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An operational amplifier is a differential electronic amplifier whose unloaded voltage is modeled in its linear regime by $v_o=A(s)(v_{+}-v_{-})$. Its large open-loop gain, large input impedance and small output impedance make negative feedback useful for setting a reproducible transfer function with external components. Supply voltages bound the available output swing; the linear model does not apply after saturation.

Two parallel branches with complex impedances $R(1+is)$ and $R(1-is)$ have a real input impedance $R(1+s^2)/2$, since their admittances have cancelling imaginary parts. For $s=\sqrt{15}$, adding a series resistance $2R$ gives total impedance $10R$. With source voltage amplitude $V_0$, each branch-resistor drop and the series-resistor drop has amplitude $V_0/5$, despite their different phases.

Use the following arrangement: an inverting operational amplifier with input resistor $R_a$ and feedback resistor $R_f$, followed by an op amp integrator with input resistor $R_i$ and feedback capacitor $C_i$. Ground both non-inverting inputs. Each ideal operational amplifier has infinite input impedance, so its inputs draw no electric current; negative feedback makes the inverting input a virtual ground.

Kirchhoff's current law at the first summing junction gives $V_1=-(R_f/R_a)V_{\mathrm{in}}$. At the second summing junction it gives $V_1/R_i+C_{i}d{V}_{\mathrm{out}}/dt=0$. Consequently the two inversions cancel:The initial output voltage records the initial capacitor charge. A reset switch across $C_i$ sets it to zero. The formula holds while both stages remain within their supply rails; a persistent input voltage otherwise produces amplifier saturation.

The non-inverting input is grounded, so the finite voltage gain gives $V_{\mathrm{OUT}}=-A{V}^{-}$, or $V^{-}=-V_{\mathrm{OUT}}/A$. Infinite input impedance makes Kirchhoff's current law at the inverting inputSubstitute $V^{-}=-V_{\mathrm{OUT}}/A$ and multiply by $R_1{R}_2$:Hence the inverting operational amplifier hasAs $A\to \infty$, the voltage gain tends to $-R_2/R_1$. Setting $V^{-}=0$ before taking that limit would discard precisely the correction being calculated.

An ideal operational amplifier has infinite input impedance, so neither input draws electric current, and zero output impedance, so its prescribed output voltage is independent of the output electric current within the ideal model. These are the two additional assumptions needed here.

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.

An ideal operational amplifier has infinite input impedance and open-loop gain, zero output impedance and no offsets or response limits. Its input currents vanish. Stable operational-amplifier negative feedback, a finite output and unsaturated linear operation then imply a virtual short, $v_{+}=v_{-}$. Real amplifiers have finite frequency-dependent gain, input bias currents, offset voltage, output loading, finite bandwidth, slew rate and supply-rail limits. Equal input voltages cannot be assumed in an open-loop or saturated circuit.

Here the noninverting input is grounded. In the ideal-feedback limit the inverting node is a virtual ground, so the resistor current charges the feedback capacitor:The initial capacitor charge fixes the integration constant. This is an op amp integrator, with $H(i\omega )=-1/(i\omega RC)$.

A tiny direct-current input, input bias current or offset can drive the output towards saturation. A parallel feedback resistor $R_f$ provides a direct-current return path, giving $H(s)=-(R_f/R)/(1+s{R}_{f}C)$; choose $\omega R_{f}C\gg 1$ over the intended integration band. A reset switch sets initial charge; low-offset components and suitable signal amplitude reduce drift and saturation. The useful band must also remain below the amplifier's bandwidth and slew-rate limits.

For the stipulated finite but constant real gain $A=100$, infinite input impedance and negligible output impedance, write the inverting-node voltage as $v=-V_{\mathrm{out}}/A$. With phasors proportional to $e^{i\omega t}$,ThereforeThe low-frequency gain is $-100$. The corner is ${\omega }_c=1/(101RC)$; above it the amplitude falls as $100/(101\omega RC)$ and the phase approaches $+\pi /2$. phase $+\pi$ and $-\pi$ describe the same low-frequency inversion. This finite-gain circuit is a low-pass response with an integrating high-frequency regime, not an ideal integrator down to zero frequency. A genuinely frequency-dependent $A(i\omega )$ would introduce additional high-frequency limitations beyond this specified model.

A representative bipolar-input 741 amplifier has $R_{\mathrm{in}}$ of order $10^6\Omega$; the LM741 specification gives a typical $2\mathrm{M\Omega }$ at $25^{\circ }\mathrm{C}$. This is an illustrative device specification, not an identification of an unspecified practical-class part. Other input architectures can have much larger electrical resistance, and finite input capacitance makes the input impedance frequency dependent.^[ref]

An ideal operational amplifier has infinite differential open-loop gain, infinite input impedance, zero output impedance and unlimited bandwidth and output capability, and zero input offset voltage and input bias current. In a stable negative-feedback circuit operating linearly, its input currents vanish and $V_{-}=V_{+}$. This equality is a feedback consequence, not a rule for a saturated or open-loop amplifier. Here the grounded non-inverting input makes the inverting node a virtual earth.

