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5.3 PID Regulyatorlar β€” Masalalar

Jami: 35 ta | Yechim bilan: βœ…


Asosiy Masalalar (1-12)​

Masala 1 ⭐⭐​

P controller: Kp=10K_p = 10. Xatolik e = 0.5. Chiqish signali u?

Yechim

u=Kpβ‹…e=10Γ—0.5=5u = K_p \cdot e = 10 \times 0.5 = 5


Masala 2 ⭐⭐​

PI controller: Kp=5K_p = 5, Ki=2K_i = 2. e(t) = 1 (doimiy), t = 3 s. Chiqish?

Yechim

uP=Kpβ‹…e=5Γ—1=5u_P = K_p \cdot e = 5 \times 1 = 5

uI=Ki∫031 dt=2Γ—3=6u_I = K_i \int_0^3 1 \, dt = 2 \times 3 = 6

u=uP+uI=5+6=11u = u_P + u_I = 5 + 6 = 11


Masala 3 ⭐⭐​

PD controller: Kp=8K_p = 8, Kd=0.5K_d = 0.5. e = 2, dedt=βˆ’4\frac{de}{dt} = -4. Chiqish?

Yechim

u=Kpe+Kddedt=8(2)+0.5(βˆ’4)=16βˆ’2=14u = K_p e + K_d \frac{de}{dt} = 8(2) + 0.5(-4) = 16 - 2 = 14


Masala 4 ⭐⭐​

Diskret PID: Kp=2K_p = 2, Ki=0.5K_i = 0.5, Kd=0.1K_d = 0.1, dt = 0.01 s. e[n] = 1, e[n-1] = 1.2, integral_sum = 5. Chiqish?

Yechim

P=Kpβ‹…e=2Γ—1=2P = K_p \cdot e = 2 \times 1 = 2

I=Kiβ‹…(integral_sum+eβ‹…dt)=0.5Γ—(5+0.01)=2.505I = K_i \cdot (integral\_sum + e \cdot dt) = 0.5 \times (5 + 0.01) = 2.505

D=Kdβ‹…eβˆ’eprevdt=0.1Γ—1βˆ’1.20.01=0.1Γ—(βˆ’20)=βˆ’2D = K_d \cdot \frac{e - e_{prev}}{dt} = 0.1 \times \frac{1 - 1.2}{0.01} = 0.1 \times (-20) = -2

u=P+I+D=2+2.505βˆ’2=2.505u = P + I + D = 2 + 2.505 - 2 = 2.505


Masala 5 ⭐⭐​

Motor tezlik kontroli: setpoint = 1000 rpm, actual = 800 rpm, Kp=0.5K_p = 0.5. PWM duty cycle (0-100%)?

Yechim

e=1000βˆ’800=200e = 1000 - 800 = 200 rpm

u=Kpβ‹…e=0.5Γ—200=100u = K_p \cdot e = 0.5 \times 200 = 100

PWM = 100% (to'yingan)


Masala 6 ⭐⭐​

Dron altitude: setpoint = 10 m, actual = 8 m. Kp=5K_p = 5. Thrust qo'shimchasi (N)?

Yechim

e=10βˆ’8=2e = 10 - 8 = 2 m

Ξ”thrust=Kpβ‹…e=5Γ—2=10\Delta thrust = K_p \cdot e = 5 \times 2 = 10 N


Masala 7 ⭐⭐​

Steady-state error: P controller Kp=20K_p = 20, plant DC gain = 1. Step input uchun SS error?

Yechim

Yopiq sikl: YR=Kpβ‹…11+Kpβ‹…1=2021\frac{Y}{R} = \frac{K_p \cdot 1}{1 + K_p \cdot 1} = \frac{20}{21}

ess=1βˆ’2021=121=0.048=4.8%e_{ss} = 1 - \frac{20}{21} = \frac{1}{21} = 0.048 = 4.8\%


Masala 8 ⭐⭐​

Integral action: xatolik e = 0.1 doimiy, Ki=2K_i = 2. 5 s da integral komponenti?

Yechim

uI=Ki∫050.1 dt=2Γ—0.1Γ—5=1u_I = K_i \int_0^5 0.1 \, dt = 2 \times 0.1 \times 5 = 1


Masala 9 ⭐⭐​

Ziegler-Nichols: Ku=50K_u = 50, Tu=0.5T_u = 0.5 s. PID parametrlari?

