EEM322 Denetim Sistemleri: Bode Kararlılık Ölçütü, Kök Yer ve PI/PID Tasarımı (Robert Edwards, ChE 450, Penn State, 2001)

Bu yazı, EEM322 Denetim Sistemleri dersi için internetten indirilip "kontrol sistemleri farklı yerlerden indirilenler" klasörüne konmuş üç üçüncü taraf PowerPoint sunumunu bir araya getiriyor: Robert Edwards'ın ChE 450 (Spring 2001) proses kontrol dersinin 36, 37 ve 38. ders slaytları. Klasördeki not dosyası kaynağı http://www.mne.psu.edu/edwards/CLASSES/CHE_SP01/lectures/lecture36.PPT olarak gösteriyor (Pennsylvania State University sunucusu). Dosyalar Ocak–Mayıs 2001 tarihli. Slaytlar İngilizce olduğundan metin özgün dilinde aktarıldı, denklemler düz metne çevrildi; üç sunum ve kaynak notu yazının sonunda ZIP olarak indirilebilir.

Bu içerik üçüncü taraf bir kaynağa aittir; arşiv ve nostalji amacıyla kaynak belirtilerek paylaşılmıştır. Hak sahibinin talebi hâlinde kaldırılır.

Notation in the slides: SP setpoint, E error, MV manipulated variable, CV controlled variable, CVm measured CV, D disturbance; G_c controller, G_v valve, G_p process, G_d disturbance, G_s sensor; G_OL = G_c·G_v·G_p·G_s is the open-loop transfer function.

Lecture 36: Bode stability criterion and sketching rules

The Bode stability criterion uses the open-loop characteristic to predict the closed-loop behavior.

Open-loop landmarks

• ω₁ is the frequency at which |G_OL(jω)| = 1 (0 dB).
• ω_π is the frequency at which ∠G_OL(jω) = −180°.
• Gain margin Gm = 1/|G_OL(jω_π)| in gain units, or −|G_OL(jω_π)| in dB.
• Phase margin Pm = 180° + ∠G_OL(jω₁).

Frequency-domain guidelines for a "good" open loop

• A larger 0 dB crossing ω₁ means larger closed-loop bandwidth and shorter rise time.
• Large |G_OL| at low frequency means small offset.
• Gain margin > 1 (> 0 dB) is required for closed-loop stability; > 2.5 (8 dB) is recommended for performance.
• Phase margin > 35° gives a small resonant peak in the closed loop, good damping and a good decay ratio.

The next question is how to add components to a feedback loop (PID, recycle, etc.) to get these desirable frequency response characteristics.

Example: three-tank mixing process

With K_c = 1, T_I = ∞, T_D = 0, i.e. G_OL(jω) = 1.0·G_v G_p G_s = 0.039/(5s + 1)³ (bode(0.039, conv([5 1],conv([5 1],[5 1])))): there is no 0 dB crossing, which implies a slow response; low gain at low ω means poor offset; the large gain margin (~50 dB) means "really stable"; the infinite phase margin means "well damped".

With PID, K_c = 21, T_I = 10, T_D = 2:

margin(0.039*21*[20 10 1],conv([10 0],conv([5 1],conv([5 1],[5 1]))))

The 0 dB crossing is ω₁ = 0.08 rad/min; the infinite gain at low ω means good (zero) offset; the gain margin is infinite (no ω_π); the phase margin of about 67° (> 35°) means a well damped response. The corresponding root locus (rlocus(conv([Ti*Td Ti 1],num),conv([Ti 0],den)) with the exact three-tank model) has open-loop poles at s = 0, −1/5, −1/5, −1/5 and open-loop zeros at −0.14 and −0.36 for T_D = 2, T_I = 10. The slides then compare setpoint and disturbance responses with the "setpoint" and "disturbance" tunings.

With the Ziegler–Nichols ultimate-period settings (ultimate gain K_c = 120), K_c = 90, T_I = 9, T_D = 2.3:

margin(0.039*90*[20.7 9 1],conv([9 0],conv([5 1],conv([5 1],[5 1]))))

the phase margin drops to about 36°. Time responses are compared for Pm = 31° (K_c = 120) and Pm = 36° (K_c = 90).

Not: Bu slaytta açıklama kutusu bir önceki slayttan kopyalanmış ("phase margin of 67° > 35°"); grafikteki faz payı ise yaklaşık 36°'dir.

