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A suitable electrical circuit has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy inductor $R_L+i\omega L$; take the output voltage across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its complex impedance,At direct current the capacitor is an open circuit and the inductor contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. ThereforeWhen $R_L$ is neglected, $Z_p=i\omega L/(1-{\omega }^{2}LC)$. At the antiresonance ${\omega }_0^{2}LC=1$ its electrical admittance vanishes, so $|H|$ reaches its maximum value of one. ConsequentlyThis value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A transfer function peak and one DC measurement do not uniquely identify every possible two-port network, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC wave amplitudes or root mean square values: the literal time averages of sinusoidal voltages are zero, whose ratio would be undefined.

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.

Write $V_{\mathrm{in}}=v(t)+V_{\mathrm{off}}$, with constant unwanted offset $V_{\mathrm{off}}$. Add an adjustable reference voltage $V_b$ through another resistor $R_b$ to the first inverting summing junction, as the dashed branch in the circuit figure shows. Kirchhoff's current law now givesChoose the reference polarity and magnitude so thatA potentiometer between positive and negative reference supplies permits either sign. The first stage then has zero output voltage when $v=0$, and the op amp integrator integrates only $v(t)$:Merely placing a large resistor across the integrating capacitor limits drift but also changes the transfer function; cancellation at the summing junction preserves ideal integration of the desired signal.

Use sinusoidal phasors with physical values $v(t)=\mathrm{Re}(\hat{V}{e}^{j\omega t})$, $i(t)=\mathrm{Re}(\hat{I}{e}^{j\omega t})$, and $j^2=-1$. The complex impedance is $Z=\hat{V}/\hat{I}=X+jY$: its modulus is the voltage-to-current amplitude ratio, and its complex argument is the voltage-minus-current phase difference. Here $X$ is the electrical resistance part and $Y$ the signed electrical reactance; the source's $X,Y$ notation differs from the common $R,X$ notation.

The component laws $v_R=Ri$, $v_L=Ldi/dt$ and $i_C=Cd{v}_C/dt$ give the complex impedances of an ideal resistor, inductor and capacitor:In a series circuit the same electric current flows through both components, so Kirchhoff's voltage law adds their voltage phasors. The resulting series RC impedance isFor positive $R,C,\omega$, the negative phase difference means that electric current leads the applied electric potential difference.

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 $\omega >0$.

As $\omega \to \infty$, both capacitors approach short circuits. Thus $Z\to 0$, $W\to R_v>0$, and the unloaded right junction tends to the source's top-terminal voltage. HenceThe connected ideal ammeter carries limiting electric current $V_0/R$ from right to left because $Z_{\mathrm{th}}\to R/2$. Reversing the output polarity reverses both nonzero open-circuit answers, without changing their magnitudes.

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 $Z_{\mathrm{th}}=R/2+R_x$.

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.

Zero capacitance removes the capacitor branches. No current enters the non-inverting input, so no current flows through $R_1$ or $R_2$ and $V_{+}=V_{\mathrm{in}}$. The inverting divider gives $V_{-}=V_{\mathrm{out}}{R}_3/(R_3+R_4)$. Equating the two input voltages in the linear solution gives

Current conservation at A, with the capacitor $C_1$ connected from A to the output as verified in the PDF, givesSubstitute $V_A=(1+s{R}_2{C}_2)V_B$ and $V_{\mathrm{out}}=K{V}_B$:Therefore, at $s=i\omega$,This algebraic transfer function alone does not establish that transients decay; the pole check in the parent solution is necessary for the requested numerical choice.

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[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.

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.

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.

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.

Let $v$ be the inverting-input potential. The non-inverting input is grounded, so the ideal operational amplifier rules give $v=0$ and no input current. Kirchhoff's current law at that node therefore givesIntegrating from an initial time $t_0$,This op amp integrator performs inverted time integration; the initial voltage across the capacitor supplies the integration constant. For zero initial conditions its transfer function is $H(s)=-1/(sRC)$. In sinusoidal steady state, with the convention $e^{i\omega t}$,For example, a cosine input produces a negative sine output after its constant component is set. The additional inversion makes the phase $+90^{\circ }$ modulo $360^{\circ }$, rather than the $-90^{\circ }$ of non-inverting integration.

Any direct-current component of the input produces an output ramp. The input offset voltage and input bias currents of a real operational amplifier therefore charge the feedback capacitor until the output saturates. Even the desired signal can exceed the available output swing when integrated for too long. Initial capacitor charge also introduces an unknown integration constant.

Place a large resistor $R_f$ in parallel with the feedback capacitor. The idealized transfer function then becomesIts finite direct-current gain $-R_f/R$ supplies a return path and bounds the response to a small offset; it approximates $-1/(sRC)$ when $\omega R_{f}C\gg 1$. The chosen gain must still keep offsets within the output range. A reset switch across the capacitor can set the initial charge, while low-offset amplifiers, bias compensation and input offset adjustment reduce drift. Useful integration also requires frequencies below the amplifier's own response limit.

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.

Choose positive electric current in the initial discharge direction, since the source diagram does not label the initial plate polarity or a current arrow. Let $q$ be the charge remaining on the initially positive plate, so $I=-\dot{q}$. Kirchhoff's voltage law for the series RLC circuit givesThe natural angular frequency, damping coefficient and quality factor areSince $\alpha <{\omega }_0$ (equivalently $Q>1/2$), this is an underdamped RLC response. With ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}=815.82\mathrm{rad}{\mathrm{s}}^{-1}$, satisfying the initial conditions givesand differentiating givesThe electric current starts at zero with $\dot{I}(0)=V_0/L=100\mathrm{A}{\mathrm{s}}^{-1}$. Its wave period is $T_d=2\pi /{\omega }_d=7.702\mathrm{ms}$, its zeros occur every $3.851\mathrm{ms}$, and its positive and negative wave envelopes decay with time constant $30.0\mathrm{ms}$. The first maximum obeys $\tan {\omega }_{d}t={\omega }_d/\alpha$ and is $0.1151\mathrm{A}$ at $1.875\mathrm{ms}$. The electric current tends to zero as the initial capacitor energy is dissipated in the resistor.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.