Yechim

Kp=0.6Ku=30K_p = 0.6 K_u = 30

Ti=Tu/2=0.25T_i = T_u/2 = 0.25 s β†’ Ki=Kp/Ti=120K_i = K_p/T_i = 120

Td=Tu/8=0.0625T_d = T_u/8 = 0.0625 s β†’ Kd=Kpβ‹…Td=1.875K_d = K_p \cdot T_d = 1.875


Masala 10 ⭐⭐​

Anti-windup: integral cheklovi Β±100. Hozirgi integral = 95, e = 0.2, Ki=50K_i = 50, dt = 0.1. Yangi integral?

Yechim

Yangi integral (cheklovsiz): 95+50Γ—0.2Γ—0.1=95+1=9695 + 50 \times 0.2 \times 0.1 = 95 + 1 = 96

96 < 100 β†’ chegarada emas

Yangi integral = 96


Masala 11 ⭐⭐​

Robot qo'li: ΞΈref=90Β°\theta_{ref} = 90Β°, ΞΈactual=85Β°\theta_{actual} = 85Β°. Kp=2K_p = 2 NΒ·m/deg. Motor moment?

Yechim

e=90βˆ’85=5Β°e = 90 - 85 = 5Β°

Ο„=Kpβ‹…e=2Γ—5=10\tau = K_p \cdot e = 2 \times 5 = 10 NΒ·m


Masala 12 ⭐⭐​

Line follower: sensor = 3 (o'rtadan chapda), Kp=20K_p = 20. Motorlar tezlik farqi?

Yechim

e=0βˆ’3=βˆ’3e = 0 - 3 = -3 (0 = markaz)

Ξ”v=Kpβ‹…e=20Γ—(βˆ’3)=βˆ’60\Delta v = K_p \cdot e = 20 \times (-3) = -60

Chap motor: base - 30 O'ng motor: base + 30

(O'ngga burilish)


O'rtacha Masalalar (13-25)​

Masala 13 ⭐⭐⭐​

Overshoot: ΞΆ=0.5\zeta = 0.5. Foiz overshoot?

Yechim

Mp=eβˆ’Ο€ΞΆ1βˆ’ΞΆ2=eβˆ’3.14Γ—0.50.866=eβˆ’1.81=0.163=16.3%M_p = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} = e^{-\frac{3.14 \times 0.5}{0.866}} = e^{-1.81} = 0.163 = 16.3\%


Masala 14 ⭐⭐⭐​

Settling time (2%) kerak: ts<2t_s < 2 s. ΞΆ=0.7\zeta = 0.7 bo'lsa, Ο‰n\omega_n minimal?

Yechim

ts=4ΞΆΟ‰n<2t_s = \frac{4}{\zeta\omega_n} < 2

Ο‰n>4ΞΆβ‹…ts=40.7Γ—2=2.86\omega_n > \frac{4}{\zeta \cdot t_s} = \frac{4}{0.7 \times 2} = 2.86 rad/s


Masala 15 ⭐⭐⭐​

Kaskadli PID: Tashqi (pozitsiya) Kp1=2K_{p1} = 2, ichki (tezlik) Kp2=5K_{p2} = 5. Pozitsiya xatoligi 0.5 m, tezlik xatoligi 0.1 m/s. Kuch chiqishi?

Yechim

Tashqi: vref=Kp1β‹…epos=2Γ—0.5=1v_{ref} = K_{p1} \cdot e_{pos} = 2 \times 0.5 = 1 m/s

Ichki: F=Kp2β‹…evel=Kp2β‹…(vrefβˆ’vactual)F = K_{p2} \cdot e_{vel} = K_{p2} \cdot (v_{ref} - v_{actual})

Agar vactualv_{actual} berilmagan, evel=0.1e_{vel} = 0.1: F=5Γ—0.1=0.5F = 5 \times 0.1 = 0.5 N


Masala 16 ⭐⭐⭐​

Derivative filter: Ο„f=0.01\tau_f = 0.01 s, raw derivative = 100. Filtrlangan qiymat (oldingi = 50)?