Bode diagram sketching rules

First, put the open-loop transfer function into Bode form, with +1 at all the constant terms:

G_OL(s) = K_zp·Π(s − z_k)/Π(s − p_i) = K_B·Π(τ_k^N s + 1)/Π(τ_i^D s + 1), K_B = K_zp·Π(−z_k)/Π(−p_i), τ_k^N = −1/z_k, τ_i^D = −1/p_i

Then substitute s = jω. Each numerator or denominator component G_k(jω) = τ_k^N jω + 1 or G_i(jω) = τ_i^D jω + 1 is a complex number with a magnitude and a phase. Collecting the exponential terms and equating magnitude and phase on each side:

∠G(jω) = ∠K_B + Σ∠G_k(jω) − Σ∠G_i(jω) 20 log₁₀|G(jω)| = 20 log₁₀|K_B| + Σ 20 log₁₀|G_k(jω)| − Σ 20 log₁₀|G_i(jω)|

Logarithms turn magnitude multiplication into addition, so the plot can be constructed graphically: numerator components are added, denominator components subtracted. Since both have the same form, only one set of sketching rules is needed. There are only 4 possible numerator components: G(s) = K_B·s^ℓ·(τ²s² + 2ζτs + 1)·Π(τ_k^N s + 1)/Π(τ_i^D s + 1).

1. Constant K_B: a real number, positive or negative. Magnitude |K_B|; phase 0° if K_B > 0, −180° if K_B < 0.
2. Repeated zeros at the origin, s^ℓ: (jω)^ℓ = ω^ℓ e^(jℓπ/2); magnitude in dB = 20ℓ log₁₀ ω, a constant slope of +20ℓ dB/decade crossing 0 dB at ω = 1; phase constant at +90ℓ°. If ℓ < 0 (in the denominator), slope and phase are negative.
3. Numerator time constants τ_k > 0: τ_k jω + 1. As ω → 0 the magnitude → 1 (0 dB) and phase → 0°; for large ω the magnitude → ωτ_k and the phase → +90°. The asymptotes intersect at the break frequency ω_b = 1/τ_k; the magnitude increases 20 dB/decade above the break. The phase is half-way between its asymptotic values (45°) at ω_b, and the slope of the phase plot extrapolates to 0° at ω_b/5 and to +90° at 5ω_b (bode([0.1 1],1)). For a denominator time constant (bode(1,[1 1])) everything is inverted: −20 dB/decade and −90°.
4. Watch out for τ_k < 0 (zero in the right half plane): the magnitude behaves the same (+20 dB/decade above the break; magnitude is just the distance from the origin), but the phase goes to −90° (bode([-0.1 1],1)).

Repeated time constants: the high-frequency asymptotic slope, the high-frequency phase and the phase at the break are multiplied by the order of the root; the 0.2ω_b and 5ω_b corners still apply.

Quadratic terms (−1 < ζ < 1) are sketched as a time constant repeated twice: for ζ > 0 the numerator slope is 40 dB/decade and the phase goes to +180°; for ζ < 0 the slope is still 40 dB/decade but the phase goes to −180°. Denominator rules have opposite signs for slope and phase.

Lecture 37: PI control examples (root locus and Bode)

Example 1: PI control of a 2-second delay

G_c(s) = K_c(T_I s + 1)/(T_I s), G_v = 1, G_s = 1, and the delay is approximated as G_p(s) ≈ (−s + 1)/(s + 1). Then

G_OL = K_c·[(T_I s + 1)/(T_I s)]·[(−s + 1)/(s + 1)] = (−K_c)·(s + 1/T_I)(s − 1)/[s(s + 1)]

so K_zp = −K_c, m = 2 zeros at s = 1 and −1/T_I, n = 2 poles at s = 0 and −1; n − m = 0, so there are no asymptotes. As |K_zp| increases from 0 to ∞, the locus leaves the poles at 0 and −1 and ends at the zeros, including the right-half-plane zero at +1.

For the Bode diagram, K_B = K_c/T_I; the numerator time constant (−s + 1) breaks at ω_b = |1/(−1)| = 1 and its phase goes to −90° (τ < 0); the denominator pole 1/s and the time constant 1/(s + 1) break at ω_b = 1. The magnitudes of (−s + 1) and 1/(s + 1) add to zero, while their phases add to twice the slope (the numerator and denominator phase contributions are the same because τ < 0 in the numerator). The PI portion (T_I s + 1)/(T_I s), e.g. T_I = 0.1, has a −20 dB/decade slope that becomes flat at ω_b = 1/T_I = 10; the phase plot has twice as many slope changes because there are two "corners" per break. Adding everything, a K_c < 1 is needed for a decent phase margin: K_c = −30 dB = 0.032 gives about 30° phase margin.

Example 2: PI control with 3 asymptotes

G_p(s) = −1/[s(s + 1)²], so

G_OL = K_c·[(T_I s + 1)/(T_I s)]·[−1/(s(s + 1)²)] = (−K_c)·(s + 1/T_I)/[s²(s + 1)²]

K_zp = −K_c (note that the gain of the plant is negative), m = 1 zero at −1/T_I, n = 4 poles at s = 0, 0, −1, −1, n − m = 3 asymptotes. The locus (T_I = 2) becomes unstable for K_c too negative; a reverse response plant needs a reverse action controller (K_c < 0). In the Bode analysis, to avoid a −180° phase contribution due to a negative Bode gain, K_c < 0 is chosen with reverse action.

