Second-Order Circuits

KVL round the single loop, using the element laws of Sets 16 and 17:

The integral is awkward, so differentiate once to clear it.

Differentiating and dividing by \(L\):

Try \(i = Ae^{st}\), as the first-order case did. Substituting and cancelling \(Ae^{st}\):

A quadratic, so there are two natural frequencies rather than one. The general solution is a combination of both, which is why two initial conditions are required.

Three cases, from the discriminant:

Why the repeated root needs a \(t\). When \(s_1 = s_2\) the two exponentials collapse into one, leaving only one arbitrary constant — not enough for two initial conditions. Substituting \(i = A_2te^{-\alpha t}\) into the equation with \(\omega_0 = \alpha\) confirms it is a second independent solution.

What the parameters mean. \(\omega_0\) is the frequency at which a lossless \(LC\) circuit would oscillate for ever (Problem 18), and \(\alpha\) measures how fast the resistance removes energy:

The envelope of any underdamped response decays as \(e^{-\alpha t}\), so \(1/\alpha\) plays the role that \(\tau\) played in Set 18.

The damping ratio is the dimensionless comparison of the two rates:

Writing the roots this way separates shape from speed: \(\zeta\) alone fixes the shape of the response, and \(\omega_0\) merely scales the time axis.

Classifying the three circuits:

Note the reversal between topologies. In a series circuit \(\alpha = R/2L\) rises with \(R\); in a parallel circuit \(\alpha = 1/2RC\) falls with \(R\):

Physically obvious once stated: a series resistor is in the oscillating loop and dissipates its current, while a parallel one is a bypass that only steals current if it is small. Problem 16 shows this is duality.

The boundary conditions in component terms:

The two critical resistances differ by a factor of four, and one is a maximum for oscillation while the other is a minimum.

Why \(\zeta\) rather than \(\alpha\). Two circuits with the same \(\zeta\) have identical response shapes, differing only in timescale. So overshoot, number of visible oscillations and settling time in units of \(1/\omega_0\) are all functions of \(\zeta\) alone — which is what makes it the design parameter, in circuits and in control systems alike. Problem 17 quantifies this.

Damping parameters:

\(\alpha > \omega_0\), so overdamped with \(\zeta = 1.25\).

The roots:

aFirst initial condition: \(i(0) = 0\) gives

bSecond initial condition — the one that needs care. Apply KVL at \(t = 0^+\):

The capacitor's stored voltage becomes the inductor's slope. This is the coupling between the two elements, and it is the whole reason a second-order circuit needs both continuity results.

Solving:

Check the shape. At small \(t\), expanding gives \(i \approx 2[(1-8t)-(1-2t)] = -12t\) — starting at zero with slope \(-12\), as required. The current goes negative, reaches a minimum, and returns to zero without ever crossing back: an overdamped response has at most one turning point and no oscillation.

Parallel damping parameters — note the different \(\alpha\):

\(\alpha < \omega_0\), so underdamped with \(\zeta = 0.447\).

The damped frequency:

The circuit oscillates at \(\omega_d = 4\), slightly below the undamped \(\omega_0 = 4.47\) — damping always slows the oscillation.

aFrom \(v(0) = 10\): \(B_1 = 10\).

bThe second condition, from KCL at the node at \(t = 0^+\). The three parallel branches must sum to zero:

Differentiating the solution and evaluating at zero:

The amplitude–phase form is often more revealing:

since \(\sqrt{10^2+5^2} = 11.18\) and \(\tan^{-1}(5/10) = 26.57°\). The envelope \(11.18e^{-2t}\) and the oscillation \(\cos(4t+26.57°)\) are now separate.

Initial conditions, from the pre-switch DC circuit:

Damping parameters for the source-free parallel circuit:

\(\alpha = \omega_0\) exactly — critically damped, with the repeated root \(s = -2\).