Yechim

Ξ±=dtdt+Ο„f\alpha = \frac{dt}{dt + \tau_f} (dt = 0.01 deb faraz qilsak)

Ξ±=0.010.02=0.5\alpha = \frac{0.01}{0.02} = 0.5

Dfiltered=Ξ±β‹…Draw+(1βˆ’Ξ±)β‹…DprevD_{filtered} = \alpha \cdot D_{raw} + (1-\alpha) \cdot D_{prev} =0.5Γ—100+0.5Γ—50=75= 0.5 \times 100 + 0.5 \times 50 = 75


Masala 17 ⭐⭐⭐​

Quadcopter roll: Kp=5K_p = 5, Kd=0.5K_d = 0.5. Ο•ref=0\phi_{ref} = 0, Ο•=10Β°\phi = 10Β°, Ο•Λ™=20Β°/s\dot{\phi} = 20Β°/s. Differential thrust?

Yechim

e=0βˆ’10=βˆ’10Β°e = 0 - 10 = -10Β° eΛ™=0βˆ’20=βˆ’20Β°/s\dot{e} = 0 - 20 = -20Β°/s

u=Kpβ‹…e+Kdβ‹…eΛ™=5(βˆ’10)+0.5(βˆ’20)=βˆ’50βˆ’10=βˆ’60u = K_p \cdot e + K_d \cdot \dot{e} = 5(-10) + 0.5(-20) = -50 - 10 = -60

(Chapga burish uchun thrust farqi -60 birlik)


Masala 18 ⭐⭐⭐​

Position vs Velocity form PID. Velocity form Ξ”u\Delta u ni toping: Kp=2K_p = 2, Ki=1K_i = 1, Kd=0.1K_d = 0.1. e[n]=5e[n] = 5, e[nβˆ’1]=6e[n-1] = 6, e[nβˆ’2]=7e[n-2] = 7.

Yechim

Velocity form: Ξ”u=Kp(enβˆ’enβˆ’1)+Kiβ‹…en+Kd(enβˆ’2enβˆ’1+enβˆ’2)\Delta u = K_p(e_n - e_{n-1}) + K_i \cdot e_n + K_d(e_n - 2e_{n-1} + e_{n-2})

=2(5βˆ’6)+1(5)+0.1(5βˆ’12+7)= 2(5-6) + 1(5) + 0.1(5 - 12 + 7) =βˆ’2+5+0=3= -2 + 5 + 0 = 3


Masala 19 ⭐⭐⭐​

Bumpless transfer: Manual mode u = 50, Auto ga o'tishda integral = ?

Yechim

uauto=Kpβ‹…e+integralu_{auto} = K_p \cdot e + integral

Transfer paytida uauto=umanualu_{auto} = u_{manual}:

integral=umanualβˆ’Kpβ‹…e=50βˆ’Kpβ‹…eintegral = u_{manual} - K_p \cdot e = 50 - K_p \cdot e

Agar e=0e = 0 (setpoint = measured): integral=50integral = 50


Masala 20 ⭐⭐⭐​

Feed-forward + PID: Reference = 100, Kff=0.8K_{ff} = 0.8, PID chiqishi = 5. Umumiy chiqish?

Yechim

u=Kffβ‹…ref+uPID=0.8Γ—100+5=85u = K_{ff} \cdot ref + u_{PID} = 0.8 \times 100 + 5 = 85


Masala 21 ⭐⭐⭐​

Dead-band: |e| < 0.5 bo'lganda PID o'chiriladi. e = 0.3, Kp=10K_p = 10. Chiqish?

Yechim

∣e∣=0.3<0.5|e| = 0.3 < 0.5 β†’ Dead-band ichida

u=0u = 0 (yoki oxirgi qiymat saqlanadi)


Masala 22 ⭐⭐⭐​

Gain scheduling: v<5v < 5 m/s β†’ Kp=2K_p = 2, vβ‰₯5v \geq 5 β†’ Kp=1K_p = 1. v = 4 m/s, e = 3. Chiqish?

Yechim

v=4<5v = 4 < 5 β†’ Kp=2K_p = 2

u=Kpβ‹…e=2Γ—3=6u = K_p \cdot e = 2 \times 3 = 6


Masala 23 ⭐⭐⭐​

Rate limiter: chiqish o'zgarishi max 10 birlik/s. Oldingi u = 50, yangi u = 80, dt = 0.5s. Haqiqiy chiqish?

Yechim

Maksimal o'zgarish: 10Γ—0.5=510 \times 0.5 = 5

Kerakli o'zgarish: 80βˆ’50=3080 - 50 = 30

30 > 5 β†’ cheklangan

uactual=50+5=55u_{actual} = 50 + 5 = 55


Masala 24 ⭐⭐⭐​

Setpoint weighting: b=0.5b = 0.5 (P uchun). r=100r = 100, y=60y = 60, Kp=2K_p = 2. P komponenti?

