Worked examples — Block diagram algebra

3.5.28 · D3 · Physics › Guidance, Navigation & Control (GNC) › Block diagram algebra

Yeh page parent note ke reduction rules ka stress-test hai. Yahan hum koi nayi rules nahi seekhte — hum har tarah ke block diagram ko dhundhte hain jo duniya aap par uchaal sakti hai, aur har ek ko ek single transfer function tak grind karte hain. Agar exam mein koi diagram dekh ke lagey "yeh shape toh maine kabhi nahi dekhi," toh neeche ka matrix check karo: dekhi hai tumne.

Shuru karne se pehle, teen symbols ki ek reminder, seedhe alfazon mein taaki koi confused na ho:

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Scenario matrix

Har block-diagram problem inhi cells mein se ek (ya ek stack) hoti hai. Aakhri column us worked example ka naam deta hai jo ise nail karta hai.

Cell	Kya special hai isme	Danger	Nail karta hai
A. Pure series	blocks sirf multiply karte hain	kuch nahi — warm-up	Ex 1
B. Parallel, minus sign	summer branches ko subtract karta hai	sign slip	Ex 2
C. Negative feedback	standard loop	$+1$ bhool jaana	Ex 3
D. Positive feedback	denominator $1-GH$ ho jaata hai	galat sign → galat poles	Ex 4
E. Zero / degenerate gain	$H=0$, ya $G\to \infty$	formula tab bhi sahi rehna chahiye	Ex 5
F. Limiting behaviour	$s\to 0$ (DC) aur $s\to \infty$	kya loop stable / finite hai?	Ex 6
G. Nested loops (inside-out)	ek loop ke andar chupta hua loop	pehle outer reduce karna = galat	Ex 7
H. Equivalence move	take-off / summer ko block ke aage-peeche slide karna	toll $G$ ya $1/G$ bhool jaana	Ex 8
I. Real-world word problem	prose mein ek physical GNC loop	words ko blocks mein translate karna	Ex 9
J. Exam twist: unity feedback + cascade	$H=1$, numeric, "poles dhundho"	loop gain aur forward gain ko mix karna	Ex 10

Signs (+/-), zero inputs, aur dono limits ($s\to 0$, $s\to \infty$) mein se har ek kam se kam ek cell mein covered hai. Kuch bhi chhoot nahi raha.

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Cell A — Pure series (warm-up)

1. Seedha multiply karo. Signal ko har block par ek baar multiply kiya jaata hai: $Y=G_3(G_2(G_{1}R))=G_1{G}_2{G}_{3}R$. Yeh step kyun? Cascade matlab time-mein-convolution = $s$-mein-multiplication; multiplication associative hai isliye order aur grouping matter nahi karte.
2. Numbers daalo: $\frac{Y}{R}=\frac{5}{s}\cdot \frac{1}{s+2}\cdot 4=\frac{20}{s(s+2)}=\frac{20}{s^2+2s}$.

Verify: Gain ke units multiply hote hain, poora dimensionless. Sanity: teen blocks, ek fraction, koi $1+GH$ nahi kyunki koi return trip nahi hai. $s=1$ par: $\frac{20}{1\cdot 3}=\frac{20}{3}\approx 6.667$. ✔

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Cell B — Subtracting summer ke saath Parallel

1. Har branch usi $R$ par act karti hai (take-off par copying karne se signal nahi badalta): top $=G_{1}R$, bottom $=G_{2}R$. Yeh step kyun? Rule 2 — split dono branches ko identical copy feed karta hai.
2. Summer subtract karta hai: $Y=(G_1-G_2)R=\left(\frac{s}{s+1}-\frac{1}{s+1}\right)R=\frac{s-1}{s+1}R$. Yeh step kyun? Same denominator hai, toh numerators directly subtract karo.

Verify: $s=1$ par numerator $s-1=0$, toh $\frac{Y}{R}=0$ — dono branches wahin exactly cancel ho jaate hain, jo ek subtracting summer ka poora point hai. ✔ $s=3$ par: $\frac{2}{4}=0.5$. ✔

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Cell C — Negative feedback (standard loop)

1. Loop gain $=GH=\frac{10}{s+1}\cdot \frac{1}{2}=\frac{5}{s+1}$. Yeh step kyun? $GH$ woh number hai jisse signal loop ke ek full trip mein multiply hota hai — woh number jiske aage "$1+$" baithta hai.
2. Closed-loop formula apply karo $\frac{Y}{R}=\frac{G}{1+GH}=\frac{\frac{10}{s+1}}{1+\frac{5}{s+1}}$. Yeh step kyun? Feedback control loops — yeh Rule 3 hai, $Y=G(R-B)$ mein $B=HY$ substitute karke derive kiya gaya.
3. Fraction clear karo (top aur bottom ko $s+1$ se multiply karo): $\frac{Y}{R}=\frac{10}{(s+1)+5}=\frac{10}{s+6}$.

