Block diagram algebra: Exercises — Physics deep dive

Intuition How to use this page

These problems climb a ladder. L1 Recognition — can you name the rule? L2 Application — can you turn the crank once? L3 Analysis — can you pick the right order and untangle? L4 Synthesis — can you combine moves you were never handed pre-arranged? L5 Mastery — can you reason about poles, stability, and edge cases?

Every solution is hidden inside a collapsible [!recall]- callout. Try first, then reveal. This page assumes the parent note Block diagram algebra and its four rules, restated here so the numbering never surprises you:

Rule 1 — Series: cascaded blocks multiply, $G_{e q} = G_{1} G_{2}$ .
Rule 2 — Parallel: split-then-summed blocks add, $G_{e q} = G_{1} \pm G_{2}$ (sign from the summer).
Rule 3 — Feedback loop: forward $G$ , feedback $H$ , negative summer $\Rightarrow \frac{Y}{R} = \frac{G}{1 + G H}$ (denominator becomes $1 - G H$ for positive feedback).
Rule 4 — Moving junctions: slide a summer or take-off point across a block $G$ and compensate — pay the toll $G$ (or take change $1/ G$ ) in the moved branch so the arithmetic still balances.

Every symbol used below is defined once here so you never meet a stranger:

The figure below is your visual dictionary of the three shapes these exercises hide inside them — glance back at it whenever you are unsure which rule a problem is asking for.

Figure 1 — The three canonical structures. Orange (top): a series cascade, whose two blocks multiply into $G_{1} G_{2}$ (Rule 1). Teal (middle): a parallel split whose branches rejoin at a summer, giving $G_{1} \pm G_{2}$ (Rule 2). Plum (bottom): a feedback loop — notice the return path re-entering the summer before the forward block $G$ , the tell-tale sign of Rule 3, collapsing to $G / (1 + G H)$ .

Level 1 — Recognition

Goal: name the rule and write the one-line result. No algebra to clear yet.

Exercise 1.1

Two blocks $G_{1} = \frac{3}{s}$ and $G_{2} = s + 4$ sit one after another (cascade), signal flowing $G_{1}$ then $G_{2}$ . Write the single equivalent block.

Recall Solution 1.1

WHAT rule: Blocks in series multiply (Rule 1). WHY: the output of $G_{1}$ becomes the input of $G_{2}$ , and "passing through a block" means "multiply by that block". This is the orange top row of Figure 1. $G_{e q} = G_{1} G_{2} = \frac{3}{s} \cdot (s + 4) = \frac{3 ( s + 4 )}{s} = \frac{3 s + 12}{s} .$

Exercise 1.2

The same input $X$ splits at a take-off point into $G_{1} = 2$ and $G_{2} = 5$ , and their outputs meet at a summer that subtracts the bottom branch from the top. Write the equivalent block.

Recall Solution 1.2

WHAT rule: Blocks in parallel add/subtract (Rule 2), sign taken from the summer. This matches the teal middle row of Figure 1. $G_{e q} = G_{1} - G_{2} = 2 - 5 = - 3.$ WHY the minus: the summer subtracts the bottom branch, so we carry that same minus into the combination.

Exercise 1.3

A forward block $G = \frac{10}{s + 1}$ has a negative unity feedback ( $H = 1$ ). Write $\frac{Y}{R}$ without simplifying.

Recall Solution 1.3

WHAT rule: The feedback loop (Rule 3), unity-feedback case $H = 1$ — the plum bottom row of Figure 1 with the $H$ block set to $1$ . $\frac{Y}{R} = \frac{G}{1 + G} = \frac{\frac{10}{s + 1}}{1 + \frac{10}{s + 1}} .$ (Simplification comes in L2 — here we only recognise the form.)

Level 2 — Application

Goal: turn the crank once and clear the fractions.

Exercise 2.1

Finish Exercise 1.3: simplify $\frac{Y}{R} = \frac{10/ ( s + 1 )}{1 + 10/ ( s + 1 )}$ to a single ratio of polynomials.

