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What is Bode plot?

A representation of the frequency response of a system as two graphs with the logarithm of frequency on the horizontal axis and the gain (dB) and phase (degrees) on the vertical axes. A product of transfer functions becomes a sum of gains (in dB) and phases, so the plot can be drawn by adding up the lines of each element. Each additional pole changes the gain slope by −20 dB/dec and adds up to 90° of phase lag. The gain margin and phase margin are read from the Bode plot of the loop transfer function to examine stability.

Japanese name: ボード線図

Bode plot appears in 4 practice questions across 1 topic, including Control Engineering Basics.

Field: Control Engineering Basics

Drawing the straight-line approximation

A first-order element 1/(1 + s/ωc) is approximated by a horizontal 0 dB line below ωc and a −20 dB/dec line above it. The actual gain at the corner is about −3 dB, and the phase is −45° at the corner and changes from about 0° to −90° between one tenth and ten times the corner frequency. A zero conversely adds a +20 dB/dec slope and phase lead, and an integrator 1/s adds a −20 dB/dec slope and −90° across the whole plot.

Reading the stability margins

The phase margin is how many degrees the phase still is from −180° at the frequency where the gain of the loop transfer function crosses 0 dB (the gain crossover frequency). The gain margin is how many decibels the gain is below 0 dB at the frequency where the phase reaches −180° (the phase crossover frequency). The gain crossover frequency also indicates the speed of the loop: the higher it is, the faster the closed-loop response.

How it differs from a Nyquist plot

Drawing the same frequency response as a single curve in the complex plane gives a Nyquist plot (polar plot). The Bode plot has an easy-to-read frequency scale and can be drawn by adding up elements, which suits design. The Nyquist plot shows stability directly from its position relative to the point −1 and can also be used when the loop transfer function itself is unstable.

Try it: questions about Bode plot

Which of the following correctly describes the Bode plot of the first-order system G(s) = 1/(1 + sT)?

  1. AAt the corner angular frequency ω = 1/T the gain is about −3 dB and the phase −45°, and well above it the gain falls at −20 dB/dec
  2. BAt the corner angular frequency ω = 1/T the gain is −6 dB and the phase −90°, and well above it the gain falls with a slope of −40 dB/dec
  3. CThe gain is 0 dB at every frequency, and only the phase lags in proportion to frequency
Show answer and explanation

Answer: A. At the corner angular frequency ω = 1/T the gain is about −3 dB and the phase −45°, and well above it the gain falls at −20 dB/dec

The magnitude of G(jω) = 1/(1 + jωT) is 1/√(1 + (ωT)²), which is 1/√2 (about −3 dB) at ω = 1/T, where the phase is −45°. Well above 1/T the magnitude approaches 1/(ωT), and the gain falls by 20 dB for every tenfold increase in frequency (−20 dB/dec). The phase changes from 0° at low frequencies to −90° at high frequencies. In the straight-line approximation of the Bode plot, the gain is drawn as a horizontal 0 dB line below the corner and a −20 dB/dec line above it.

Which of the following correctly describes the Bode plot of an integrator G(s) = 1/s?

  1. AThe gain is 0 dB at every frequency, and the phase is −90° at every frequency
  2. BThe gain is a +20 dB/dec line passing through 0 dB at ω = 1 rad/s, and the phase is +90° at every frequency
  3. CThe gain is a −20 dB/dec line passing through 0 dB at ω = 1 rad/s, and the phase is −90° at every frequency
Show answer and explanation

Answer: C. The gain is a −20 dB/dec line passing through 0 dB at ω = 1 rad/s, and the phase is −90° at every frequency

The magnitude of G(jω) = 1/(jω) is 1/ω, so it is 0 dB at ω = 1 rad/s and falls by 20 dB for every tenfold increase in frequency. Since 1/j = −j, the phase is −90° at every frequency. Because the gain grows without limit at low frequencies, a system containing an integrator can bring the steady-state error to a step setpoint to zero. A +20 dB/dec line with a phase of +90° belongs to a differentiator s.

FAQ

What does −20 dB/dec mean on a Bode plot?

It is the slope at which the gain falls by 20 dB, that is, the amplitude falls to one tenth, for every tenfold (one decade) increase in frequency. Each additional pole makes the slope steeper by −20 dB/dec. It is the same slope as −6 dB/oct, a drop of about 6 dB for every doubling of frequency.

How is a Bode plot measured?

A sine wave is applied while its frequency is swept, and the amplitude and phase of the output are measured. For power-supply control loops, a common method is to insert a small resistor into the loop, inject a signal across it and measure the loop transfer function with a frequency response analyzer.

Bode plot in the explanations

Questions that mention Bode plot (4)

  1. Which of the following correctly describes the Bode plot of the first-order system G(s) = 1/(1 + sT)?
  2. Which of the following correctly describes the Bode plot of an integrator G(s) = 1/s?
  3. For G(s) = 10/((1 + s/100)(1 + s/1000)), which of the following is the gain at the angular frequency ω = 10⁴ rad/s, found with the straight-line approximation of the Bode plot?
  4. Which of the following correctly describes the gain margin and phase margin read from the Bode plot of the loop transfer function?

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Bode plot calculatorDraw magnitude and phase plots from a transfer function and find phase and gain margins

Related terms

First-order system Steady-state error Gain margin Phase margin Transfer function Block diagram Feedback control Feedforward control Integral windup Laplace transform

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