Block diagram algebra: Exercises — Hinglish deep dive

3.5.28 · D4 · Physics › Guidance, Navigation & Control (GNC) › Block diagram algebra

Intuition Is page ko kaise use karein

Ye problems ek seedhi chadhai ki tarah hain. L1 Recognition — kya aap rule ka naam bata sakte hain? L2 Application — kya aap ek baar crank ghuma sakte hain? L3 Analysis — kya aap sahi order chun ke ulajhan suljha sakte hain? L4 Synthesis — kya aap woh moves combine kar sakte hain jo pehle se saji hui nahi thi? L5 Mastery — kya aap poles, stability, aur edge cases ke baare mein reason kar sakte hain?

Har solution ek collapsible [!recall]- callout ke andar chhupa hua hai. Pehle khud try karo, phir reveal karo. Ye page parent note Block diagram algebra aur uske chaar rules ko assume karta hai, jo yahan dobara likhe gaye hain taaki numbering kabhi surprise na kare:

Rule 1 — Series: cascaded blocks multiply hote hain, $G_{e q} = G_{1} G_{2}$ .
Rule 2 — Parallel: split-then-summed blocks add hote hain, $G_{e q} = G_{1} \pm G_{2}$ (sign summer se lena hai).
Rule 3 — Feedback loop: forward $G$ , feedback $H$ , negative summer $\Rightarrow \frac{Y}{R} = \frac{G}{1 + G H}$ (positive feedback ke liye denominator $1 - G H$ ho jaata hai).
Rule 4 — Moving junctions: ek summer ya take-off point ko block $G$ ke aage-peeche slide karo aur compensate karo — toll $G$ chukao (ya change $1/ G$ lo) moved branch mein, taaki arithmetic balance rahe.

Neeche use hone waale har symbol ki definition ek baar yahan di gayi hai taaki koi symbol anjaan na lage:

Neeche di gayi figure ek visual dictionary hai un teen shapes ki jo ye exercises andar chhupati hain — jab bhi shak ho kaunsa rule kaam aa raha hai, ise dobara dekh lo.

Figure 1 — Teen canonical structures. Orange (upar): ek series cascade, jiske do blocks $G_{1} G_{2}$ mein multiply ho jaate hain (Rule 1). Teal (beech): ek parallel split jiske branches ek summer par milte hain, giving $G_{1} \pm G_{2}$ (Rule 2). Plum (neeche): ek feedback loop — dhyan do ki return path summer mein re-enter kar raha hai forward block $G$ se pehle, jo Rule 3 ki pehchan hai, yeh $G / (1 + G H)$ mein collapse hota hai.

Level 1 — Recognition

Goal: rule ka naam lo aur ek-line result likho. Abhi koi algebra clear nahi karni.

Exercise 1.1

Do blocks $G_{1} = \frac{3}{s}$ aur $G_{2} = s + 4$ ek ke baad ek (cascade) hain, signal $G_{1}$ phir $G_{2}$ se flow kar raha hai. Single equivalent block likho.

Recall Solution 1.1

WHAT rule: Series mein blocks multiply hote hain (Rule 1). WHY: $G_{1}$ ka output $G_{2}$ ka input ban jaata hai, aur "kisi block se guzarna" matlab hai "us block se multiply karna". Yeh Figure 1 ki orange top row hai. $G_{e q} = G_{1} G_{2} = \frac{3}{s} \cdot (s + 4) = \frac{3 ( s + 4 )}{s} = \frac{3 s + 12}{s} .$

Exercise 1.2

Same input $X$ ek take-off point par $G_{1} = 2$ aur $G_{2} = 5$ mein split hota hai, aur unke outputs ek aise summer par milte hain jo bottom branch ko top se subtract karta hai. Equivalent block likho.

Recall Solution 1.2

WHAT rule: Parallel blocks add/subtract hote hain (Rule 2), sign summer se lena hai. Yeh Figure 1 ki teal middle row se match karta hai. $G_{e q} = G_{1} - G_{2} = 2 - 5 = - 3.$ WHY minus: summer bottom branch ko subtract karta hai, isliye hum wahi minus combination mein laate hain.

