Lead, Lag, and Lead–Lag Compensator Design
A single gain moves the closed-loop poles only along a locus the plant has already fixed, so when accuracy and damping make demands that no point on that locus can satisfy at once, the cure is to change the locus itself — which is what a compensator does, by adding one zero and one pole and choosing whether they buy phase near crossover or gain near the origin.
- Why a plant's root locus is fixed and what adding a pole–zero pair does to it.
- The lead network, its parameter \(\alpha\), and the derivation of \(\sin\phi_m=(1-\alpha)/(1+\alpha)\) at \(\omega_m=1/T\sqrt{\alpha}\).
- The complete Bode design procedure for a lead compensator, including why a safety allowance of \(5^\circ\) to \(12^\circ\) is always added.
- The angle-deficiency method that places a lead compensator from a root-locus specification.
- How a lag network multiplies the error constant by \(\beta\) while contributing almost no phase at crossover.
- When one network cannot do the job, and how a lead–lag compensator splits the two tasks between its two halves.
- The passive \(RC\) realisations, their loading and attenuation penalties, and how lead and lag relate to the PD and PI actions of Chapter 19.
Why Gain Alone Is Not Enough
Chapter 12 built the root locus as the set of points satisfying the angle condition, and that set depends only on the open-loop poles and zeros — not on the gain. Turning the gain knob slides the closed-loop poles along the locus; it cannot move them off it. If the locus never passes through a region where the transient specification is met, no value of \(K\) will meet it.
The frequency-domain statement is the same fact in different clothing. Raising \(K\) lifts the whole magnitude curve without touching the phase curve, so the gain crossover slides to a higher frequency where the plant's phase is more negative, and the phase margin of Chapter 17 falls. Chapter 9 closed with the resulting deadlock: accuracy wants a large \(K\), damping wants a small one, and a plant of any complexity soon reaches a point where the two demands cannot both be met.
A compensator breaks the deadlock by adding dynamics of its own — in the classical case one zero and one pole, cascaded with the plant.
Both are one zero at \(-1/T\) and one pole. In a lead the pole lies further left than the zero, so the network contributes positive phase. In a lag the pole lies closer to the origin than the zero, so it contributes negative phase — and a useful attenuation at high frequency.
The Lead Network and Its Maximum Phase
Write the lead network in its two equivalent forms, one showing the corner frequencies, the other the pole and zero locations:
The zero sits at \(-1/T\) and the pole at \(-1/(\alpha T)\); since \(\alpha<1\) the pole is a factor \(1/\alpha\) further from the origin. Between the two corner frequencies the magnitude rises at \(20\) dB/decade and the phase is positive. The phase contribution is the difference of two arctangents,
which is zero at both extremes and therefore has a maximum somewhere between. Differentiating and setting the derivative to zero locates it.
The maximum falls at the geometric mean of the two corner frequencies — the midpoint on a logarithmic axis, which is why the phase bump looks symmetric on a Bode plot. Substituting \(\omega_mT=1/\sqrt{\alpha}\) into the phase expression gives the peak value.
at \(\omega_m=1/(T\sqrt{\alpha})\), where the magnitude of \(K_c\alpha(Ts+1)/(\alpha Ts+1)\) is \(K_c\sqrt{\alpha}\) — that is, \(10\log_{10}(1/\alpha)\) decibels above its low-frequency value. The phase and the gain are not independent: a network that supplies more phase necessarily supplies more gain at the same frequency, and this is the fact that drives the whole design procedure.
| \(\alpha\) | \(\phi_m\) | High-frequency gain \(1/\alpha\) |
|---|---|---|
| 0.5 | \(19.5^\circ\) | 2 (6 dB) |
| 0.25 | \(36.9^\circ\) | 4 (12 dB) |
| 0.1 | \(54.9^\circ\) | 10 (20 dB) |
| 0.05 | \(64.8^\circ\) | 20 (26 dB) |
| 0.01 | \(78.6^\circ\) | 100 (40 dB) |
The last column is the reason a single lead network is rarely pushed past about \(60^\circ\). The factor \(1/\alpha\) is a high-frequency gain applied to everything the sensor delivers, and a sensor delivers noise as well as signal. Chapter 19 raised this objection against derivative action; the lead network is that same derivative action with the amplification deliberately limited, and \(\alpha\) is the limit. Taking \(\alpha\) below \(0.05\) trades an unreliable measurement for a marginal gain in phase. When more than \(60^\circ\) is genuinely needed, two lead sections in cascade — each supplying half — cost far less in noise than one section supplying all of it.
