Circuit Analysis in the s-Domain

Transform each element law, taking zero initial conditions for now:

Each is now \(V = \mathbf{Z}I\) — Ohm's law with a complex-valued coefficient. The calculus has vanished.

The impedance table:

Setting \(s = j\omega\) recovers Set 20's phasor impedances exactly. The \(s\)-domain impedance is the more general object; the phasor impedance is its restriction to the imaginary axis, valid under the stability condition of Set 31, Problem 16.

Why every Part 1 method transfers. The techniques of Sets 3 to 12 relied on exactly three facts:

Nothing else was ever used. Mesh analysis, nodal analysis, Thévenin, Norton, superposition and source transformation therefore all carry over verbatim — Problem 19 audits this claim properly.

Combination rules are unchanged:

A series \(RLC\) branch, for instance:

A ratio of polynomials — every impedance of a lumped network is rational in \(s\), which is why every response is a sum of exponentials (Set 31, Problem 14).

Reading the limiting behaviour straight off the impedances:

By the initial and final value theorems these correspond to \(t\to0^+\) and \(t\to\infty\) — the familiar rules of Set 18, now derived from impedance rather than remembered. Problem 14 uses this as a checking method.

The inductor, from the differentiation theorem (Set 31, Problem 5):

Read as a KVL statement: an impedance \(sL\) in series with a voltage source of value \(Li(0^-)\), oriented to oppose the assumed current direction. Note the source is a constant in \(s\), hence an impulse in time.

The Norton form follows by rearranging for \(I\):

Admittance \(1/sL\) in parallel with a current source \(i(0^-)/s\) — a step of current in the time domain, which is the more intuitive picture: the inductor was already carrying that current.

Check they are equivalent by source transformation (Set 8):

The capacitor, from the integration theorem (Set 31, Problem 6):

Check: \(I_N = V_{Th}/\mathbf{Z} = [v(0^-)/s](sC) = Cv(0^-)\) ✓.

The complete table, which is all one needs to draw a transformed circuit:

Note the pleasing duality: the inductor's series source is an impulse and its parallel source a step; the capacitor's are the other way round. Sets 8 and 18's dualities, appearing once more.

Which form to use — the same reasoning as Set 8:

Getting the polarity right is the one place errors creep in. The safe procedure:

Step 4 is decisive: if the initial value comes out as \(-i(0^-)\), a source is reversed. Problem 4 demonstrates the check in use.

1Transform the circuit. With no stored energy there are no extra sources:

The natural frequencies are already visible — at \(-2\) and \(-3\), before any response has been computed. They are the zeros of the impedance.

2Apply Ohm's law:

The \(s\) cancels. The pole at the origin from the step is removed by the capacitor's zero — physically, no DC current flows through a capacitor, so there is no steady-state term.

3Invert by partial fractions:

4The capacitor voltage, from \(V_C = I/sC\):

Check every endpoint:

All four correct with no extra work — the initial and final conditions were never imposed, they simply emerged.

The current peak, by differentiating:

Well below \(V/R = 2\) A, because the capacitor begins charging before the inductor has finished building current.

1Draw the transformed circuit with both stored-energy sources in Thévenin form, since this is a single loop:

The signs are the crux. The inductor source pushes current in the direction it was already flowing; the capacitor's existing voltage opposes further charging.

2Apply KVL round the single loop:

3Solve — and something remarkable happens:

The \((s+3)\) factor cancelled. A second-order circuit has produced a purely first-order current — the natural mode at \(s = -3\) is entirely absent. This is not an arithmetic accident, and Problem 13 explains it.

4The capacitor voltage:

Also purely first-order, and note the initial-condition term \(4/s\) had to be added back — \(I/sC\) alone gives only the voltage change.

Verify by KVL in the time domain, the strongest available check:

The exponentials cancel exactly, leaving the source voltage. KVL holds at every instant, confirming both the solution and the source polarities of step 1.

Endpoint checks:

Both match the given data and physical expectation. Had a source been reversed, \(i(0^+)\) would have come out as \(-2\) A — which is precisely why the check is worth two lines.