Solve for the inductor current (the state variable), using the repeated-root form:

The second condition, from the inductor's own element law:

Hence both quantities:

Check the two against each other. Differentiating the current:

Consistent with \(v = L\,di/dt\) at every instant, not merely at \(t=0\) — a worthwhile check whenever both quantities have been written out.

Reading the response. The voltage has zero initial slope, since

which vanishes at \(t=0\) and is negative thereafter. So \(v\) decays monotonically from 2 V, and the current climbs monotonically from \(-2\ \text{A}\) towards zero. A critically damped response never crosses its final value — the \(te^{-\alpha t}\) term can produce at most one hump, and here the constant term suppresses even that.

Initial conditions from the DC steady state before switching:

Both carry across by continuity: \(v(0^+) = 16\ \text{V}\) and \(i(0^+) = 2\ \text{A}\).

Damping parameters after switching:

Overdamped, \(\zeta = 1.67\).

aFrom \(i(0) = 2\):

bSecond condition by KVL at \(t = 0^+\). The capacitor drives the loop against the resistance:

Note that both initial conditions were needed here — the capacitor voltage and the inductor current — which is exactly what distinguishes second order from first.

Solving:

The requested quantities:

Check: \(v_R(0) = 16\ \text{V} = 8 \times 2\) ✓, consistent with the initial current.

The difficulty. Continuity gives \(v_C(0^+)\) and \(i_L(0^+)\) directly — two numbers. But if the unknown is \(v_C\), the general solution needs \(v_C(0^+)\) and \(\dot{v}_C(0^+)\), and the derivative is not given. It must be manufactured from the other element's condition.

The bridge is always an element law:

So the derivative of one state variable is set by the other quantity at the same element — and finding \(i_C(0^+)\) or \(v_L(0^+)\) is an ordinary resistive problem at the single instant \(t = 0^+\).

The procedure. Draw the circuit at \(t = 0^+\) with each storage element replaced by a source of its known value:

This is the substitution theorem of Set 14 applied at one instant. The resulting network is purely resistive and yields to KCL or KVL immediately.

The standard results, which follow from applying that procedure to the two canonical topologies:

The first is KVL round the loop; the second is KCL at the node. Neither should be memorised — both are one line of Kirchhoff at \(t = 0^+\).

Three ways this goes wrong:

A worked instance. Take Problem 3: unknown \(i\), with \(i(0^+) = 0\) and \(v_C(0^+) = 12\ \text{V}\). Replacing the capacitor by a 12 V source and the inductor by a 0 A source (an open circuit):

The resistor contributes nothing because no current flows yet. Had \(i(0^+)\) been non-zero, its \(Ri\) drop would have entered.

Initial conditions:

Damping parameters:

Underdamped, \(\zeta = 0.6\), \(\omega_d = 4\ \text{rad/s}\).

The final value. In DC steady state the capacitor is open and no current flows, so the whole source voltage appears across it:

aFrom \(v(0) = -12\):

bFrom the capacitor's element law, with zero initial current:

The response:

Or in amplitude–phase form, with \(\sqrt{62^2+46.5^2} = 77.5\):

The circuit overshoots. Starting at \(-12\ \text{V}\) and heading for 50 V, it does not stop there — Problem 17 computes the peak as 55.9 V, an overshoot of 9.5% of the 62 V step. No first-order circuit can do this, and Set 18, Challenge C3 proved why.

Initial conditions after converting the current source to a voltage source (Set 12):

Damping parameters:

Heavily overdamped: \(\zeta = 5\). The large capacitance makes \(\omega_0\) very small.

The roots are widely separated, which is characteristic of large \(\zeta\):

aFrom \(v(0) = 10\):

bFrom the capacitor's element law:

The response:

Read what the numbers say. The fast root's amplitude is a thousandth of the slow root's, so to any practical accuracy

— a first-order response with \(\tau = 19.8\ \text{s}\). When \(\zeta \gg 1\) the two roots separate widely, the fast one contributes almost nothing, and the circuit behaves as though the inductor were absent. Setting \(L = 0\) gives \(\tau = RC = 20\ \text{s}\), close to the 19.8 s found.