Recall Solution 2.1

WHAT we do: multiply top and bottom by $(s + 1)$ to clear the inner fraction. WHY it is valid: multiplying a fraction's numerator and denominator by the same non-zero quantity is multiplying by $1$ , so the value cannot change. WHY it helps: it removes the "fraction inside a fraction", leaving a clean ratio of polynomials whose poles you can read off. $\frac{Y}{R} = \frac{10}{( s + 1 ) + 10} = \frac{10}{s + 11} .$ Sanity check: a single pole at $s = - 11$ , further left than the open-loop pole at $s = - 1$ — feedback pushed the pole leftward (faster response). More on that in L5.

Exercise 2.2

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

Recall Solution 2.2

Step 1 (series): $G = G_{1} G_{2} = \frac{4}{s ( s + 2 )}$ . WHY: Rule 1 turns the cascade into one forward block so Rule 3 applies. 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): multiply top and bottom by $s (s + 2)$ . WHY it is valid: $s (s + 2)$ is exactly the common denominator of every fraction in sight, so multiplying top and bottom by it is again multiplying by $1$ — the value is untouched. WHY it helps: it wipes out all the nested $s (s + 2)$ denominators at once, collapsing the mess into a single polynomial ratio. $\frac{Y}{R} = \frac{4}{s ( s + 2 ) + 12} = \frac{4}{s ^{2} + 2 s + 12} .$

Exercise 2.3

Two blocks $G_{1} = \frac{s}{s + 1}$ and $G_{2} = \frac{s + 1}{s}$ in series. Simplify.

Recall Solution 2.3

$G_{e q} = \frac{s}{s + 1} \cdot \frac{s + 1}{s} = 1.$ The two blocks are inverses — they cancel exactly, leaving a pure pass-through of gain $1$ . WHY it matters: this is precisely the " $1/ G$ " correction you insert when moving a take-off point past a block (Rule 4): the extra block undoes the block it crossed.

Level 3 — Analysis

Goal: choose the right order and untangle nesting.

Exercise 3.1 (two loops, inside-out)

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

The figure below shows this exact two-loop layout; the plum dashed box marks the inner loop you must collapse first, before the outer feedback formula can even be written.

Figure 2 — Nested loops for Exercise 3.1. Teal is the inner loop ( $G_{2}$ with feedback $H_{2}$ ), boxed by the plum dashed line and labelled "reduce THIS first". Orange is the outer loop ( $G_{1}$ with feedback $H_{1}$ ) that wraps everything. Two plum take-off dots feed the two return paths.

Recall Solution 3.1

Step 1 — reduce the innermost loop first. WHY: the outer feedback formula needs the forward path to be one block; it cannot see through a hidden loop (the plum dashed box in Figure 2). $G_{inn er} = \frac{G _{2}}{1 + G _{2} H _{2}} = \frac{5/ s}{1 + 5/ s} = \frac{5}{s + 5} .$ Step 2 — series with $G_{1}$ : $G_{f w d} = G_{1} G_{inn er} = 2 \cdot \frac{5}{s + 5} = \frac{10}{s + 5} .$ Step 3 — outer loop: apply Rule 3, then multiply top and bottom by $(s + 5)$ . WHY multiply by $(s + 5)$ : it is the common denominator of the fractions stacked inside the big fraction, so multiplying top and bottom by it is multiplying by $1$ — the value stays put. WHY it helps: it clears the nested fraction in one stroke, leaving a single first-order transfer function whose pole ( $s = - 15$ ) is immediately visible. $\frac{Y}{R} = \frac{G _{f w d}}{1 + G _{f w d} H _{1}} = \frac{10/ ( s + 5 )}{1 + 10/ ( s + 5 )} = \frac{10}{( s + 5 ) + 10} = \frac{10}{s + 15} .$

Exercise 3.2 (move a take-off point, then reduce)

Forward: $R \to [-] \to G \to Y$ with $G = \frac{6}{s + 2}$ . The feedback take-off currently sits after $G$ and returns through $H = 1$ . You are asked to move the take-off before $G$ and give the new feedback block, then confirm $\frac{Y}{R}$ is unchanged.