Lead Design on the Bode Plot
The design problem is usually posed as a pair of specifications: an error constant, which fixes the low-frequency gain, and a phase margin, which fixes the behaviour at crossover. The procedure below satisfies the first exactly and the second by construction.
Its one subtlety is the reason \(\phi_m\) is not simply the shortfall in phase margin. Adding the lead network raises the magnitude curve near \(\omega_m\), which pushes the gain crossover to a higher frequency, where the plant contributes more phase lag than it did before. Some of the phase we just bought is therefore immediately spent. An allowance of \(5^\circ\) for a gently sloping plant, up to \(12^\circ\) for a steep one, covers the loss.
| Step | Action |
|---|---|
| 1 | Choose the open-loop gain \(K\) so that the error-constant specification (\(K_p\), \(K_v\) or \(K_a\) from Chapter 9) is met exactly. Call the result \(G_1(s)\). |
| 2 | Draw the Bode plot of \(G_1\) and read its phase margin \(\gamma_1\). |
| 3 | Required lead \(\phi_m=\gamma_{\text{spec}}-\gamma_1+\varepsilon\), with \(\varepsilon=5^\circ\) to \(12^\circ\). |
| 4 | \(\alpha=\dfrac{1-\sin\phi_m}{1+\sin\phi_m}\). |
| 5 | Find \(\omega_m\), the frequency at which \(|G_1(j\omega)|=\sqrt{\alpha}\), i.e. \(10\log_{10}\alpha\) dB. This becomes the new gain crossover, because the network adds exactly \(10\log_{10}(1/\alpha)\) dB there. |
| 6 | \(T=\dfrac{1}{\omega_m\sqrt{\alpha}}\); the zero is at \(-1/T\) and the pole at \(-1/(\alpha T)\). |
| 7 | Set \(K_c=1/\alpha\) so that \(G_c(0)=K_c\alpha=1\) and the error constant of step 1 is untouched. |
| 8 | Verify the phase margin of \(G_cG_1\). If it falls short, increase \(\varepsilon\) and repeat. |
Lead Design on the Root Locus
When the specification is given as overshoot and settling time rather than as a phase margin, Chapter 8 converts it into a pair of desired dominant poles \(s_d=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}\), and the design question becomes geometric: how must the locus be bent so that it passes through \(s_d\)?
The answer comes straight from the angle condition. Evaluate the open-loop transfer function at the desired point; if the angle is not \(-180^\circ\), the compensator must make up the difference.
If \(\phi_c\) is positive the point lies to the left of the existing locus and a lead network is required; if it is negative the point lies to the right and a lag — or a reduction in gain — is called for. A single lead section can supply any \(\phi_c\) up to roughly \(60^\circ\); beyond that, use two.
Knowing how much angle is needed does not by itself fix the pole and the zero: infinitely many pairs supply the same angle at \(s_d\). Three choices are standard. The first is cancellation — put the compensator zero on top of an existing plant pole, which removes that pole from the picture and leaves only the compensator pole to place. It is the simplest choice and the one that gives the cleanest algebra, but it must never be used on an unstable or very lightly damped plant pole, because an imperfect cancellation leaves that pole in the closed-loop response where it is no longer controllable by the gain.
The second is the bisector rule. Draw the line from \(s_d\) to the origin and the horizontal line through \(s_d\); bisect the angle between them, then place the zero and the pole on the real axis at \(\pm\phi_c/2\) from that bisector, as seen from \(s_d\). This construction can be shown to maximise \(\alpha\) — that is, to place the pole and zero as close together as possible for the required angle — which minimises the high-frequency gain \(1/\alpha\) and therefore the noise penalty.
The third is simply to fix the zero somewhere convenient, just to the left of the plant's slowest pole, and solve for the pole from the angle condition. Whichever placement is used, the gain follows from the magnitude condition \(|G_c(s_d)G(s_d)|=1\), and the design is not finished until two checks have been made: that the remaining closed-loop poles are far enough left for \(s_d\) to dominate, and that the error constant has not been ruined — a lead network multiplies \(K_v\) by \(\alpha\) at fixed \(K_c\), which is a reduction.
