Transfer Functions and Bode Plots

The definition. For a linear circuit driven at one frequency, the output is a sinusoid of the same frequency with a different amplitude and phase. The ratio of the two phasors is

Both magnitude and phase are functions of \(\omega\), and together they are the complete description.

Why one function suffices. Two facts from earlier sets combine:

Knowing the response to every sinusoid therefore determines the response to every input. That is a remarkably strong claim, and it rests entirely on linearity.

Four common transfer functions, distinguished by what is divided by what:

All four are "transfer functions"; a plot is meaningless without saying which.

A worked instance. For a series \(RC\) with the output across \(C\), voltage division gives

Every method of Set 21 produces transfer functions directly — the only change is to leave \(\omega\) symbolic instead of substituting a number.

What it cannot describe:

The third exclusion is the one most easily forgotten. A transfer function describes the circuit long after switching, exactly as the phasor method has assumed since Set 20.

Set 27's transformer is a good example of the need. Its passband was bounded below by magnetising inductance and above by leakage — a statement about \(H(j\omega)\) that no single-frequency calculation could express.

Definitions. Writing \(H(s) = N(s)/D(s)\):

For the given function: one zero at \(s = -1\), and poles at \(s = 0\), \(-10\) and \(-100\).

Convert to standard Bode form, in which every factor reads \((1+s/\omega_c)\):

The leading constant collapsed from 100 to \(100/(10\times100) = 0.1\). Failing to extract those factors is the commonest error in Bode work — it shifts the whole plot vertically by the amount missed, here 60 dB.

Why standard form matters. Each factor now equals 1 at low frequency, so the low-frequency behaviour is read straight off the leading term:

Slope \(-20\) dB/dec, crossing 0 dB where \(0.1/\omega = 1\), i.e. at \(\omega = 0.1\) rad/s. That single point anchors the entire sketch.

What the pole locations tell you. Poles are the natural frequencies of the circuit — the exponents of its transient response, from Sets 18 and 19:

A pole in the right half-plane means a growing exponential and hence instability — Set 19's damping criterion, restated in the language of poles. All poles here are on or left of the imaginary axis.

Counting poles and zeros. Including those at infinity, the two counts must be equal. Here there are three finite poles and one finite zero, so there are two zeros at infinity — which is why the high-frequency slope is \(-40\) dB/dec:

A quick check on any completed sketch, and one that catches a dropped factor immediately.

The definition for a voltage or current ratio:

The factor is 20 for amplitude ratios and 10 for power ratios, because power goes as the square: \(10\log(P_2/P_1) = 10\log(V_2/V_1)^2 = 20\log(V_2/V_1)\). The two give the same number for the same physical situation, which is the point of the convention.

Reason one: products become sums. A transfer function is a product of factors, and

so the individual factors' plots simply add. Phase adds too, without needing logarithms. This is what makes graphical assembly possible at all.

Reason two: power laws become straight lines. With a logarithmic frequency axis, \(|H| \propto \omega^n\) plots as

a straight line of slope \(20n\) dB per decade. Every asymptote in Bode work is such a power law, so every asymptote is a straight line — the whole basis of the sketching method.

Reason three: enormous ranges fit on one page. An audio system spans \(10^3\) in frequency and perhaps \(10^6\) in gain. Linear axes would make most of the plot unreadable.

Conversions worth memorising:

From 6 dB and 20 dB almost anything can be built: 12 dB is \(\times4\), 26 dB is \(\times20\), 34 dB is \(\times50\).

Why \(-3\) dB defines the bandwidth. At that point the amplitude ratio is \(1/\sqrt2\) and hence the power ratio is exactly one half:

The convention is a statement about power, not about amplitude, which is why the same 3 dB appears in every discipline that measures bandwidth.

