Worked examples — Control system fundamentals — plant, actuator, sensor, controller

3.5.26 · D3 · Physics › Guidance, Navigation & Control (GNC) › Control system fundamentals — plant, actuator, sensor, contr

Prerequisites jinpe hum rely karte hain (sab parent topic mein bane hain): Transfer Functions and Laplace Domain, Poles Zeros and Stability, Second-order System Response, PID Control, aur Actuator Saturation and Anti-Windup.

Scenario matrix

Kuch bhi work karne se pehle, chaliye har distinct case class ki list banate hain jo ek closed-loop problem de sakti hai. Har row ek problem ka "shape" hai; neeche har worked example ko us cell ke saath tag kiya gaya hai jise woh fill karta hai.

Neeche har numeric answer machine-checked verify block mein hai.

Forecast: Pehle guess karo — ek plant pole aur positive gain ke saath, kya aap expect karte hain ki yeh blow up hoga ya settle hoga? Step ka error zero hoga ya kuch leftover rahega?

1. Forward path $G=CAP$. Yeh step kyun? Cascaded Laplace blocks multiply karte hain, isliye poori forward chain ek product hai. $G(s)=2\cdot 1\cdot \frac{4}{s+1}=\frac{8}{s+1}$

2. Boxed formula apply karo. Kyun? $H=1$, isliye $T=\frac{G}{1+G}$. $T(s)=\frac{8/(s+1)}{1+8/(s+1)}=\frac{8}{(s+1)+8}=\frac{8}{s+9}$

3. Pole padho. Kyun? Denominator ka root closed-loop pole hai; uska sign stability decide karta hai. $s+9=0\Rightarrow s=-9(<0)\Rightarrow \text{stable}$

4. Step ke liye steady-state error. Kyun? $e_{ss}=\frac{1}{1+G(0)}$, valid hai kyunki loop stable hai (final value theorem apply hota hai). $G(0)=\frac{8}{0+1}=8,e_{ss}=\frac{1}{1+8}=\frac{1}{9}\approx 0.111$

Verify: Pole $-1$ (open loop) se $-9$ (closed loop) par move hua — feedback ne use speed up kiya, jo "zyada gain, faster" se match karta hai. Error chhota hai but nonzero hai kyunki koi integrator nahi hai. ✓

Forecast: Ek value slow no-overshoot motion deti hai, doosri snappy-but-ringing. Guess karo aage padhne se pehle ki kaun sa $K$ zyada ring karta hai.

1. Closed loop. Kyun? Same procedure jaise parent ka Ex 1. $T(s)=\frac{K}{s(s+2)+K}=\frac{K}{s^2+2s+K}$

2. Standard second-order form se match karo. Kyun? $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ se compare karne par (dekho Second-order System Response) damping seedha padh sakte hain. ${\omega }_n=\sqrt{K},2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{1}{\sqrt{K}}$

3. $K=1$ plug karo. Kyun? Low-gain corner test karo. ${\omega }_n=1,\zeta =1\Rightarrow \text{critically damped, no overshoot.}$

4. $K=16$ plug karo. Kyun? High-gain corner test karo. ${\omega }_n=4,\zeta =\frac{1}{4}=0.25\Rightarrow \text{underdamped, big overshoot.}$

Figure dekho: jaise $K$ badhta hai, dono poles real axis se slide karke complex plane mein upar jaate hain — woh vertical climb hi ringing hai.

5. $K=16$ ke liye overshoot. Kyun? Percent overshoot sirf $\zeta$ par depend karta hai: $M_p=e^{-\pi \zeta /\sqrt{1-{\zeta }^2}}$. $M_p=e^{-\pi (0.25)/\sqrt{1-0.0625}}=e^{-0.811}\approx 0.444=44.4\%$

Verify: $\zeta =1$ exactly overshoot aur no-overshoot ke boundary par baith ta hai — consistent hai $K=1$ ke critically damped hone se. $K$ ko 1 se 16 tak badhane par ${\omega }_n$ chaar guna ho gaya aur $\zeta$ chauthai ho gayi. ✓

Forecast: Feedback mein minus sign ka matlab hai "aap target ki taraf jaane ke bajaye usse door jaate ho." Guess karo pole zero ke left mein aayega ya right mein.

