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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.

A wirewound resistor uses a metal alloy chosen for stable electrical resistivity. Compared with a typical carbon composition resistor, it can offer tighter resistor tolerance, a smaller temperature coefficient of resistance, lower excess electrical noise, and greater continuous electrical dissipation when its construction provides an adequate heat path. These are design-dependent advantages, rather than guarantees for every alloy or package.

The winding has parasitic inductance and interturn capacitance. Its electrical impedance therefore ceases to be a nearly real electrical resistance at sufficiently high frequencies; an approximate model is a series circuit $R+i\omega L$ shunted by a small parasitic capacitance. A resonance can occur, making a wirewound resistor unsuitable where a nearly frequency-independent electrical resistance is required. Noninductive winding patterns reduce this problem. For comparable small-signal duties it is usually larger and more expensive. A carbon composition resistor is compact and has relatively small parasitic inductance, but commonly has poorer stability, larger resistor tolerance, and more excess electrical noise.

Use phasors with time dependence $e^{i\omega t}$ and electrical impedance $Z=\tilde{V}/\tilde{I}$. The element laws $V=IR$, $I=CdV/dt$ and $V=LdI/dt$ giveA complex number records both amplitude and phase, turning the linear differential equations of a sinusoidally driven electrical circuit into algebraic equations. A resistor has in-phase voltage and electric current and dissipates average power. For an ideal capacitor, the electric current leads the voltage by $90^{\circ }$; for an ideal inductor, it lags by $90^{\circ }$. Their imaginary electrical impedances express reversible energy storage and return, with zero net average power over a period. Using $e^{-i\omega t}$ would conjugate the displayed reactive signs without changing predictions.

In the illustrated Maxwell bridge, the left and right vertices are the supply terminals and the upper and lower vertices are detector terminals. Insert a sensitive AC null detector between the latter and drive the former with a sinusoidal source. The upper-left arm is $Z_x=R_x+i\omega L_x$, the upper-right arm is $R_3$, the lower-left arm is $R_2$, and the lower-right arm is the parallel combinationAt a null no detector electric current flows, so the two voltage dividers giveMultiplying by the parallel-arm electrical admittance yieldsEquating real and imaginary parts therefore givesTwo independent adjustments are needed to annul the in-phase and quadrature detector signals. For example, change $R_2$ to set $L_x$ and $R_1$ to set $R_x$, iterating until the detector reads zero. Alternatively, with fixed $R_2,R_3$, independently vary $R_1$ and $C$.

The Maxwell bridge compares an unknown lossy inductor with a stable standard capacitor, avoiding the need for an accurately known reference inductor. The ideal balance equations contain no frequency, and a null measurement need not rely on absolute detector calibration. In practice choose a suitable frequency below the coil's self-resonance, where the series $R_x,L_x$ model and the standard capacitor remain accurate.

If the initial setting is far from balance, the bridge as drawn needs adjustable components and a detector, not merely a more precise readout. Fit resistance boxes or switched ranges for $R_1$ and a ratio arm, and a selectable standard $C$. Choose ranges aroundThen approach the null with coarse adjustments followed by finer ones. A large required $R_1$ can be brought into range by increasing $C$ and decreasing $R_2{R}_3$ proportionally. Start with reduced excitation or reduced detector sensitivity to avoid overload, then increase sensitivity near the null. Changing frequency alone cannot repair an impossible component range, since the ideal balance conditions are frequency-independent. Stray capacitance and lead electrical resistance set practical accuracy limits and may require shielding and short connections.

In the parallel LC circuit both components have the same voltage phasor. Adding their current phasors by Kirchhoff's current law gives the total electrical admittance:ThusAt the ideal parallel resonance, ${\omega }_0=1/\sqrt{LC}$, the two branch current phasors cancel. The net electrical admittance is zero and the magnitude of the complex impedance tends to infinity: the parallel combination behaves as an open circuit at that frequency. Its complex argument is undefined exactly at the pole. Finite component losses replace the infinite peak by a large finite electrical impedance. This conclusion concerns a voltage-driven sinusoidal response; an ideal resonant current drive does not have a bounded sinusoidal steady-state voltage.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to $R_x$. With the detector removed, no electric current flows down that branch, so $V_R=V_0$ and $V_L=V_0/2$. ThereforeFor the connected ideal ammeter, the resulting electric current from right to left is $V_0/(R+2R_x)$, despite its zero terminal voltage. This follows from $I_A=V_{\mathrm{oc}}/Z_{\mathrm{th}}$ with $Z_{\mathrm{th}}=R/2+R_x$.

