Fourier Series

The obstacle. Set 31, Problem 17 gave the transform of any periodic waveform:

and driving a circuit \(H(s)\) gives

This is not a rational function. The factor \(1-e^{-sT_p}\) has infinitely many zeros, so there is no finite partial-fraction expansion and no finite sum of exponentials to invert to. The transform is correct and entirely useless for computation.

Where the difficulty really lies. Those infinitely many zeros are at

All on the imaginary axis, evenly spaced at multiples of the fundamental. That is the harmonic structure, already present in the transform — the problem is not that the information is missing but that it is packed into a form no inversion technique can unpack.

The change of strategy. Rather than seeking one closed-form answer, decompose and superpose:

Every ingredient already exists. The only new thing is the first step, and even that was implicit in Set 28, Problem 1's claim that \(H(j\omega)\) determines the response to any input.

What is given up, and what is gained:

The loss of the transient is deliberate, not accidental. Fourier series describes the steady state only — the condition after the circuit has settled into repeating its input's period. Set 31, Problem 16 established exactly when that is legitimate: the circuit must be stable.

Why this is the right trade in practice. Periodic excitation in power and electronics is almost always a steady-state question:

None of these asks what happened in the first few cycles. The transient that Fourier discards is the part nobody was asking about.

The claim. Over one period \(T = 2\pi/\omega_0\):

The last is the strongest: a sine and a cosine are orthogonal even at the same frequency.

Proof of the first, by the product-to-sum identity:

Both terms are cosines at integer multiples of \(\omega_0\), and the integral of any such cosine over a whole period is zero — an integer number of complete cycles. The only exception is \(m = n\), where \(\cos(m-n)\omega_0t = \cos 0 = 1\) contributes \(T/2\).

Extracting a coefficient. Multiply the series by \(\cos m\omega_0t\) and integrate over a period:

Orthogonality annihilates every term but one. All the sine products vanish; all the cosine products with \(n \ne m\) vanish; the \(a_0\) term vanishes. What survives is \(a_m(T/2)\):

and identically for \(b_m\) with a sine.

Why \(a_0/2\) rather than \(a_0\). Setting \(m = 0\) in the same formula:

so the DC term in the series is \(a_0/2 = \) the mean. The awkward factor of two exists purely so that one formula covers every \(n\) including zero — a convention worth remembering rather than deriving each time.

The geometric picture. Orthogonality makes the harmonics behave like perpendicular axes:

Finding a Fourier coefficient is projecting onto an axis — which is why each is computed independently of all the others, and why truncating the series is a genuine best approximation rather than a crude one.

The physical consequence, which Problem 13 develops: since different harmonics are orthogonal, a voltage harmonic of one order and a current harmonic of another produce no average power. Only matched pairs contribute — orthogonality is the reason power adds harmonic by harmonic.

The waveform: \(f(t) = +A\) for \(0 < t < T/2\) and \(-A\) for \(T/2 < t < T\).

Two symmetry arguments have eliminated three quarters of the work before any integration. Problem 4 justifies them.

Compute \(b_n\). Using the odd symmetry, integrate over the half period and double:

With \(\omega_0T/2 = \pi\):

Evaluate the bracket:

confirming the half-wave symmetry prediction independently.

The harmonic amplitudes, for the running example \(A = 100\) V at 50 Hz:

Note the fundamental exceeds the square wave's own amplitude — 127.3 V peak against 100 V. The higher harmonics subtract near the middle of each half-cycle to flatten the top, so the fundamental must overshoot for their sum to come out flat.

Check by evaluating the series at \(t = T/4\), where \(f = A\). There \(\sin(n\pi/2)\) alternates \(+1, -1, +1,\dots\) for \(n = 1,3,5,\dots\):

The Leibniz series for \(\pi\), obtained as a by-product. This is a genuine check: had the coefficient been wrong by any factor, the identity would fail.

The \(1/n\) decay is the slowest of any common waveform, and it is the direct consequence of the discontinuity:

Problem 10 makes the connection between smoothness and decay rate general, and Problem 9 shows the discontinuity never disappears no matter how many terms are kept.

aEven symmetry, \(f(-t) = f(t)\). The integrand of \(b_n\) is

and an odd function integrates to zero over a symmetric interval. Hence \(b_n = 0\) for all \(n\) — the series contains only cosines and a constant.

bOdd symmetry, \(f(-t) = -f(t)\). By the same argument with the roles exchanged:

including \(a_0\) — an odd function has zero mean automatically.

cHalf-wave symmetry, \(f(t+T/2) = -f(t)\). Split the coefficient integral at the half period:

Substituting \(t = \tau+T/2\) in the second integral, and using \(\cos n\omega_0(\tau+T/2) = (-1)^n\cos n\omega_0\tau\):

The bracket vanishes for even \(n\). So half-wave symmetry kills every even harmonic — including the DC term — and the same argument applies to \(b_n\).