1. Forward path. Kyun? Same product, ab negative $K$ ke saath. $G(s)=\frac{-8}{s+1}$

2. Closed loop. Kyun? Formula apply karo aur dekho sign denominator mein kaise land karta hai. $T(s)=\frac{-8/(s+1)}{1-8/(s+1)}=\frac{-8}{(s+1)-8}=\frac{-8}{s-7}$

3. Pole. Kyun? $s-7=0$ ka root. $s=+7(>0)\Rightarrow \text{UNSTABLE — right-half-plane pole.}$

Verify: Open-loop pole $-1$ par tha; positive feedback ne use right push kiya imaginary axis ke paar. Physically: ek error controller ko galat direction mein push karata hai, error ko grow karta hai — exponential blow-up $e^{7t}$. Isliye sensor/actuator par sign convention life-or-death hai. ✓ (Ex 1 se compare karo — same plant, sahi sign → pole at $-9$.)

Forecast: Denominator mein ek akela $s$ hai ($s=0$ par ek pole). Guess karo step error finite hoga ya exactly zero.

1. Forward path. Kyun? Multiply karo. $G(s)=\frac{18}{s(s+4)}$

2. DC gain. Kyun? Step error ko $G(0)$ chahiye. $G(0)={\lim }_{s\to 0}\frac{18}{s(s+4)}=+\infty (\text{kyunki }1/s\text{ hai})$

3. Steady-state error. Kyun? $e_{ss}=\frac{1}{1+G(0)}$. $e_{ss}=\frac{1}{1+\infty }=0$

Verify: $s=0$ par ek pole (ek "free integrator") ka matlab hai loop error accumulate karta rehta hai jab tak woh exactly wipe out na ho jaaye. Yahi reason hai ki PID Control mein "I" steady-state error khatam karta hai — woh deliberately $1/s$ insert karta hai. Ex 1 se compare karo (koi integrator nahi → $e_{ss}=1/9\ne 0$). ✓

Forecast: Agar sensor aadha report karta hai, toh kya controller bahut jaldi ruk jaayega ya true target ko overshoot karega? Galti ki direction guess karo.

1. $H$ ke saath closed loop. Kyun? Ab full $T=\frac{G}{1+GH}$ use karo. $T(s)=\frac{10/(s+2)}{1+0.5\cdot 10/(s+2)}=\frac{10}{(s+2)+5}=\frac{10}{s+7}$

2. $y$ ki final value. Kyun? $y_{ss}={\lim }_{s\to 0}sT(s)R(s)=T(0)$ unit step ke liye. $y_{ss}=T(0)=\frac{10}{7}\approx 1.429$

3. Interpret karo. Kyun? Loop measured signal $y_m=Hy$ ko reference ki taraf drive karta hai, $y$ ko nahi. $y_m=0.5\times 1.429=0.714\text{(1 ke kareeb, lekin true }y\text{ ne 1.43 tak overshoot kiya)}$

Verify: Kyunki sensor reading half kar deta hai, controller ko true output double karni padti hai taaki measurement match ho — isliye true $y$ target se upar end hoti hai. "1.0" command ne physical 1.43 produce kiya. Yahi woh galti hai jiske baare mein parent warn karta hai: ek "accurate-looking" lekin mis-scaled $H$ loop ko silently bias karta hai. ✓

Forecast: Ek limit loop ko "off" kar deta hai, doosra usse utna hard crank karta hai jitna physics allow kare. Guess karo poles har extreme par kya karte hain.

1. $K\to 0^{+}$. Kyun? Controller ko kuch bhi nahi par le aao. $s^2+2s+0=s(s+2)=0\Rightarrow s=0,-2$ Poles open-loop locations par wapas aa jaate hain — $0$ par pole ke saath, output track nahi karta (loop effectively open ho gaya).