Exercise 1.3

Ek forward block $G = \frac{10}{s + 1}$ ke saath negative unity feedback hai ( $H = 1$ ). $\frac{Y}{R}$ ko simplify kiye bina likho.

Recall Solution 1.3

WHAT rule: Feedback loop (Rule 3), unity-feedback case $H = 1$ — Figure 1 ki plum bottom row jisme $H$ block $1$ set hai. $\frac{Y}{R} = \frac{G}{1 + G} = \frac{\frac{10}{s + 1}}{1 + \frac{10}{s + 1}} .$ (Simplification L2 mein aayegi — yahan hum sirf form pahchante hain.)

Level 2 — Application

Goal: ek baar crank ghunao aur fractions clear karo.

Exercise 2.1

Exercise 1.3 finish karo: $\frac{Y}{R} = \frac{10/ ( s + 1 )}{1 + 10/ ( s + 1 )}$ ko polynomials ke single ratio mein simplify karo.

Recall Solution 2.1

WHAT we do: inner fraction clear karne ke liye upar aur neeche $(s + 1)$ se multiply karo. WHY it is valid: kisi fraction ke numerator aur denominator ko same non-zero quantity se multiply karna, $1$ se multiply karna hai, isliye value badal nahi sakti. WHY it helps: yeh "fraction ke andar fraction" hataa deta hai, ek saaf polynomial ratio chhod deta hai jiske poles seedhe padhe ja sakte hain. $\frac{Y}{R} = \frac{10}{( s + 1 ) + 10} = \frac{10}{s + 11} .$ Sanity check: $s = - 11$ par ek single pole hai, open-loop pole $s = - 1$ se zyada left mein — feedback ne pole ko left mein push kiya (faster response). Iske baare mein L5 mein aur baat hogi.

Exercise 2.2

Forward path ek cascade hai $G_{1} = \frac{4}{s}$ , $G_{2} = \frac{1}{s + 2}$ ; feedback $H = 3$ , negative. $\frac{Y}{R}$ nikalo.

Recall Solution 2.2

Step 1 (series): $G = G_{1} G_{2} = \frac{4}{s ( s + 2 )}$ . WHY: Rule 1 cascade ko ek forward block mein badal deta hai taaki Rule 3 apply ho sake. Step 2 (feedback): $\frac{Y}{R} = \frac{G}{1 + G H} = \frac{\frac{4}{s ( s + 2 )}}{1 + \frac{4}{s ( s + 2 )} \cdot 3} .$ Step 3 (clear): upar aur neeche $s (s + 2)$ se multiply karo. WHY it is valid: $s (s + 2)$ saamne dikhne wali har fraction ka common denominator hai, isliye upar aur neeche se multiply karna phir se $1$ se multiply karna hai — value unchanged hai. WHY it helps: yeh ek saath saare nested $s (s + 2)$ denominators ko mita deta hai, gande expression ko single polynomial ratio mein collapse kar deta hai. $\frac{Y}{R} = \frac{4}{s ( s + 2 ) + 12} = \frac{4}{s ^{2} + 2 s + 12} .$

Exercise 2.3

Do blocks $G_{1} = \frac{s}{s + 1}$ aur $G_{2} = \frac{s + 1}{s}$ series mein hain. Simplify karo.

Recall Solution 2.3

$G_{e q} = \frac{s}{s + 1} \cdot \frac{s + 1}{s} = 1.$ Ye dono blocks inverses hain — ye exactly cancel ho jaate hain, ek pure pass-through of gain $1$ chhod kar. WHY it matters: yeh bilkul wohi " $1/ G$ " correction hai jo aap tab insert karte ho jab koi take-off point kisi block ke paas se move kiya jaata hai (Rule 4): extra block us block ko undo karta hai jise usne cross kiya.

Level 3 — Analysis

Goal: sahi order chuno aur nesting suljhao.