Initial conditions from the mesh equations:

Damping parameters:

Exactly critical: \(\zeta = 1\), repeated root \(s = -4\).

Solve for the capacitor voltage, whose final value is the source voltage:

The second condition. The capacitor current is the loop current, here flowing so as to discharge it:

The \(t\,e^{-4t}\) term vanishes — a legitimate outcome, not an error. Critical damping guarantees the form, not that both constants are non-zero.

Hence the capacitor voltage and the current:

Two checks. At \(t = 0\), \(i = 5\ \text{A}\) ✓ matching the initial condition; and as \(t \to \infty\), \(i \to 0\) and \(v \to 10\ \text{V}\) ✓ — the capacitor blocks DC and takes the full source voltage.

Initial conditions:

Current division sends the larger share through the smaller resistance — the 6 Ω on top divides into the inductor's branch.

Damping parameters:

Critically damped again, \(s = -5\) repeated.

Solve for the capacitor voltage:

The second condition, from the capacitor current — which in a series loop is the inductor current:

The current follows by differentiating:

The current changes sign at \(t = \tfrac13\ \text{s}\), where the bracket vanishes. Physically: the capacitor charges past 12 V, then discharges back towards it. That single reversal is the most a critically damped response can manage — it approaches the final value from one side and never crosses.

The capacitor voltage does exceed 12 V — but that is not overshoot of the current's final value, which is zero and approached monotonically after the reversal.

Damping parameters:

Very lightly damped: \(\zeta = 0.01\). This is a resonant circuit rather than a damped one.

The damped frequency is barely distinguishable from \(\omega_0\):

A 1% damping ratio shifts the frequency by only 0.005% — which is why \(\omega_d \approx \omega_0\) is a safe approximation for any lightly damped circuit.

The final value is the whole source current, since the inductor is a DC short:

aFrom \(i(0) = 0\): \(A = -6\ \text{mA}\).

bFrom \(v(0) = L\,\dot{i}(0) = 0\):

\(B\) is a hundredth of \(A\) — for light damping the sine term is nearly negligible and the response is almost a pure cosine.

The response:

The current rings around 6 mA, reaching nearly 12 mA at its first peak — an overshoot approaching 100%.

How long does the ringing last? The envelope decays with \(1/\alpha = 20\ \text{ms}\), while one oscillation takes \(2\pi/\omega_d = 1.26\ \text{ms}\). So roughly

before settling. Problem 19 shows this count is essentially the quality factor.

Damping parameters:

Critically damped, \(s = -2\) repeated.

The final value is the source current, since the inductor shorts in DC:

aFrom \(i(0) = 6\):

bFrom \(v(0) = 0\), using the inductor's element law:

The response:

Trace its shape. The current starts at 6 A with zero slope, must reach 3 A, and cannot overshoot. Differentiating:

Zero at \(t=0\) and negative thereafter — a flat start followed by a monotonic fall to 3 A. The \(6t\) term does not create a hump here; it holds the response up briefly before the exponential takes over.

Initial conditions. Before switching, the two halves are independent. The 4 A source drives the inductor, and the capacitor sits at the divider's output:

Recall \(30u(-t)\) equals 30 for \(t<0\) and zero for \(t>0\) — a source that switches off.

Damping parameters after switching:

Overdamped, \(\zeta = 2.5\).

The final value is the source current, since the inductor is a DC short:

aFrom \(i(0) = 4\): \(A_1 + A_2 = 0\), so \(A_1 = -A_2\).

bFrom the inductor's element law:

The inductor current:

The transient is tiny — 65 mA against a 4 A steady value — because the initial current already equals the final one. Only the capacitor's 15 V disturbs it.