What each saves:

The square wave of Problem 3 has both, which is why only \(b_n\) for odd \(n\) needed computing — one integral instead of four families.

An important caution: symmetry depends on the time origin. Shifting the square wave by \(T/4\) makes it even instead of odd:

Choose the origin to create symmetry before integrating. Problem 18 shows the physical content is unaffected — only the split between sine and cosine terms changes, never the harmonic magnitudes.

Half-wave symmetry is different in kind. Unlike the other two it cannot be created or destroyed by shifting the origin:

and it has real physical significance. Any waveform whose positive and negative half-cycles are identical in shape contains no even harmonics — which is why a symmetric inverter output or a balanced magnetic circuit produces only odd harmonics, and why the appearance of even harmonics signals an asymmetry such as DC offset or half-cycle saturation.

Substitute Euler's formulas into the trigonometric series:

Collecting terms in \(e^{jn\omega_0t}\) and \(e^{-jn\omega_0t}\) gives

The relationships:

For a real \(f(t)\), \(c_{-n} = c_n^*\) always. This conjugate symmetry is not a coincidence — it is exactly the condition that the sum comes out real.

Why negative frequencies. They are a bookkeeping device, not a physical claim:

The pair combines into one real sinusoid. Each rotating phasor is complex; only their sum is physical. This is precisely the "take the real part" convention of Set 20, made symmetric — and the reward is that the algebra involves only exponentials, which multiply and differentiate trivially.

The square wave in exponential form. From Problem 3, \(a_n = 0\) and \(b_n = 4A/n\pi\) for odd \(n\):

Each is half the trigonometric amplitude, the other half sitting at the mirror frequency. Adding the pair recovers the 127.3 V peak of Problem 3.

Why this form is preferred for everything beyond hand computation:

The last is decisive. The Fourier transform of Set 34 is the exponential series with the sum replaced by an integral, and that passage is invisible in the trigonometric form.

aThe sawtooth, rising linearly from \(-A\) to \(+A\) over one period and dropping instantly. It is odd, so only \(b_n\) survives:

Integrating by parts:

All harmonics present, both odd and even, with alternating signs. There is no half-wave symmetry — the waveform's two half-cycles are not mirror images, since one contains the discontinuity.

bThe triangular wave, which rises and falls linearly with no jump. Odd and half-wave symmetric:

Compare the three waveforms met so far, normalised to unit fundamental:

The triangle's harmonics collapse far faster. Its seventh harmonic is 2% of the fundamental where the square wave's is 14% — a sevenfold difference, growing with \(n\).

The reason is the discontinuity. Square and sawtooth both jump; the triangle does not:

Problem 10 generalises this into a rule. The bandwidth a waveform demands is set by its sharpest feature, not its amplitude.

A useful relationship. The triangle is the integral of the square wave, and integrating divides each coefficient by \(jn\omega_0\):

Integration in time is division by \(n\) in the spectrum — Set 31, Problem 6's theorem applied harmonic by harmonic. Differentiating a triangle gives a square wave, and correspondingly multiplies each coefficient by \(n\).

Why the sawtooth has even harmonics but the square wave does not. Both are discontinuous, so both decay as \(1/n\) — the difference is symmetry, not smoothness:

Decay rate is governed by smoothness; which harmonics appear at all is governed by symmetry. The two questions are independent, and confusing them is a common error.

aHalf-wave rectified sine — \(A\sin\omega_0t\) for the positive half-cycle, zero otherwise.

The DC term:

The fundamental survives at half amplitude:

The even harmonics:

Odd harmonics above the first are entirely absent — an unusual spectrum, and one that follows from the waveform being a sine multiplied by a square wave of the same period.

bFull-wave rectified sine — \(A|\sin\omega_0t|\). The period halves, so the "fundamental" of the rectified wave is at \(2\omega_0\):

No fundamental at all, and twice the DC output. Both facts make full-wave rectification the better choice.

