Step 3, second-order systems: from the physical model to the transfer function, characteristic equation, symbols, s-plane, lab, specifications, comparison of five systems and exercises.
CDSSC · Step 3 (complete) · builds on Step 1 (K, first order) and Step 2 (poles → shape)
Second-order systems (sistemas de segundo orden)
Formulas and names in English, Spanish in brackets as on the slides. Goal: go from the physical equation to G(s), read \omega_n,\ \zeta,\ K from it, draw the poles and, without simulating, say how much it will overshoot, when it reaches its peak and when it settles.
1 From physics to G(s) · 2 Characteristic equation · 3 Symbols · 4 Other applications · 5 s-plane step by step · 6 Lab · 7 Specifications · 8 Five systems · 9 Pitfalls · 10 MATLAB · 11 Test yourself
1 · From physics to G(s): model → transfer function (función de transferencia)
One example runs through the whole step: a quarter car on one wheel. Mass m (weight of the body), a spring with stiffness k (how many newtons per metre of compression), a damper (amortiguador) with coefficient c (how many newtons it brakes with at a speed of 1 m/s). Someone gets in: a force F pushes down. Output y = how far the body sinks.
Derivation step by step
General recipe: ODE → G(s) (SS-05 p.2, SS-04 p.13)
Write the differential equation (Newton for mechanics, Kirchhoff for circuits). Everything with the output y on the left, the input u on the right.
Laplace with zero initial conditions: every derivative = multiplying by s. \ddot y\to s^2Y,\ \dot y\to sY,\ y\to Y.
Take Y(s) out as a common factor and divide: G(s)=Y(s)/U(s).
Divide the numerator and denominator by the coefficient of s^2 and compare with the standard form.
Is c our zero? No. Zeros (ceros) are the roots of the numerator, poles (polos) are the roots of the denominator (SS-05 p.3). Here the numerator is a constant, so there are no zeros at all. All three parameters m,\ c,\ k sit in the denominator, so together they set the poles. A zero would only appear if the input also acted through a derivative (e.g. the road pushes the wheel through the damper: numerator cs+k).
2 · A way without rearranging the formula: solve the characteristic equation (ecuación característica)
Yes, you can do it this way. It is a fully valid method.
The poles are simply the solutions of denominator = 0 (the characteristic equation, polinomio característico, SS-05 p.3). You solve an ordinary quadratic equation, get a complex number -\sigma\pm j\omega_d (SS-10 p.3), and read \omega_n and \zeta from the geometry: length and angle (SS-10 p.4–5). No need to match anything to the standard form.
The car step by step: 250s^2+3000s+25000=0
Discriminant (discriminante): \Delta=b^2-4ac=3000^2-4\cdot250\cdot25000=-16\cdot10^6. Negative, so the roots are complex.
You don't have to divide by a. Dividing the whole equation by 250 does not change its solutions, so the poles come out the same.
One thing the roots won't tell you: K. Poles only describe shape and speed. Get K separately: K=G(0)=\text{numerator}/\text{constant term}=1/25000.
When Δ ≥ 0, you get two real roots p_1,\ p_2: the system is overdamped (sobreamortiguado, ζ ≥ 1). Then \omega_n=\sqrt{p_1p_2} and \zeta=-\dfrac{p_1+p_2}{2\omega_n}.
Why do both ways give the same result? For the equation as^2+bs+c=0 the product of the roots is c/a and the sum is -b/a (Vieta's formulas). For the pair -\sigma\pm j\omega_d: product =\sigma^2+\omega_d^2=\omega_n^2, sum =-2\sigma=-2\zeta\omega_n. So "matching coefficients" is the same geometry, just in shorthand.
Way A: compare with the standard form
Way B: roots of the equation
you do
divide by a, read \omega_n^2 and 2\zeta\omega_n
Δ, quadratic formula, read \sigma,\ \omega_d, then Pythagoras and cosine
good when
G(s) already looks like the standard form (fast in the exam)
you want to understand where the poles are; any polynomial, also higher order
MATLAB
damp(G)
roots([250 3000 25000])
Calculator: enter the denominator coefficients
a\,s^2+b\,s+c=0a b c
Examples:
The quadratic formula is ordinary school maths. Reading σ, ωd from the pole: SS-10 p.3. ωn = magnitude, ζ = cos θ: SS-10 p.4–5. Vieta and the calculator are my own additions, not from the slides.