Write the impedance matrix exactly as in Set 4, but with polynomials:

The matrix is symmetric, as it must be for a network of passive bilateral elements — the same check applied in Set 26, Problem 8.

Solve by Cramer's rule. The determinant:

The natural frequencies are the roots of the common denominator:

Underdamped — both mesh currents will oscillate at \(\sqrt2/3 = 0.471\) rad/s while decaying with \(\tau = 1.5\) s. The natural frequencies belong to the network, not to either mesh, which is why the same pair appears in both currents.

Check the endpoints without inverting:

At DC the inductor short-circuits mesh 2's contribution and the capacitor blocks it, so all the current flows round mesh 1 through \(R_1\) and the inductor. The \(s\)-domain result agrees with elementary reasoning.

Where the initial conditions would go. Had the inductor carried \(i_L(0^-)\), its Thévenin source \(Li_L(0^-)\) would appear in both mesh equations — with opposite signs, since the branch is shared:

Exactly the treatment a shared voltage source would receive in Set 4. No new rule is needed.

What carried over from Set 4, item by item:

Write KCL at the single node — the admittance form is natural here:

Note that the natural frequencies come from the numerator of \(\mathbf{Y}\), whereas in Problem 3's series circuit they came from the numerator of \(\mathbf{Z}\). Duality again.

Solve:

Complete the square (Set 31, Problem 4):

Check both endpoints:

A common trap is to expect \(v(\infty) = IR = 8\) V. That would be right for a resistor alone, but the inductor short-circuits it at DC. The final voltage is zero, and the interesting question is where the 4 A goes.

Follow the current. The inductor current is \((1/L)\int v\,dt\), i.e. \(V(s)/sL\):

The inductor takes the entire source current and holds it with no voltage across it — a perfect DC short. This is the dual of Problem 3's capacitor holding the entire source voltage with no current through it.

Reading the damping. Comparing with the standard form of Set 29:

\(\zeta = 1/\sqrt2\) exactly — a second-order Butterworth (Set 30, Problem 5). The damping ratio is read from the same denominator that gave the poles, so no separate classification step is needed.

For a parallel circuit, use \(\mathbf{Y}\), not \(\mathbf{Z}\). Computing \(\mathbf{Z} = 1/\mathbf{Y}\) first and then dividing would work, but inverting a three-term sum is needless labour. The rule from Set 5 stands:

One expression covers all three cases:

Only \(R\) changes. The damping case is decided entirely by whether \(s^2+Rs+6\) has real, repeated or complex roots — i.e. by the sign of \(R^2-24\).

aOverdamped, \(R = 5\ \Omega\). Discriminant \(25-24 = 1 > 0\):

Two distinct real poles, two exponentials, no oscillation — Problem 3's result.

bCritically damped, \(R = 2\sqrt6 = 4.899\ \Omega\). Discriminant zero:

The \(t\) factor is a repeated pole — Set 31, Problem 10. Nothing special had to be done: the partial-fraction rule for a double pole produced it automatically.

cUnderdamped, \(R = 2\ \Omega\). Discriminant \(4-24 = -20 < 0\):

Poles at \(-1\pm j\sqrt5\): decay rate 1, ringing at \(\sqrt5 = 2.236\) rad/s.

The three cases side by side:

using \(\zeta = R/(2\sqrt{L/C}) = R/(2\sqrt6)\). The natural frequency \(\omega_0 = \sqrt6 = 2.449\) rad/s is the same in all three — only the damping differs, which is Set 29's reading of pole geometry.

What has been avoided. Set 19 required identifying the case first, then selecting the matching general solution:

The three "cases" are three ways a quadratic can factor, not three physical situations requiring different theory.

The open-circuit voltage, by the divider rule of Set 3:

The Thévenin impedance, with the source killed (short-circuited, as in Set 9):

Not a resistance but a function of \(s\) — which is the whole difference from Part 1. It still behaves exactly as a Thévenin impedance should.

What it means, read at the two extremes:

So the equivalent source has zero output impedance the instant a load is connected, rising to 4 Ω once everything settles — a fair description of a capacitor's stiffening effect on a supply, and a result no purely resistive equivalent could express.