Compare the two, as fractions of the DC output:

Full-wave wins three times over: double the DC, less relative ripple, and the ripple that remains is at twice the frequency.

Why the ripple frequency matters most. A smoothing capacitor's attenuation goes as \(1/\omega\), so doubling the ripple frequency halves the required capacitance for the same ripple:

Combining both effects, a full-wave rectifier needs roughly a fifth of the smoothing capacitance — which is why half-wave rectification survives only in the least demanding applications.

The \(1/n^2\) decay is worth noticing. Both rectified waves have coefficients falling as \(1/n^2\) rather than \(1/n\):

because neither waveform is discontinuous — the full-wave output has a corner at each zero crossing but no jump. By Problem 10's rule that gives \(1/n^2\), and it means the high harmonics are easily filtered.

Definition. A line spectrum plots \(|c_n|\) against frequency \(n\omega_0\) — a set of discrete lines, one per harmonic:

The last point is what makes a periodic signal special. A periodic waveform occupies discrete frequencies only, which is why a spectrum analyser shows a comb rather than a continuum.

The square wave's magnitude spectrum (\(A = 100\) V, 50 Hz):

A \(1/n\) envelope with every even line missing. On a log–log plot the envelope is a straight line of slope \(-20\) dB/decade — the same slope as a single pole, and for the same reason.

The phase spectrum plots \(\angle c_n\). For the odd square wave every \(c_n\) is purely imaginary:

Constant phase. Shifting the time origin would change this entirely while leaving the magnitude spectrum untouched — Problem 18.

What each spectrum tells you:

The last is important and often forgotten. Two waveforms with identical magnitude spectra can look completely different, because rearranging the phases redistributes where the harmonics reinforce. A square wave and a narrow pulse train can share an envelope and share nothing else.

The bandwidth question, made concrete. To reproduce a 50 Hz square wave with harmonics down to 1% of the fundamental:

A 50 Hz square wave needs 5 kHz of bandwidth — a hundred times its own repetition rate. For the triangle wave of Problem 6, \(1/n^2 < 0.01\) gives \(n > 10\), so 500 Hz suffices. The edges, not the period, set the bandwidth.

The experiment. Sum \(N\) harmonics of a \(\pm1\) square wave and record the peak value near the jump:

The overshoot does not shrink. It converges to a fixed 17.90% and stays there however many terms are added — which is the phenomenon.

The exact value. The limit is

involving the sine integral \(\text{Si}\). Quoted as a fraction of the total jump — here \(2\), from \(-1\) to \(+1\) — this is

The 8.949% figure is the standard one, and the two numbers are frequently confused: 8.949% of the jump, which for a \(\pm A\) square wave reads as 17.9% of \(A\).

Why this does not contradict convergence. The series does converge — but not uniformly:

The ripple gets narrower but never shorter, and slides ever closer to the discontinuity. Pick any point away from the jump and the series converges there perfectly well; the overshoot simply moves out of the way rather than diminishing.

At the discontinuity itself the series converges to the midpoint:

For the \(\pm A\) square wave, the series gives exactly zero at the jump. It splits the difference — a general property, not a special case.

Practical consequences, which are real rather than academic:

A sharp-cut filter applied to a square wave produces 9% overshoot regardless of how good the filter is — the overshoot comes from the truncation, not from any defect in the design.

The cure is to truncate gently. Instead of cutting the series off abruptly, taper the coefficients with a window:

This is exactly the trade of Set 30: sharpness against ringing. A Bessel filter's gentle roll-off is a taper in disguise, which is why it shows almost no overshoot on a step.

The rule:

Why it holds. Integration by parts, applied to the coefficient integral, transfers a derivative onto \(f\) and produces a factor \(1/jn\omega_0\):

Each integration by parts gains one power of \(n\) in the denominator — but only if the boundary term vanishes, which requires continuity. A jump makes the boundary term survive and stops the process, fixing the decay rate at that point.

Bandwidth for 1% accuracy, the practical consequence:

A twentyfold difference in bandwidth for waveforms of the same period and amplitude. This is why data transmission shapes its pulses — the raised-cosine pulse is used precisely because its \(1/n^3\) decay minimises the spectrum it occupies.

The reverse reading is equally useful. Given a spectrum, the decay rate identifies the sharpest feature:

The last row is the limiting case: an impulse train has coefficients that never decay, since \(\mathcal{L}\{\delta\} = 1\) is flat (Set 31, Problem 13). Infinite bandwidth is the price of an infinitely sharp feature.