The Lag Network
Reverse the ordering of pole and zero and the network changes character completely.
Now the pole at \(-1/(\beta T)\) is nearer the origin than the zero at \(-1/T\). The magnitude starts at \(K_c\beta\) at low frequency, falls at \(-20\) dB/decade between the two corners, and settles at \(K_c\) — a factor \(\beta\) lower. The phase is negative throughout, with a minimum of \(-\phi_m\) at the geometric mean, where \(\sin\phi_m=(\beta-1)/(\beta+1)\) by the same algebra as before.
Negative phase is exactly what a stability-limited loop does not need, so the lag network is not used for its phase at all. It is used for the ratio \(\beta\) between its low- and high-frequency gains, and the design consists of arranging for the phase penalty to be paid somewhere harmless.
The trick that makes this possible is to place both corner frequencies well below the intended gain crossover. Put the zero one decade below the new crossover \(\omega_2\), so \(1/T=\omega_2/10\). At \(\omega_2\) the zero contributes \(\arctan 10=84.3^\circ\) and the pole contributes very nearly \(90^\circ\) of lag, so the network's net phase there is only about \(-5.7^\circ\) — an amount easily covered by the same safety allowance \(\varepsilon\) used in the lead procedure. Meanwhile the full attenuation \(1/\beta\) is already in effect at \(\omega_2\), because \(\omega_2\) lies a decade above the upper corner.
Equivalently: keep the low-frequency gain fixed and the lag attenuates the loop near crossover by \(\beta\), dragging the crossover down to a frequency where the plant's own phase lag is smaller. Either reading is correct — the network improves the error constant at fixed margin, or the margin at fixed error constant.
Lag Design in Both Domains
The frequency-domain procedure mirrors the lead procedure but reads the Bode plot in the opposite order: the phase curve is consulted first to choose the new crossover, and the magnitude curve then tells us how much attenuation is needed there.
| Step | Action |
|---|---|
| 1 | Set \(K\) from the error-constant specification, giving \(G_1(s)\). |
| 2 | On the phase curve of \(G_1\), find \(\omega_2\) where \(\angle G_1=-180^\circ+\gamma_{\text{spec}}+\varepsilon\), with \(\varepsilon=5^\circ\) to \(12^\circ\). This will be the new gain crossover. |
| 3 | Read \(|G_1(j\omega_2)|\). The lag must remove exactly this much: \(\beta=|G_1(j\omega_2)|\), or \(20\log_{10}\beta=|G_1(j\omega_2)|_{\mathrm{dB}}\). |
| 4 | Place the zero a decade below: \(1/T=\omega_2/10\). |
| 5 | Place the pole at \(1/(\beta T)\). |
| 6 | Set \(K_c=1/\beta\) so that \(G_c(0)=K_c\beta=1\) and the error constant survives. |
| 7 | Verify the phase margin; the \(-6^\circ\) contributed by the network is already covered by \(\varepsilon\). |
On the root locus the lag plays a different-looking but identical role. Suppose the uncompensated locus already passes through an acceptable \(s_d\) — the transient specification is met — but the gain there gives an error constant that is too small by a factor \(\beta\). Add a zero at \(-z\) and a pole at \(-z/\beta\), both very close to the origin, with \(z\) chosen small compared with \(|s_d|\).
The pair is called a dipole. Seen from a point \(s_d\) far away, the pole and the zero are almost at the same place: their angles nearly cancel and their magnitudes nearly cancel, so the locus in the region of interest is essentially unchanged and the gain required at \(s_d\) is essentially unchanged. Seen from the origin, where \(s\) is comparable with \(z\), they are far apart, and the ratio of their distances is exactly \(\beta\). The error constant is multiplied by \(\beta\) for almost nothing.
"Almost" deserves a word. The dipole adds a slow closed-loop pole near \(s=-z\), close to the compensator zero but not on it. Its residue is small, so it contributes a small-amplitude term to the step response — but that term decays with time constant \(1/z\), which is deliberately long. The result is a long, low tail on the settling response, familiar to anyone who has watched a lag-compensated servo creep into its final position. Making \(z\) smaller reduces the tail's amplitude and lengthens its duration; there is no setting that removes it.