Yechim

eP=bβ‹…rβˆ’y=0.5Γ—100βˆ’60=βˆ’10e_P = b \cdot r - y = 0.5 \times 100 - 60 = -10

uP=Kpβ‹…eP=2Γ—(βˆ’10)=βˆ’20u_P = K_p \cdot e_P = 2 \times (-10) = -20

(Oddiy PID: e=rβˆ’y=40e = r - y = 40, uP=80u_P = 80)


Masala 25 ⭐⭐⭐​

Back-calculation anti-windup: u=120u = 120, usat=100u_{sat} = 100, Tt=0.5T_t = 0.5, dt = 0.01. Integral tuzatish?

Yechim

Ξ”integral=dtTt(usatβˆ’u)=0.010.5(100βˆ’120)=0.02Γ—(βˆ’20)=βˆ’0.4\Delta integral = \frac{dt}{T_t}(u_{sat} - u) = \frac{0.01}{0.5}(100 - 120) = 0.02 \times (-20) = -0.4

Integral 0.4 ga kamayadi har qadamda.


Murakkab Masalalar (26-35)​

Masala 26 ⭐⭐⭐⭐​

Ziegler-Nichols reaction curve: L = 0.5 s (delay), T = 2 s (time constant). PID?

Yechim

Kp=1.2TL=1.2Γ—20.5=4.8K_p = \frac{1.2T}{L} = \frac{1.2 \times 2}{0.5} = 4.8

Ti=2L=1T_i = 2L = 1 s β†’ Ki=KpTi=4.8K_i = \frac{K_p}{T_i} = 4.8

Td=0.5L=0.25T_d = 0.5L = 0.25 s β†’ Kd=Kpβ‹…Td=1.2K_d = K_p \cdot T_d = 1.2


Masala 27 ⭐⭐⭐⭐​

Auto-tuning relay: Relay amplitude d = 5, oscillation amplitude a = 2, period TuT_u = 1 s. KuK_u va PID?

Yechim

Ku=4dΟ€a=4Γ—53.14Γ—2=206.28=3.18K_u = \frac{4d}{\pi a} = \frac{4 \times 5}{3.14 \times 2} = \frac{20}{6.28} = 3.18

Kp=0.6Ku=1.91K_p = 0.6 K_u = 1.91 Ti=0.5T_i = 0.5 s β†’ Ki=3.82K_i = 3.82 Td=0.125T_d = 0.125 s β†’ Kd=0.24K_d = 0.24


Masala 28 ⭐⭐⭐⭐​

SIMC tuning: K=2K = 2, Ο„=5\tau = 5 s, ΞΈ=1\theta = 1 s (FOPDT model). Closed-loop Ο„c=2\tau_c = 2 s uchun PI?

Yechim

SIMC PI: Kp=Ο„K(Ο„c+ΞΈ)=52(2+1)=56=0.83K_p = \frac{\tau}{K(\tau_c + \theta)} = \frac{5}{2(2 + 1)} = \frac{5}{6} = 0.83

Ti=min⁑(Ο„,4(Ο„c+ΞΈ))=min⁑(5,12)=5T_i = \min(\tau, 4(\tau_c + \theta)) = \min(5, 12) = 5 s

Ki=KpTi=0.835=0.17K_i = \frac{K_p}{T_i} = \frac{0.83}{5} = 0.17


Masala 29 ⭐⭐⭐⭐​

Lambda tuning: Plant = 25s+1eβˆ’s\frac{2}{5s+1}e^{-s}. Lambda = 3 s uchun PI?

Yechim

K=2K = 2, Ο„=5\tau = 5, ΞΈ=1\theta = 1

Kp=Ο„K(Ξ»+ΞΈ)=52(3+1)=0.625K_p = \frac{\tau}{K(\lambda + \theta)} = \frac{5}{2(3 + 1)} = 0.625

Ti=Ο„=5T_i = \tau = 5 s

Ki=0.125K_i = 0.125


Masala 30 ⭐⭐⭐⭐​

2-DOF PID: Kp=2K_p = 2, b=0.5b = 0.5 (reference weight), c=0c = 0 (derivative weight). Step response overshoot kamayadi, isbotlang.