The connection to filter roll-off. The envelope slopes are exactly those of Set 28:

and the reason is the same in both cases — repeated integration. A waveform's spectrum falls off for the same mathematical reason a filter's response does, which is why filtering a waveform and smoothing it are the same operation.

1Note the circuit at the fundamental:

A convenient choice: the circuit is at 45° at the fundamental, and becomes progressively more inductive at higher harmonics.

2The impedance at the \(n\)-th harmonic. Only the reactance scales:

This is the entire method — evaluate \(\mathbf{Z}(s)\) at \(s = jn\omega_0\), which is legitimate by Set 31, Problem 16 because the circuit is stable.

3Divide each voltage harmonic by its own impedance:

Each harmonic is treated as a completely separate problem — a different source, a different impedance, a different phase. Nothing couples them, because the circuit is linear.

4Assemble the answer:

in peak values (RMS \(\times\sqrt2\)). Convergence is rapid — the eleventh harmonic contributes barely 1% of the fundamental.

The two attenuation mechanisms compound, which is why convergence is so much faster than the input's:

The source falls as \(1/n\); the impedance rises roughly as \(n\); the current therefore falls as \(1/n^2\). The inductor has smoothed the current without any deliberate filtering.

Why the transient is missing, and when that matters. The method finds only the steady state:

so the circuit settles within about one cycle and the steady state is reached almost immediately. For a circuit with \(\tau \gg T\) the steady state might take hundreds of cycles to establish, and Fourier says nothing about what happens meanwhile — that is Laplace's territory.

The theorem. The mean square of a waveform is the sum of the mean squares of its harmonics:

Squares add, not amplitudes. This follows directly from orthogonality (Problem 2): squaring the series produces cross-terms, and every one of them integrates to zero.

Verify on the square wave, whose RMS is obviously \(A\) since \(|f| = A\) always:

Using the known sum \(\sum_{n \text{ odd}}1/n^2 = \pi^2/8\):

Exact. And read backwards, this is a derivation of the series \(1+\frac19+\frac1{25}+\cdots = \pi^2/8\) — a second number-theoretic by-product, after Problem 3's Leibniz series.

Convergence in practice. Truncating the sum:

The fundamental alone accounts for 90% of the RMS despite the waveform's severe distortion — because the harmonics enter as squares, and \(0.333^2 = 0.111\) is already small.

The current in the running example:

The harmonics add only 1.3% to the RMS current, even though they add 48% to the RMS voltage — because the inductor has removed most of them (Problem 11).

Why this matters for measurement. Two instruments give different readings on a distorted waveform:

The second type is calibrated assuming a sine wave, where the form factor is 1.111. On a square wave it over-reads by 11%; on a peaky rectifier current it can under-read by 40%. Only a true-RMS instrument is trustworthy on a distorted waveform.

The average power is the mean of \(vi\) over a period. Substituting both series and multiplying out gives terms of the form

By orthogonality (Problem 2) this vanishes unless \(m = n\). A 150 Hz voltage and a 250 Hz current transfer no net energy, however large both may be.

Hence power adds harmonic by harmonic:

with \(V_n, I_n\) RMS and \(\phi_n\) the phase angle at that harmonic. Each harmonic behaves as an independent single-frequency circuit — Set 21's power theory applied term by term.

Compute for the running example, method 1 — sum \(V_nI_n\cos\phi_n\):

Method 2 — all the power is dissipated in the resistor, so

Agreeing exactly. The second method is far quicker when the circuit has a single resistor, and it also confirms Parseval — the two are the same statement.

Where the power actually goes:

The harmonics carry only 2.6% of the power despite the input being 48% distorted — because the inductor limits their current, and power goes as current squared.

The inductor consumes no average power at any harmonic, as expected:

so it exchanges reactive power at every harmonic while dissipating none — Set 21's result, holding independently at each frequency.

A caution about reactive power. Unlike \(P\), the definition of \(Q\) with harmonics is not settled:

The shortfall is called distortion power \(D\), with \(S^2 = P^2+Q^2+D^2\). The familiar power triangle of Set 21 is a two-dimensional picture of what is really a three-dimensional situation, and it applies only to sinusoidal operation.

The definition — harmonic content relative to the fundamental:

using RMS values. A pure sinusoid has THD = 0, and the measure grows without bound as distortion increases.