Exercise 3.1 (do loops, inside-out)

Inner loop: forward $G_{2} = \frac{5}{s}$ , feedback $H_{2} = 1$ (negative). Outer loop: forward $G_{1} = 2$ phir inner block, feedback $H_{1} = 1$ (negative). $\frac{Y}{R}$ nikalo.

Neeche di gayi figure yeh exact two-loop layout dikhati hai; plum dashed box inner loop ko mark karta hai jise aapko pehle collapse karna hai, iske baad hi outer feedback formula likhi ja sakti hai.

Figure 2 — Exercise 3.1 ke liye nested loops. Teal inner loop hai ( $G_{2}$ with feedback $H_{2}$ ), plum dashed line se box karke "reduce THIS first" label kiya gaya hai. Orange outer loop hai ( $G_{1}$ with feedback $H_{1}$ ) jo sab kuch wrap karta hai. Do plum take-off dots do return paths ko feed karte hain.

Recall Solution 3.1

Step 1 — pehle sabse inner loop reduce karo. WHY: outer feedback formula ko forward path ek single block ki zarurat hai; woh ek hidden loop ke andar nahi dekh sakti (Figure 2 mein plum dashed box). $G_{inn er} = \frac{G _{2}}{1 + G _{2} H _{2}} = \frac{5/ s}{1 + 5/ s} = \frac{5}{s + 5} .$ Step 2 — $G_{1}$ ke saath series: $G_{f w d} = G_{1} G_{inn er} = 2 \cdot \frac{5}{s + 5} = \frac{10}{s + 5} .$ Step 3 — outer loop: Rule 3 apply karo, phir upar aur neeche $(s + 5)$ se multiply karo. WHY multiply by $(s + 5)$ : yeh badi fraction ke andar stacked fractions ka common denominator hai, isliye upar aur neeche se multiply karna $1$ se multiply karna hai — value unchanged rehti hai. WHY it helps: yeh ek hi stroke mein nested fraction clear kar deta hai, ek single first-order transfer function chhod kar jiska pole ( $s = - 15$ ) seedha dikhta hai. $\frac{Y}{R} = \frac{G _{f w d}}{1 + G _{f w d} H _{1}} = \frac{10/ ( s + 5 )}{1 + 10/ ( s + 5 )} = \frac{10}{( s + 5 ) + 10} = \frac{10}{s + 15} .$

Exercise 3.2 (ek take-off point move karo, phir reduce karo)

Forward: $R \to [-] \to G \to Y$ with $G = \frac{6}{s + 2}$ . Feedback take-off filhaal $G$ ke baad baitha hai aur $H = 1$ se guzar kar wapas aata hai. Aapko take-off ko $G$ se pehle move karna hai, naya feedback block dena hai, phir confirm karna hai ki $\frac{Y}{R}$ unchanged hai.

Recall Solution 3.2

Step 1 — tap abhi kya read kar raha hai: $G$ ke baad, tapped signal $Y = G E$ hai. Fed-back signal hai $B = H Y = H G E$ . Step 2 — tap ko $G$ se pehle move karo: ab yeh $E$ read karta hai, $GE$ nahi — humne $G$ ka ek factor khoya. $B$ ko same rakhne ke liye hum moved branch mein $G$ insert karte hain: new feedback block $= G H = G \cdot 1 = G$ . (Rule 4: "block cross karo, toll $G$ chukao".) Step 3 — confirm karo ki closed loop unchanged hai. Loop gain phir bhi $G \cdot H = G$ hai dono taraf se, isliye $\frac{Y}{R} = \frac{G}{1 + G} = \frac{6/ ( s + 2 )}{1 + 6/ ( s + 2 )} = \frac{6}{s + 8} .$ Junction move karne se sirf drawing badla, transfer function kabhi nahi. ✔

Level 4 — Synthesis

Goal: aisi kai moves assemble karo jo pehle se saji hui nahi thi.

Exercise 4.1 (parallel forward, single feedback)

Forward path ek parallel combination hai: input $G_{a} = \frac{1}{s}$ aur $G_{b} = 2$ mein split hota hai, aur unke outputs forward output banane ke liye ek summer par add hote hain. Yeh forward output phir ek negative unity feedback loop drive karta hai. $\frac{Y}{R}$ nikalo.