Test it. Connect a load \(R_L = 4\ \Omega\) and find \(v_L(t)\):

Check directly: the loaded circuit is a 10 V step behind 4 Ω charging \(C = \tfrac12\) F in parallel with 4 Ω. Final value \(10 \times 4/8 = 5\) V ✓; time constant \((4\parallel4)C = 2 \times 0.5 = 1\) s ✓.

The theorems that carry over, all of Part 1's:

The one requiring care is maximum power transfer, which is inherently a steady-state, single-frequency idea — it needs \(s = j\omega\) and the conjugate, so it does not generalise to arbitrary \(s\).

With stored energy present, the initial-condition sources are simply additional sources when finding \(V_{Th}\) — and are killed like any other source when finding \(\mathbf{Z}_{Th}\). Voltage sources short, current sources open, including the IC sources.

Three sources act on the transformed circuit of Problem 4:

Since the loop impedance is the same in every case, each contribution is that source divided by \(\mathbf{Z}(s) = (s+2)(s+3)/s\).

Contribution of each:

Sum the three:

Matching Problem 4 exactly.

What the decomposition reveals. The \(e^{-3t}\) column is the interesting one: \(-10+6+4 = 0\). Each source individually excites the \(s = -3\) mode; their contributions cancel exactly.

The pole–zero cancellation of Problem 4 was not a single source failing to excite a mode — it was three sources exciting it in amounts that sum to zero. Problem 13 shows this is a designable condition.

Zero-state and zero-input, in Set 31, Problem 19's language:

The zero-state response is precisely Problem 3's answer — the same circuit starting from rest. Superposition therefore says: a circuit with stored energy behaves as the at-rest circuit plus a source-free discharge.

When superposition is worth using here. For a single loop it is more work than direct KVL, but it earns its place when:

Definition. A transfer function is defined with zero initial conditions:

This restriction is essential. With stored energy present the output is not proportional to the input, so no single ratio describes the circuit — the IC contribution of Problem 9 would be there regardless of the input.

Apply the divider rule — the circuit is a single loop, so this is one line:

Multiply top and bottom by \(s/L\):

Compare with the standard form of Set 29:

Set 29's \(Q\) formulas, derived rather than quoted. With \(R = 5\) the circuit is slightly overdamped, consistent with the real poles found in Problem 3.

Substituting the running values:

The output terminals decide the filter type. Same circuit, three different transfer functions:

with \(D(s) = s^2+(R/L)s+1/LC\) throughout. The poles are a property of the network; the zeros are a property of where you measure. This is the single most useful structural fact in the set, and Problem 18 develops it.

Sanity check the band-pass claim via element behaviour:

Zero at both ends, peak in the middle — a band-pass, exactly as Set 29 described series resonance.

The impulse response is \(\mathcal{L}^{-1}\{H(s)\}\), since \(\mathcal{L}\{\delta\} = 1\):

Check: \(h(0) = 0\) ✓ — an impulse of voltage cannot instantaneously change the capacitor voltage, because the series inductor limits the current.

The step response is \(\mathcal{L}^{-1}\{H(s)/s\}\):

This is Problem 3's \(v_C(t)\) divided by 10 — as it must be, since that was the response to a 10 V step.

The two are related by calculus, in either direction:

Verify by differentiating:

In the \(s\)-domain this is just the factor \(1/s\) of Set 31, Problem 6 — integration in time is division by \(s\), and the step is the integral of the impulse.

Useful checks on both:

The area under the impulse response equals the DC gain. This follows directly from \(H(0) = \int_0^\infty h\,e^{0}dt\) and is one of the quickest sanity checks available.

Why the impulse response is the fundamental one. By the convolution theorem (Set 31, Problem 12), \(h(t)\) determines the response to every input:

In practice, though, the step response is what gets measured — an impulse is hard to generate and the resulting signal is small. Measure the step response, differentiate to get \(h\), transform to get \(H\).

