Rule of thumb · Control SystemsNº 75 / 167

Each pole costs 20 dB a decade and 90° of phase

Every additional real pole in a transfer function adds another −20 dB/decade of rolloff and, eventually, another −90° of phase lag.

Why it works

It’s why stacking poles (extra filtering stages, extra lag elements) steadily erodes phase margin — each one you add brings the phase closer to the −180° instability threshold.

When it fails

A zero does the opposite and can cancel the pole's contribution entirely, which is the whole point of lead compensation. And a right-half-plane zero adds the gain of a zero with the phase lag of a pole — the counting rule breaks exactly where it matters most for stability.

How wrong is it?

At the rule says and the exact answer is 0.7071× — 41% high. At the point most people use it the shortcut is already outside its own 10% tolerance. It only holds between 0.1× and 0.45×, which is not where it gets used.

+10%0−10%0.15.0510Frequency, relative to the corner (×)

The straight-line Bode plot is two asymptotes, and the truth is the curve between them. They meet at the corner, which is precisely where the approximation is at its worst: 41% high, the famous 3 dB. A decade either side it is within half a per cent, so the rule is excellent everywhere except the one frequency you care about.

The rule against the exact answer, computed across the range. Inside the shaded band the shortcut is close enough to use; outside it, reach for the calculator.

Do it exactly

Estimate with the rule, then check it against the calculator that models it properly.

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What does each pole do to a Bode plot?

Every additional real pole in a transfer function adds another −20 dB/decade of rolloff and, eventually, another −90° of phase lag. It’s why stacking poles (extra filtering stages, extra lag elements) steadily erodes phase margin — each one you add brings the phase closer to the −180° instability threshold.

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