Verify: Open-loop pole $s=-1$ par tha; feedback ne ise $s=-6$ tak move kar diya — feedback ne pole ko left half-plane mein aur gehrayi mein kheench liya (tez, zyada stable), exactly yahi negative feedback karta hai. $s=0$ par (DC): $\frac{10}{6}\approx 1.667$. ✔

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Cell D — Positive feedback (sign palat jaata hai)

1. Positive-feedback formula use karo: $\frac{Y}{R}=\frac{G}{1-GH}=\frac{\frac{10}{s+1}}{1-\frac{5}{s+1}}$. Yeh step kyun? $E=R+B$ ke saath, $Y$ collect karne par $Y(1-GH)=GR$ milta hai — minus bake in ho jaata hai.
2. Clear karo: $\frac{Y}{R}=\frac{10}{(s+1)-5}=\frac{10}{s-4}$.

Verify: Pole ab $s=+4$ par hai — right half-plane mein. Matlab response unbounded grow karta hai: positive feedback yahan unstable hai. Ex 3 ka pole $-6$ par tha se compare karo. Isliye Stability & characteristic equation denominator $1\pm GH=0$ dekhta hai: sign zindagi ya maut decide karta hai loop ki. ✔

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Cell E — Zero / degenerate gain

1. Case (a), $H=0$: $\frac{Y}{R}=\frac{G}{1+G\cdot 0}=\frac{G}{1}=G=\frac{7}{s+3}$. Yeh step kyun? $H=0$ return path kaat deta hai — diagram ek single open-loop block mein degenerate ho jaata hai. Formula gracefully $G$ return karta hai.
2. Case (b), $H=1$ (unity feedback): $\frac{Y}{R}=\frac{G}{1+G}=\frac{\frac{7}{s+3}}{1+\frac{7}{s+3}}=\frac{7}{s+10}$. Yeh step kyun? Unity feedback sabse common special case hai; pole $-3\to -10$ move hota hai.
3. Case (c), $G\to \infty$: $\frac{Y}{R}=\frac{G}{1+GH}\stackrel{G\to \infty }{\to }\frac{1}{H}=\frac{1}{1/3}=3$. Yeh step kyun? Top aur bottom ko $G$ se divide karo: $\frac{1}{1/G+H}\to \frac{1}{H}$. Bahut bada open-loop gain ek clean $1/H$ ban jaata hai — yahi exact trick Op-amp gain ke peeche hai.

Verify: (a) $s=0$ par: $\frac{7}{3}\approx 2.333$. (b) $s=0$ par: $\frac{7}{10}=0.7$. (c) limit $=3$ exactly, $s$ se independent. ✔

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Cell F — Limiting behaviour ($s\to 0$ aur $s\to \infty$)

1. $s\to 0$ (DC): $s=0$ substitute karo: $\frac{20}{0+0+20}=1$. Yeh step kyun? DC par "frequency knob" off hai; yeh number steady-state gain hai — loop constant inputs ko gain 1 par pass karta hai.
2. $s\to \infty$: top degree 0 hai, bottom degree 2; bottom tezi se explode karta hai, toh $\frac{20}{s^2+4s+20}\to 0$. Yeh step kyun? Physical systems infinitely fast wiggles par respond nahi kar sakte — output khatam ho jaata hai. Yeh loop ka built-in low-pass character hai.

Verify: DC $=1$ exactly. Ek mid value $s=2$ par (size check ke liye real treat kiya): $\frac{20}{4+8+20}=\frac{20}{32}=0.625$, $1$ aur $0$ ke beech. ✔ Monotone squeeze confirmed.

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Cell G — Nested loops (inside-out)

1. Inner loop reduce karo: $G_{inner}=\frac{G_2}{1+G_2{H}_2}=\frac{1/s}{1+(1/s)(2)}=\frac{1}{s+2}$. Yeh step kyun? Outer feedback formula ko forward path mein ek single block chahiye; hum ise ek hidden loop ke across apply nahi kar sakte. Inside-out kaam karo.
2. $G_1$ ke saath series: forward $=G_1{G}_{inner}=\frac{3}{s+2}$. Yeh step kyun? Rule 1 — outer forward path ab $G_1$ cascaded with ek clean inner block hai.
3. Outer feedback apply karo ($H_1=1$): $\frac{Y}{R}=\frac{\frac{3}{s+2}}{1+\frac{3}{s+2}}=\frac{3}{s+5}$.