Left: standard form, SS-10 p.3 (SS-10 writes ξ instead of ζ, it is the same letter). Right: the car after dividing by m. Compare term by term.
\omega_n
natural frequency (frecuencia natural no amortiguada) =\sqrt{k/m} How fast the car would bounce without a damper. Stiff spring or light mass: fast. It is the "speed" of the whole system. It does not depend on c.
In the s-plane (plano s): distance of the pole from the point (0, 0). Dashed circle = all poles with the same ωn.
\zeta
damping ratio (coeficiente de amortiguamiento) =\dfrac{c}{2\sqrt{km}} How hard the damper brakes compared with the amount that is "just enough not to overshoot" (\zeta=1). \zeta=0: bounces forever. \zeta=0.6: overshoots once, a little. \zeta\ge1: does not overshoot. No unit.
In the s-plane: the pole angle, \zeta=\cos\theta.
K
static gain (ganancia estática) =G(0)=1/k in this example Where the motion stops, per unit of input. At the end nothing moves, so the spring alone holds the force: y_\infty=F/k (Hooke's law). Does changing k change K? Here yes, but only because the input is a force. In an RLC circuit or in an arm with a controller (part 4) K=1 whatever the stiffness. Besides, k also changes \omega_n and \zeta.
In the s-plane: not there. Poles tell you shape and speed, not the final value.
\sigma
decay rate (factor de amortiguamiento o decrecimiento) =\zeta\omega_n=\dfrac{c}{2m} How fast the bouncing dies out: the funnel of the envelope (envolvente) e^{-\sigma t}. Unit 1/s.
In the s-plane: how far left of the vertical axis the pole lies.
\omega_d
damped frequency (frecuencia amortiguada) =\omega_n\sqrt{1-\zeta^2} How fast the car really bounces, the one you count on the plot. Unit rad/s.
In the s-plane: height of the pole above the horizontal axis.
m
mass (masa): inertia, i.e. resistance to speeding up. Heavier car: slower (smaller ωn) and smaller ζ with the same damper. It is a parameter of this example: in other systems a different quantity plays the same role (part 4).
Why is \omega_d<\omega_n? In the s-plane it is a right triangle: \sigma to the left, \omega_d up, \omega_n is the hypotenuse. Pythagoras: \omega_n^2=\sigma^2+\omega_d^2, and the hypotenuse is always longer. Physically: the damper does not only kill the bouncing, it also slows it down. Only two numbers are independent (SS-10 p.3): usually \omega_n and \zeta, you calculate the rest.
4 · The same thing in other applications
Mass, spring and damper are just one "costume". Every second-order system (sistema de segundo orden) has the same three roles: inertia (what does not want to speed up), stiffness (what pulls back) and damping (what brakes the motion). The formulas are always the same:
Robot arm: the P controller pushes the arm with a force proportional to the error, so it behaves like a spring. Larger K_p = stiffer "spring" = faster, but smaller ζ, so more overshoot. The same problem comes back when tuning controllers (DC-02).
RLC: the inductor dislikes changes in current (inertia), the resistor turns energy into heat (damping), the capacitor "pushes back" like a spring. Without R the circuit rings forever.
K depends on where the input enters: a force on the spring gives 1/k, while a set position or a supply voltage gives 1, because in the end the output catches up with the input.
The models and numbers in parts 1, 4 and 6 are mine (Newton / Kirchhoff), not from the slides. ODE → G(s) procedure: SS-05 p.2, derivative rule: SS-04 p.13.
5 · The s-plane (plano s) step by step
Reminder from Step 2: every point of the s-plane is one way of moving (mode). Horizontal axis (Re): dies out or grows. Vertical axis (Im): how fast it bounces. We draw the poles (polos) of the car: s=-6\pm8j. The last step answers the question of which way the angle is measured.