The resistor current, obtained from the node voltage \(v = L\,di/dt\):

Check: \(i_R(0) = 0.785 - 0.034 = 0.75\ \text{A} = 15/20\) ✓, the current the 15 V initially drives through the 20 Ω.

The standard formulas do not apply here, because the circuit fits neither canonical topology. The general method is to derive the differential equation from Kirchhoff's laws and read \(\alpha\) and \(\omega_0\) from its coefficients.

Mesh equations, with \(i_0\) in the source loop and \(i\) in the other:

Here \(1/C = 25\), and the capacitor is common to both loops.

Eliminate \(i_0\). Differentiating (2):

Substituting into (1), together with \(25\int(i_0+i)dt = -4i\):

Read the parameters off the coefficients:

Underdamped, but only just: \(\zeta = 0.957\). Neither \(R/2L\) nor \(1/2RC\) would have given 15.125.

The final value from the forcing term, or equally from the DC circuit:

Check: at steady state the capacitor is open, so the two loops form one path of \(6+4 = 10\ \Omega\) carrying \(20/10 = 2\ \text{A}\), which is \(-2\) in the second loop's reference direction ✓.

Fitting the constants, with \(i(0) = 0\) and \(\dot{i}(0) = 0\) (from \(v(0) = 0\) through equation 2):

The two governing equations. For the series loop, KVL in terms of current; for the parallel node, KCL in terms of voltage:

Identical under the substitutions \(i \leftrightarrow v\), \(L \leftrightarrow C\), \(R \leftrightarrow G = 1/R\) — the dual dictionary of Set 17, Problem 20.

The parameters map accordingly:

\(\omega_0\) being self-dual is the reason both topologies resonate at the same frequency for the same \(L\) and \(C\) — a fact Set 29 relies on.

Converting Problem 3. There a series circuit with \(R = 10\ \Omega\), \(L = 1\ \text{H}\), \(C = \tfrac1{16}\ \text{F}\) gave \(\alpha = 5\), \(\omega_0 = 4\) and

The dual circuit needs \(G = 10\ \text{S}\) (so \(R = 0.1\ \Omega\)), \(C = 1\ \text{F}\), \(L = \tfrac{1}{16}\ \text{H}\). Check:

with \(v(0) = 0\) and \(i_L(0) = 12\ \text{A}\) as the dual initial conditions. Not one line of algebra was repeated.

Why the critical resistances differ by four. Setting \(\alpha = \omega_0\) in each case:

Both involve the characteristic impedance \(\sqrt{L/C}\), and the factor of 4 is the accumulated effect of the two factors of 2 sitting on opposite sides.

The normalised step response. For a step from 0 to 1 with zero initial slope:

Differentiate to find the peak. The cosine terms cancel, leaving

which vanishes whenever \(\omega_dt = n\pi\). The first maximum is at

Substituting back, with \(\sin\pi = 0\) and \(\cos\pi = -1\):

using \(\alpha/\omega_d = \zeta/\sqrt{1-\zeta^2}\). The overshoot depends on \(\zeta\) alone — not on \(\omega_0\), nor on the component values individually.

Applying to Problem 8 (\(\alpha = 3\), \(\omega_d = 4\), \(\zeta = 0.6\)):

The step there ran from \(-12\) to 50 V, a span of 62 V, so the predicted peak is \(50 + 0.0948(62) = 55.9\ \text{V}\) — which is exactly what evaluating the full expression gives at \(t = 0.785\ \text{s}\).

The design table:

Settling time is governed by the envelope, so to within 2% takes

For Problem 8, \(t_s = 4/3 = 1.33\ \text{s}\). Note that reducing \(\zeta\) below 1 makes settling slower, not faster — the ringing has to die away.