Recall Solution 3.2

Step 1 — what the tap reads now: after $G$ , the tapped signal is $Y = G E$ . The fed-back signal is $B = H Y = H G E$ . Step 2 — move the tap before $G$ : now it reads $E$ , not $GE$ — we lost a factor of $G$ . To keep $B$ identical we insert $G$ in the moved branch: new feedback block $= G H = G \cdot 1 = G$ . (Rule 4: "cross a block, pay the toll $G$ ".) Step 3 — check the closed loop is unchanged. The loop gain is still $G \cdot H = G$ either way, so $\frac{Y}{R} = \frac{G}{1 + G} = \frac{6/ ( s + 2 )}{1 + 6/ ( s + 2 )} = \frac{6}{s + 8} .$ Moving the junction changed the drawing, never the transfer function. ✔

Level 4 — Synthesis

Goal: assemble several moves you were never handed pre-arranged.

Exercise 4.1 (parallel forward, single feedback)

The forward path is a parallel combination: input splits into $G_{a} = \frac{1}{s}$ and $G_{b} = 2$ , their outputs added at a summer to form the forward output. This forward output then drives a negative unity feedback loop. Find $\frac{Y}{R}$ .

Recall Solution 4.1

Step 1 — collapse the parallel forward path (Rule 2, sign $+$ ). WHY combine into one fraction: Rule 3 needs a single forward block $G$ , so we first add the two parallel branches over a common denominator $s$ . $G = G_{a} + G_{b} = \frac{1}{s} + 2 = \frac{1 + 2 s}{s} .$ Step 2 — apply unity feedback (Rule 3, $H = 1$ ), then clear the compound fraction by multiplying top and bottom by $s$ . WHY multiply by $s$ : $s$ is the shared denominator of the pieces inside the big fraction, so this is multiplication by $1$ — value unchanged. WHY it helps: it flattens the fraction-inside-a-fraction into a plain ratio of first-degree polynomials. $\frac{Y}{R} = \frac{G}{1 + G} = \frac{( 1 + 2 s ) / s}{1 + ( 1 + 2 s ) / s} = \frac{1 + 2 s}{s + 1 + 2 s} = \frac{2 s + 1}{3 s + 1} .$

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

$R \to [-] \to G_{1} \to ∙ \to G_{2} \to Y$ , where the take-off point $∙$ (between $G_{1}$ and $G_{2}$ ) feeds back through $H$ into the summer. So the loop wraps only $G_{1}$ , and $G_{2}$ is a plain cascade after the loop. With $G_{1} = \frac{4}{s + 1}$ , $G_{2} = \frac{1}{s}$ , $H = 1$ , find $\frac{Y}{R}$ .

Recall Solution 4.2

Step 1 — identify the loop scope. The tap sits between $G_{1}$ and $G_{2}$ , so the fed-back signal is the output of $G_{1}$ only. The loop encloses just $G_{1}$ with feedback $H$ . Step 2 — reduce that inner loop (Rule 3), clearing the fraction by multiplying top and bottom by $(s + 1)$ for the same "multiply by $1$ " reason as before: $G_{l oo p} = \frac{G _{1}}{1 + G _{1} H} = \frac{4/ ( s + 1 )}{1 + 4/ ( s + 1 )} = \frac{4}{s + 5} .$ Step 3 — cascade the leftover $G_{2}$ (Rule 1, $G_{2}$ is outside the loop): $\frac{Y}{R} = G_{l oo p} \cdot G_{2} = \frac{4}{s + 5} \cdot \frac{1}{s} = \frac{4}{s ( s + 5 )} = \frac{4}{s ^{2} + 5 s} .$ WHY this order: $G_{2}$ is downstream of the take-off, so no feedback wraps it — treat the loop first, then multiply the clean cascade block.

Level 5 — Mastery

Goal: reason about poles, stability, sign of feedback, and degenerate inputs.

