Hardware-in-the-Loop (HIL) simulation — real hardware, simulated plant: Exercises

5.5.21 · D4 · Coding › Embedded Systems & Real-Time Software › Hardware-in-the-Loop (HIL) simulation — real hardware, simul

Intuition Is page ko kaise use karein

Ye problems ek ladder climb karti hain. L1 Recognition sirf check karta hai ki tum sahi idea spot kar sako. L5 Mastery tak aate-aate tum sab kuch ek full test plan mein jodoge aur woh timing arithmetic karoge jo decide karti hai ki ek real HIL rig kaam bhi karega ya nahi. Har solution ek collapsible box ke andar chhupa hua hai — pehle khud try karo, phir reveal karo. Numbers sab machine-checked hain.

Prerequisite ideas parent mein hain: Hardware-in-the-Loop (HIL). Jab koi neighbour concept aata hai to hum use link karte hain taaki tum branch kar sako: Software-in-the-Loop (SIL), Model-in-the-Loop (MIL), Processor-in-the-Loop (PIL), Real-time Operating Systems (RTOS), PID Control, PWM (Pulse Width Modulation), CAN Bus Protocol, Fault Injection Testing, FPGA for Real-Time Simulation।

Notation jo hum baar-baar use karenge (ek baar yahan define ki gayi hai)

Kisi bhi problem se pehle, aao har symbol ko earn karein taaki baad mein kuch bhi surprise na kare.

Definition Is page ke symbols

$Δ t$ — simulation time step: woh fixed slice of pretend-time jitni simulator har cycle mein physics ko aage badhata hai। Agar $Δ t = 1 ms$ hai to simulator real time ke har millisecond mein duniya ko redraw karta hai।
$t_{compute}$ — wall-clock time jo simulator actually ek step crunch karne mein spend karta hai। Real stopwatch se measure hota hai।
$f = 1/Δ t$ — update rate (hertz mein, cycles per second)। $Δ t = 1 ms \Rightarrow f = 1000 Hz = 1 kHz$ ।
$D$ — PWM duty cycle, $0$ se $1$ tak ka ek fraction। Ye woh slice hai jitni har on/off cycle mein signal "on" rehta hai। Dekho PWM (Pulse Width Modulation)।
$V_{batt}$ — battery voltage। Applied motor voltage hai $V = D \cdot V_{batt}$ ।
$ω$ — angular velocity (rad/s), motor shaft kitni tezi se ghoomta hai।
RPM — revolutions per minute। Conversion: $RPM = ω \cdot \frac{60}{2 π}$ ।
$J, b, K_{t}, K_{e}, R$ — motor constants (inertia, damping, torque constant, back-EMF constant, resistance), ye sab parent note ke DC-motor model se hain।

L1 — Recognition

Exercise 1.1 (L1)

In mein se kaunsa HIL ki defining feature hai jo ise Software-in-the-Loop (SIL) se alag karti hai? (a) Plant simulated hota hai। (b) Controller real target hardware par run karta hai। (c) Ek PID loop use hoti hai। (d) Test automated hota hai।

Recall Solution

(b). SIL aur HIL dono mein plant simulated hota hai (isliye (a) dono par true hai — distinguishing feature nahi hai)। PID aur automation kai test styles mein aate hain। HIL ki unique signature ye hai ki production ECU/microcontroller real firmware execute karta hai, real electrical signals exchange karte hue। Ladder compare karo: Model-in-the-Loop (MIL) (sab model), Software-in-the-Loop (SIL) (code on PC), Processor-in-the-Loop (PIL) (target CPU par code, real I/O nahi), phir HIL (target + real I/O)।

Exercise 1.2 (L1)

Ek HIL simulator $f = 2 kHz$ par run karta hai। Milliseconds mein uska time step $Δ t$ kya hai?