Rise time, using Set 28, Problem 19:

With the dominant pole at \(-2\), a first-order circuit of the same time constant would give \(2.2\tau = 1.1\) s. The true rise time is 28% longer, because the second pole at \(-3\) adds its own lag. A dominant-pole estimate is optimistic about the rise even when it is accurate about the tail (Set 31, Problem 14) — the extra poles cannot speed anything up.

aThe stage-by-stage method, as in Set 18.

Stage 1 \((0 \le t < 2)\) — charging from zero:

Stage 2 \((t \ge 2)\) — this value becomes the new initial condition, and the source is now zero:

Correct, but note what was required: solve, evaluate at the switching instant, restart the problem. With three or four switching events this bookkeeping becomes the dominant labour and the main source of error.

bThe one-shot method, using the delay theorem (Set 31, Problem 7). Describe the whole input as a single function:

Invert the exponential-free part first:

Then apply the delay theorem to the second piece:

One expression, valid for all \(t\). The switching instant never had to be treated as a separate problem.

Verify the two agree:

Identical to four decimals. The delay theorem has absorbed the stage-two initial condition automatically — \(-g(t-2)\) encodes exactly the discharge from whatever level had been reached.

Why the second stage decays from 8.65 V and not 10 V. The pulse lasts only \(2\tau\), so the capacitor reaches \(1-e^{-2} = 86.5\%\) of the source:

When each method wins:

The last row is the real limitation. The delay theorem handles a changing input to a fixed network. If the switch alters the topology, each configuration has its own \(H(s)\) and the stages must be solved separately, carrying the state across.

Write the current with general initial conditions. For the running circuit with source \(V_0u(t)\):

The numerator is a first-order polynomial in \(s\) whose coefficients are set entirely by the initial conditions and the source.

Locate its zero:

The initial conditions place a zero. If that zero lands on a pole, the corresponding natural mode is cancelled and never appears in the response.

The condition for silencing each mode. With \(L = 1\), \(V_0 = 10\), \(v_0 = 4\):

Problem 4 used \(i_0 = 2\) A, so it silenced the fast mode. With \(i_0 = 3\) A instead, the response is \(3e^{-3t}\) — a purely fast decay from a second-order circuit.

Verify the second case from scratch:

Check: \(i(0^+) = 3\) A ✓, matching the stated initial current.

What this means physically. A natural mode is a particular pattern of energy exchange between \(L\) and \(C\):

The special initial conditions are eigenvector-like: they set the circuit going in a state that decays without changing shape. Every other starting state is a mixture of the two.

A caution — this is cancellation, not removal. The pole is still there:

The cancellation is a property of this particular excitation, not of the network. Perturb the initial current by 1% and the \(e^{-3t}\) term reappears in proportion.

Where this matters in practice. The same mathematics appears whenever a zero is placed on a pole:

All are fragile in the same way: cancellation requires exact matching, and component tolerance or drift resurrects the mode — Set 30, Challenge C3's lesson in a different guise.

The correspondence. The initial and final value theorems relate \(s\) limits to \(t\) limits, and the impedances have simple limits:

These are Set 18's rules, now derived. An inductor opposes sudden current change (open at \(t=0^+\)) but passes DC freely (short at \(t=\infty\)); the capacitor does the reverse.

The important qualification. These apply to elements with no stored energy. With initial conditions:

The uncharged cases are the special instances with source value zero — an inductor carrying no current is an open circuit, and an uncharged capacitor is a short.

Apply to Problem 3 (at rest, 10 V step):

Apply to Problem 4 (energised):

Note the final values are the same in Problems 3 and 4. Initial conditions affect how a circuit gets there, never where it ends up — the final state is set by the sources and the topology alone.

Apply to Problem 6 (parallel, 4 A step) — the case that catches people out:

Two seconds of inspection would have prevented the natural but wrong guess that \(v(\infty) = IR = 8\) V.

Use it as a working discipline:

Doing step 1 first matters. Predictions made after seeing the answer tend to agree with it.

The time-domain equations, from Set 26, with the sign set by the dot convention:

Transforming term by term:

The mutual term brings its own initial-condition source. A current in coil 2 contributes flux linkage to coil 1, so it appears in coil 1's initial-condition term — an effect with no counterpart in uncoupled circuits.

