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Bode Plot of Second Order System

Resonant peak, damping effect on magnitude.

Mohith N
Updated: 19 March 2026
5 min read

The Bode plot of a second-order system introduces phenomena that are absent in first-order analysis: a resonant peak in the magnitude plot, a steeper -40 dB/decade rolloff above the natural frequency, and the effect of the damping ratio on both the peak magnitude and the sharpness of the phase transition. Understanding these characteristics is essential for designing control systems that avoid excessive resonance or undesirable transient behavior.

Second Order System Bode Plot: Effect of Damping RatioMagnitude Plotlog ωdB+200-40ωnζ=0.1 (large peak)ζ=0.5ζ=0.7070dB flat regionPhase Plotlog ω°0-90-180ωnζ=0.1 (steep)ζ=0.707-90° at ωnStandard Second Order Transfer FunctionG(s) = ωn² / (s² + 2ζωns + ωn²)ωn = natural frequency    ζ = damping ratioResonant PeakMr = 1/(2ζsqrt(1-ζ²)) for ζ < 0.707Peak occurs at ωr = ωn*sqrt(1-2ζ²)No Resonant PeakWhen ζ ≥ 0.707 (= 1/sqrt(2))Magnitude monotonically decreases
Figure 1: Second-order system Bode plot — resonant peak appears for damping ratio below 0.707; rolloff is -40dB/decade above natural frequency

Core Concept Explanation

The standard second-order transfer function is written as G(s) = omega_n squared divided by (s squared plus 2 zeta omega_n s plus omega_n squared), where omega_n is the undamped natural frequency in radians per second and zeta is the damping ratio, a dimensionless parameter that controls how quickly oscillations decay. This form ensures the DC gain is unity (0 dB at omega approaching zero), which is a standard convention for analyzing the shape of the frequency response.

At frequencies well below omega_n, both the magnitude and phase are approximately unchanged from their DC values. As frequency approaches omega_n, the denominator of G(j omega) can become very small if zeta is small, causing the magnitude to rise sharply above 0 dB. This rise is the resonant peak, and it occurs only when zeta is less than 1/sqrt(2) = 0.707. For larger damping ratios, the magnitude decreases monotonically and no peak is observed.

At frequencies well above omega_n, the s squared term dominates the denominator. The magnitude of s squared grows as omega squared, so the overall magnitude decreases as 1/omega squared. In decibels, this corresponds to a slope of -40 dB per decade, which is twice the -20 dB/decade rolloff of a first-order pole. This is because a second-order system has two poles, each contributing -20 dB/decade at high frequencies.

The phase of the second-order system transitions from 0 degrees at low frequency to -180 degrees at high frequency. At exactly omega = omega_n, the phase is always -90 degrees regardless of the value of zeta. The rate of the phase transition, however, depends strongly on zeta: small damping produces a very steep phase transition near omega_n, while large damping produces a gradual, spread-out transition.

Mathematical Expression

For G(s) = omega_n squared / (s squared + 2 zeta omega_n s + omega_n squared), evaluating at s = j omega gives:

  • Magnitude: |G(j omega)| = omega_n squared / sqrt[ (omega_n squared minus omega squared) squared plus (2 zeta omega_n omega) squared ]
  • Phase: angle G = - arctan[ (2 zeta omega_n omega) / (omega_n squared minus omega squared) ]
  • Resonant peak magnitude: Mr = 1 / (2 zeta sqrt(1 minus zeta squared)) — valid only for zeta less than 0.707
  • Resonant frequency: omega_r = omega_n times sqrt(1 minus 2 zeta squared) — valid only for zeta less than 0.707

The resonant peak Mr is obtained by differentiating the magnitude expression with respect to omega and setting the derivative to zero. The result shows that as zeta approaches zero, the resonant peak tends to infinity, and as zeta approaches 0.707, the resonant frequency approaches zero and the peak disappears. For zeta equal to 0.707, the magnitude rolls off most smoothly, which is the maximally flat (Butterworth) condition for a second-order low-pass filter.

Practical Understanding

The resonant peak in the frequency response directly corresponds to oscillatory behavior in the time domain. A large resonant peak (small zeta) means the system will exhibit significant overshoot and ringing in its step response. This is why feedback control system design typically aims for a damping ratio between 0.4 and 0.8 in the closed-loop poles, balancing speed of response against undesirable oscillation.

In GATE problems, second-order Bode plots are tested through identification of omega_n and zeta from a given plot, calculation of the resonant peak in dB, and determination of phase margin from the open-loop Bode plot of a second-order system. Recognizing the -40 dB/decade slope above omega_n and the -90 degree phase at omega_n are essential shortcuts.

Example
Given:
G(s) = 100 / (s^2 + 2s + 100)
omega_n^2 = 100 → omega_n = 10 rad/s
2*zeta*omega_n = 2 → zeta = 2/(2*10) = 0.1

Why this formula applies:
Since zeta = 0.1 < 0.707, a resonant peak exists.
Use: Mr = 1/(2*zeta*sqrt(1-zeta^2))

Formula:
Mr = 1 / (2 * zeta * sqrt(1 - zeta^2))
omega_r = omega_n * sqrt(1 - 2*zeta^2)

Substitution:
Mr = 1 / (2 * 0.1 * sqrt(1 - 0.01))
   = 1 / (0.2 * sqrt(0.99))
   = 1 / (0.2 * 0.995)
   = 1 / 0.199 = 5.025

omega_r = 10 * sqrt(1 - 2*(0.1)^2)
        = 10 * sqrt(1 - 0.02)
        = 10 * sqrt(0.98)
        = 10 * 0.9899 = 9.899 rad/s

Final Answer:
Resonant peak Mr = 5.025 (= 20*log(5.025) = 14.02 dB)
Resonant frequency = 9.899 rad/s (just below omega_n = 10 rad/s)
Exam Tip: In GATE, if a second-order system has zeta = 0.5, the resonant peak Mr = 1/(2*0.5*sqrt(1-0.25)) = 1/(0.866) = 1.155, which is only about 1.25 dB above 0 dB — quite small. Resonant peaks become significant only for zeta below 0.3. Also, always check if zeta < 0.707 before applying the resonant peak formula; applying it for zeta >= 0.707 gives an incorrect imaginary result.

Quick Revision

  • Standard form: G(s) = omega_n squared / (s squared + 2 zeta omega_n s + omega_n squared); DC gain = 0 dB.
  • Resonant peak exists only when zeta < 0.707; magnitude is monotonically decreasing for zeta >= 0.707.
  • Peak magnitude: Mr = 1 / (2 zeta sqrt(1 - zeta squared)); peak frequency: omega_r = omega_n sqrt(1 - 2 zeta squared).
  • High-frequency rolloff: -40 dB/decade (two poles); phase transitions from 0 to -180 degrees.
  • At omega = omega_n: phase is always exactly -90 degrees, independent of zeta.
  • zeta = 0.707 gives maximally flat (Butterworth) response — no resonant peak, smoothest rolloff.
  • Exam trap: The resonant frequency omega_r is slightly less than omega_n for underdamped systems — they are not equal.

Second Order Bode Quiz

Test your understanding of resonant peaks, damping ratio effects, and Bode plots for second-order systems.

Question 1 of 3

Q1.For a standard second-order system G(s) = wn^2 / (s^2 + 2*zeta*wn*s + wn^2), the resonant peak in the Bode magnitude plot occurs at the resonant frequency wr. What is wr?