A caution on absolute levels. A decibel is a ratio, so an absolute level needs a stated reference — dBm relative to 1 mW, dBV relative to 1 V, dBSPL relative to 20 µPa. "A signal of 30 dB" alone means nothing.

aThe constant \(K\). Magnitude and phase are both independent of frequency:

A horizontal line. Its only role is to slide the whole magnitude plot up or down — but getting it wrong shifts everything, which is why Problem 2 insisted on extracting it correctly.

bA pole at the origin, \(1/j\omega\):

A straight line of slope \(-20\) dB/dec passing through 0 dB at \(\omega = 1\). There is no break and no curvature — the asymptote is the exact curve.

cA zero at the origin, \(j\omega\): the mirror image, \(+20\) dB/dec and \(+90°\).

Repeated factors multiply the effect:

The circuits that produce them:

The connection is exact: \(v_C = \frac{1}{C}\int i\,dt\) transforms to \(\mathbf{V} = \mathbf{I}/j\omega C\), so a capacitor is an integrator. Set 20 established this and it is now visible as a plot.

Where a constant plus an origin factor anchors the sketch. Combining them:

For the running example's \(0.1/j\omega\), the low-frequency asymptote crosses 0 dB at \(\omega = 0.1\) — the single most useful number for starting a sketch, because everything else is built by changing slope at the break frequencies.

The phase contribution is constant, which is unusual: every other factor has a phase that varies with frequency. An integrator's \(-90°\) is present at DC and at a gigahertz alike.

The exact expressions:

The two asymptotes. Well below the break:

Well above it:

The high-frequency asymptote is a power law \(\omega^{-1}\), hence a line of slope \(-20\) dB/dec, and the two asymptotes meet exactly at \(\omega = \omega_c\). That intersection is what makes \(\omega_c\) "the corner".

The exact curve around the break:

Three numbers carry the whole correction: 3 dB at the break, 1 dB an octave either side, negligible a decade away. Sketching those three and joining smoothly gives a curve indistinguishable from the exact one at plotting accuracy.

The phase asymptotes. The standard construction takes phase as 0° up to \(\omega_c/10\), then a straight line to \(-90°\) at \(10\omega_c\):

The phase approximation is worse than the magnitude one — 5.7° of error at the ends, and phase influences a decade either side of the break rather than being confined near it.

Phase spreads much further than magnitude. At a decade below the break the magnitude error is 0.04 dB — invisible — but the phase is already 5.7° from zero. This asymmetry matters for stability margins (Problem 18), where a distant pole can erode phase margin while barely touching the gain.

The zero is the exact reciprocal of the pole, so its Bode contribution is the mirror image about the horizontal axis:

Every sign flips; every magnitude is identical. That single fact halves the work of learning the building blocks.

What a zero does physically. It marks a frequency at which some path through the circuit begins to help rather than hinder:

A zero cannot make gain rise forever. Every physical circuit has at least as many poles as zeros, because at high enough frequency stray capacitance always intervenes:

A transfer function with more zeros than poles would have \(|H| \to \infty\) as \(\omega \to \infty\) — an infinite output from a finite input, which no passive or finite-gain circuit can produce.

The phase is the reason zeros are placed deliberately. A zero adds up to \(+90°\) of phase, and it does so over two decades centred on \(\omega_c\). Placing a zero just below a troublesome frequency lifts the phase there:

That is the whole mechanism of lead compensation, and it is why Problem 18's phase margin can be improved without changing the gain crossover much.

A worked check. For \(H = (1+j\omega/10)\) at \(\omega = 10\):

The break of a zero is where its real and imaginary parts are equal — exactly as for a pole.

1Standard form, from Problem 2:

2List the factors and their breaks:

3Anchor the low-frequency asymptote. Below all breaks, \(H \to 0.1/j\omega\), which crosses 0 dB at \(\omega = 0.1\) with slope \(-20\) dB/dec. At \(\omega = 0.01\) it is therefore at \(+20\) dB.

4Accumulate the slopes, changing at each break:

Note the flat section between 1 and 10: the zero's \(+20\) exactly cancels the integrator's \(-20\) there.

5Check the final slope against the pole–zero count of Problem 2:

Compare with the exact values:

The errors are within 3 dB everywhere and essentially zero away from the breaks — the asymptotic sketch is a good plot, not merely a rough one.

