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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 positive angular frequency, the squared amplitude response of the series-parallel RC band-pass filter isIts denominator is minimized exactly when $w=w^{-1}$, soThis is a band-pass filter maximum produced entirely by the two RC arms; the drawn electrical circuit contains no inductor. The low- and high-frequency limits areThus the output amplitude tends to zero linearly with $\omega$ at low frequency and as $1/\omega$ at high frequency. In real time,The upper series capacitor suppresses low frequencies, while the lower shunt capacitor suppresses high frequencies.

The precise bandwidth depends on the output: this is the distinction between displacement and current bandwidths of a series RLC circuit. For current output, or voltage across the resistor, the electrical impedance givesThe half-power points satisfy $|\omega L-1/(\omega C)|=R$. Their positive roots areChanging capacitance shifts the centre without changing this absolute width. For the charge response in part (d), the exact half-height points instead obeywhen both exist their separation only approaches $\gamma =R/L$ at large quality factor. Therefore the printed capacitance-independence claim is exact for current or resistor-voltage output and an approximation for the preceding charge resonance.

For a narrow band-pass filter, use current or resistor-voltage output and make $R/L$ small, counting source, load and inductor losses in the total electrical resistance. Increase $Q=\sqrt{L/C}/R$; if a specified centre frequency is to remain fixed, adjust $C$ with $L$ to keep $LC$ fixed. Capacitor-voltage output alone is not a band-pass response because it has a nonzero DC limit.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.

Drive a series RLC circuit with electric potential difference $V(t)=V_0\cos (\omega t)$, where $V_0>0$ and $\omega >0$. After the free transient decays, the electric current has the same angular frequency as the source:For fixed positive $R,L,C,V_0$, the electric current amplitude is largest exactly at $\omega ={\omega }_0=1/\sqrt{LC}$. The two electrical reactance contributions cancel there, leaving only the drop from electrical resistance. In particular,Every solution approaches this periodic response because its difference from the displayed solution is a decaying free response. The peak remains at ${\omega }_0$ even for an overdamped RLC response; free ringing is not required for a maximum of driven electric current. ^[ref].

For example, stipulate $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$, $R=20\Omega$, and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4{\mathrm{s}}^{-1}$, $f_0={\omega }_0/(2\pi )\approx 1591.55\mathrm{Hz}$, and the peak electric current amplitude is $50\mathrm{mA}$. Holding the source amplitude fixed is essential when comparing responses at different frequencies.

Three frequencies must be distinguished. The free underdamped RLC response rings at ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$. Driven electric current peaks at ${\omega }_0$. Driven electric charge has amplitudewhose peak occurs at $\sqrt{{\omega }_0^2-2{\alpha }^2}$ only if $\alpha <{\omega }_0/\sqrt{2}$. Otherwise it decreases from its zero-frequency limit $C{V}_0$, with no positive-frequency peak. The different peak locations follow because electric current is the time derivative of electric charge, multiplying a sinusoidal amplitude by $\omega$. The limiting responses are $I_{\mathrm{amp}}\sim V_{0}C\omega$ as $\omega \to 0^{+}$ and $I_{\mathrm{amp}}\sim V_0/(L\omega )$ as $\omega \to \infty$.

The Q factor measures selectivity:The average Joule heating power is $R{I}_{\mathrm{amp}}^2/2$. Its half-maximum angular frequencies and their separation are exactlyThese follow by setting $|\omega L-1/(\omega C)|=R$. In the numerical example, $Q=5$ and $\Delta \omega =2000{\mathrm{s}}^{-1}$. Reducing $R$ makes a taller, narrower electric current peak; varying $C$ tunes the selected frequency. Measuring the drop $v_R=Ri$ gives a band-pass filter that suppresses very low and very high frequencies while passing a range near resonance. This is useful for selecting a radio signal, though the range must be broad enough to pass its modulation. ^[ref].

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

MIT's laboratory guide mounts resistors on an aluminum box beneath a metal beaker shield, with short twisted leads. Its SR560 amplifier uses nominal gain $1000$ before a Krohn-Hite $1$–$50\mathrm{kHz}$ band-pass filter; an oscilloscope reads root mean square voltage. Calibration sweeps an attenuated sinusoid and measures the actual gain $g(f)$. Students alternate the resistor and a short circuit, subtracting squared readings $V_R^2-V_S^2$, and disconnect the ohmmeter during noise acquisition. Lead and input capacitance enter the gain-squared integral. With settings unchanged, varying electrical resistance or temperature tests the predicted mean square. The guide prescribes repeated readings and warns against averaging noise waveforms to zero. This is a documented teaching protocol, not a supplied dataset of measured outcomes.^[ref]

The subtraction requires care: shorting the input changes the source electrical impedance. An amplifier's input current noise can therefore produce different voltage noise in the two configurations; a shorted baseline alone does not remove that contribution. Gain drift, interference and insufficient observation time also limit the inference. These are limitations of interpreting the measurement, not corrections that can be inferred from the nominal signal bandwidth alone. In an independent analysis one would compare the measured difference withwhere $C$ represents the measured shunt capacitance and $g$ excludes that separately modeled shunt. If calibration already includes the same shunt response, applying the denominator again counts the attenuation twice. The thermal noise worked example and both original figures illustrate ideal models; they claim no measured agreement or measured value of the Boltzmann constant.

Use the equal-component Wien bridge with upper-left parallel arm and lower-left series arm. Let $A$ be the left arm junction, $C$ the upper input node, and $V_i$ the upper-minus-lower input electric potential difference. With $u=\omega RC$, the unloaded transfer ratio $H=(V_A-V_C)/V_i$ is the displayed expression. Its band-pass filter magnitude is $[9+(u-u^{-1}{)}^2{]}^{-1/2}$, maximal at $u=1$ with $H=-1/3$. One continuous wave phase is $\varphi =-\pi -\arctan [(u-u^{-1})/3]$, tending from $-\pi /2$ at low positive frequency to $-3\pi /2$ at high frequency. Reversing the measured output polarity multiplies $H$ by $-1$ and shifts its phase by $\pi$ without changing its magnitude.