In matrix form, the transformed circuit is

The impedance matrix is symmetric, as Set 26, Problem 8 required — and the check applies just as forcefully here. An asymmetric off-diagonal pair means an error, since \(M_{12} = M_{21}\) for any passive coupled network.

Reflected impedance. With a load \(\mathbf{Z}_L\) on the secondary and no initial energy, the secondary loop gives

Substituting into the primary equation:

The sign of \(M\) has disappeared — it is squared. Set 26 found the same: the dot convention affects individual voltages but never the input impedance, because the coupling is traversed twice.

A worked case. Take \(L_1 = L_2 = 1\) H, \(M = 0.5\) H, \(\mathbf{Z}_L = 1\ \Omega\):

At high frequency the secondary current is limited by \(sL_2\) rather than \(\mathbf{Z}_L\), so the reflected term becomes the constant \(M^2/L_2\). Note this makes the effective inductance larger, because the reflected impedance adds here.

The T-equivalent of Set 26 transfers unchanged:

This removes the coupling entirely, leaving three ordinary inductors — after which mesh and nodal analysis apply without any special treatment of the mutual term. Usually the easier route for a circuit with more than two meshes.

The general inverting configuration. With an ideal op-amp the inverting input is a virtual earth, so Set 12's result applies with impedances in place of resistances:

One formula covers every inverting op-amp circuit. Choosing \(\mathbf{Z}_f\) and \(\mathbf{Z}_i\) is the whole design problem.

aThe integrator — feedback capacitor, input resistor:

A pole at the origin, gain falling at \(-20\) dB/dec — the ideal integrator of Set 28, Problem 4. But the pole is exactly on the imaginary axis, which by Set 31, Problem 15 is marginally stable: any DC input, or the op-amp's own input offset, integrates without limit until the output saturates.

bThe practical version puts \(R_f\) across the capacitor to bound the DC gain:

The pole has moved off the axis into the left half-plane, which is exactly what makes the circuit usable. It is a first-order low-pass that happens to integrate at high frequencies.

cThe design. For gain 10 and \(f_c = 159\) Hz:

Choose \(R_f = 100\) kΩ (a convenient value that keeps currents small):

Check: \(H(0) = -10\) ✓, pole at \(-1000\) rad/s \(= 159.2\) Hz ✓.

Swapping the elements gives the differentiator — input capacitor, feedback resistor:

A zero at the origin and a pole at \(-1/R_iC_i\) — a high-pass. The series \(R_i\) is essential: without it the gain would rise indefinitely with frequency, amplifying noise and inviting instability. Every practical differentiator is band-limited.

What the ideal op-amp assumption costs. A real device has finite gain-bandwidth product, adding a pole of its own:

The second is the one that bites. It is also what limits Set 30's Sallen–Key sections at high \(Q\) — the amplifier's own pole conspires with the intended ones.

A network function is any ratio of a transformed response to a transformed excitation, with zero initial conditions. They divide into two families:

The four transfer types are voltage gain, current gain, transfer impedance and transfer admittance. All are dimensionally distinct and none should be called simply "the transfer function" without saying which.

The driving-point impedance of the running circuit, seen from the source:

A zero of a driving-point impedance is a short-circuit natural frequency. At \(s = -2\) the loop can sustain a current of the form \(e^{-2t}\) with the terminals shorted — which is precisely what Problems 4 and 13 observed.

The admittance is the reciprocal, so poles and zeros exchange:

Its poles are now at \(-2\) and \(-3\). This is why Problem 6's parallel circuit took its natural frequencies from the numerator of \(\mathbf{Y}\) while Problem 3's series circuit took them from the numerator of \(\mathbf{Z}\) — the same physical frequencies, appearing in whichever function the topology makes convenient.

Properties of driving-point functions of passive networks, which act as strong checks:

A function satisfying all of these is called positive real, and it is exactly the condition for a rational function to be realisable as a passive network — the foundation of network synthesis.