Recall Solution

$Δ t = \frac{1}{f} = \frac{1}{2000 Hz} = 0.0005 s = 0.5 ms .$

L2 — Application

Exercise 2.1 (L2)

ECU $12 V$ battery par ek motor mein duty cycle $D = 0.7$ command karta hai। HIL plant model kaunsa voltage $V$ apply karta hai?

Recall Solution

$V = D \cdot V_{batt} = 0.7 \times 12 = 8.4 V .$ Kyun: PWM ka average delivered voltage duty cycle times supply hota hai — ye exactly woh pehla step hai jo simulator ECU ki PWM pin read karne par karta hai।

Exercise 2.2 (L2)

Aage jaari rakhte hain: motor resistance $R = 2 Ω$ , back-EMF constant $K_{e} = 0.05 V\cdots/rad$ , current shaft speed $ω = 20 rad/s$ । Quasi-static current equation $I \approx (V - K_{e} ω) / R$ use karke current $I$ nikalo।

Recall Solution

$I = \frac{V - K _{e} ω}{R} = \frac{8.4 - ( 0.05 ) ( 20 )}{2} = \frac{8.4 - 1.0}{2} = \frac{7.4}{2} = 3.7 A .$ Kyun quasi-static: humne $L d I / d t$ inductance term drop ki kyunki wo $R I$ ke comparison mein tiny hai, jisse simulator $Δ t$ ke andar finish ho sakta hai।

Exercise 2.3 (L2)

Torque constant $K_{t} = 0.05 N\cdotm/A$ । Woh current kaunsa motor torque $τ_{motor}$ produce karta hai?

Recall Solution

$τ_{motor} = K_{t} \cdot I = 0.05 \times 3.7 = 0.185 N\cdotm .$

L3 — Analysis

Exercise 3.1 (L3)

Upar se $τ_{motor} = 0.185 N\cdotm$ lo। Load torque $τ_{load} = 0.05 N\cdotm$ , damping $b = 0.001 N\cdotm\cdots/rad$ , inertia $J = 0.01 kg\cdotm^{2}$ , current $ω (t_{k}) = 20 rad/s$ , step $Δ t = 1 ms$ । $ω (t_{k + 1})$ nikalne ke liye ek Euler update karo।

Recall Solution — figure follow karo

Parent note se Euler update: $ω (t_{k + 1}) = ω (t_{k}) + \frac{Δ t}{J} (τ_{motor} - τ_{load} - b ω (t_{k})) .$ Net torque: $τ_{net} = 0.185 - 0.05 - (0.001) (20) = 0.185 - 0.05 - 0.02 = 0.115 N\cdotm .$ Acceleration $\times$ step: $\frac{Δ t}{J} τ_{net} = \frac{0.001}{0.01} \times 0.115 = 0.1 \times 0.115 = 0.0115 rad/s .$ $ω (t_{k + 1}) = 20 + 0.0115 = 20.0115 rad/s .$ Kaisa dikhta hai: figure mein shaft speed thodi upar nudge hoti hai kyunki net torque positive hai — ek $Δ t$ par woh tiny slope segment।

Exercise 3.2 (L3)

$ω (t_{k}) = 20 rad/s$ aur updated $ω (t_{k + 1}) = 20.0115 rad/s$ dono ko RPM mein convert karo।

Recall Solution

$RPM = ω \cdot \frac{60}{2 π} .$ $20 \cdot \frac{60}{2 π} = \frac{1200}{2 π} \approx 190.986 RPM .$ $20.0115 \cdot \frac{60}{2 π} \approx 191.096 RPM .$

Exercise 3.3 (L3)

Ek step mein $Δ t = 1 ms$ par $t_{compute} = 0.85 ms$ laga। Budget ka kaunna fraction use hua, aur kya rig real-time-valid hai?

