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Second-order systems: transfer function, damping ratio and natural frequency

Turning a mass-spring-damper's equation of motion into a transfer function, then identifying its damping ratio and natural frequency.

Almost every second-order system a control systems module throws at you — electrical, mechanical, or a mix of both — reduces to the same two numbers once you have its transfer function: natural frequency and damping ratio. Between them, they tell you whether the system oscillates, how fast, and how quickly it settles. Here's a full derivation from a physical system through to both numbers.

Problem. A mass m = 2 kg is connected to a fixed wall by a spring of stiffness k = 50 N/m and a damper with damping coefficient b = 8 N·s/m, arranged in parallel. A horizontal force F(t) applied to the mass produces a displacement x(t). Find the transfer function X(s)/F(s), then identify the system's natural frequency and damping ratio.

mkbF(t)x(t)

Spring and damper act in parallel between the wall and the mass.

From equation of motion to transfer function

mx¨(t)+bx˙(t)+kx(t)=F(t)m\ddot x(t) + b\dot x(t) + kx(t) = F(t)

Sum of forces on the mass: the spring, the damper, and the applied force F(t).

(ms2+bs+k) X(s)=F(s)(ms^2 + bs + k)\,X(s) = F(s)

Each time-derivative becomes a factor of s in the Laplace domain, assuming the mass starts at rest.

G(s)=X(s)F(s)=1ms2+bs+kG(s) = \dfrac{X(s)}{F(s)} = \dfrac{1}{ms^2+bs+k}

The transfer function is output over input, with zero initial conditions.

G(s)=12s2+8s+50G(s) = \dfrac{1}{2s^2+8s+50}

m = 2 kg, b = 8 N·s/m, k = 50 N/m.

Identifying ζ and ωₙ

The standard second-order form is s2+2ζωns+ωn2s^2 + 2\zeta\omega_n s + \omega_n^2 in the denominator — dividing through by m puts the transfer function into that shape directly.

G(s)=1/ms2+(b/m)s+(k/m)=0.5s2+4s+25G(s) = \dfrac{1/m}{s^2+(b/m)s+(k/m)} = \dfrac{0.5}{s^2+4s+25}

Dividing every term by m normalises the s² coefficient to 1.

ωn2=km=25 ⇒ ωn=5 rad/s\omega_n^2 = \dfrac{k}{m} = 25 \ \Rightarrow\ \omega_n = 5\ \text{rad/s}

The natural frequency depends only on stiffness and mass, never on damping.

2ζωn=bm=4 ⇒ ζ=42(5)=0.42\zeta\omega_n = \dfrac{b}{m} = 4 \ \Rightarrow\ \zeta = \dfrac{4}{2(5)} = 0.4

ζ < 1, so the system is underdamped: it overshoots and oscillates before settling.

ωd=ωn1−ζ2=51−0.16≈4.58 rad/s\omega_d = \omega_n\sqrt{1-\zeta^2} = 5\sqrt{1-0.16} \approx 4.58\ \text{rad/s}

This is the frequency actually seen in the step response — slightly below the natural frequency, because damping slows the oscillation.

With ζ = 0.4, this mass will overshoot its final position and ring for a few cycles before settling — the same underdamped behaviour shows up in suspension systems, sensor housings and closed-loop control designs, which is exactly why ζ and ωₙ are the two numbers a control systems module keeps asking for.

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