Lead–Lag Compensation
Specifications frequently demand both things at once: a large error constant and a fast, well-damped transient. A lag alone would meet the first and slow the system down; a lead alone would meet the second and might even reduce the error constant. Cascading the two gives a network with two zeros and two poles that does both jobs, each in its own frequency band.
The condition on the corner frequencies is what keeps the two halves from interfering: the lag section does all its work one or two decades below crossover, where its \(-6^\circ\) is harmless, while the lead section does all of its work at crossover, where its phase bump is worth the most. Read the magnitude curve from left to right and the compensator lifts the low frequencies by \(\beta\), drops back down, then lifts again by \(\gamma\) around crossover.
The design is done sequentially, and the order matters. Start with the lead half, because it is the lead that determines where the crossover ends up: choose the dominant poles or the phase margin, design the lead section against the plant alone, and compute the error constant the lead-compensated loop achieves. Then take the ratio of the required error constant to the achieved one — that ratio is \(\beta\) — and design a lag dipole to supply it, placed far enough below the new crossover to leave the lead's work undisturbed. A single verification pass at the end confirms both specifications.
The lead half sets the crossover frequency and the phase margin; the lag half then multiplies the error constant by whatever factor is still missing. Neither half is asked to do the other's work, which is why a lead–lag design usually needs less phase from its lead section than a lead-only design of the same plant would.
A single passive network can realise the whole compensator if the two ratios are linked, \(\gamma=\beta\); this is the classical lag–lead network of the analogue era, and its restriction is the price of using one \(RC\) section instead of two. With an operational amplifier, or in software, the two ratios are independent and the more general form above applies.
Realisation and Comparison
Both networks are two-resistor, one-capacitor circuits, which is why they dominated control hardware for forty years before digital implementation made the question moot. Their transfer functions follow from a voltage divider with one frequency-dependent arm.
The lag network is the same exercise with the capacitor moved into the lower arm: \(E_o/E_i=(1+R_2Cs)/\bigl(1+(R_1+R_2)Cs\bigr)\), giving \(T=R_2C\) and \(\beta=(R_1+R_2)/R_2\).
Two practical points follow from these expressions. The passive lead has a DC gain of \(\alpha\), an attenuation that must be made up by an amplifier of gain \(1/\alpha\) somewhere in the loop — the \(K_c=1/\alpha\) of the design procedure is not a bookkeeping device but a real amplifier. The passive lag, by contrast, has unity DC gain and needs no such amplifier, which is one reason lag networks were popular in the era of expensive amplification. And because both are passive dividers, cascading two of them directly makes the second load the first and changes both transfer functions; a buffer between the sections, or a single active realisation, is required.
| Property | Lead | Lag | Lead–lag |
|---|---|---|---|
| Pole–zero order | zero then pole | pole then zero | both pairs |
| Phase at crossover | \(+\phi_m\) added | about \(-6^\circ\) | \(+\phi_m\) added |
| Bandwidth | increased | reduced | increased |
| Rise / settling time | faster | slower | faster |
| Error constants | slightly reduced | multiplied by \(\beta\) | multiplied by \(\beta\) |
| Noise sensitivity | worse by \(1/\alpha\) | improved | worse by \(1/\alpha\) |
| PID counterpart | filtered PD | approximate PI | approximate PID |
| Use it when | margin or speed is short | accuracy is short | both are short |
Worked Examples
Problem. For \(G_c(s)=\dfrac{1+0.2s}{1+0.05s}\), find \(\alpha\), the maximum phase lead, the frequency at which it occurs, and the gain there.
Solution. Compare with \(K_c\alpha(Ts+1)/(\alpha Ts+1)\): the zero corner gives \(T=0.2\), and the pole corner gives \(\alpha T=0.05\).
Checking directly: \(G_c(j10)=(1+j2)/(1+j0.5)\), whose magnitude is \(2.236/1.118=2\) and whose phase is \(63.43^\circ-26.57^\circ=36.87^\circ\). The network's DC gain is unity, so \(K_c=1/\alpha=4\).
Problem. For \(G(s)=\dfrac{K}{s(s+2)}\) with unity feedback, design a lead compensator giving \(K_v=20\ \mathrm{s^{-1}}\) and a phase margin of at least \(50^\circ\).
Solution. Step 1 fixes the gain: \(K_v=\lim_{s\to0}sG=K/2=20\), so \(K=40\) and \(G_1(s)=40/\bigl(s(s+2)\bigr)\).