Recall Solution

$utilization = \frac{t _{compute}}{Δ t} = \frac{0.85}{1.0} = 0.85 = 85%.$ Kyunki $0.85 ms < 1 ms$ hai, law $t_{compute} < Δ t$ hold karta hai — valid hai, lekin sirf $15%$ headroom ke saath। Kaafi close hai: ek slower branch (jaise ek fault-injection path) deadline blow kar sakti hai। Dekho Real-time Operating Systems (RTOS)।

L4 — Synthesis

Exercise 4.1 (L4)

Ek engine-idle HIL test ko plant $Δ t = 0.5 ms$ par run karna hai। Har step mein teen tasks run hoti hain: combustion model $0.18 ms$ , crankshaft dynamics $0.09 ms$ , aur I/O generation (encoder pulses + CAN Bus Protocol frames) $0.15 ms$ । (a) Total compute? (b) Valid hai? (c) Is workload ke saath maximum update rate $f_{m a x}$ kya hai?

Recall Solution

(a) Total: $t_{compute} = 0.18 + 0.09 + 0.15 = 0.42 ms .$ (b) $0.42 < 0.5$ → valid, headroom $0.08 ms$ ( $16%$ )। (c) Sabse fast step jo tum sustain kar sakte ho usmein $Δ t = t_{compute}$ hoga, isliye $f_{m a x} = \frac{1}{0.42 ms} = \frac{1}{0.00042 s} \approx 2380.95 Hz \approx 2.38 kHz .$ Isse upar, compute step overrun karta hai aur ECU frozen sensors dekhta hai।

Exercise 4.2 (L4)

Words mein ek fault-injection test plan sketch karo (+ figure reference karo) jo verify kare ki ECU safely limp karta hai jab ek throttle-position sensor open-circuit read kare। Closed-loop cycle aur PASS criterion list karo।

Recall Solution — loop figure dekho

Plan (ye Fault Injection Testing hai HIL par layered):

Plant model mein engine ko steady $800 RPM$ idle par lao।
Fault inject karo: I/O box ke DAC ko command karo ki TPS line ko $0 V$ par drive kare (ya float kare) — ek broken sensor wire simulate karte hue।
ECU ka diagnostic ek implausible/out-of-range TPS read karta hai।
ECU ko ek fault code set karke limp-home mein enter karna chahiye (fixed safe throttle map)।
Plant respond karta hai; engine speed aur throttle-motor PWM measure karo। PASS criterion: injection ke $500 ms$ ke andar ECU (a) diagnostic flag raise kare aur (b) engine speed ek safe band mein rakhe (na stall, na runaway)। Figure closed loop ko red fault-injection tap ke saath dikhata hai jo sensor path mein cut karta hai। Kyun SIL nahi balki HIL: fault electrical signal path mein rehta hai (open circuit, wrong voltage)। Sirf real I/O hardware hi wo reproduce kar sakta hai; ek pure-software Software-in-the-Loop (SIL) rig wire nahi tod sakta।

L5 — Mastery

Exercise 5.1 (L5)

Full idle-control loop। Start: $ω_{0} = 0 rad/s$ । Motor params: $R = 2 Ω$ , $K_{e} = K_{t} = 0.05$ , $J = 0.01$ , $b = 0.001$ , $τ_{load} = 0.05 N\cdotm$ , $V_{batt} = 12 V$ , $Δ t = 1 ms$ । Ek PID Control loop sirf proportional term ke saath setpoint $ω^{⋆} = 20 rad/s$ ki taraf speed error se duty cycle drive karta hai: $D (t_{k}) = clip (K_{p} (ω^{⋆} - ω (t_{k})), 0, 1), K_{p} = 0.1 (per rad/s) .$ Pehle do Euler steps compute karo: har step par $D$ , $I$ , $τ_{motor}$ , $ω$ do। (Quasi-static current use karo, negative current ko $0$ par clip karo।)

Recall Solution — step by step

Step k = 0 ( $ω_{0} = 0$ ):

