Linear filter

A linear filter obeys the superposition principle: filtering $\alpha x+\beta z$ gives $\alpha T[x]+\beta T[z]$. Many familiar examples are also linear time-invariant systems, represented by an impulse response and convolution. Linearity alone does not imply time invariance: multiplying $x[n]$ by the index $n$ is linear but changes its behavior when the same input is shifted.

The moving average is a finite impulse response example implemented by arithmetic; a band-pass filter can instead be built from a series RLC circuit. Their frequency responses explain which oscillations pass through. Choosing a filter means choosing how much amplitude, timing and detail may change, not simply removing every visible fluctuation.

For a linear time-invariant system, a sinusoidal wanted signal at ${\omega }_s$ and an unwanted sinusoidal component at ${\omega }_n$ acquire their separate frequency response factors. The ratio of their output amplitudes, divided by their input amplitude ratio, is $\eta =|H({\omega }_n)|/|H({\omega }_s)|$. An amplitude attenuation factor of $F=1/\eta$ corresponds to a factor $F^2$ in squared-amplitude power ratios under a common impedance convention. The unwanted component's own absolute attenuation is $|H({\omega }_n)|$, which need not equal its attenuation relative to the wanted signal.

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.

A maximally flat passband magnitude design. A normalized second-order low-pass denominator is $1+\sqrt{2}s/{\omega }_0+s^2/{\omega }_0^2$. An equal-component Sallen-Key low-pass filter realizes this damping when $K=3-\sqrt{2}$, with the corresponding non-inverting passband gain.

A second-order active low-pass architecture using a non-inverting amplifier and a resistor-capacitor network with output feedback into the network. For the convention with $R_1$ from source to A, $R_2$ from A to the non-inverting input, $C_2$ from that input to ground and $C_1$ from A to output, $a=C_1{R}_1(1-K)+C_2(R_1+R_2)$ and $b=R_1{R}_2{C}_1{C}_2$. The poles, not merely finite values of $G(i\omega )$, determine whether a settled sinusoidal response exists.^[ref]

For equal frequency-setting resistors $R$ and capacitors $C$, write $\tau =RC$. The denominator is $1+(3-K)s\tau +s^2{\tau }^2$. With positive $\tau$, asymptotic stability requires $K<3$; at $K=3$ the ideal poles are imaginary, and larger gain gives right-half-plane poles. In particular $K=5$ gives $(1-s\tau {)}^2$, a repeated unstable pole. A finite formal gain curve must not be confused with stable filter operation.

An RL circuit with the output measured across its resistor is a first-order low-pass filter. For an ideal inductor, an ideal voltage source and a high-resistance output instrument, the electrical impedances $i\omega L$ and $R$ giveThe gain approaches one at low angular frequency and $R/(\omega L)$ at high angular frequency; at ${\omega }_c$ it is $1/\sqrt{2}$. The time-domain equation is $L\dot{I}+RI=V_{\mathrm{in}}$ and $V_{\mathrm{out}}=RI$, so the time constant is $L/R$. Instrument loading, source resistance and nonideal inductance can change the response.

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.

For an equal-duty square wave alternating between $\pm A$ with period $T$, the steady periodic capacitor voltage has extrema $\pm v_p$. A positive half-period maps $-v_p$ to $A-(A+v_p)e^{-T/(2RC)}=v_p$, proving the title formula. The negative half-period is its sign-reversed copy. Short periods give a small nearly triangular output; long periods give input-like plateaux with exponential transitions.

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 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 real electrical load can draw electric current from the output. For a load resistor $R_L>0$ across the capacitor, Kirchhoff's current law instead givesTherefore the loaded continuous-time frequency response isLoading reduces the eventual gain and shortens the time constant. With the constructed values and $R_L=10\mathrm{k}\Omega$, $K_L\approx 0.81967$, ${\tau }_L\approx 84.754\mu \mathrm{s}$, and the corner relative to the loaded low-frequency gain moves to $1.8778\mathrm{kHz}$. A $3\mathrm{V}$ step now approaches $2.4590\mathrm{V}$. A quicker rise alone would conceal the error in the final voltage.