Yechim

Oddiy PID: Reference step β†’ katta P signal β†’ overshoot

2-DOF: eP=brβˆ’ye_P = br - y, eD=crβˆ’ye_D = cr - y

b=0.5b = 0.5: P ga reference faqat 50% ta'sir qiladi c=0c = 0: D faqat output ga javob beradi

Step paytida:

  • P komponenti kichikroq (b < 1)
  • D smooth (c = 0, reference derivative yo'q)

Natija: Kamroq overshoot, smooth javob.


Masala 31 ⭐⭐⭐⭐​

IMC-PID: Plant G(s)=32s+1G(s) = \frac{3}{2s+1}, filter Ξ»=1\lambda = 1 s. PID parametrlari?

Yechim

IMC controller: Q(s)=Gβˆ’1(s)1+Ξ»s=2s+13(1+s)Q(s) = \frac{G^{-1}(s)}{1 + \lambda s} = \frac{2s+1}{3(1+s)}

PID approximation: Kp=Ο„KΞ»=23Γ—1=0.67K_p = \frac{\tau}{K\lambda} = \frac{2}{3 \times 1} = 0.67

Ti=Ο„=2T_i = \tau = 2 s

Td=0T_d = 0 (FOPDT uchun)


Masala 32 ⭐⭐⭐⭐​

Discrete PID - Tustin transform: KdsK_d s β†’ diskret. dt = 0.01 s.

Yechim

Tustin (bilinear): s=2dtzβˆ’1z+1s = \frac{2}{dt}\frac{z-1}{z+1}

Kdsβ†’Kd2dtzβˆ’1z+1=200Kd(zβˆ’1)z+1K_d s \to K_d \frac{2}{dt}\frac{z-1}{z+1} = \frac{200K_d(z-1)}{z+1}

Differentsiya tenglamasi: D[n]=βˆ’D[nβˆ’1]+200Kd(e[n]βˆ’e[nβˆ’1])D[n] = -D[n-1] + 200K_d(e[n] - e[n-1])


Masala 33 ⭐⭐⭐⭐​

Model Reference Adaptive PID: Reference model Ο‰n=10\omega_n = 10, ΞΆ=0.7\zeta = 0.7. Actual response Ο‰=8\omega = 8, ΞΆ=0.5\zeta = 0.5. Gains qaysi yo'nalishda o'zgartiriladi?

Yechim

Ο‰\omega past (8 < 10):

  • Bandwidth kam β†’ KpK_p va/yoki KiK_i oshirish

ΞΆ\zeta past (0.5 < 0.7):

  • Damping kam, overshoot ko'p β†’ KdK_d oshirish yoki KpK_p kamaytirish

Optimal: Kd↑K_d ↑, KpK_p biroz ↑↑


Masala 34 ⭐⭐⭐⭐​

Windup example: Setpoint = 100, output stuck at 50 (physical limit). Kp=1K_p = 1, Ki=0.5K_i = 0.5, dt = 0.1 s. 10 s da integral?

Yechim

e=100βˆ’50=50e = 100 - 50 = 50 (doimiy)

Integralsiz: integral=Kiβˆ‘eβ‹…dt=0.5Γ—50Γ—10=250integral = K_i \sum e \cdot dt = 0.5 \times 50 \times 10 = 250

Bu katta windup! Anti-windup kerak.


Masala 35 ⭐⭐⭐⭐​

Parallel vs Ideal PID: Parallel Kp=2K_p = 2, Ki=1K_i = 1, Kd=0.5K_d = 0.5. Ideal formaga o'tkazing.

Yechim

Parallel: u=Kpe+Ki∫e+KdeΛ™u = K_p e + K_i \int e + K_d \dot{e}

Ideal: u=Kp(e+1Ti∫e+TdeΛ™)u = K_p(e + \frac{1}{T_i}\int e + T_d \dot{e})

Kp=2K_p = 2 (bir xil)

Ti=KpKi=21=2T_i = \frac{K_p}{K_i} = \frac{2}{1} = 2 s

Td=KdKp=0.52=0.25T_d = \frac{K_d}{K_p} = \frac{0.5}{2} = 0.25 s


βœ… Tekshirish Ro'yxati​

  • 1-12: Asosiy PID
  • 13-25: Tuning va amaliy
  • 26-35: Murakkab usullar

Keyingi Qadam​

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