Error $= 20 - 0 = 20$ । $D = clip (0.1 \times 20, 0, 1) = clip (2.0, 0, 1) = 1.0$ (saturated)।
$V = 1.0 \times 12 = 12 V$ ।
$I = (12 - 0.05 \cdot 0) /2 = 6.0 A$ ।
$τ_{motor} = 0.05 \times 6.0 = 0.30 N\cdotm$ ।
$τ_{net} = 0.30 - 0.05 - 0.001 \cdot 0 = 0.25 N\cdotm$ ।
$ω_{1} = 0 + \frac{0.001}{0.01} (0.25) = 0.1 \times 0.25 = 0.025 rad/s$ ।

Step k = 1 ( $ω_{1} = 0.025$ ):

Error $= 20 - 0.025 = 19.975$ । $D = clip (0.1 \times 19.975, 0, 1) = 1.0$ (abhi bhi saturated)।
$V = 12 V$ ।
$I = (12 - 0.05 \cdot 0.025) /2 = (12 - 0.00125) /2 = 5.999375 A$ ।
$τ_{motor} = 0.05 \times 5.999375 = 0.29996875 N\cdotm$ ।
$τ_{net} = 0.29996875 - 0.05 - 0.001 \cdot 0.025 = 0.24994375 N\cdotm$ ।
$ω_{2} = 0.025 + 0.1 \times 0.24994375 = 0.025 + 0.024994375 = 0.049994375 rad/s$ ।

Isko padhna: shuruaat mein error bahut bada hai, isliye P-controller duty cycle ko ceiling par slam karta hai ( $D = 1$ , saturated)। Shaft rest se dheere-dheere upar crawl karta hai — realistic hai, kyunki inertia $J$ sudden speed ka resist karti hai। Sirf bahut baad (jab $ω \to 20$ ) $D$ rail se neeche aayega aur loop settle hoga।

Exercise 5.2 (L5)

Mastery synthesis question (concept + arithmetic): Tumhe ek stiff electrical sub-model (fast $L d I / d t$ dynamics) add karna hai jisko stability ke liye $Δ t \leq 0.1 ms$ chahiye, lekin tumhara current mechanical loop $Δ t = 1 ms$ par khush hai। Har electrical step $0.06 ms$ compute cost karta hai; mechanical step $0.30 ms$ cost karta hai lekin sirf har $1 ms$ mein run karna chahiye। (a) Agar sab kuch fast rate par chale, to us slice ka worst-case compute per $0.1 ms$ slice kya hoga jo mechanics bhi karta hai, aur kya ye valid hai? (b) Architectural fix aur ek hardware option batao।

Recall Solution

(a) Unlucky slice par dono run karte hain: $t_{compute} = 0.06 + 0.30 = 0.36 ms$ , lekin slice budget sirf $0.1 ms$ hai। Kyunki $0.36 > 0.1$ hai, ye $3.6 \times$ overrun karta hai — invalid। Sirf electrical slices bhi theek hain ( $0.06 < 0.1$ ); mechanical burst hi use maarta hai। (b) Multi-rate / multi-tasking simulation: electrical model ko fast $0.1 ms$ task mein aur mechanical model ko slower $1 ms$ task mein run karo, separate cores par ya hard-real-time scheduler (Real-time Operating Systems (RTOS)) se, taaki heavy mechanical work kabhi ek single fast slice mein na aaye। Hardware option: stiff, deterministic electrical ODE ko ek FPGA for Real-Time Simulation par offload karo jo parallel mein sub-microsecond rates par run kare, CPU ko free kare। Exactly isliye FPGAs high-end HIL rigs mein appear karte hain।

Recall Self-test — one-line recalls

SIL ke comparison mein HIL ki defining feature ::: controller real target hardware par real I/O ke saath run karta hai Real-time validity law ::: $t_{compute} < Δ t$ Max sustainable update rate ::: $f_{m a x} = 1/ t_{compute}$ rad/s ko RPM mein convert karo ::: $60/ (2 π)$ se multiply karo Stiff fast sub-model jo overrun kare uska fix ::: multi-rate scheduling aur/ya ek FPGA co-processor Back-EMF kyun subtract karni chahiye ::: current sirf leftover voltage $V - K_{e} ω$ feel karta hai