The inverting two-pole band-pass filter has a series resistor and capacitor input, while the feedback arm is a parallel resistor and capacitor. With Laplace variable $s$,Kirchhoff's current law at virtual earth gives the minus sign. Substituting $s=i\omega$ yieldsIt vanishes at zero and infinite frequency, with low-frequency $|H{|}^2\simeq (\omega R_2{C}_1{)}^2$ and high-frequency $|H{|}^2\simeq 1/(\omega R_1{C}_2{)}^2$. Differentiation gives the single maximumFor ${\tau }_1\gg {\tau }_2$, the broad interval $1/{\tau }_1\ll \omega \ll 1/{\tau }_2$ has $|H|\simeq R_2/R_1$. For ${\tau }_2\gg {\tau }_1$, the interval $1/{\tau }_2\ll \omega \ll 1/{\tau }_1$ instead has $|H|\simeq C_1/C_2$. The same rising and falling asymptotes apply outside either plateau.

Writing $a={\tau }_1+{\tau }_2$ and $b={\tau }_1{\tau }_2$, the half-power frequencies and quality factor areEqual time constants give the narrowest fractional pass-band achievable here, with ${\omega }_{\pm }=(\sqrt{2}\pm 1){\omega }_0$ andThis topology cannot make a genuinely sharp, high-$Q$ band-pass. Its poles are real, and the selectivity bound for a two-real-pole band-pass prevents arbitrarily narrow selection. A more selective resonant filter needs additional circuit structure. The gain above answers the component-ratio question for the sharpest available setting without assuming an impossible limit.

Real amplifiers have finite frequency-dependent gain and phase lag, so virtual earth becomes inaccurate, the response changes and stability must be checked. Gain-bandwidth limits, slew rate for large signals, output loading and amplifier saturation also matter. Component tolerances move the corners; capacitor parasitics, stray capacitance, resistor noise and input noise become significant. The ideal high-frequency tail should not be extrapolated beyond the amplifier and lumped-component operating ranges.

The ideal voltage-feedback Operational amplifier has zero input currents and, while stable negative feedback holds it in its linear operating range, equal input voltages: $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$. The first rule follows from infinite input impedance; the second is the large-open-loop-gain limit of $v_o=A(v_{+}-v_{-})$. Equal input electric potential is a virtual short, not an electrically conducting connection.

For an inverting amplifier with the noninverting input grounded, $v_{-}=0$ and node-current balance gives $v_i/R_i=-v_o/R_f$, hence $v_o=-R_f{v}_i/R_i$. For a noninverting amplifier, the feedback divider gives $v_{-}=v_o{R}_g/(R_f+R_g)=v_i$, hence $v_o=(1+R_f/R_g)v_i$.

One should check output swing, input common-mode range, slew rate, Gain-bandwidth product and stability before using these limits in a real circuit. Saturation, positive feedback or operation beyond bandwidth invalidates the equal-voltage rule; finite input bias currents and offset voltage can also matter.^[ref]

Use an inverting operational amplifier. Ground the non-inverting input, connect the source to the inverting input through $R_{\mathrm{in}}$, and connect $R_f$ from the output back to that same input.

In linear operation the ideal operational amplifier draws no input electric current, and feedback holds the inverting input at the same electric potential as the grounded non-inverting input. This node is a virtual ground. Kirchhoff's current law there givesThe source supplies current $I_i=V_i/R_{\mathrm{in}}$, so its input impedance is $R_{\mathrm{in}}$. The required design is thereforeOnly these two resistors and the single ideal operational amplifier are needed in the signal circuit; its power supplies are implicit. The result assumes an input small enough to keep the output within its linear operating range.

A systematic error is a reproducible bias in the measurement model or procedure, such as a zero offset, incorrect gain, background contamination, misalignment, loading or temperature-dependent calibration. Repeating the same biased procedure more often reduces independent random scatter but does not remove the bias. An unknown correction also has a measurement uncertainty, even when the mechanism is systematic.

A simple instrument model is $y=(1+g)x+a+d(t)+ϵ$, with gain error $g$, offset $a$, drift $d(t)$ and random noise $ϵ$. Comparison with calibrated standards at several input values separates an offset from a scale error and can test nonlinearity. Use standards that cover the operating range, verify units and reference values, and propagate uncertainty in the standards into the corrected result. Calibration changes the model; it does not make the corrected measurement exact.