The components 1/s and 1/(s + 1)² add to a slope of −20 − 40 = −60 dB/decade after ω = 1. With T_I = 10 (1/T_I = 0.1) the phase margin is maximised by choosing K_c = 0.1 (−20 dB). With K_c = 0.25, T_I = 10:

margin(0.25*[10 1],conv([10 0],[1 2 1 0]))   % Gm = 16 dB, Pm = 40 deg

The Simulink test with reverse action selected in the controller confirms the response. With T_I = 5 the reset time is too short and the maximum phase margin is only 23°:

margin(0.25*[5 1],conv([5 0],[1 2 1 0]))     % Pm = 23 deg

Not: Yanıt grafiklerinin başlığında süreç "1/[s(s² + 1)]" diye yazılmış; incelenen süreç 1/[s(s + 1)²]'dir (Simulink bloğunda da −1/(s³ + 2s² + s)).

Lecture 38: PID control of a third order plant

Lecture 38 begins by repeating Example 2 (as "Example 4") and then continues with PID control.

Example 5: PID control with a 3rd order plant

G_c(s) = K_c(T_I T_D s² + T_I s + 1)/(T_I s), G_p(s) = 1/[s(s + 1)²]:

G_OL = K_c·(T_I T_D/T_I)·[s² + s/T_D + 1/(T_I T_D)]/s · 1/[s(s + 1)²]

K_zp = K_c T_D; m = 2 zeros at s = [−T_I ± √(T_I² − 4T_I T_D)]/(2T_I); n = 4 poles at s = 0, 0, −1, −1; n − m = 2 asymptotes. With T_I = 4, T_D = 1 the asymptotes are at σ = −1/2 and the response is stable for all K_c; adding T_D requires a change in T_I.

Bode example 3: choose T_I and T_D to get a repeated numerator root: T_D = T_I/4 gives the repeated time constant T_I/2, G_OL = K_c(T_I s/2 + 1)²/(T_I s) · 1/[s(s + 1)²]. With T_I = 20 (T_D = 5): |1/20| = −26 dB; the PID components added together rise to 40 − 26 = 14 dB at high frequency, giving an ultimate phase improvement of +90°, but increasing the gain at large ω creates a noise response problem. Combined with the plant (−60 dB/decade), the phase margin is about 30° in the asymptotic sketch.

margin([5*20 20 1],conv([20 0],[1 2 1 0]))   % Pm = 47 deg, Gm infinite, Kc = 1
rlocus([5*20 20 1],conv([20 0],[1 2 1 0]))   % Kc = 0.28

Placing the PID zeros close to the origin forces a slow closed-loop response (compare K_c = 1 and K_c = 0.28).

Why frequency response is generally more capable: with root locus, choosing K_c to get a desired damping factor on "one" set of roots does not account for the presence of additional poles and zeros; frequency response accounts for all the poles and zeros and their relative positions as well.

Moving the zeros away from the origin speeds the response

rlocus([10 11 1],conv([11 0],[1 2 1 0]))
margin(2*[10 11 1],conv([11 0],[1 2 1 0]))        % Kc = 2, PM = 35 deg
rlocus([2 3 1],conv([3 0],[1 2 1 0]))
margin(2*[3 2 1],conv([3 0],[1 2 1 0]))           % PM = 45 deg (slaytta yazıldığı gibi)
margin(0.6*[6.25 6.25 1],conv([6.25 0],[1 2 1 0])) % PM = 47 deg
rlocus([6.25 6.25 1],conv([6.25 0],[1 2 1 0]))

Moving the zeros to the left "pulls" the locus to faster pole locations. With T_I = 11, T_D = 0.9 the PID zero at −1 cancels a pole of the plant, and the high-frequency gain increase starts at a higher frequency, which makes the loop less susceptible to noise; the asymptotic sketch shows about 60° phase margin, and K_c was increased until the PM was reduced to 35°.

Not: T_I = 3, T_D = 0,67 için PID payı T_I·T_D·s² + T_I·s + 1 = 2s² + 3s + 1, yani [2 3 1] olmalıdır; margin satırında [3 2 1] yazılmış (aynı slayt grubundaki rlocus satırında doğru biçim kullanılmış).

The lecture ends by comparing the three PID tunings: K_c = 2, T_I = 11, T_D = 0.9; K_c = 0.6, T_I = 6.25, T_D = 1; and K_c = 1, T_I = 20, T_D = 5.