Bode Plot - Control Systems - Electrical Engineering (EE)

Introduction

The Bode plot is a frequency-response representation of a linear time-invariant system that consists of two separate plots sharing a common logarithmic frequency axis: a magnitude plot (usually in decibels) and a phase-angle plot (in degrees). It is commonly drawn for the open-loop transfer function because the open-loop Bode plot conveniently indicates stability margins and helps in controller design.

• The magnitude plot is usually expressed as 20 log10 |G(jω)|, where the logarithm is base 10 and the unit is decibel (dB).
• The horizontal axis is logarithmic in frequency (ω), while the vertical axes for magnitude and phase are linear.
• Because logarithms convert products into sums, the magnitude in decibels allows contribution of individual factors in a transfer function to be added algebraically. For example, if G(jω) = K / [jω (1 + jωT)], then:

|G(jω)| = K / [ω √(1 + ω²T²)]

∠G(jω) = -90° - tan⁻¹(ωT)

20 log10 |G(jω)| = 20 log10 K - 20 log10 ω - 20 log10 √(1 + ω²T²)

Thus each multiplicative factor in G(s) contributes an additive term in the total magnitude measured in dB. This property is the basis of the straight-line (asymptotic) sketching method for Bode plots.

Constant gain K

• A constant gain factor is G(s) = K → G(jω) = K.
• The phase of a positive constant gain is 0°; for a negative gain the phase is 180° (or -180° depending on the chosen branch).
• The magnitude in decibels is 20 log10 K.
• Behaviour by ranges:
  • K > 1: 20 log10 K > 0 dB
  • K = 1: 20 log10 K = 0 dB
  • 0 < K < 1: 20 log10 K < 0 dB
• On a Bode magnitude plot a constant gain appears as a horizontal straight line.

Integral (pole at origin) factor

An integral factor (pole at the origin) of order one is given by G(s) = K / s, so G(jω) = K / (jω).

• Magnitude: |G(jω)| = K / ω.
• Phase: ∠G(jω) = -90° (for each pole at origin, -90° per order).
• Magnitude in dB: 20 log10 |G(jω)| = 20 log10 K - 20 log10 ω.
• The asymptotic slope is -20 dB/decade for a single pole at origin. For G(s) = K / sⁿ the slope is -20n dB/decade and the phase is -90n°.

Derivative (zero at origin) factor

A derivative factor (zero at the origin) of order one is given by G(s) = K s, so G(jω) = K jω.

• Magnitude: |G(jω)| = K ω.
• Phase: ∠G(jω) = +90° (for each zero at origin, +90° per order).
• Magnitude in dB: 20 log10 |G(jω)| = 20 log10 K + 20 log10 ω.
• The asymptotic slope is +20 dB/decade for a single zero at origin. For G(s) = K sⁿ the slope is +20n dB/decade and the phase is +90n°.

First-order factor in numerator (zero)

Consider a first-order zero: G(s) = 1 + sT → G(jω) = 1 + jωT.

• Magnitude: |G(jω)| = √(1 + ω²T²).
• Phase: ∠G(jω) = tan⁻¹(ωT).
• Magnitude in dB: 20 log10 √(1 + ω²T²).
• Asymptotic approximation:
  • For ω ≪ 1/T, |G(jω)| ≈ 1 → 0 dB (flat line).
  • For ω ≫ 1/T, |G(jω)| ≈ ωT → slope +20 dB/decade.
• The frequency ω = 1/T is the corner (break) frequency; the exact magnitude differs from the asymptotes by about +3 dB at the corner for a first-order zero.

First-order factor in denominator (pole)

Consider a first-order pole: G(s) = 1 / (1 + sT) → G(jω) = 1 / (1 + jωT).

• Magnitude: |G(jω)| = 1 / √(1 + ω²T²).
• Phase: ∠G(jω) = -tan⁻¹(ωT).
• Magnitude in dB: 20 log10 [1 / √(1 + ω²T²)] = -20 log10 √(1 + ω²T²).
• Asymptotic approximation:
  • For ω ≪ 1/T, |G(jω)| ≈ 1 → 0 dB (flat line).
  • For ω ≫ 1/T, |G(jω)| ≈ 1 / (ωT) → slope -20 dB/decade.
• Corner at ω = 1/T gives about -3 dB deviation of the exact curve from the asymptotes.

Note: Zeros produce positive slopes in the magnitude plot; poles produce negative slopes.

Second-order (quadratic) factor in denominator

A standard second-order denominator (normalised) is

G(s) = ωn² / (s² + 2ζωn s + ωn²)

Writing s = jω and normalising by ωn gives

G(jω) = 1 / [1 - (ω/ωn)² + j 2ζ (ω/ωn)]

• Magnitude:

  |G(jω)| = 1 / √{[1 - (ω/ωn)²]² + [2ζ (ω/ωn)]²}

• Phase:

  ∠G(jω) = -tan⁻¹{ [2ζ (ω/ωn)] / [1 - (ω/ωn)²] }

• Asymptotic behaviour:
  • For ω ≪ ωn, |G(jω)| ≈ 1 → 0 dB.
  • For ω ≫ ωn, |G(jω)| ≈ (ωn/ω)² → slope -40 dB/decade.
• At ω = ωn the phase is -90° (phase passes from 0° toward -180° as ω increases). At ω → ∞ the phase tends to -180°.
• If the damping ratio ζ is small, the magnitude curve shows a resonant peak near ω ≈ ωn; the peak height increases as ζ decreases (peak height is inversely related to ζ).

