Frequency response and Bode plots

Sinusoidal steady-state response, Bode magnitude and phase plots built from standard factors, corner frequencies and slopes, resonant peak, minimum-phase systems, and identifying a transfer function from an asymptotic plot.

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Why it matters

Vibration, sensor noise and periodic loads all reach a machine as sinusoids, so how a system treats each frequency decides whether it tracks commands, rejects disturbances or rings. Frequency response can also be measured on real hardware with a signal generator, even when no model exists. The Bode plot turns that response into straight-line sketches you can draw by hand, read margins from, and use to design compensators.

Key ideas

Frequency response. If a stable LTI system with transfer function G(s) is driven by u(t) = A·sin(ωt), then after the transients die out the output is y(t) = A·|G(jω)|·sin(ωt + φ), where φ = ∠G(jω). The output has the same frequency; only amplitude and phase change. It requires stability — an unstable system has no steady-state sinusoidal response, although G(jω) can still be computed and used for Nyquist and Bode stability tests.

Bode plot. Two graphs against log₁₀ω: magnitude in decibels, 20·log₁₀|G(jω)|, and phase in degrees. Because logarithms turn products into sums, the plot of a product of factors is the sum of the plots of each factor — the key to hand sketching.

Standard factors (time-constant form):

  • Constant K: flat line at 20·log₁₀K dB; phase 0° (K > 0) or −180° (K < 0).
  • Integrator 1/s (or 1/sᴺ): −20N dB/decade line through 0 dB at ω = 1 rad/s; phase −90°N. Differentiator s: +20 dB/decade, +90°.
  • First-order pole 1/(1 + sT): 0 dB below the corner ω_c = 1/T, −20 dB/decade above it; exact value at the corner is −3 dB. Phase from 0° to −90°, −45° at the corner (asymptote: 0° below 0.1ω_c, −90° above 10ω_c).
  • First-order zero (1 + sT): the mirror image, +20 dB/decade and 0° to +90°.
  • Quadratic pole 1/(1 + 2ζ(s/ωn) + (s/ωn)²): corner at ωn, −40 dB/decade above; phase 0° to −180°, −90° at ωn. Near ωn the true curve peaks above the asymptote when ζ < 0.707; the resonant peak is Mr = 1/(2ζ√(1 − ζ²)) at ωr = ωn√(1 − 2ζ²).

Sketching procedure. (1) Put G in time-constant form and find K. (2) List the corner frequencies in increasing order. (3) Start the low-frequency line: for type N it has slope −20N dB/decade and passes through 20·log₁₀K at ω = 1 rad/s (equivalently, for type 1 it crosses 0 dB at ω = K). (4) At each corner, change the slope by ±20 dB/decade per first-order factor (±40 for a quadratic). (5) Add the phase of each factor. (6) Apply corrections at corners if needed.

Minimum phase. A system with no poles or zeros in the right half-plane (and no delay) is minimum-phase: its phase is fixed by its magnitude curve, so the transfer function can be identified from the magnitude plot alone. Right-half-plane zeros and transport delay e^(−sT_d) (magnitude 1, phase −ωT_d rad) add extra lag.

Frequency-domain specifications. Gain crossover frequency ω_gc (|G| = 0 dB), phase crossover ω_pc (∠G = −180°), gain and phase margins (next topic), bandwidth (closed-loop magnitude 3 dB below its low-frequency value) and resonant peak Mr. A larger bandwidth means a faster response; a larger Mr means more overshoot.

Formulas

|G(jω)|_dB = 20·log₁₀|G(jω)| φ(ω) = ∠G(jω) (degrees)

|1/(1 + jωT)| = 1/√(1 + ω²T²), ∠ = −tan⁻¹(ωT) — first-order pole; T in s, ω in rad/s.

ω_c = 1/T — corner frequency (rad/s).

Slope change at each corner: −20 dB/decade per simple pole, +20 dB/decade per simple zero, −40 dB/decade per quadratic pole pair.

Mr = 1/(2ζ√(1 − ζ²)), ωr = ωn·√(1 − 2ζ²) — resonant peak and frequency; valid for ζ < 0.707.