The phase has an interesting feature. It starts at \(-90°\) from the integrator, rises to about \(-37°\) as the zero takes effect, then falls to \(-180°\) once both poles act. A non-monotonic phase like this is characteristic of a zero placed below the poles, and it is exactly what lead compensation exploits.

The 3 dB rule is exact for an isolated break. Each factor contributes exactly \(\pm3.01\) dB of error at its own corner and essentially nothing a decade away. Errors add, so the total error at any break is

At \(\omega = 1\): the zero contributes \(+3.01\) dB. The pole at 10 is one decade away and contributes

and the pole at 100 contributes \(-0.0004\) dB. Total:

At \(\omega = 100\): the pole at 100 contributes \(-3.01\) dB, the pole at 10 contributes \(-0.043\) dB, and the zero at 1 contributes \(+0.0004\) dB:

Both discrepancies accounted for exactly. The rule is not approximate — the other factors simply were not counted.

How the error grows as breaks approach each other:

Two coincident poles give 6 dB of error at the break, not 3. A repeated factor \((1+s/\omega_c)^2\) is 6.02 dB below its asymptote at \(\omega_c\) and its phase is \(-90°\) there rather than \(-45°\).

When the asymptotic method fails badly. Two situations:

The second is far more serious. At \(\zeta = 0.05\) the true curve peaks 20 dB above an asymptote that predicts 0 dB — an error of a factor of ten in amplitude, which no amount of care with the corners will reveal.

A practical rule. The asymptotic sketch is trustworthy to about 1 dB provided

Outside those conditions, sketch the asymptotes to get the shape and then compute the few points that matter.

The standard form, with \(u = \omega/\omega_n\):

This is Set 19's second-order system seen in the frequency domain: \(\omega_n\) the natural frequency, \(\zeta\) the damping ratio.

The asymptotes are simple:

The \(-90°\) at \(\omega_n\) is independent of \(\zeta\) — a reliable landmark, and the practical way to locate \(\omega_n\) on a measured plot.

Why the asymptotes are insufficient. The value at the break depends entirely on \(\zeta\), which the asymptotes do not contain:

A 32 dB spread at one frequency, all with identical asymptotes. The asymptotic sketch simply cannot represent a quadratic pair without the damping ratio being supplied separately.

The critically useful case \(\zeta = 0.5\), where \(|H(j\omega_n)| = 1\) exactly — the curve passes through the corner of the asymptotes, as a simple pole never does.

Two real poles versus a complex pair. When \(\zeta \ge 1\) the quadratic factorises into two real poles:

and the ordinary two-break sketch applies. Only for \(\zeta < 1\) is the quadratic irreducible, and only for \(\zeta < 1/\sqrt2\) does it peak — Problem 10.

The phase transition sharpens as damping falls. At \(\zeta = 1\) the phase swings from 0° to \(-180°\) over about three decades; at \(\zeta = 0.05\) it does so over a fraction of a decade — almost a step. That abruptness is what makes lightly damped systems difficult to stabilise.

Maximise \(|H|\) by minimising the quantity under the root:

The non-trivial root:

The peak is below \(\omega_n\), and moves further below as damping increases.

The condition for a peak to exist. A real \(\omega_r\) requires

For \(\zeta \ge 0.707\) the only stationary point is \(u = 0\) and the response falls monotonically. Note this differs from the time-domain threshold: overshoot in the step response appears for \(\zeta < 1\), but frequency-domain peaking only for \(\zeta < 0.707\). A system with \(\zeta = 0.8\) overshoots in time yet shows no resonant peak.

The peak value. Substituting \(u^2 = 1-2\zeta^2\):

Tabulated and verified numerically:

As \(\zeta \to 0.707\) the peak both shrinks to nothing and migrates to zero frequency, so it disappears smoothly rather than abruptly.