6 · Lab: pick a system and turn the physical parameters
The sliders change real parts (mass, spring, damper, inductor...). Next to them you see the step response and the poles straight away. Numbers below the plots.
System:
Set damping to:
nowpinned (dashed)final value and ±5 % bandpeak ▲ and exact ts
Poles in the s-plane (fixed scale for each system)
What to notice
Move only the damping (c, b or R): the circle stays, the pole moves along it. The speed \omega_n does not depend on damping.
Move the stiffness (k, Kp or C): the circle grows or shrinks, and the angle changes too.
Move the inertia (m, J or L): heavier = smaller circle (slower) and larger angle (more bouncing).
In the car, changing k also changes the final value (F/k). In the arm and the RLC circuit the final value always equals the input (K = 1).
Second-order waves sit inside an envelope (envolvente) that shrinks like a first-order system with time constant T=1/\sigma (SS-10 p.6). In the s-plane: the further left the pole lies, the faster everything dies out, in both cases.
0.30
second orderenvelope e^(−σt)first order with T = 1/σ (dashed)±5 % band
This inversion of the formula from FR-06 p.4 is my own algebra (solved for ζ). The slide only shows the shapes of the regions (FR-06 p.5).
10 %
1.5 s
Pole map: green = meets both specifications
Which systems pass
Quiz on the five systems
9 · Pitfalls
\xi (SS-10) and \zeta (FR-06, Franklin, MATLAB) are the same damping ratio (coeficiente de amortiguamiento).
Angle: the mathematical argument of a pole is measured from the positive Re axis, counter-clockwise (−6 + 8j: 126.9°). SS-10 uses the supplementary angle \theta=180^\circ-\arg p (53.1°), measured from the negative Re axis, so that \zeta=\cos\theta (SS-10 p.4). FR-06 p.4 and Franklin measure from the Im axis: \sin^{-1}\zeta (36.9°), so sin and cos swap (SS-10 p.5, footnote 1).
c is not a zero. Zeros are roots of the numerator, and c sits in the denominator, so it moves the poles (SS-05 p.3).
K is not the numerator. \frac{50}{s^2+6s+25} has K = 2, not 50, because K=\text{numerator}/\omega_n^2=G(0).
First divide by the coefficient of s^2 (m, J, L...). 250s^2+3000s+25000 → s^2+12s+100. Otherwise ωn and ζ come out wrong.
Mp is measured relative to the final value, not to 1 or to the step size. K = 2 and peak 2.66: M_p=0.66/2=33\% (SS-10 p.9).
t_s\approx3/\sigma is an estimate from the envelope (SS-10 p.12). The exact ts can be a bit smaller or larger and jumps when ζ changes (the "exact" column in the lab). MATLAB stepinfo uses 2 % by default.
The Mp and tp formulas only work for 0 < ζ < 1 and a pure second-order system without zeros. Extra poles or zeros close to the dominant ones change the response (SS-10 p.2, FR-06 p.8).
What you see as bouncing has frequency ωd, not ωn: t_p=\pi/\omega_d.
10 · MATLAB
s = tf('s');
m = 250; c = 3000; k = 25000;
G = 1/(m*s^2 + c*s + k) % model -> transfer function, straight from m, c, k
damp(G) % Pole -6 +/- 8i, Damping 0.6, Frequency 10
pzmap(G) % pole map
step(800*G) % response to F = 800 N, final value 800/k = 0.032 m
S = stepinfo(G, 'SettlingTimeThreshold', 0.05) % 5 % band like the course
% S.Overshoot in %, S.PeakTime, S.SettlingTime, S.RiseTime (10-90 %)
step(G, 1/(m*s^2 + 1000*s + k), 1/(m*s^2 + 8000*s + k)); legend('c=3000','c=1000','c=8000')
Expected: Overshoot ≈ 9.5, PeakTime ≈ 0.39, SettlingTime close to 0.5 (estimate 3/σ).
11 · Test yourself
E1. Model → G(s). Robot arm: J\ddot\theta+b\dot\theta+K_p\theta=K_p\theta_{ref}, J = 0.05 kg·m², b = 0.5, Kp = 5. Find ωn, ζ, K.