Procedure to draw the Bode magnitude and phase curves (asymptotic sketch)

• Express the transfer function in time-constant (normalised) form so corner frequencies are explicit.
• List all corner (break) frequencies in increasing order. Prepare a table of terms, corner frequencies and change in slope at each corner.
• Choose a frequency lower than the lowest corner frequency and evaluate the asymptotic contribution of each term at that frequency.
• Use the rule that between two corner frequencies the slope is constant; to find magnitude at a higher corner ωy from a lower corner ωx use:
  
• The incremental form (asymptotic) is:
  
• Mark the calculated points on semilog graph paper (logarithmic x-axis for ω) and join them with straight lines (the asymptotes). For a closer approximation, adjust each corner by ±3 dB for first-order elements (±3 dB at ω = corner) and account exactly for resonant peaks of second-order terms when required.

Worked sketch example (combination of factors)

Consider the transfer function

G(s) = K (1 + sT1)² / [s² (1 + sT2) (1 + sT3)]

Assume T2 < T3 < T1 so the corner frequencies are ωc2 = 1/T2, ωc3 = 1/T3, ωc1 = 1/T1 in increasing order.

• Decompose into four elementary terms: K / s², (1 + sT1)², 1/(1 + sT2), 1/(1 + sT3).
• Slope contributions:
  • K / s² → slope -40 dB/decade (two poles at origin)
  • (1 + sT1)² → slope +40 dB/decade for ω > 1/T1 (double zero)
  • 1/(1 + sT2) → slope -20 dB/decade for ω > 1/T2
  • 1/(1 + sT3) → slope -20 dB/decade for ω > 1/T3
• The complete magnitude plot is the algebraic sum (in dB) of these contributions; between breakpoints add the slopes numerically. For example, in the region between ωc1 and ωc3 the net slope/value equals (-40 dB/dec + 40 dB/dec + 0 + 0) giving a flat net contribution in that region.

Phase margin and gain margin

• Gain crossover frequency (ωgc) is the frequency where the magnitude |L(jω)| = 1 (0 dB) for the open-loop transfer L(s).
• Phase margin (PM)is defined at ωgc by

  PM = 180° + ∠L(jωgc)

  Here ∠L(jωgc) is typically a negative number; PM is the additional phase required to bring the loop to -180° at the gain crossover. A larger PM indicates more phase reserve and generally better relative stability.

• Phase crossover frequency (ωpc) is the frequency where ∠L(jω) = -180°.
• Gain margin (GM)is determined at ωpc as

  GM = 1 / |L(jωpc)|

  In decibels: GM(dB) = 20 log10 GM. If |L(jωpc)| < 1 then GM > 1 (positive dB), indicating margin in gain before instability.

Example: determine K when gain crossover is 5 rad/s

Problem: For the open-loop transfer function G(s) = K s² / [(1 + 0.2 s)(1 + 0.02 s)], determine K such that the gain crossover frequency is ωgc = 5 rad/s. Also indicate how magnitude and phase are obtained.

Solution (step-wise calculation):

Write G(jω) and evaluate the magnitude without K at ω = 5:
G(jω) = K (jω)² / [(1 + j0.2ω)(1 + j0.02ω)]
Magnitude without K at ω = 5:
Compute |1 + j0.2·5| = |1 + j1| = √(1² + 1²) = √2 ≈ 1.4142
Compute |1 + j0.02·5| = |1 + j0.1| = √(1² + 0.1²) = √1.01 ≈ 1.00499
Compute ω² = 5² = 25
|G(j5)| (without K) = 25 / (1.4142 × 1.00499) ≈ 25 / 1.4214 ≈ 17.585
Magnitude in dB (without K): 20 log10(17.585) ≈ 24.90 dB
For the gain crossover at ω = 5 we require total magnitude to be 0 dB, therefore 20 log10 K + 24.90 = 0
Hence 20 log10 K = -24.90
Therefore log10 K = -24.90 / 20 = -1.245
K = 10^(-1.245) ≈ 0.0569
Thus the numerical value of K that makes the gain crossover frequency 5 rad/s is approximately 0.0569.
Phase expression for the open-loop transfer (general):
∠G(jω) = ∠(jω)² - tan⁻¹(0.2ω) - tan⁻¹(0.02ω) = 180° - tan⁻¹(0.2ω) - tan⁻¹(0.02ω)

Use this expression to sketch the phase plot versus ω (evaluating at selected frequencies). The magnitude and phase plots are drawn on the same logarithmic frequency scale for convenience.

Exact versus asymptotic plots

• Asymptotic (straight-line) Bode sketches are based on adding the dB contributions of each pole/zero and approximating corner transitions by straight lines. They are simple and useful for hand analysis and design.
• The exact plot requires evaluating the exact magnitude and phase expressions at frequencies of interest. Around corner frequencies the exact magnitude deviates from the straight-line asymptotes (first-order corners give about ±3 dB difference at the corner; second-order shapes may show resonant peaks depending on damping).

Advantages of Bode plots

• They show behaviour over many decades of frequency (low and high frequency) on a single graph.
• They directly provide gain margin and phase margin (measures of relative stability).
• They allow simple sketching and hand-analysis by superposition of elementary factors (poles and zeros) since magnitudes in dB add.
• They separate magnitude and phase information on two plots with a common frequency axis, reducing confusion.
• Asymptotic straight-line sketches are quick and give useful design insight with small information loss for many practical problems.

Limitations of Bode plots

• Asymptotic sketches are approximations and can miss fine features (exact evaluation is needed when accuracy matters, e.g., near resonances).
• Bode plots give frequency-domain information but do not directly show time-domain response; further analysis is required for time-response specifications.
• Some systems with closely spaced poles/zeros or nonminimum-phase behaviour need careful exact evaluation rather than straight-line addition.