Exact THD of a square wave. Since \(V_{\text{rms}} = A\) and \(V_1 = 4A/(\pi\sqrt2)\):

A closed form, using the same \(\pi^2/8\) that appeared in Parseval. The relation \(\text{THD} = \sqrt{(F_{\text{rms}}/F_1)^2-1}\) is the quickest route whenever the true RMS is known.

Typical values, for calibration:

Note the two rectifier entries measure different things. The ripple factor compares all AC content with the DC output and comes to \(\sqrt{\pi^2/8-1} = 48.3\%\) — numerically identical to the square wave's THD, since both reduce to the same expression. The THD proper compares the harmonics with the \(2\omega_0\) fundamental and is 22.7%.

THD can exceed 100% — it is referenced to the fundamental, not the total, so a waveform with more harmonic content than fundamental is entirely possible. A capacitor-input rectifier draws exactly such a current.

The running example, input against output:

The inductor has cut the distortion by a factor of three without any deliberate filter design — simply because \(|\mathbf{Z}_n|\) rises with \(n\). Problem 16 pursues this.

An alternative definition exists and is used in power engineering:

Always state which is meant. \(\text{THD}_F\) is the usual convention and is assumed throughout here; \(\text{THD}_R\) is sometimes preferred because it is bounded.

What THD does not tell you. It is a single number summarising a whole spectrum:

Two waveforms with equal THD can have entirely different effects. Standards therefore specify individual harmonic limits as well as a total, precisely because the total hides the distribution.

Power factor is always defined the same way, harmonics or not:

This definition never fails. It is the shortcuts that fail.

Compute directly from Problems 12 and 13:

Now the standard formula. It splits PF into displacement and distortion factors:

0.6978 against a true value of 0.6451 — an error of 8%. The formula is not approximately right here; it is wrong.

Why it fails. The derivation assumes the voltage is a pure sinusoid:

— all harmonic currents meet zero voltage and carry no power. Here the voltage is the square wave, so every harmonic has both voltage and current and every one contributes:

The contrast makes it clear. Take the identical circuit with a pure 100 V rms sinusoid:

The same RMS voltage delivers 17% less power when it is a square wave, because the harmonic voltage meets a high impedance and achieves little.

When each approach is safe:

The first row covers most real installations — the supply is reasonably sinusoidal and the load draws distorted current — which is exactly why the formula became standard, and why it is applied outside its range so often.

The mechanism. The transfer function from source voltage to current is

a first-order low-pass with corner at \(\omega = R/L\). Here \(R/L = 314.2\) rad/s \(= \omega_0\) exactly — the corner sits at the fundamental, which is why the fundamental is at 45°.

Harmonic by harmonic:

The attenuation factor is \(|\mathbf{Z}_n|/|\mathbf{Z}_1|\), growing roughly linearly with \(n\). Combined with the source's own \(1/n\) fall, the current harmonics decay as \(1/n^2\).

The overall effect on distortion:

A reduction by a factor of 2.96. A single inductor has done most of the work, which is why series line reactors are the standard first remedy for harmonic problems in industrial installations.

Improving it. To cut distortion further, move the corner frequency down:

The difficulty is that a first-order filter cannot separate the fundamental from the third harmonic sharply. They are only a factor of three apart, and 20 dB/decade gives just 9.5 dB over that span.

This is precisely Set 30's problem. The specification would be:

A transition ratio of 3 is generous by Set 30's standards — Challenge C1 there faced 1.2 — so a modest order suffices. For 40 dB of rejection at the third harmonic with 1 dB of passband loss, Set 30's formula gives \(n \ge 4.8\), hence a 5th-order Butterworth.

The alternative used in practice: tuned traps. Rather than a low-pass, place a series \(LC\) resonant at the offending harmonic (Set 29):

A trap tuned to 150 Hz shunts the third harmonic to earth while barely loading the 50 Hz fundamental. It is far more economical than a high-order low-pass when the spectrum is known and stable — and Set 29, Problem 20 gave the detuning-reactor design that keeps it from resonating with the supply.

Set up the phases. In a balanced system phase \(b\) lags \(a\) by 120° and \(c\) by 240°. At the \(n\)-th harmonic that delay becomes

because a fixed time delay corresponds to \(n\) times the phase angle at the \(n\)-th harmonic — Problem 18's time-shift result.

