Block diagram algebra: Worked examples — Physics deep dive

This page is a stress-test of the reduction rules from the parent note. We do not learn new rules here — we hunt down every kind of block diagram the world can throw at you and grind each one to a single transfer function. If you meet a diagram in an exam and think "I've never seen this shape," check the matrix below: you have.

Before anything, one reminder of the three symbols we will lean on, stated in plain words so nobody is lost:

The scenario matrix

Every block-diagram problem is one (or a stack) of these cells. The final column names the worked example that nails it.

Cell	What makes it special	Danger	Nailed by
A. Pure series	blocks only multiply	none — warm-up	Ex 1
B. Parallel, minus sign	summer subtracts branches	sign slip	Ex 2
C. Negative feedback	the standard loop	forget the $+ 1$	Ex 3
D. Positive feedback	denominator flips to $1 - G H$	wrong sign → wrong poles	Ex 4
E. Zero / degenerate gain	$H = 0$ , or $G \to \infty$	formula must still make sense	Ex 5
F. Limiting behaviour	$s \to 0$ (DC) and $s \to \infty$	is the loop stable / finite?	Ex 6
G. Nested loops (inside-out)	a loop hiding inside a loop	reducing outer first = wrong	Ex 7
H. Equivalence move	slide a take-off / summer past a block	forget the toll $G$ or $1/ G$	Ex 8
I. Real-world word problem	a physical GNC loop in prose	translating words → blocks	Ex 9
J. Exam twist: unity feedback + cascade	$H = 1$ , numeric, "find poles"	mixing up loop gain and forward gain	Ex 10

Signs (+/-), zero inputs, and both limits ( $s \to 0$ , $s \to \infty$ ) are each covered by at least one cell. Nothing is left uncovered.

Cell A — Pure series (warm-up)

Multiply straight through. The signal is multiplied once per block: $Y = G_{3} (G_{2} (G_{1} R)) = G_{1} G_{2} G_{3} R$ . Why this step? A cascade is convolution-in-time = multiplication-in- $s$ ; multiplication is associative so order and grouping don't matter.
Plug numbers: $\frac{Y}{R} = \frac{5}{s} \cdot \frac{1}{s + 2} \cdot 4 = \frac{20}{s ( s + 2 )} = \frac{20}{s ^{2} + 2 s}$ .

Verify: Units of gain multiply, dimensionless throughout. Sanity: three blocks, one fraction, no $1 + G H$ because there is no return trip. At $s = 1$ : $\frac{20}{1 \cdot 3} = \frac{20}{3} \approx 6.667$ . ✔

Cell B — Parallel with a subtracting summer

Each branch acts on the same $R$ (copying at a take-off does not change a signal): top $= G_{1} R$ , bottom $= G_{2} R$ . Why this step? Rule 2 — the split feeds an identical copy to both branches.
Summer subtracts: $Y = (G_{1} - G_{2}) R = (\frac{s}{s + 1} - \frac{1}{s + 1}) R = \frac{s - 1}{s + 1} R$ . Why this step? Same denominator, so subtract numerators directly.

Verify: At $s = 1$ the numerator $s - 1 = 0$ , so $\frac{Y}{R} = 0$ — the two branches exactly cancel there, which is the whole point of a subtracting summer. ✔ At $s = 3$ : $\frac{2}{4} = 0.5$ . ✔

Cell C — Negative feedback (the standard loop)

Loop gain $= G H = \frac{10}{s + 1} \cdot \frac{1}{2} = \frac{5}{s + 1}$ . Why this step? $G H$ is what a signal is multiplied by after one full trip around the loop — the number that " $1 +$ " sits in front of.
Apply the closed-loop formula $\frac{Y}{R} = \frac{G}{1 + G H} = \frac{\frac{10}{s + 1}}{1 + \frac{5}{s + 1}}$ . Why this step? Feedback control loops — this is Rule 3, derived by substituting $B = H Y$ into $Y = G (R - B)$ .
Clear the fraction (multiply top and bottom by $s + 1$ ): $\frac{Y}{R} = \frac{10}{( s + 1 ) + 5} = \frac{10}{s + 6}$ .

