Exercise 12 Illustrations

What each element does to the loop shape

Loop shaping modifies L = KG on the Bode plot. In series, controller elements add their dB and degrees to the plant. Pick an element and watch where it acts.

nominal stability
gain margin
phase margin
|L(0)| ≫ 1, small stationary error
crossover frequency, bandwidth

P K = kshifts gain, (1)–(5) PI K = k(1 + 1/(T_Is))(4): |K(0)| = ∞ PD K = k(1 + T_Ds)(3): +90° lead lead k(Ts+1)/(αTs+1), α<1(3): filtered PD lag α(Ts+1)/(αTs+1), α>1(4): finite PI

k = 1

T = 1 s corner at 1/T = 1 rad/s

α = 0.2

K(iω)corner frequencies 1/T and 1/(αT)ω_max, φ_max

Task 12.1 · Lead compensator

G(s) = s + 10s (s + 1) K(s) = k · 1(1 + 0.01s)² · 1 + s/ω_z1 + s/ω_p

Specs: ramp error ≤ 0.02, slope −60 dB/dec above 100 rad/s, PM ≥ 70°, GM ≥ 20 dB. Build the controller step by step and watch the margins.

k = 5 ramp error 1/(10k) = 0.02

Roll-off filter 1/(1 + 0.01s)² part iii Lead compensator part iv

zero ω_z = 5 rad/s

pole ω_p = 20 rad/s

Rule of thumb: put the lead zero a little below the crossover (it lifts the phase there) and the pole above it (it limits the high-frequency gain, which protects the gain margin and keeps noise down).

plant G aloneL = KGω_c, PMω₁₈₀, GMclosed-loop step response

Task 12.2 · Lead–lag design

G(s) = 1s (1 + s/10)(1 + s/200) k = 100 for e_ramp = 1/k ≤ 0.01

The lead adds phase at ω_c but also gain, so k₁ < 1 restores the crossover and the ramp error grows. The lag puts the low-frequency gain back without touching the phase at ω_c, if its corner is far below ω_c.

i) L₀ = kG ii) L₁ = k·K_lead·G iii) L₂ = k·K_lead·K_lag·G

lead: φ_max = 30.6° α₁ = (1 − sin φ)/(1 + sin φ) = k₁ chosen so that ω_c stays at ω_max

lag: α₂ = 1.75 lag: T₂ = 1 s corner 1/T₂ = 1 rad/s

L₀ = kGL₁ with leadL₂ with lead and lagω_c, PM of the selected stageclosed-loop step response

Task 12.3 · Internal model control gives a PI

G(s) = abs + 1 T(s) = 1T_fs + 1 Q = G⁻¹T K = Q1 − ĜQ

With a perfect model, T = GQ and S = 1 − GQ are affine in Q: pick the closed loop you want, then read off K. For this plant K is exactly a PI controller with k = b/(aT_f) and T_I = b.

plant gain a = 1 plant time constant b = 1 s desired T_f = 0.3 s

model mismatch: true gain = 1 × a model mismatch: true time constant = 1 × b

T = 1 is not achievable: a causal controller cannot make the closed loop faster than order 0 for a first-order plant. For a second-order plant the same recipe with T = 1/(T_fs+1)² gives a PID with a first-order filter 1/(T_fs/2 + 1).

designed T: y = 1 − e^−t/T_fy(t) with K on the true plantcontrol input u(t)closed-loop poles (×), PI zero (○)