∠e^(−jωT_d) = −ω·T_d (rad) — pure delay; magnitude 0 dB.

1 decade = factor of 10 in ω; 1 octave = factor of 2; 20 dB/decade ≈ 6 dB/octave.

Worked examples

Example 1 (standard). For G(s) = 100/(s(s + 10)), sketch the asymptotic Bode plot and find the exact gain crossover frequency and the phase there.

  1. Time-constant form: G(s) = 10/(s(1 + 0.1s)). K = 10, one integrator, one corner at 1/0.1 = 10 rad/s.
  2. Low-frequency line: −20 dB/decade, value 20·log₁₀10 = 20 dB at ω = 1 rad/s.
  3. At ω = 10 rad/s the line reaches 20 − 20 = 0 dB; the slope then becomes −40 dB/decade.
  4. Exact crossover: |G| = 10/(ω√(1 + 0.01ω²)) = 1 → 0.01ω⁴ + ω² − 100 = 0 → ω² = 61.8 → ω_gc = 7.86 rad/s (the asymptote's 10 rad/s overestimates it).
  5. Phase: −90° − tan⁻¹(0.786) = −90° − 38.2° = −128.2° (so the phase margin will be 51.8°).

Example 2 (GATE level, identification). An asymptotic magnitude plot of a minimum-phase system starts with a −20 dB/decade slope whose extension crosses 0 dB at 5 rad/s. The slope becomes −40 dB/decade at 2 rad/s and −60 dB/decade at 20 rad/s. Find G(s), the asymptotic magnitude at 10 rad/s, and the exact magnitude and phase there.

  1. Initial −20 dB/decade means one integrator; crossing 0 dB at ω = 5 means K = 5.
  2. Each −20 dB/decade change is a simple pole: G(s) = 5/(s(1 + s/2)(1 + s/20)).
  3. Asymptote at ω = 2: 20·log₁₀(5/2) = 7.96 dB. From 2 to 10 rad/s (log₁₀5 = 0.699 decade) at −40 dB/decade: 7.96 − 27.96 = −20.0 dB.
  4. Exact: |G(j10)| = 5/(10 × √(1 + 25) × √(1 + 0.25)) = 5/(10 × 5.099 × 1.118) = 0.0877 → −21.1 dB.
  5. Phase: −90° − tan⁻¹(5) − tan⁻¹(0.5) = −90° − 78.7° − 26.6° = −195.3°. The phase has passed −180° before 10 rad/s.

Common mistakes

  • Not converting to time-constant form, so K is read wrongly (100/(s(s + 10)) has K = 10, not 100).
  • Drawing the integrator line through 0 dB at ω = 1 instead of through 20·log₁₀K.
  • Forgetting the −3 dB correction at a simple corner, or ignoring the resonant peak for a lightly damped quadratic.
  • Using log₁₀ for the frequency axis but ln for dB.
  • Assuming the phase can be read from the magnitude plot for a system with a right-half-plane zero or a time delay.
  • Mixing rad/s and Hz (ω = 2πf).

For GATE ME

Expect: identifying a transfer function from an asymptotic magnitude plot; magnitude or phase at a given frequency; slope after several corners; corner frequency and −3 dB point of a first-order system; resonant peak and bandwidth of a second-order system; and the effect of a delay on phase. Practise working with decades quickly (log₁₀2 = 0.301, log₁₀5 = 0.699).

Quick check

  1. A gain of 10 is how many dB?
  2. What is the slope above the corner of 1/(1 + 0.5s), and where is the corner?
  3. What is the exact magnitude and phase of 1/(1 + sT) at ω = 1/T?
  4. Which ζ is the limit below which a quadratic pole shows a resonant peak?
  5. For G(s) = K/s, where does the magnitude line cross 0 dB?

Answers: 1. 20 dB. 2. −20 dB/decade, above 2 rad/s. 3. −3 dB and −45°. 4. ζ = 0.707. 5. At ω = K rad/s.

Interview questions

All Control Systems interview questions

Try answering each one aloud before you open it.