Work through the harmonics:

The pattern repeats with period 3. Harmonics divisible by 3 — the triplens — have zero phase shift, meaning they are identical in all three lines at every instant.

The consequence at the neutral. The neutral carries the sum of the three line currents:

Triplens do not cancel — they add arithmetically. Everything else cancels perfectly in a balanced system.

The practical hazard. Consider a balanced load drawing 100 A of fundamental plus 30 A of third harmonic per phase:

A neutral sized on the assumption that a balanced load needs none is now carrying 86% of the line current. This is a genuine and frequent cause of overheating — and it is invisible to a clamp meter reading only the fundamental.

Where the third harmonic comes from. Almost every non-linear single-phase load produces it:

Note the common feature: all are half-wave symmetric, so they produce odd harmonics only — and the third is the largest of them. An office full of computers is a large, well-balanced, heavily triplen-producing load.

The remedies, all standard practice:

The delta winding is why Dy is the standard distribution connection — Set 27 noted this as "harmonic trapping" without explaining it. The delta provides a closed loop in which the in-phase triplen currents can circulate, keeping them off the supply side entirely.

Derive the rule. For \(g(t) = f(t-t_0)\):

Substituting \(\tau = t-t_0\) and using periodicity:

Magnitude unchanged, phase rotated by \(-n\omega_0t_0\) — linear in \(n\), which is Set 31, Problem 7's delay theorem restricted to harmonic frequencies.

The key consequence:

This resolves Problem 4's puzzle. Whether a waveform looks even or odd depends on where the clock started, and shifting merely redistributes each harmonic between its sine and cosine parts — the magnitude \(\sqrt{a_n^2+b_n^2}\) is invariant.

Demonstrate on the square wave. Shifting by a quarter period, \(t_0 = T/4\):

All phases are now 0° or 180°, meaning every coefficient is real — the shifted wave is even, and its series contains only cosines. Same waveform, same spectrum, entirely different-looking series.

Why linear phase is a delay. The shift adds phase proportional to \(n\):

This is Set 28, Problem 17's constant group delay, and Set 30, Problem 9's Bessel criterion, seen from the series side. If a filter shifted every harmonic by the same angle instead of the same time, the waveform would be distorted — which is exactly what a non-linear-phase filter does.

The practical moral for computation:

If only magnitudes are wanted — for RMS, THD or heating — step 3 can be skipped entirely. Most power-quality calculations need nothing more.

The limitation to remove. Everything so far has required strict periodicity — but a single pulse, a decaying transient or a speech waveform is not periodic. The trick is to treat such a signal as periodic with \(T \to \infty\).

What happens to the spectrum. Line spacing is \(\omega_0 = 2\pi/T\):

The lines crowd together until they merge into a continuum. A discrete spectrum becomes a continuous one — which is the essential difference between series and transform.

What happens to the coefficients. Each is

and the \(1/T\) drives every coefficient to zero. The individual lines vanish — which is why the transform cannot simply be the limit of the coefficients.

The fix: rescale before taking the limit. Define

Multiplying by \(T\) exactly cancels the shrinking factor. The Fourier transform of Set 34 — and the reconstruction becomes an integral:

The correspondence, item by item:

The units change is the conceptual step. \(F(\omega)\) is not "the amount at \(\omega\)" — no single frequency carries finite energy in a continuous spectrum — but a density that must be integrated over a band.

A worked illustration. Take a rectangular pulse of width \(\tau\) repeating with period \(T\). Its coefficients are

The envelope was the transform all along. The series samples it at multiples of \(\omega_0\); removing the periodicity simply fills in between the samples.

What Set 34 gains from the change:

The third is the one Set 28, Problem 19 has been owing since \(t_rf_c = 0.35\) was first observed.

The method, complete:

What the harmonic view explained that the time-domain view could not:

Each is invisible in the time domain and obvious in the frequency domain.

The recurring warnings, all of one kind — a shortcut applied outside its range:

Every one is a sinusoidal-operation result quietly assumed to hold generally. Problem 15's 8% power-factor error is the sharpest instance.

What the method cannot do:

The first two are the price paid in Problem 1 for computability, and they were worth paying. The third is a genuine gap — the reciprocity between rise time and bandwidth needs a continuum to state at all.

The remaining programme:

Set 34 follows directly from Problem 19: stretch the period to infinity and rescale. Set 35 addresses the one structural gap that has survived every method so far — that a two-stage circuit's transfer function is not the product of its stages'.