With the ideal input voltage source set to zero, the output impedance seen by the electrical load isThis follows from Thévenin's theorem: the resistor and capacitor are then in parallel. A sufficient small-loading condition is $|Z_L(\omega )|\gg |Z_{\mathrm{out}}(\omega )|$ over the frequencies of interest. An oscilloscope is itself an electrical load: its input electrical resistance and cable or probe capacitance can both change the response. Pure extra parallel capacitance increases $C$ and lowers the corner, whereas a finite parallel electrical resistance gives the different gain-and-corner changes above.

The definition of input also matters. If voltage is prescribed inside a source with series electrical resistance $R_s$, the unloaded response to that internal source uses $R+R_s$. If the actual voltage at the filter input terminal is measured, the ratio of output to that measured input still uses $R$. Measuring both terminals avoids mistaking a source setting for the voltage delivered to the filter.

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.

A causal filter uses only present and past input values to produce its current output. The trailing moving average is causal; a centered average that uses future samples is not. Future samples can be available when processing a finished recording, but they are unavailable to a live measurement at the current time.

A linear time-invariant system has finite impulse response when its impulse response is nonzero at only finitely many indices. For a causal filter supported at indices $0,\dots ,M-1$,This is a finite discrete convolution. The coefficients $h[j]$ specify the complete operation; they need not be positive or equal. Such a filter has a simple bounded-input estimate, $|y[n]|\le \Vert x{\Vert }_{\infty }{\sum }_j|h[j]|$, so bounded inputs remain bounded. The moving average chooses $h[j]=1/M$.

A moving average replaces each value by an average over a moving window of neighboring values. The simple trailing version with positive integer window length $M$ is the causal filterIt is a finite impulse response linear time-invariant system. A constant input remains unchanged because the weights sum to one. Weighted and centered averages are other choices; the definition of the window determines both the weights and whether future values are needed.

For an exact five-sample example, letdefined at every integer index. The first component completes one cycle in forty samples and the second in five. Averaging five consecutive samples cancels the second component exactly. The first becomesThe remaining component is slightly smaller and delayed by two samples; the filter does not recover the original component unchanged.

The example is in steady operation: its early output values use the specified negative-index samples. Starting a device at $n=0$ with an unknown past requires an initialization rule. Treating earlier samples as zero, shortening the initial window, and postponing output until a full window is available produce different initial values.

For zero-mean uncorrelated random variables $\epsilon [n]$ with common variance ${\sigma }^2$, the averaged disturbance hasThe standard deviation decreases by $\sqrt{M}$, but successive filtered errors share samples and are generally correlated. This variance calculation also assumes the wanted component is not damaged by averaging. A rapidly changing feature can be suppressed along with the disturbance.

The frequency response explains the cancellation and limits of this method. A moving average preserves low frequencies comparatively well, yet its side lobes make it a poor substitute for a sharply selective frequency cutoff. Smith develops this time-domain versus frequency-domain distinction in The Scientist and Engineer's Guide to Digital Signal Processing, chapter 15: moving-average filters.

A band-pass filter preferentially transmits signals within a range of frequencies and attenuates sufficiently lower and higher frequencies. Taking the output voltage across the resistor in a series RLC circuit gives one example: the resonant series RLC response selects a band around $1/\sqrt{LC}$. ^[ref].

A series resistor $R$ and capacitor $C$ feed a parallel pair with the same values. Measuring the parallel pair's voltage relative to the return gives this unloaded transfer function for phasors proportional to $e^{i\omega t}$. The voltage divider applied to $Z_s=R+1/(i\omega C)$ and $Z_p=R/(1+i\omega RC)$ proves the formula. The amplitude peaks at $\omega =1/(RC)$ with ratio $1/3$ and zero relative phase. At low and high positive frequencies, the transfer function tends respectively to $i\omega RC$ and $-i/(\omega RC)$. Reversing output polarity changes its sign, while loading changes the response itself.

A chosen boundary in a frequency response, commonly where the voltage amplitude gain falls to $1/\sqrt{2}$ of its pass-region value. The convention must specify the reference gain.

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