A blank or source-off measurement estimates additive background. For independent source-plus-background count $K_s$ in time $t_s$ and background count $K_b$ in time $t_b$, the net-rate estimate and its counting variance areSubtracting a measured background removes its estimated mean, not its fluctuations or its uncertainty. A long background exposure can reduce the latter, while drifting backgrounds require monitoring or a joint model.

Reversal can distinguish a wanted signal from an unchanged offset. For example, $V(+I)=IR+V_{\mathrm{th}}$ and $V(-I)=-IR+V_{\mathrm{th}}$ give $R=[V(+I)-V(-I)]/(2I)$ when the thermoelectric offset is stable between readings. Interchanging sample and reference, reversing field direction or reversing an apparatus orientation can similarly expose sign-dependent biases, but only if the wanted and unwanted contributions transform differently.

Track slow instrumental drift using frequent references, interleaved readings or an ABBA measurement sequence. Equal-time readings in the order A, B, B, A cancel a common linear drift in the difference estimator $(y_1+y_2-y_0-y_3)/2$. Nonlinear drift or unequal sample conditions can survive. Randomizing run order helps prevent a time drift from imitating dependence on a controlled experimental variable.

Control physical couplings directly: stabilize temperature, minimize gradients, shield unwanted fields, align geometry, and use high input impedance when loading is the concern. Differential and bridge measurements suppress common changes when their channels are well matched. Varying an apparatus parameter or using an independent method can reveal an unsuspected bias. Report residual corrections, their uncertainties and correlations; adding more identical readings cannot justify discarding a shared calibration uncertainty.

An ideal operational amplifier has infinite differential open-loop gain, infinite input impedance, zero output impedance, unlimited bandwidth and slew rate, and zero input offset voltage, input bias current and intrinsic noise. The idealized model has no finite output-swing or output-current limit.

The usual circuit rules areThe first follows from infinite input impedance. The second follows because a finite output requires $v_{+}-v_{-}=v_o/A\to 0$ as the open-loop gain tends to infinity. It is a virtual equality of potentials, not a conducting connection between the inputs. It cannot be imposed for open-loop operation or after an amplifier reaches its supply rails.

A real operational amplifier has large but finite, frequency-dependent open-loop gain; the gain falls and its phase changes at higher frequencies. Finite bandwidth and slew rate constrain small-signal and large-signal responses respectively. There are also finite input impedance and output impedance, nonzero input bias currents and input offset voltage, and internally generated noise. Offset and bias can drift with temperature.

The output is limited by the supply voltages and output-current capability, and the inputs have a permitted common-mode range. Rejection of common-mode and supply changes is finite. Consequently $v_{+}-v_{-}$ need not vanish exactly, and neither the virtual-ground approximation nor linear gain formulas apply after saturation. For the calculation below the stipulated infinite input impedance removes the input loading currents, while finite gain and optionally finite output resistance are retained.

A non-ideal amplifier is not specified by infinite input impedance alone: its open-loop gain and output impedance must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit finite-gain operational-amplifier integrator model.

Writing Kirchhoff's current law at the input and output nodes givesThe second equation gives $y=(sC{r}_o-A)v/(1+sC{r}_o)$. Substitution in the first gives the general transfer functionFor sinusoidal phasors, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the amplitude ratio and their arguments give the relative phase.

The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It givesThus below ${\omega }_l=[(1+A_0)RC{]}^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that frequency it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq 1/(\omega RC)$ and phase $+90^{\circ }$. Constant finite gain by itself predicts no further high-frequency turnover.

For a more realistic sketch, use the standard dominant-pole open-loop gain $A(s)=A_0/(1+s/{\omega }_a)$, still with $r_o=0$. ThenThe exact frequency response isAssuming $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, the turnover frequencies are well separated: ${\omega }_l\simeq 1/(A_{0}RC)$ and ${\omega }_h\simeq A_0{\omega }_a$, the approximate unity-gain angular frequency of the amplifier. The regimes arewith magnitude slopes $0,-20,-40\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^{\circ }$ and $-180^{\circ }$ denote the same phase.

If finite $r_o$ is retained and $A(i\omega )\to 0$, the general expression instead approaches $r_o/(R+r_o)$ at very high frequency: the capacitor couples the input to the output through a passive voltage divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,{\omega }_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.

Digital acquisition converts a continuous voltage signal into timed numerical samples. A practical chain is a suitably buffered and amplified sensor signal, an analogue anti-aliasing filter, a sampling or sample-and-hold stage, an analog-to-digital converter, and a clocked data-storage system. Choose the gain so that the interesting variation uses the available input range without clipping. Finite input impedance can load a sensor, so an appropriate buffer or operational amplifier may be needed.