Let the source's top terminal have voltage phasor $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left resistors are equal, so at balance $V_L=V_0/2$. Zero electric current through the ammeter requires $V_R=V_L$, and therefore equal upper and lower right-arm electrical impedances:Write $x=\omega R_x{C}_x$. Separating real parts and imaginary parts givesThe second equation implies $1+x^2={\omega }^2{R}_x^{2}C{C}_x$. Subtract $x^2={\omega }^2{R}_x^2{C}_x^2$ to obtainA finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity byFor a detector of finite electrical impedance $Z_A$, its terminal voltage is insteadHere $Z_{\mathrm{th}}$ is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit $Z_A\to 0$ has zero terminal voltage even when the bridge is unbalanced.

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.

Electrical noise reduction starts by identifying what is unwanted in the measured signal and where it enters. Increasing gain amplifies both signal and input noise; it does not by itself improve the input signal-to-noise ratio. Separate intrinsic component fluctuations, amplifier noise and external pickup before choosing a remedy.

Physical sources. A resistor at temperature $T$ has classical one-sided open-circuit thermal electrical noise density $S_V=4k_{B}TR$ per hertz. In a flat measurement bandwidth $B$, its rms voltage is $\sqrt{4k_{B}TRB}$. Lowering an appropriate resistance, reducing temperature or narrowing the useful bandwidth can reduce this noise, with loading and signal-gain tradeoffs considered. These are power spectral densities; an amplitude spectral density is their square root.

Discrete independent charge transfers give shot noise, with one-sided current density $S_I=2eI$ in the simple Poisson process model. Reducing leakage or unwanted bias current can help, but reducing a photocurrent may also reduce the desired signal. Charge correlations can change the simple shot-noise law. At low frequency, flicker noise and drift often dominate and are not described by a white spectrum.

Amplifier and impedance choice. An operational amplifier has equivalent input voltage and current noise. In a simplified uncorrelated model, a resistive source contributes the input-referred voltage densityHere $e_n,i_n$ are the voltage and current amplitude spectral densities and $Z_s$ is the source electrical impedance; feedback components add further terms in a full circuit. Low current noise matters particularly for a high-impedance source, while voltage noise can dominate for a low-impedance source. Minimize needless source impedance, leakage and long sensitive connections without sacrificing the required transduction gain or bandwidth. Bias-current offsets and thermal drift also require calibration; they are not removed simply by quoting a random-noise density.

Pickup and bandwidth. Shield sensitive circuits against electric-field coupling, minimize loop area to reduce changing magnetic-flux pickup, and use twisted signal/return pairs or suitable coaxial connections. Differential readout rejects common-mode interference when impedances and amplifier responses are adequately matched. Plan the signal returns to avoid unwanted circulating ground currents, and decouple supply rails locally so supply fluctuations do not enter the signal path. Mechanical vibration, cable motion and temperature changes may be converted into electrical noise as well.

A low-pass filter removes out-of-band noise before digitization; an anti-aliasing filter is needed because digital filtering cannot undo already aliased content. Avoid a narrower passband than the physical signal requires. For a known periodic signal, modulation and a lock-in amplifier select a narrow frequency interval, often moving the measurement above a strong flicker noise region. A notch can reject a narrow interference line if that does not remove essential signal information.

Independent noise contributions add in variance, not directly in rms amplitude. Averaging $N$ independent equal-variance readings reduces the standard error by $N^{-1/2}$, but slow correlated drift and systematic errors need separate controls. Good noise reduction preserves the information-bearing signal while reducing unwanted coupling, intrinsic fluctuations or the admitted noise bandwidth.

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.

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.

Mutual inductance expresses how a changing electric current in one circuit induces an electromotive force in another. Choose winding orientations and define their linked magnetic fluxes byFor a reciprocal linear magnetic system, the same $M$ appears in both expressions. Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$ when the secondary electric current does not itself change. The sign depends on winding orientation and is consistent with Lenz's law.