For light damping the two useful quantities nearly coincide:

At \(\zeta = 0.1\), \(1/2\zeta = 5.00\) against a true 5.025 — half a percent. This is also the quality factor \(Q = 1/2\zeta\) that Set 29 develops.

The design consequence. Peaking is usually unwanted:

\(\zeta = 0.707\) appears throughout engineering for exactly this reason: it is the boundary at which peaking first appears.

Read the slope changes, working upward in frequency:

Every slope change of \(-20\) is a pole, of \(+20\) a zero. The initial slope gives the behaviour at the origin.

Write the form:

The shape is now fixed; only \(K\) remains, and it comes from any single level on the plot.

Find \(K\) from the flat region. Between the two breaks, the numerator's \(s\) is cancelled by the first pole:

Check the answer by evaluating at several frequencies:

The plateau never quite reaches 26 dB — it peaks at 25.57 dB. The two breaks are only a factor of 20 apart, so by Problem 8 each still contributes about \(-0.2\) dB at the midpoint. Reading the flat level off a measured plot therefore underestimates \(K\) slightly.

The symmetry is worth noticing. This response is a band-pass, and its magnitude is symmetric about the geometric mean:

The values at 0.2 and 400 are equal, as are those at 2 and 40. Band-pass responses are symmetric on a logarithmic axis — a fact that Set 29 uses heavily and that provides a free check on any such plot.

What cannot be recovered from magnitude alone. Problem 16 shows that a right-half-plane zero at \(+\omega_z\) has exactly the same magnitude as one at \(-\omega_z\). This reconstruction therefore assumes a minimum-phase system — the standard assumption, and one that must be verified against the measured phase.

Voltage division gives the transfer function directly:

One pole, no zeros — the simplest non-trivial transfer function there is, and the factor of Problem 5 in concrete form.

The cut-off frequency:

Note \(RC = 1\ \text{ms}\) is exactly the time constant of Set 18. Cut-off frequency and time constant are reciprocals — the same physical fact viewed in two domains.

The response at three points:

A factor of ten above cut-off attenuates by 20 dB — the \(-20\) dB/dec roll-off, confirmed.

The physical explanation at each extreme:

The cut-off is where the two impedances are equal — which is why \(\omega_c = 1/RC\) and not something requiring calculation.

What it does to a square wave. A square wave is a sum of odd harmonics (Set 33). If the fundamental is well below \(\omega_c\), the harmonics are attenuated progressively:

Edges are made of high-frequency content, so removing high frequencies rounds them. Problem 19 makes this quantitative through the bandwidth–rise time relation.

The roll-off is gentle. Only 20 dB per decade means a signal at ten times \(\omega_c\) is still 10% of its input amplitude. Sharper selectivity needs more poles, which is the subject of Set 30.

The high-pass transfer function:

A zero at the origin and a pole at \(1/RC\) — the two factors of Problems 4 and 5 combined.

The two responses are complementary:

They sum to unity at every frequency — as they must, since the two voltages are the two parts of the same divider and KVL applies. Note this is a sum of complex quantities: the magnitudes do not sum to 1, and at cut-off both are 0.707.

The same cut-off:

Exact mirror images in magnitude, and negated in phase. The high-pass phase is positive — the output leads the input, because a capacitor's current leads its voltage.

The four first-order circuits:

The output element swaps between the two families. Taking the output across the reactive element gives low-pass for \(C\) but high-pass for \(L\) — because their impedances vary oppositely with frequency.

Why \(RC\) is preferred in practice. Both families give identical responses, yet \(RC\) dominates:

The last is decisive for modern electronics: inductors do not integrate well, so active \(RC\) filters (Set 30) replaced \(LC\) designs wherever an amplifier was available.

The naive expectation. If the second section drew no current from the first:

The actual result. Analysing the full four-element network — by mesh analysis, or by noting that the first section is loaded by the second's input impedance:

That single change is the entire loading effect, and it arises because the second capacitor draws current through the first resistor.

What the "3" does to the poles. The buffered version has a repeated pole at \(1/T = 1000\); the actual one factorises into two distinct real poles:

The poles have separated — one moved down by a factor of 2.6, the other up by the same factor. Their product is unchanged at \(1/T^2\), since the constant term did not change.