Verify: The open-loop pole was at $s = - 1$ ; feedback moved it to $s = - 6$ — feedback pulled the pole deeper into the left half-plane (faster, more stable), exactly what negative feedback does. At $s = 0$ (DC): $\frac{10}{6} \approx 1.667$ . ✔

Cell D — Positive feedback (sign flips)

Use the positive-feedback formula: $\frac{Y}{R} = \frac{G}{1 - G H} = \frac{\frac{10}{s + 1}}{1 - \frac{5}{s + 1}}$ . Why this step? With $E = R + B$ , collecting $Y$ gives $Y (1 - G H) = GR$ — the minus is baked in.
Clear: $\frac{Y}{R} = \frac{10}{( s + 1 ) - 5} = \frac{10}{s - 4}$ .

Verify: The pole is now at $s = + 4$ — in the right half-plane. That means the response grows without bound: positive feedback here is unstable. Compare Ex 3's pole at $- 6$ . This is why Stability & characteristic equation watches the denominator $1 \pm G H = 0$ : the sign decides life or death of the loop. ✔

Cell E — Zero / degenerate gain

Case (a), $H = 0$ : $\frac{Y}{R} = \frac{G}{1 + G \cdot 0} = \frac{G}{1} = G = \frac{7}{s + 3}$ . Why this step? $H = 0$ cuts the return path — the diagram degenerates to a single open-loop block. The formula gracefully gives back $G$ .
Case (b), $H = 1$ (unity feedback): $\frac{Y}{R} = \frac{G}{1 + G} = \frac{\frac{7}{s + 3}}{1 + \frac{7}{s + 3}} = \frac{7}{s + 10}$ . Why this step? Unity feedback is the most common special case; pole moves $- 3 \to - 10$ .
Case (c), $G \to \infty$ : $\frac{Y}{R} = \frac{G}{1 + G H} G \to \infty \frac{1}{H} = \frac{1}{1/3} = 3$ . Why this step? Divide top and bottom by $G$ : $\frac{1}{1/ G + H} \to \frac{1}{H}$ . Huge open-loop gain becomes a clean $1/ H$ — this is the exact trick behind Op-amp gain.

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

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

$s \to 0$ (DC): substitute $s = 0$ : $\frac{20}{0 + 0 + 20} = 1$ . Why this step? At DC the "frequency knob" is off; this number is the steady-state gain — the loop passes constant inputs through at gain 1.
$s \to \infty$ : the top is degree 0, the bottom degree 2; the bottom explodes faster, so $\frac{20}{s ^{2} + 4 s + 20} \to 0$ . Why this step? Physical systems can't respond to infinitely fast wiggles — the output dies. This is the loop's built-in low-pass character.

Verify: DC $= 1$ exactly. At a mid value $s = 2$ (treated as real for a size check): $\frac{20}{4 + 8 + 20} = \frac{20}{32} = 0.625$ , between $1$ and $0$ . ✔ Monotone squeeze confirmed.

Cell G — Nested loops (inside-out)

Reduce the inner loop: $G_{inn er} = \frac{G _{2}}{1 + G _{2} H _{2}} = \frac{1/ s}{1 + ( 1/ s ) ( 2 )} = \frac{1}{s + 2}$ . Why this step? The outer feedback formula needs a single block in the forward path; we can't apply it across a hidden loop. Work inside-out.
Series with $G_{1}$ : forward $= G_{1} G_{inn er} = \frac{3}{s + 2}$ . Why this step? Rule 1 — the outer forward path is now $G_{1}$ cascaded with one clean inner block.
Apply outer feedback ( $H_{1} = 1$ ): $\frac{Y}{R} = \frac{\frac{3}{s + 2}}{1 + \frac{3}{s + 2}} = \frac{3}{s + 5}$ .