Transfer functions are less constrained. Their zeros may sit anywhere, including the right half-plane:

Right-half-plane zeros give the non-minimum-phase behaviour of Set 28, Challenge C3 — extra phase lag with no magnitude penalty, and an initial undershoot in the step response. A passive network can have them; a passive driving-point impedance cannot.

Why the poles are shared. Every network function of a given network has the same denominator, because it comes from the same determinant:

as Problem 5's mesh determinant showed. Poles belong to the network; zeros belong to the particular input–output pair chosen. Problem 18 turns this into a design observation.

The three transfer functions, all with denominator \(D(s) = s^2+(R/L)s+1/LC\):

Add them:

Exactly unity, at every frequency. The reason is KVL: the three element voltages must sum to the source voltage at every instant, so their transfer functions must sum to 1. The numerators are forced to be the terms of the denominator.

Verify with the running values (\(R=5\), \(L=1\), \(C=\tfrac16\)):

At DC the capacitor takes the whole source; at high frequency the inductor does; and the resistor takes it only in between. The three filters partition the source voltage.

The structural lesson, which Set 30 relied on without stating:

This is why Set 30's low-pass-to-high-pass transformation left the order unchanged: it moved zeros, not poles. Choosing a filter family is choosing poles; choosing a filter type is choosing zeros.

What zeros do to the response. Poles alone give a sum of decaying exponentials; zeros determine how those exponentials are weighted:

The fourth row is the practical one: a zero near a pole makes that mode's contribution small without eliminating it, which is how a real compensator works when exact cancellation is unattainable.

Counting poles from the circuit. The number of poles equals the number of independent energy-storage elements:

"Independent" matters. Two capacitors directly in parallel combine into one, so a circuit's order can be lower than its element count suggests — which is why the order is properly read from the denominator degree rather than counted on the diagram.

What the techniques actually rest on. Every method in Sets 3 to 12 was built from exactly three ingredients:

KVL and KCL are statements about charge conservation and the single-valuedness of potential; they hold instant by instant, and the transform is linear, so they hold for transformed quantities too. Problem 1 supplied the third. Therefore the methods must transfer — the only question is whether any of them used something extra.

The audit:

The genuine qualification: maximum power transfer. Set 24 established \(\mathbf{Z}_L = \mathbf{Z}_{Th}^*\), and the conjugate is the difficulty:

Conjugation is an operation on the imaginary axis. Average power is inherently a steady-state, single-frequency concept — it requires a settled sinusoid, so it belongs to \(s = j\omega\) and does not generalise to arbitrary \(s\). For a transient, one computes energy by integrating \(vi\), and there is no simple matching condition.

The star–delta caveat is subtler and worth noting. The transformation is algebraically valid:

but the resulting \(\mathbf{Z}_a(s)\) may not correspond to any physical \(R\), \(L\) or \(C\) combination — it can require a negative element. As a computational device it is fine; as a physical equivalent it may not be buildable. The same reservation applied in Set 3 whenever a negative resistance appeared.

What is genuinely new, as against merely inherited:

Every one of these is a consequence of the element models, not of a new circuit principle. The physics did not change; the algebra did.

The practical upshot. A student who can analyse a resistive network can analyse any linear circuit:

with the difficulty concentrated in the last step — partial fractions — rather than in the circuit analysis. That is a considerable simplification over Sets 18 and 19, where each new topology required its own differential equation.

The method, complete:

Only step 4 involves anything the earlier sets did not already contain.

What the two sets together have replaced:

Four separate bodies of technique, reduced to one. They were never different subjects — they were the same rational function examined on different parts of the \(s\)-plane.

The recurring structural facts, worth carrying forward:

What remains unaddressed. Two gaps, both genuine:

The first is a computational obstacle: Laplace establishes that a square wave has poles at the odd harmonics but gives no practical route to the response. The second is worse — it means \(H(s)\) for a two-stage circuit cannot be found by multiplying the stages.

The remaining programme:

Set 33 attacks the first gap by abandoning the closed form: a periodic input becomes a sum of sinusoids, each handled by the phasor method that Set 31, Problem 16 has now justified. Set 35 closes the second by characterising a network with four parameters instead of one, so that loading is accounted for rather than assumed away.