The bandwidth penalty:

Loading costs 42% of the bandwidth. At \(\omega = 1000\) the buffered pair gives \(-6.02\) dB against the cascaded pair's \(-9.54\) dB — 3.5 dB of unexpected loss.

The equivalent damping. Comparing with the standard quadratic form:

Overdamped, hence the two real poles. The buffered version has \(\zeta = 1\) — critically damped, with the repeated pole. Loading always increases the damping of a passive cascade, which is why passive \(RC\) networks alone can never produce a peaked or sharp response.

The remedies:

The second is common where an amplifier is unavailable: scaling the impedances by ten reduces the loading error to a few percent without changing either section's own cut-off.

The definition. A transfer function is minimum-phase if all its poles and all its zeros lie in the left half of the \(s\)-plane.

Stability concerns only the poles. A right-half-plane zero does not cause instability — it causes extra phase lag, which is a different and subtler problem.

The Bode gain–phase relation. For a minimum-phase system the phase is determined by the magnitude:

The relation is exact only where the slope has held for several decades; near a break it is approximate. Problem 7's example illustrates both: at \(\omega = 0.01\) the slope has been \(-20\) for a long way and the phase is \(-89.5°\), but at \(\omega = 3.16\), in a flat region between two nearby breaks, the phase is \(-36.9°\) rather than 0°.

Why this is useful. Three practical consequences:

The second is the classical design rule: arrange for the gain to cross 0 dB at a slope of \(-20\) dB/dec, and adequate phase margin follows automatically. Crossing at \(-40\) would give near \(-180°\) and no margin at all.

The name explained. Among all systems with a given magnitude response, the minimum-phase one has the least phase lag. Problem 16 constructs a counterexample with identical magnitude and up to 180° more lag.

What makes a system non-minimum-phase. Right-half-plane zeros arise from competing paths through a circuit whose contributions cancel and then reverse:

The common signature is an initial response in the wrong direction — the output first moves opposite to where it will settle.

The magnitude. Put \(s = j\omega\) and let \(u = \omega/\omega_0\):

Numerator and denominator are complex conjugates, so their magnitudes are identical. The magnitude plot is a flat line at 0 dB — no break, no slope, nothing.

The phase, however, sweeps a full 180°:

The zero at \(s = +\omega_0\) and the pole at \(s = -\omega_0\) each contribute \(-45°\) at \(\omega_0\) — both lagging, because a right-half-plane zero lags where a left-half-plane one leads.

The consequence for reconstruction. Take any minimum-phase \(H_{mp}(s)\) and multiply by this all-pass:

The two are indistinguishable on a magnitude plot, and differ by up to 180° in phase. Problem 11's reconstruction was therefore only valid under the minimum-phase assumption, and no amount of magnitude data can verify it.

A concrete pair. Take \(H_{mp} = 10(1+s/10)/[(1+s)(1+s/100)]\) and the same with the zero reflected to \(+10\):

Magnitudes agree to every decimal place; phases diverge to nearly 180°. A control loop closed around the second would behave utterly differently.

What all-pass networks are used for. Since they change phase without changing magnitude, they are the tool of choice wherever delay must be adjusted:

The name "minimum phase" justified. Any non-minimum-phase system factorises as a minimum-phase part times an all-pass, and the all-pass only ever adds lag. So among all systems with a given magnitude, the minimum-phase one really does have the least phase lag — the term is a theorem, not a label.

The definition:

It is the delay experienced by the envelope of a narrow band of frequencies near \(\omega\) — the time taken for information, as distinct from the phase delay \(-\angle H/\omega\) of an individual carrier cycle.

Why a constant group delay means no distortion. Suppose \(\angle H = -\omega t_0\), a straight line through the origin. Then every component is delayed by

and the output is the input shifted bodily in time:

Linear phase, constant delay, no waveform distortion — the ideal a filter designer aims for and rarely achieves.