Sampling makes time discrete. If the conditioned signal is confined below frequency $f_{\max }$, choose a sampling rate $f_s>2f_{\max }$, allowing a transition band for the physical filter. The Nyquist–Shannon sampling theorem permits ideal reconstruction of a suitably bandlimited function from exact samples. Without filtering, components separated by integer multiples of $f_s$ have identical samples; in particular real sinusoids at $f$ and $f_s-f$ can be indistinguishable. This aliasing cannot generally be undone by later digital filtering. Oversampling can ease filter design, but does not replace the requirement to suppress unwanted high-frequency content before sampling.

Amplitude quantization is separate from sampling. A uniform $b$-bit converter covering voltage span $V_{\mathrm{span}}$ has step $q\simeq V_{\mathrm{span}}/2^b$. Rounding an in-range input gives the uniform-quantization error bound $|ϵ|\le q/2$. If the rounding errors can additionally be modeled as uniform and uncorrelated, their standard deviation is $q/\sqrt{12}$. This statistical estimate is not a guarantee for a constant input or a coherent periodic signal. More bits increase resolution, but accuracy also depends on voltage-reference calibration, offset, gain error, nonlinearity and electronic noise. Out-of-range signals clip and invalidate the rounding bound.

Clock quality matters too. Timing jitter $\delta t$ gives a voltage error approximately $(dV/dt)\delta t$; for a sinusoid its scale grows with $2\pi f{V}_p{\sigma }_t$. A clock calibration error biases measured frequencies and time intervals. Simultaneous multichannel signals need known relative timing, and the sampling and analogue settling times must be appropriate to the signal bandwidth.

For $n_c$ channels the raw recording rate is approximately $n_c{f}_{s}b$ bits per second, before overhead, so storage and sustained transfer speed must prevent missing data. Store units, channel identities, timestamps, sampling rate, input ranges and calibration information along with the readings. Preserve the original samples when applying digital filters or other processing. Averaging can improve precision for independent fluctuating errors, but cannot restore aliased frequencies, clipped peaks or a fixed rounding bias. Digital recording makes reproducible analysis and transmission possible when its sampling, amplitude and timing limitations are all controlled.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

The reflected impedance is the contribution of a coupled secondary load to a primary winding's input impedance. For two ideal mutually coupled inductors, choose winding signs so that positive electric currents reinforce their magnetic fluxes. If the secondary loop has complex impedance $Z_s$, its Kirchhoff's voltage law gives ${\tilde{I}}_s=-i\omega M{\tilde{I}}_p/Z_s$. Substitution in ${\tilde{V}}_p=i\omega L_p{\tilde{I}}_p+i\omega M{\tilde{I}}_s$ gives $Z_{\mathrm{in}}=i\omega L_p+{\omega }^2{M}^2/Z_s$. For $Z_s=R+i\omega L_s$, the reflected real part is ${\omega }^2{M}^{2}R/[R^2+(\omega L_s{)}^2]$, and primary mean input power equals secondary mean Joule heating.

The square-root collapse of capacitive bridging bandwidth near the DC error floor describes how the usable frequency interval disappears when a resistive divider has almost exhausted a complex-error allowance. Use the model and exact formula in capacitive bridging complex-error bandwidth: $g=R_s/R_{\mathrm{in}}>0$, ${\tau }_s=R_s{C}_{\mathrm{in}}$, and the DC floor ${\eta }_0=g/(1+g)$. Write the available extra tolerance as $\Delta =\eta -{\eta }_0>0$. The exact bandwidth becomesAt fixed positive $g$, the leading numerator is linear in $\Delta$, so $f_{\eta }$ is proportional to $\sqrt{\Delta }$. Thus demanding four times less extra tolerance approximately halves the bandwidth; it does not reduce it by a factor of four.

For an original example, choose $R_s=20\mathrm{k\Omega }$, $R_{\mathrm{in}}=1\mathrm{M\Omega }$, and $C_{\mathrm{in}}=50\mathrm{pF}$. Then $g=0.02$, ${\tau }_s=1\mu \mathrm{s}$, and ${\eta }_0\approx 0.019608$. Taking $\Delta =0.00020$ gives about $456\mathrm{Hz}$ from the exact formula. Taking $\Delta =0.00005$ gives about $228\mathrm{Hz}$. The small deviation from an exact two-to-one ratio comes from the retained quadratic term and denominator.

The qualification fixed positive $g$ matters. If the receiving resistance is infinite, $g={\eta }_0=0$ andwhich is linear in small $\eta$. For a small but nonzero floor, the square-root approximation needs $\Delta$ small compared with ${\eta }_0$ as well as with the remaining distance to unit error. The crossover from square-root to linear capacitive bridging bandwidth scaling prevents using the fixed-floor asymptotic expression uniformly as $g$ also tends to zero.

This is a limit of the chosen passive input impedance model, not an empirical limit for every amplifier. The manufacturer's shunt-capacitance model motivates the resistive/capacitive paths; an actual bench comparison must include both amplitude and phase shift and the probe's added capacitance.^[ref]