2. $K\to \infty$. Kyun? Gain ko ceiling tak push karo. ${\omega }_n=\sqrt{K}\to \infty ,\zeta =\frac{1}{\sqrt{K}}\to 0$ Undamped oscillation: poles imaginary direction mein rocket karte hain, $M_p\to 100\%$. Response forever ring karta hai (instability ki taraf marginally).

3. Middle ground. Kyun? Best trade $\zeta =1/\sqrt{2}\approx 0.707$ par. $\frac{1}{\sqrt{K}}=\frac{1}{\sqrt{2}}\Rightarrow K=2$

Verify: Jaise $K$, $0\to \infty$ run karta hai, $\zeta$ monotonically $\infty \to 0$ run karta hai — har intermediate damping exactly ek baar hit hota hai, aur $K=2$ textbook "nicest" damping deta hai. Ex 2 ke consistent ($K=1$: $\zeta =1$; $K=16$: $\zeta =0.25$). ✓

Forecast: Plant mein phir ek integrator hai. Guess karo ki hover height exactly 10 m par lock hogi ya drift karegi.

1. Words → blocks translate karo. Kyun? Reference $r=10$ (height 10 ka ek step), $R(s)=10/s$.

2. DC gain & step error. Kyun? Integrator ⇒ $G(0)=\infty$ ⇒ zero error (Case D logic). $G(s)=\frac{2K}{s(s+3)},G(0)=\infty \Rightarrow e_{ss}=0\Rightarrow \text{exactly 10 m hold karta hai.}$

3. $K=4.5$ ke liye damping. Kyun? Loop close karo aur second-order form se match karo. $T(s)=\frac{2K}{s^2+3s+2K}\Rightarrow {\omega }_n=\sqrt{2K},2\zeta {\omega }_n=3\Rightarrow \zeta =\frac{3}{2\sqrt{2K}}$ $K=4.5:{\omega }_n=\sqrt{9}=3,\zeta =\frac{3}{2\cdot 3}=0.5$

Verify: $\zeta =0.5<1$ → underdamped: drone 10 m se upar jaata hai, dip karta hai, phir exactly 10 m par settle karta hai (zero steady error integrator ki wajah se). Units: ${\omega }_n$ rad/s mein, $\zeta$ dimensionless. ✓

Forecast: Integrator add karne se zero steady error milta hai — lekin integrators destabilizing hote hain. Guess karo ki $K$ ki koi upper limit hai ya nahi.

1. Characteristic equation. Kyun? Stability $1+GH=0$ ke roots mein rehti hai. $G=\frac{K}{s(s+1)(s+2)},H=1\Rightarrow 1+\frac{K}{s(s+1)(s+2)}=0$ $\Rightarrow s(s+1)(s+2)+K=0\Rightarrow s^3+3s^2+2s+K=0$

2. $s^3+3s^2+2s+K$ ke liye Routh array. Kyun? Routh–Hurwitz cubic solve kiye bina sign changes = right-half poles batata hai.

3. Sab first-column entries $>0$ honi chahiye. Kyun? Koi bhi sign change ⇒ right-half-plane pole ⇒ instability. $K>0\text{aur}\frac{6-K}{3}>0\Rightarrow K<6$ $\boxed{0<K<6}$

4. Boundary check $K=6$. Kyun? Edge par $s^1$ row vanish ho jaati hai → poles imaginary axis par baith jaate hain (sustained oscillation). $s^3+3s^2+2s+6=(s+3)(s^2+2)\Rightarrow s=-3,\pm j\sqrt{2}$ Purely imaginary pair ⇒ marginal, $\omega =\sqrt{2}$ rad/s par oscillate karta hai.

Verify: $K=6$ par exactly $\pm j\sqrt{2}$ milta hai (marginal) aur ek stable real pole $-3$ par; $K$ ke 6 se thoda neeche hone par woh complex poles left move karte hain (stable), 6 se thoda upar hone par right cross karte hain (unstable). Isliye $0<K<6$ poori stable window hai. ✓

Active recall