Recall Solution 4.1

Step 1 — parallel forward path collapse karo (Rule 2, sign $+$ ). WHY ek fraction mein combine karo: Rule 3 ko ek single forward block $G$ chahiye, isliye hum pehle dono parallel branches ko common denominator $s$ ke upar add karte hain. $G = G_{a} + G_{b} = \frac{1}{s} + 2 = \frac{1 + 2 s}{s} .$ Step 2 — unity feedback apply karo (Rule 3, $H = 1$ ), phir compound fraction ko upar aur neeche $s$ se multiply karke clear karo. WHY multiply by $s$ : $s$ badi fraction ke andar pieces ka shared denominator hai, isliye yeh $1$ se multiply karna hai — value unchanged. WHY it helps: yeh fraction-inside-a-fraction ko plain first-degree polynomials ke ratio mein flatten kar deta hai. $\frac{Y}{R} = \frac{G}{1 + G} = \frac{( 1 + 2 s ) / s}{1 + ( 1 + 2 s ) / s} = \frac{1 + 2 s}{s + 1 + 2 s} = \frac{2 s + 1}{3 s + 1} .$

Exercise 4.2 (cascade with feedback around only part of it)

$R \to [-] \to G_{1} \to ∙ \to G_{2} \to Y$ , jahan take-off point $∙$ ( $G_{1}$ aur $G_{2}$ ke beech) $H$ se guzar kar summer mein wapas aata hai. Toh loop sirf $G_{1}$ ke around wrap karta hai, aur $G_{2}$ loop ke baad ek plain cascade hai. $G_{1} = \frac{4}{s + 1}$ , $G_{2} = \frac{1}{s}$ , $H = 1$ ke saath $\frac{Y}{R}$ nikalo.

Recall Solution 4.2

Step 1 — loop scope identify karo. Tap $G_{1}$ aur $G_{2}$ ke beech baitha hai, isliye fed-back signal sirf $G_{1}$ ka output hai. Loop sirf $G_{1}$ ko feedback $H$ ke saath enclose karta hai. Step 2 — woh inner loop reduce karo (Rule 3), fraction ko pehle ki tarah "multiply by $1$ " reason se upar aur neeche $(s + 1)$ se multiply karke clear karo: $G_{l oo p} = \frac{G _{1}}{1 + G _{1} H} = \frac{4/ ( s + 1 )}{1 + 4/ ( s + 1 )} = \frac{4}{s + 5} .$ Step 3 — bacha hua $G_{2}$ cascade karo (Rule 1, $G_{2}$ loop ke baahir hai): $\frac{Y}{R} = G_{l oo p} \cdot G_{2} = \frac{4}{s + 5} \cdot \frac{1}{s} = \frac{4}{s ( s + 5 )} = \frac{4}{s ^{2} + 5 s} .$ WHY yeh order: $G_{2}$ take-off ke downstream hai, isliye koi feedback usse wrap nahi karta — pehle loop treat karo, phir clean cascade block multiply karo.

Level 5 — Mastery

Goal: poles, stability, feedback ka sign, aur degenerate inputs ke baare mein reason karo.

Exercise 5.1 (characteristic equation → stability)

Exercise 2.2 ke system ke liye, $\frac{Y}{R} = \frac{4}{s ^{2} + 2 s + 12}$ , characteristic equation denominator ko zero set karne par milti hai, $s^{2} + 2 s + 12 = 0$ . Poles nikalo aur batao ki closed loop stable hai ya nahi.

Recall Solution 5.1

WHAT rule: poles characteristic equation $1 + G H = 0$ ke roots hain, jo exactly denominator $= 0$ hai. $s = \frac{- 2 \pm 4 - 48}{2} = \frac{- 2 \pm - 44}{2} = - 1 \pm j 11 .$ Yahan $j = - 1$ (imaginary unit) aur $11 \approx 3.317$ . Dono poles ki real part $- 1 < 0$ hai, yaani woh complex plane ke left half mein hain. Verdict: stable. Oscillation (imaginary part $\pm j 11$ se) decay hoti hai kyunki envelope $e^{- 1 \cdot t}$ shrink hoti hai.