The stored magnetic energy isIts nonnegativity for arbitrary electric currents requires $M^2\le L_1{L}_2$. Strong coupling makes $|M|$ approach $\sqrt{L_1{L}_2}$; flux leakage makes it smaller.

An ideal electrical transformer has two windings linked by the same time-varying core magnetic flux. The induced winding voltages are proportional to their turn counts. For appropriately chosen terminal polarities,The electric current convention in the power relation takes primary power into the device and secondary power out to the load. Consequently a load electrical impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_1/N_2{)}^2{Z}_L$. A larger secondary voltage therefore comes with a smaller available secondary electric current, rather than extra energy.

A steady direct electric current produces no sustained transformer action after its switching transient. Practical departures from the ideal include winding electrical resistance, leakage magnetic flux, finite magnetizing electric current, core saturation, eddy currents and magnetic hysteresis losses.

In a homogeneous Ohmic conductor, $\mathbf{J}=\sigma \mathbf{E}$. Combining this with Maxwell's equations givesFor a good conductor, $\sigma \gg \omega ϵ$, so conduction dominates displacement current. With time dependence $e^{-i\omega t}$ and depth $z>0$, take $E\propto e^{ikz}$. ThenThe field is consequently proportional to $e^{-z/\delta }{e}^{iz/\delta }$: it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by $e^{-1}$; power density falls as $e^{-2z/\delta }$, so its attenuation length is $\delta /2$.

Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys ${\partial }_t\mathbf{B}=(\mu \sigma {)}^{-1}{\nabla }^2\mathbf{B}$. Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces $\delta$, provided the relevant material parameters remain applicable.

For a thick conductor the tangential surface electrical impedance is $Z_s=(1-i)/(\sigma \delta )$ with the stated time convention and consistent field orientation. Its real part $1/(\sigma \delta )$ represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when $\delta$ is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When $\sigma /\gg \omega ϵ$, the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.

Use $e^{i\omega t}$ phasors. The relation $V=L\dot{I}$ gives $Z_L=i\omega L$. The three branches form a parallel RLC circuit and share the source voltage. The source-current phasor and its real time dependence areConsequently its amplitude is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L){)}^2}$ and its phase relative to the voltage is ${\tan }^{-1}[R(\omega C-1/(\omega L))]$. At low frequency it is inductive and lags toward $-\pi /2$; at high frequency it is capacitive and leads toward $+\pi /2$. The ideal amplitude diverges at either extreme and has a finite minimum between them; real component losses and source electrical impedance limit those asymptotes.

Zero phase requires cancellation of electrical susceptances:The inductor and capacitor currents are nonzero, equal in magnitude and opposite in phase, while the resistor supplies the in-phase source current. energy oscillates between the reactive elements; dissipation occurs in the shunt resistor. A source-current maximum would describe a different, series-resonant circuit.

The specified ratio isIt exceeds $1/\sqrt{2}$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and ${\omega }_0=1/\sqrt{LC}$. The two positive endpoints are ${\omega }_{\pm }=[\sqrt{b^2+4{\omega }_0^2}\pm b]/2$, soThe endpoints have geometric mean ${\omega }_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a series RLC circuit. Equivalently, $Q={\omega }_0{U}_{\max }/P_{\mathrm{loss}}$, with $U_{\max }=C{V}_0^2/2$ and $P_{\mathrm{loss}}=V_0^2/(2R)$. The parallel RLC bandwidth here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.

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.

The original diagram places the output across the $10\Omega$ resistor, in series with an ideal $1\mathrm{mH}$ inductor. Treat the voltages as phasors with time dependence $e^{i\omega t}$. The electrical impedances are $Z_R=R$ and $Z_L=i\omega L$, so the same electric current flows through both:This is an RL low-pass filter, with corner angular frequencyThe frequency response tends to one for $\omega \ll {\omega }_c$, has magnitude $1/\sqrt{2}$ at the corner, and tends to ${\omega }_c/\omega$ for $\omega \gg {\omega }_c$. The requested sketch uses a logarithmic horizontal axis and a linear voltage-gain axis:

At the two specified angular frequencies,The relative attenuation by a linear filter is the ratio of these gains:Thus the noise amplitude is attenuated by a factor of 10 relative to the signal. A squared-voltage ratio would decrease by a factor of $100$; the question's voltage transfer function gives the amplitude comparison.

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

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.