Verify: General two-loop result use karte hue $\frac{G_1{G}_2}{1+G_2{H}_2+G_1{G}_2{H}_1}=\frac{3/s}{1+2/s+3/s}=\frac{3}{s+5}$ — Step 3 se exactly match karta hai. $s=0$ par: $\frac{3}{5}=0.6$. ✔

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Cell H — Equivalence move (toll bharo)

1. Move se pehle branch $A$ carry karta hai. Baad mein tap downstream baithta hai aur $GA$ read karta hai. Yeh step kyun? Ek take-off us point par jo bhi value hoti hai usse copy karta hai; point move karne se value badal jaati hai.
2. $\frac{1}{G}$ insert karo moved branch mein: new value $=\frac{1}{G}\cdot GA=A$. Toll pay ho gaya. Yeh step kyun? Rule 4 — take-off ke saath ek block ke downstream cross karo, extra multiply cancel karne ke liye $1/G$ pay karo.

Verify (dummy input): $A=2$, $s=0\Rightarrow G=\frac{1}{4}$. Moved-branch reading $=GA=\frac{1}{4}\cdot 2=0.5$; $\times \frac{1}{G}=\times 4$ ke baad: $0.5\times 4=2=A$. ✔ Original se identical.

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Cell I — Real-world word problem

1. Prose ko translate karo: forward block $G=\frac{4}{s(s+2)}$, feedback block $H=0.5$, negative summer. Yeh step kyun? "Plant" = forward block; "sensor in feedback path" = $H$; "subtracts" = $E=R-B$ minus sign.
2. Loop gain: $GH=\frac{4}{s(s+2)}\cdot 0.5=\frac{2}{s(s+2)}$.
3. Closed loop: $\frac{Y}{R}=\frac{G}{1+GH}=\frac{\frac{4}{s(s+2)}}{1+\frac{2}{s(s+2)}}=\frac{4}{s^2+2s+2}$. Yeh step kyun? $s(s+2)$ se multiply karke fraction clear karo — pure algebra.

Verify: Denominator $s^2+2s+2$ ke roots $s=-1\pm j$ hain (dono left half-plane mein), toh yeh satellite loop stable hai — ek physically sensible design. DC gain: $s=0$ par, $\frac{4}{0}$ → $\infty$, matlab zero steady-state angle error ($G$ mein ek free integrator). $s=1$ par: $\frac{4}{1+2+2}=\frac{4}{5}=0.8$. ✔

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Cell J — Exam twist: unity feedback + cascade, "poles dhundho"

1. Forward path ko series-combine karo: $G=G_1{G}_2=\frac{6}{s(s+5)}$. Yeh step kyun? Rule 1 cascade ko ek block mein collapse karta hai loop rule apply karne se pehle.
2. Unity feedback: $\frac{Y}{R}=\frac{G}{1+G}=\frac{\frac{6}{s(s+5)}}{1+\frac{6}{s(s+5)}}=\frac{6}{s^2+5s+6}$. Yeh step kyun? $H=1$ hai toh $1+GH=1+G$; $s(s+5)$ se multiply karke clear karo.
3. Denominator factor karo (characteristic equation): $s^2+5s+6=(s+2)(s+3)$, toh poles $s=-2$ aur $s=-3$ par. Yeh step kyun? $1+GH=0$ set karne se poles milte hain — woh numbers jo stability aur speed govern karte hain (Stability & characteristic equation).

Verify: $(s+2)(s+3)=s^2+5s+6$ ✔; dono poles negative → stable. $s=0$ par DC gain: $\frac{6}{6}=1$. ✔

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Recall KISI BHI diagram ke liye one-line recipe

Series collapse karo (×), parallel collapse karo (±), phir loops inside-out mein $\frac{\text{forward}}{1\pm \text{loop gain}}$ se; $G$ ya $1/G$ ka toll tab do jab koi junction kisi block cross kare.

Quick self-test

Connections

• Transfer functions — yahan har $G$, $H$ yahi hai.
• Feedback control loops — Cells C, D, E, G, I, J.
• Op-amp gain — Cell E case (c), $1/H$ limit.
• Stability & characteristic equation — Cells D, I, J denominator padhte hain.
• Signal flow graphs & Mason's gain formula — Cell G karne ka ek algebra-free tarika.
• Laplace transform — isliye Cell A multiply karta hai.