3.5.41 · D5Guidance, Navigation & Control (GNC)
Question bank — Bode plot — magnitude and phase vs frequency
True or false — justify
Every prompt below is a full statement. Decide true or false, then say why in one breath before revealing.
The Bode plot works for any system at all, linear or not.
False — only linear time-invariant (LTI) systems have the "sine in → same-frequency sine out" property; a nonlinear system can create new frequencies, so a single gain-and-phase pair per is meaningless.
A first-order pole drops the magnitude by exactly 3 dB at its corner frequency.
True — at we get , and dB, which is the defining "−3 dB corner."
At the corner frequency the phase of a single pole is , since that is where the pole "starts."
False — the phase is already at the corner (halfway through its swing); the corner is a magnitude landmark, not a phase-starting line.
A standard Bode slope of dB/decade is the same as dB/octave.
False — a decade is and an octave is ; dB/decade is only about dB/octave, so the two numbers differ by a factor of .
Multiplying two transfer functions means adding their Bode magnitude curves in dB.
True — , and taking turns that product into a sum; phase adds too. This is the entire reason Bode sketching is fast.
A pure integrator has a magnitude that depends on the DC gain of the plant.
False — has magnitude with no free constant; it always passes through dB at rad/s and slopes dB/dec regardless of anything else.
If a plant sits at exactly phase where its gain is ( dB), the closed loop is comfortably stable.
False — that is exactly the danger point defined above: zero phase margin means the loop is marginally unstable and any extra lag tips it into sustained oscillation.
For a minimum-phase system, you can shape the magnitude curve freely without affecting phase.
False — the Bode gain–phase relation ties them: a dB/dec slope forces about of phase, so steep roll-off always costs you phase lag.
A first-order zero is the mirror image of a pole: dB/dec and phase rising to .
True — a zero multiplies where a pole divides, so every sign in the asymptote flips; that is exactly how a lead network buys back phase.
The factor in is because we always use for decibels.
False — power decibels use ; the appears because gain is an amplitude ratio and power amplitude, so .
Spot the error
Each line contains one flawed claim. Name the flaw in a sentence.
"A double integrator has slope dB/decade because it is still just an integrator."
The error: two integrators each contribute , so the slope is dB/decade and the phase is , not .
"Since phase is a separate graph, a phase lag changes how much the system amplifies a signal at that frequency."
The error: at a fixed frequency the magnitude curve alone sets amplification — the phase shift only says how much the output lags, not how big it is, so phase does not change amplification at that frequency (though gain and phase together decide loop stability).
"The gain margin is read at the frequency where the phase is ."
The error: gain margin is read at the ==phase-crossover == where phase (the danger-point phase), not .
"To find phase margin, subtract the phase at gain crossover from ."
The error: it is evaluated at the gain-crossover — you add the (negative) phase, since phase there is typically like , giving a positive margin.
"A low-pass filter with a sharper cutoff is always the safer choice inside a control loop."
The error: a sharper low-pass rolls off faster, which by the gain–phase relation dumps more phase lag near crossover — often destabilizing the loop rather than helping it.
"On a Bode plot the horizontal axis is frequency in linear scale so you can read directly."
The error: the frequency axis is logarithmic (, with in rad/s); that is what makes the simple factors appear as straight-line asymptotes across many decades.
"A constant gain tilts the low-frequency asymptote by dB per decade."
The error: a constant gain is a flat horizontal shift of dB with zero slope and phase — it changes the level, not the tilt.
"The corner frequency of is , so it lives on the real axis."
The error: the corner is a frequency value rad/s evaluated at ; we walk up the imaginary axis, and the corner is where equals the pole's magnitude.
Why questions
Answer each in one or two sentences of mechanism. Recall from the definitions box that is the loop transfer function (plant × controller × sensor), and walks up the imaginary axis.
Why do we evaluate specifically at rather than any other point?
Because is the eigenfunction of an LTI system, so (the imaginary axis, ) extracts exactly the steady-state gain and phase applied to a pure sine of frequency (in rad/s).
Why does taking the logarithm of magnitude make hand-sketching possible?
Logarithms convert the products of factors in a Transfer function into sums, so each pole and zero contributes an additive straight-line piece you can stack by eye.
Why does a pole's phase transition spread across roughly a decade on each side of the corner?
Because changes gradually — it is near a decade below and near a decade above, so the swing is smooth, not a jump.
Why do GNC engineers care about delay and not just gain?
Because delay (phase lag) plus loop gain is what turns negative feedback into positive feedback: enough lag flips a correcting signal into a reinforcing one, causing oscillation.
Why does a lead compensator help a loop sitting near ?
A lead network's zero adds positive phase near crossover, restoring phase margin so the corrective action arrives before the loop response has lagged all the way to the danger point.
Why is the Nyquist plot sometimes preferred over Bode for judging stability?
Bode margins assume the danger point is a single crossover of , but for loops that cross multiple times the Nyquist encirclement count gives the rigorous stability verdict that separate margins can misread.
Why does raising loop gain shrink both margins?
Lifting the whole magnitude curve of pushes gain crossover to a higher frequency where phase is more negative (less phase margin) and moves the level closer to dB at the point (less gain margin).
Edge cases
The degenerate and limiting scenarios the topic hides.
What happens to the phase of as versus ?
Nothing changes — the integrator's phase is a constant at every frequency; only its magnitude sweeps from down through dB at rad/s.
What is the magnitude of a first-order pole exactly at DC ()?
It is exactly (i.e. dB above the DC-gain level), because the term equals when , so the pole has not "kicked in" yet.
What if two poles share the same corner frequency ?
Their contributions simply double: the slope becomes dB/dec above the corner and the phase heads to , since each pole adds its own dB/dec and .
What does a pure time-delay element do to the two Bode curves?
Its magnitude is (= dB, flat, unchanged at every frequency), yet its phase is radians — a lag that grows without bound as rises. So a pure delay costs you nothing in gain but eats phase margin faster and faster, a classic Loop shaping trap that a straight-line Bode sketch (with no corner to warn you) completely hides.
If a transfer function has a right-half-plane (non-minimum-phase) zero, does the "sharper slope means more lag" intuition still hold?
No — a non-minimum-phase zero adds magnitude slope like a normal zero but subtracts phase like a pole, so the gain–phase link that holds for minimum-phase systems breaks down.
What does a right-half-plane (unstable) pole do to the Bode phase compared with an ordinary pole?
An RHP pole keeps the same dB/dec magnitude roll-off but its phase moves the wrong way — it contributes instead of — so the magnitude looks normal while the phase drifts opposite to what a minimum-phase reader expects, a classic trap in non-minimum-phase analysis.
What does the Bode magnitude do at a resonant (lightly damped) complex-pole pair near its natural frequency?
Instead of a gentle dB corner it forms a sharp peak rising well above dB whose height grows as damping falls; the straight-line asymptote misses this, so Loop shaping must account for the peak explicitly.
What is the gain margin if the phase never reaches at any finite frequency?
The gain margin is infinite — with no phase-crossover there is no gain level that drives the loop unstable through that mechanism, so the margin is unbounded (though other criteria may still bind).
What happens to phase margin if the gain-crossover frequency is pushed to where phase is exactly ?
Phase margin becomes , the danger point — now coincides with and the loop sustains a steady oscillation, the boundary between a stable and an unstable closed loop.
Recall One-line self-check
Cover the answers above and re-derive three of them from scratch — pick one "true/false," one "spot the error," and one "edge case." If any reason came out as a bare yes/no, revisit the parent note's section that owns it.