Why \(\zeta = 0.707\). It is the compromise: 4.3% overshoot is usually tolerable, and settling is markedly faster than at \(\zeta = 1\). It also gives the flattest possible frequency response with no peaking — the Butterworth condition of Set 30. The same number appearing in both the time and frequency domains is not a coincidence, as Set 28 will show.

With \(R = 0\), \(\alpha = 0\) and the roots are purely imaginary:

Undamped: the oscillation never decays, because there is nothing to dissipate energy.

Track the energy:

That is the mechanism. Energy moves completely from the electric field to the magnetic field and back, twice per cycle, with the total conserved. The oscillation is not something the circuit is driven to do — it is the only way two conservative stores connected together can behave.

Why two different kinds are needed. Two capacitors connected together also exchange charge, but they reach a common voltage and stop (Set 18, Challenge C2). The exchange is one-way because the driving quantity — voltage difference — is destroyed by the transfer itself. With an \(L\) and a \(C\):

Each element's state drives the other's rate of change, so neither can settle while the other is charged. That 90° lag is what turns exchange into oscillation.

Adding resistance takes a share of the energy on every pass:

— the envelope of the energy, decaying at twice the rate of the amplitude, as Set 18, Problem 8 found for the first-order case. When \(\alpha \ge \omega_0\) the resistance removes the energy faster than one exchange can complete, and no oscillation is visible at all. That is precisely what overdamping is.

Definition:

A high \(Q\) means light damping. The two quantities carry the same information; \(\zeta\) is preferred for transient work and \(Q\) for resonance, where Set 29 will show it also equals \(\omega_0/\text{bandwidth}\).

Component forms, obtained by substituting each \(\alpha\):

Again inverse in \(R\), as duality requires. Both involve \(\sqrt{L/C}\), so \(Q\) compares the resistance with the circuit's characteristic impedance.

Checking against earlier problems:

Verify the last directly: \(Q = R\sqrt{C/L} = 2000\sqrt{5\times10^{-6}/8\times10^{-3}} = 2000(0.025) = 50\) ✓.

The energy interpretation, which is where the name comes from:

Since the stored energy decays as \(e^{-2\alpha t}\), the fraction lost in one period \(T = 2\pi/\omega_d\) is about \(2\alpha T = 4\pi\alpha/\omega_0 = 2\pi/Q\) for light damping — which inverts to the definition above.

Counting the ringing. The response is essentially over after \(5/\alpha\), and each cycle takes \(2\pi/\omega_d\), so

For Problem 12 with \(Q = 50\) that predicts about 80 visible cycles — matching the direct calculation there exactly. A useful rule: a circuit rings for roughly \(Q\) cycles.

The comparison:

What is genuinely new is the last row. A first-order circuit's stored energy has only one destination — the resistance. A second-order circuit has somewhere else to put it, and the round trip takes time, which is what a frequency is.

What is not new is the method. Kirchhoff's laws, continuity, DC analysis for initial and final values, and the split into natural and forced response all work exactly as in Set 18. Even the procedure is the same shape — find the state, find where it is going, find how fast — with "how fast" now requiring two numbers instead of one.

The limitation. Problem 15 showed what happens when the topology is not canonical: the differential equation must be assembled by hand, integrals differentiated away, and one variable eliminated between two equations. For third order and beyond this becomes impractical, and there are three further difficulties:

What the Laplace transform does about it. Set 31 converts differentiation into multiplication by \(s\), so the circuit's equations become algebraic. Initial conditions enter as sources rather than as separate conditions, the natural and forced responses emerge together, and Set 32 shows the whole analysis reduces to writing impedances and applying Set 6's nodal analysis unchanged.

The characteristic equation of this set becomes the pole polynomial, and the three damping cases become three arrangements of poles in the complex plane.

The immediate next step, though, is different. Sets 20 onwards ask what happens when the source is a sustained sinusoid rather than a step. The transient dies away and only the forced response remains — and for that special case there is a shortcut that makes the whole of Part 1 apply again, unchanged, to AC circuits.