For the single pole, \(\angle H = -\tan^{-1}(\omega/\omega_c)\):

At DC the delay equals the time constant \(RC = 1\) ms exactly — a satisfying link to Set 18 — and it halves at cut-off.

What non-constant delay does. Different frequency components arrive at different times, so a pulse spreads and its shape changes even though no component has been attenuated unequally:

An all-pass has group delay but no attenuation. For \((1-s/\omega_0)/(1+s/\omega_0)\):

Exactly twice the single pole's, because both the pole and the RHP zero contribute lag. A network can distort a signal in time while leaving every amplitude untouched — the practical meaning of Problem 16.

The design trade-off. Filters that are sharpest in magnitude are worst in delay:

Set 30 develops these. The choice is governed by whether the application cares about separating frequencies or about preserving waveshape — and it can rarely have both.

The criterion. A negative-feedback loop oscillates if the signal returns in phase and undiminished:

The margins measure how far the loop is from that condition, in the two independent directions.

The phase crossover is where \(\angle L = -180°\), and it does not depend on \(K\):

Two angles summing to 90° have reciprocal tangents, so \(\omega = 10/\omega\):

A clean closed form — worth noting that for \(K/[s(s+a)(s+b)]\) the phase crossover is always at \(\sqrt{ab}\).

The critical gain. At \(\omega_{pc}\) with \(K = 1\):

Above \(K = 110\) the closed loop is unstable. Note the neat coincidence: \(K_{\text{crit}} = ab(a+b)/1 = 1\times10\times11 = 110\), the Routh result for this cubic.

The margins tabulated:

Gain margin depends on \(K\) in the simplest possible way: GM in dB is \(20\log(110/K)\), since raising \(K\) lifts the whole magnitude curve without touching the phase.

Typical design targets and what they buy:

Phase margin also predicts the transient response: \(\text{PM} \approx 100\zeta\) degrees is a serviceable rule, so 45° corresponds to \(\zeta \approx 0.45\) and about 20% overshoot — connecting directly to Problem 10.

Why both margins are needed. Each measures one direction, and a system can be comfortable in one and dangerous in the other. A large gain margin with 10° of phase margin will ring badly; a large phase margin with 2 dB of gain margin will oscillate as soon as a component drifts.

The step response of a single-pole low-pass, from Set 18:

Find the 10% and 90% times:

Express in terms of the bandwidth in hertz:

The famous constant — and it is \(\ln9/2\pi\), not an empirical fit.

Worked numbers for the running \(RC\):

How general is it? The constant varies a little with the response shape:

Close enough across the useful range that 0.35 is used universally. The relation is essentially independent of the details — bandwidth and rise time are two measurements of one quantity.

The practical uses:

The last is why an instrument should have three to five times the bandwidth of the signal: at \(3\times\), the observed rise time is \(\sqrt{1+1/9} = 1.054\) times the true one — a 5% error.

What was gained. Making \(\omega\) a variable replaced a family of single-frequency calculations by one function:

The three recurring numbers, worth having by heart:

Almost every result in the set is an application or a correction of these.

The recurring caution. Three separate problems found the same failure mode — a plot that looks reasonable but is wrong:

In each case the sketch is internally consistent and looks entirely plausible. The safeguard is knowing which assumption each step relies on — exactly the lesson of Set 26's Problem 15 and Set 27's Challenge C2.

The standing limitation. \(H(j\omega)\) is a steady-state description. It says what a circuit does long after switching, and nothing about the transient. Yet Problem 19 found that the bandwidth and the rise time are rigidly linked, and Problem 2 found that the poles are simultaneously the slope changes and the transient exponents.

The same numbers govern both domains, but the present machinery cannot say why — because \(s\) has been used only as shorthand for \(j\omega\).

What comes next.

Set 31 is the pivotal one. Letting \(s = \sigma + j\omega\) rather than \(j\omega\) alone makes the pole–transient correspondence a theorem instead of an observation, and it retires the steady-state restriction that has held since Set 20.