  1. 1.What is frequency response in control systems?Concept

    Frequency response is the steady-state response of a system to a sinusoidal input signal as a function of frequency. It describes how the amplitude and phase of the output signal change with respect to the input signal's frequency. This analysis helps in understanding the behavior of the system over a range of frequencies.

  2. 2.Explain what a Bode plot is and its components.Concept

    A Bode plot is a graphical representation of a system's frequency response. It consists of two plots: the magnitude plot, which shows the gain of the system in decibels (dB) versus frequency on a logarithmic scale, and the phase plot, which shows the phase shift in degrees versus frequency. These plots help in analyzing the stability and performance of control systems.

  3. 3.Why are Bode plots used in control systems analysis?Application

    Bode plots are used because they provide a clear and concise way to analyze the frequency response of a system. They help engineers assess system stability, gain margin, and phase margin, which are critical for designing stable and robust control systems. Additionally, Bode plots make it easier to visualize the effects of adding controllers or compensators.

  4. 4.What is gain margin and phase margin in the context of Bode plots?Concept

    Gain margin is the amount by which the gain of a system can be increased before it becomes unstable, measured at the phase crossover frequency where the phase angle is -180 degrees. Phase margin is the additional phase lag required to bring the system to the verge of instability, measured at the gain crossover frequency where the gain is 1 (0 dB). Both margins are indicators of system stability.

  5. 5.How does the Nyquist criterion relate to Bode plots?Application

    The Nyquist criterion is a graphical method for determining the stability of a control system by analyzing its frequency response. While the Nyquist plot provides a complete picture of stability, Bode plots offer a more intuitive and easier-to-interpret view. The gain and phase margins derived from Bode plots can be used to infer stability, similar to the Nyquist criterion.

  6. 6.What happens to the Bode plot if a pole is added to the system?Application

    Adding a pole to a system typically decreases the slope of the magnitude plot by -20 dB/decade for each pole and introduces an additional phase lag of -90 degrees. This change affects the system's stability and performance, potentially reducing the gain and phase margins.

  7. 7.What is the effect of adding a zero to a system on its Bode plot?Application

    A left-half-plane zero (1 + sT) adds +20 dB/decade to the magnitude slope above its corner 1/T and adds phase lead, rising from 0° to +90° (+45° at the corner). Placed near the gain crossover, the lead raises the phase margin, which is the idea behind PD and lead compensation, although the extra high-frequency gain amplifies noise. A right-half-plane zero has the same magnitude effect but adds phase lag, which reduces the margins.

  8. 8.Calculate the gain margin and phase margin for a unity-feedback loop with open-loop transfer function G(s) = 10/(s² + 2s + 10).Numerical

    Gain crossover: |G(jω)| = 10/√((10 − ω²)² + 4ω²) = 1 gives ω⁴ − 16ω² = 0, so ω_gc = 4 rad/s. The phase there is −tan⁻¹ of (2ω)/(10 − ω²) with 10 − ω² = −6 < 0, so ∠G = −(180° − 53.1°) = −126.9° and the phase margin is 180° − 126.9° = 53.1°. A second-order system's phase only approaches −180° as ω → ∞, so there is no finite phase crossover and the gain margin is infinite.

  9. 9.Explain how you would sketch a Bode plot for a first-order system with a transfer function G(s) = 1 / (τs + 1).Numerical
    1. Identify the corner frequency, ω_c = 1/τ.
    2. For frequencies much lower than ω_c, the magnitude plot is approximately 0 dB, and the phase is near 0°.
    3. At ω_c, the magnitude decreases by 3 dB, and the phase is -45°.
    4. For frequencies much higher than ω_c, the magnitude decreases at a rate of -20 dB/decade, and the phase approaches -90°.
    5. Plot these characteristics on logarithmic scales for frequency.
  10. 10.What is the significance of the corner frequency in a Bode plot?Concept

    The corner frequency, also known as the cutoff frequency, is the frequency at which the magnitude of the system's response begins to change significantly. For a first-order system, it is the frequency where the magnitude drops by 3 dB from its low-frequency value. It marks the transition between the flat and sloped regions of the magnitude plot and is crucial for understanding the system's bandwidth and response characteristics.

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