. At $V_{\mathrm{th}}=0.5\mathrm{V}$, both crossing delays equal $\tau \log (4/3)$, so $t_{\mathrm{low}}=D\approx 0.693147\mu \mathrm{s}$. The reset margin is a positive $1/6\mathrm{V}$, yet a stipulated requirement of at least $0.8\mu \mathrm{s}$ low time is not met.

The strict crossing assumptions are essential. At $V_{\mathrm{th}}=V_b$, the waveform only touches the threshold and the low interval vanishes. At or above the peak there is no finite high interval to separate. A Schmitt trigger uses distinct switching thresholds, so its output-state interval requires the corresponding falling and rising crossings rather than substituting one threshold indiscriminately.

A proposed measurement records the pulse generator, capacitor and electronic comparator outputs on an oscilloscope. Measure the low interval between actual output edges while also locating both capacitor-level crossings. Their difference can expose switching delays; the capacitor formula alone does not specify a real device's minimum pulse duration.

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

The threshold-reset margin for a periodic RC pulse train is the separation between the reset threshold of an electronic comparator and the largest baseline that the periodic RC low-pass filter may leave between pulses. A positive margin excludes missing comparator reset in an RC pulse train in the ideal threshold model; its size measures tolerance to additional electric potential difference errors.

Let $V_0>0$ be the input high level, $T$ the high duration, $D$ the quiet duration and $\tau >0$ the time constant. From the periodic solution in missing comparator reset in an RC pulse train, writeThis defines a margin in volts, rather than in units of standard deviation. If a nominal baseline ${\bar{V}}_b$ has a known absolute error bound $b_b$ and the reset threshold ${\bar{V}}_{\mathrm{th}}$ has bound $b_{\mathrm{th}}$, a sufficient worst-case condition isThese are simultaneous bounds on the actual baseline and threshold. Adding their standard uncertainties would not by itself provide a deterministic guarantee; a probabilistic statement requires a specified joint probability distribution.^[ref]

For an original example, let $V_0=2\mathrm{V}$, $\tau =2\mu \mathrm{s}$ and $T=D=\tau \log 2$. Then $\alpha =\beta =1/2$, $V_b=2/3\mathrm{V}$ and the peak is $4/3\mathrm{V}$. A threshold of $0.800\mathrm{V}$ gives $m=0.133333\dots \mathrm{V}$. Bounds $b_b=0.040\mathrm{V}$ and $b_{\mathrm{th}}=0.020\mathrm{V}$ leave $m_{\mathrm{guaranteed}}=0.073333\dots \mathrm{V}$. Raising the baseline bound to $0.120\mathrm{V}$ would remove this guarantee even though the nominal waveform still crosses the threshold.

A positive minimum separation does not specify the reset dwell time below threshold in an RC pulse train. A real electronic comparator needs enough time to switch, and the peak must separately exceed the acceptance threshold. For a Schmitt trigger, use the lower switching threshold in $m$, while checking the upper one against the peak.

A focused bench test records the pulse generator output, capacitor output and electronic comparator output on an oscilloscope, then varies $D$ near the predicted boundary. Compare the measured minimum with the reset threshold and record the low interval as well as the minimum. The underlying exponential decay can be checked by the natural-response experiment in the MIT laboratory; the margin calculation above is an original application of that model.^[ref]