Step 2 finds the uncompensated crossover from \(|G_1|=1\):
Step 3: the shortfall is \(32^\circ\); adding a \(6^\circ\) allowance for the crossover shift gives \(\phi_m=38^\circ\). Steps 4 and 5:
With \(K_c=1/\alpha=4.20\) the compensator is
Verification at the new crossover \(\omega=8.95\): the plant contributes \(-90^\circ-\arctan(4.47)=-167.4^\circ\) and the network contributes \(+38^\circ\), giving a phase margin of \(50.6^\circ\). Since \(G_c(0)=1\), \(K_v\) remains \(20\). Both specifications are met.
Problem. The same plant \(G(s)=K/\bigl(s(s+2)\bigr)\) must have dominant closed-loop poles with \(\zeta=0.5\) and \(\omega_n=4\). Design a lead compensator and find the resulting \(K_v\).
Solution. The desired poles are \(s_d=-2\pm j3.464\). Without compensation the characteristic equation is \(s^2+2s+K\), whose roots always have real part \(-1\); no gain can produce a real part of \(-2\), so compensation is unavoidable. Compute the angle at \(s_d\):
Choose the compensator zero to cancel the plant pole at \(-2\). Its angle is \(+90^\circ\), so the compensator pole at \(-p\) must satisfy
So \(G_c(s)=K_c(s+2)/(s+4)\), and the open loop collapses to \(K_cK/\bigl(s(s+4)\bigr)\) with characteristic equation \(s^2+4s+K_cK\). Matching \(s^2+2\zeta\omega_ns+\omega_n^2=s^2+4s+16\) gives \(K_cK=16\), and the closed-loop poles are exactly \(-2\pm j3.464\). The velocity constant is
The transient specification is met exactly, but \(K_v=4\) is modest. If the specification also demanded \(K_v=20\), a lag section would have to be added — which is Example 6.
Problem. For \(G(s)=\dfrac{K}{s(s+1)(0.5s+1)}\), design a lag compensator giving \(K_v=5\ \mathrm{s^{-1}}\) and a phase margin of at least \(40^\circ\).
Solution. \(K_v=K=5\), so \(G_1(s)=5/\bigl(s(s+1)(0.5s+1)\bigr)\). Check the uncompensated loop first: the characteristic equation \(s^3+3s^2+2s+10=0\) has a Routh \(s^1\) entry of \((6-10)/3=-1.33\), so the closed loop is unstable. Its gain crossover is at \(1.80\) rad/s where the phase is \(-193^\circ\) — a phase margin of \(-13^\circ\).
Step 2: find where the phase equals \(-180^\circ+40^\circ+12^\circ=-128^\circ\).
Step 3: the magnitude there must be removed entirely.
Verification: with the network's factor of \(0.1\) in place, the new crossover is where \(|G_1|=10\), namely \(\omega=0.446\) rad/s. There the plant contributes \(-126.6^\circ\) and the network \(\arctan(8.91)-\arctan(89.1)=-5.8^\circ\), giving a phase margin of \(47.7^\circ\). A loop that was unstable now has a comfortable margin and the full \(K_v=5\), at the cost of a bandwidth reduced from \(1.80\) to \(0.45\) rad/s.
Problem. For \(G(s)=\dfrac{K}{s(s+1)(s+2)}\) the gain is set so that the dominant poles have \(\zeta=0.5\). Find that gain and the resulting \(K_v\), then design a lag network raising \(K_v\) to at least \(5\) without disturbing the transient.
Solution. Write the dominant pair as \(-0.5\omega_n\pm j0.866\omega_n\) and the third root as \(-p\). Matching \(s^3+3s^2+2s+K\) coefficient by coefficient:
The required boost is \(5/0.519=9.6\), so take \(\beta=10\). Place the dipole far inside \(|s_d|=0.667\): zero at \(-0.02\), pole at \(-0.002\). Its angular contribution at \(s_d\) is
Well under the \(5^\circ\) usually tolerated, so the dominant poles are essentially unmoved and the gain \(K=1.037\) is retained. The compensated loop has \(K_v\approx 10\times0.519=5.19\ \mathrm{s^{-1}}\). The price is a closed-loop pole near \(-0.02\), which adds a small tail decaying with a time constant of about \(50\) s.