Verify: Using the general two-loop result $\frac{G _{1} G _{2}}{1 + G _{2} H _{2} + G _{1} G _{2} H _{1}} = \frac{3/ s}{1 + 2/ s + 3/ s} = \frac{3}{s + 5}$ — matches Step 3 exactly. At $s = 0$ : $\frac{3}{5} = 0.6$ . ✔

Cell H — Equivalence move (pay the toll)

Before the move the branch carries $A$ . After the tap sits downstream and reads $G A$ . Why this step? A take-off copies whatever value exists at that point; moving the point changes the value.
Insert $\frac{1}{G}$ in the moved branch: new value $= \frac{1}{G} \cdot G A = A$ . Toll paid. Why this step? Rule 4 — cross a block downstream with a take-off, pay $1/ G$ to cancel the extra multiply.

Verify (dummy input): $A = 2$ , $s = 0 \Rightarrow G = \frac{1}{4}$ . Moved-branch reading $= G A = \frac{1}{4} \cdot 2 = 0.5$ ; after $\times \frac{1}{G} = \times 4$ : $0.5 \times 4 = 2 = A$ . ✔ Identical to the original.

Cell I — Real-world word problem

Translate the prose: forward block $G = \frac{4}{s ( s + 2 )}$ , feedback block $H = 0.5$ , negative summer. Why this step? "Plant" = forward block; "sensor in feedback path" = $H$ ; "subtracts" = the $E = R - B$ minus sign.
Loop gain: $G H = \frac{4}{s ( s + 2 )} \cdot 0.5 = \frac{2}{s ( s + 2 )}$ .
Closed loop: $\frac{Y}{R} = \frac{G}{1 + G H} = \frac{\frac{4}{s ( s + 2 )}}{1 + \frac{2}{s ( s + 2 )}} = \frac{4}{s ^{2} + 2 s + 2}$ . Why this step? Clear the fraction by multiplying by $s (s + 2)$ — pure algebra.

Verify: Denominator $s^{2} + 2 s + 2$ has roots $s = - 1 \pm j$ (both in the left half-plane), so this satellite loop is stable — a physically sensible design. DC gain: at $s = 0$ , $\frac{4}{0}$ → $\infty$ , meaning zero steady-state angle error (a free integrator in $G$ ). At $s = 1$ : $\frac{4}{1 + 2 + 2} = \frac{4}{5} = 0.8$ . ✔

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

Series-combine the forward path: $G = G_{1} G_{2} = \frac{6}{s ( s + 5 )}$ . Why this step? Rule 1 collapses the cascade into one block before the loop rule can apply.
Unity feedback: $\frac{Y}{R} = \frac{G}{1 + G} = \frac{\frac{6}{s ( s + 5 )}}{1 + \frac{6}{s ( s + 5 )}} = \frac{6}{s ^{2} + 5 s + 6}$ . Why this step? $H = 1$ so $1 + G H = 1 + G$ ; clear by multiplying by $s (s + 5)$ .
Factor the denominator (the characteristic equation): $s^{2} + 5 s + 6 = (s + 2) (s + 3)$ , so poles at $s = - 2$ and $s = - 3$ . Why this step? Setting $1 + G H = 0$ gives the poles — the numbers that govern stability and speed (Stability & characteristic equation).

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

Recall One-line recipe for ANY diagram

Collapse series (×), collapse parallel (±), then loops inside-out with $\frac{forward}{1 \pm loop gain}$ ; pay the toll $G$ or $1/ G$ whenever a junction crosses a block.

Quick self-test

Ex 3 negative-feedback answer

\frac{10}{s + 1}

H = \frac{1}{2}

\frac{10}{s + 6}

Ex 4 positive-feedback pole location

s = + 4

(unstable, right half-plane)

Ex 5c limit of

\frac{G}{1 + G H}

G \to \infty

\frac{1}{H}

Ex 7 nested-loop closed-loop TF

\frac{3}{s + 5}

Ex 10 poles of

\frac{6}{s ^{2} + 5 s + 6}

s = - 2

and

s = - 3

Connections

Transfer functions — every $G$ , $H$ here is one.
Feedback control loops — Cells C, D, E, G, I, J.
Op-amp gain — Cell E case (c), the $1/ H$ limit.
Stability & characteristic equation — Cells D, I, J read the denominator.
Signal flow graphs & Mason's gain formula — an algebra-free way to do Cell G.
Laplace transform — why Cell A multiplies.