Exercise 5.2 (positive feedback aur sign flip)

Forward $G = \frac{k}{s + 1}$ lo positive feedback ke saath, $H = 1$ . Closed-loop denominator $1 - G H$ ho jaata hai. Kis value of $k$ par closed loop ka ek pole exactly $s = 0$ par hoga (instability ki boundary)?

Recall Solution 5.2

WHAT positive feedback ke liye badalta hai: denominator $1 - G H$ hai, $1 + G H$ nahi. $\frac{Y}{R} = \frac{G}{1 - G H} = \frac{k / ( s + 1 )}{1 - k / ( s + 1 )} = \frac{k}{( s + 1 ) - k} = \frac{k}{s + 1 - k} .$ Pole $s = k - 1$ par hai. $s = 0$ par pole ke liye $k - 1 = 0$ chahiye, isliye $k = 1 .$ Interpretation: $k < 1$ ke liye pole $s = k - 1 < 0$ par hai (stable); $k = 1$ par yeh imaginary axis par baithta hai (marginal); $k > 1$ ke liye yeh right half-plane mein cross kar jaata hai aur system blow up ho jaata hai. Positive feedback pole ko rightward push karta hai — Exercise 2.1 mein dekhe stabilising leftward push se bilkul ulta.

Exercise 5.3 (degenerate case — loop gain vanish ho jaata hai)

Ek negative feedback loop $\frac{Y}{R} = \frac{G}{1 + G H}$ mein, maan lo feedback path broken hai (ek open wire), jise hum $H = 0$ se model karte hain. Closed loop kya ban jaata hai, aur physically matlab kya hai?

Recall Solution 5.3

$H = 0$ set karo: $\frac{Y}{R} = \frac{G}{1 + G \cdot 0} = \frac{G}{1} = G .$ Matlab: koi feedback return nahi ho raha, isliye koi round trip nahi, loop gain $G H = 0$ hai aur sirf direct pass-through (woh " $1$ ") bachta hai. System sirf raw open-loop block $G$ hai — exactly woh op-amp jo apne full untamed gain par chal raha hai. Yeh $1 + G H$ mein $1$ ki physical reading confirm karta hai: yeh woh signal hai jo seedha ek baar guzarta hai, bina kisi feedback trip ke.

Exercise 5.4 (limiting behaviour — bahut bada loop gain)

Abhi bhi $\frac{Y}{R} = \frac{G}{1 + G H}$ use karo, dekhte hain kya hota hai jab forward gain $G \to \infty$ (bahut bada) ho jaata hai. Us limit mein closed-loop expression simplify karo.

Recall Solution 5.4

Upar aur neeche $G$ se divide karo. WHY divide by $G$ : yeh $1/ G$ term ko isolate karta hai, jo akela change hota hai jab $G$ badhta hai — limit padhna trivial ho jaata hai. $\frac{Y}{R} = \frac{1}{\frac{1}{G} + H} G \to \infty \frac{1}{0 + H} = \frac{1}{H} .$ Matlab: jab forward gain enormous ho, closed-loop behaviour sirf feedback path $H$ par depend karta hai, $G$ par nahi. Yeh precision op-amps aur feedback control ka golden principle hai: $G$ ko huge aur sasta banao, aur ek precise, stable $H$ ko actual gain $1/ H$ set karne do.

Connections

Parent: Block diagram algebra — woh rules jo yahan drill hote hain.
Transfer functions — upar har block ek hai.
Laplace transform — $s$ kyun aata hai aur blocks kyun multiply karte hain.
Feedback control loops — Rule 3 ki physical setting.
Signal flow graphs & Mason's gain formula — inside-out reduction ka ek alternative.
Op-amp gain — Exercises 5.3 aur 5.4 disguise mein.
Stability & characteristic equation — Exercise 5.1 aur 5.2.