Problem. For \(G(s)=\dfrac{K}{s(s+1)(s+2)}\), design a compensator giving dominant poles with \(\zeta=0.5\) and \(\omega_n=1.5\), together with \(K_v=10\ \mathrm{s^{-1}}\).
Solution. The desired poles are \(s_d=-0.75\pm j1.299\). Example 5 showed that \(\zeta=0.5\) alone gives \(\omega_n=0.667\); asking for \(\omega_n=1.5\) as well as a twenty-fold increase in \(K_v\) puts both specifications out of reach of gain adjustment. Design the lead half first.
Cancel the plant pole at \(-1\) with the lead zero. That zero contributes \(79.1^\circ\), so the lead pole at \(-p_1\) must contribute
The lead-compensated loop is \(K/\bigl(s(s+2)(s+6)\bigr)\), whose characteristic equation \(s^3+8s^2+12s+K\) passes through \(s_d\) when \(K=14.6\). That gives \(K_v=14.6/12=1.22\), short of \(10\) by a factor
Place the lag zero at \(-0.05\) and its pole at \(-0.05/8.2=-0.0061\). The pair contributes \(118.3^\circ-119.8^\circ=-1.5^\circ\) at \(s_d\) and reduces the magnitude there by \(1.4\%\), so the gain is raised slightly to \(K_c=14.8\). The complete compensator is
The lead half moved the dominant poles from \(\omega_n=0.667\) to \(\omega_n=1.5\) at the same damping; the lag half then multiplied the velocity constant by \(8.2\). Neither section interferes with the other, because their corner frequencies are two decades apart.
Chapter Summary
Gain moves poles along a fixed locus; a compensator changes the locus itself.
\(\sin\phi_m=(1-\alpha)/(1+\alpha)\) at \(\omega_m=1/T\sqrt{\alpha}\), with gain \(1/\sqrt{\alpha}\) there.
Set \(K\), find the shortfall, add \(5^\circ\)–\(12^\circ\), get \(\alpha\), centre \(\omega_m\) on the new crossover.
\(\phi_c=-180^\circ-\angle G(s_d)\); positive means a lead, negative means a lag.
A dipole near the origin multiplies the error constant by \(\beta\) for about \(6^\circ\) of phase.
Lead first to set the crossover, lag second to supply the missing \(\beta\) — a filtered PID in disguise.
Practice Problems
Decide first which specification the uncompensated system fails — margin or accuracy — because that decides the network. Assume unity feedback throughout. Difficulty rises down the list.
- For \(G_c(s)=\dfrac{1+0.5s}{1+0.1s}\), find \(\alpha\), \(\phi_m\), \(\omega_m\) and the gain in decibels at \(\omega_m\).
- What value of \(\alpha\) is needed for a maximum phase lead of \(45^\circ\), and what high-frequency gain does that imply?
- A lag network has \(\beta=8\) and its zero at \(-0.4\). Where is its pole, and what is its phase contribution at \(\omega=4\) rad/s?
- For \(G(s)=\dfrac{K}{s(s+4)}\), find \(K\) for \(K_v=10\) and the resulting phase margin.
- Continuing Problem 4, design a lead compensator that raises the phase margin to \(45^\circ\) while keeping \(K_v=10\).
- For \(G(s)=\dfrac{K}{s(s+1)}\), design a lead compensator giving dominant poles with \(\zeta=0.707\) and \(\omega_n=2\), using cancellation of the plant pole.
- Explain why a lead compensator reduces the velocity error constant when \(K_c\) is held fixed, and how the design procedure compensates for that.
- A system meets its transient specification with \(K_v=2\) but requires \(K_v=25\). Design a suitable lag dipole and state the resulting settling-tail time constant.
- Show that placing the lag zero one decade below the new gain crossover limits the network's phase contribution there to about \(-6^\circ\), whatever the value of \(\beta\).
- For \(G(s)=\dfrac{K}{s(s+2)(s+8)}\), design a lead–lag compensator for \(\zeta=0.5\), \(\omega_n=3\) and \(K_v=15\).
- Two lead sections, each supplying \(30^\circ\), are cascaded. Compare the total high-frequency gain with that of a single section supplying \(60^\circ\), and explain the practical significance.
- A plant has a lightly damped pole pair at \(-0.1\pm j5\). Explain why cancelling it with compensator zeros is a poor design choice, and what the root locus of the imperfectly cancelled system looks like.