Exercise 5.1 (characteristic equation → stability)

For the system of Exercise 2.2, $\frac{Y}{R} = \frac{4}{s ^{2} + 2 s + 12}$ , the characteristic equation is the denominator set to zero, $s^{2} + 2 s + 12 = 0$ . Find the poles and state whether the closed loop is stable.

Recall Solution 5.1

WHAT rule: poles are the roots of the characteristic equation $1 + G H = 0$ , which is exactly the denominator $= 0$ . $s = \frac{- 2 \pm 4 - 48}{2} = \frac{- 2 \pm - 44}{2} = - 1 \pm j 11 .$ Here $j = - 1$ (the imaginary unit) and $11 \approx 3.317$ . Both poles have real part $- 1 < 0$ , i.e. they lie in the left half of the complex plane. Verdict: stable. The oscillation (from the $\pm j 11$ imaginary part) decays because the envelope $e^{- 1 \cdot t}$ shrinks.

Exercise 5.2 (positive feedback and the sign flip)

Take forward $G = \frac{k}{s + 1}$ with positive feedback, $H = 1$ . The closed-loop denominator becomes $1 - G H$ . For what value of $k$ does the closed loop have a pole exactly at $s = 0$ (the boundary of instability)?

Recall Solution 5.2

WHAT changes for positive feedback: the denominator is $1 - G H$ , not $1 + G H$ . $\frac{Y}{R} = \frac{G}{1 - G H} = \frac{k / ( s + 1 )}{1 - k / ( s + 1 )} = \frac{k}{( s + 1 ) - k} = \frac{k}{s + 1 - k} .$ The pole is at $s = k - 1$ . A pole at $s = 0$ requires $k - 1 = 0$ , so $k = 1 .$ Interpretation: for $k < 1$ the pole is at $s = k - 1 < 0$ (stable); at $k = 1$ it sits on the imaginary axis (marginal); for $k > 1$ it crosses into the right half-plane and the system blows up. Positive feedback pushes the pole rightward — the opposite of the stabilising leftward push we saw in Exercise 2.1.

Exercise 5.3 (degenerate case — the loop gain vanishes)

In a negative feedback loop $\frac{Y}{R} = \frac{G}{1 + G H}$ , suppose the feedback path is broken (an open wire), which we model as $H = 0$ . What does the closed loop become, and what does that mean physically?

Recall Solution 5.3

Set $H = 0$ : $\frac{Y}{R} = \frac{G}{1 + G \cdot 0} = \frac{G}{1} = G .$ Meaning: with no feedback returning, there is no round trip, so the loop gain $G H = 0$ and only the direct pass-through (the " $1$ ") survives. The system is just the raw open-loop block $G$ — exactly the op-amp running at its full untamed gain. This confirms the physical reading of the $1$ in $1 + G H$ : it is the signal going straight through, once, with no feedback trip.

Exercise 5.4 (limiting behaviour — huge loop gain)

Still using $\frac{Y}{R} = \frac{G}{1 + G H}$ , examine what happens as the forward gain $G \to \infty$ (very large). Simplify the closed-loop expression in that limit.

Recall Solution 5.4

Divide top and bottom by $G$ . WHY divide by $G$ : it isolates the $1/ G$ term, which is the only thing that changes as $G$ grows — making the limit trivial to read. $\frac{Y}{R} = \frac{1}{\frac{1}{G} + H} G \to \infty \frac{1}{0 + H} = \frac{1}{H} .$ Meaning: when the forward gain is enormous, the closed-loop behaviour depends only on the feedback path $H$ , not on $G$ . This is the golden principle of precision op-amps and feedback control: make $G$ huge and cheap, and let a precise, stable $H$ set the actual gain $1/ H$ .

Connections

Parent: Block diagram algebra — the rules these drill.
Transfer functions — every block above is one.
Laplace transform — why $s$ appears and why blocks multiply.
Feedback control loops — the physical setting of Rule 3.
Signal flow graphs & Mason's gain formula — an alternative to inside-out reduction.
Op-amp gain — Exercises 5.3 and 5.4 in disguise.
Stability & characteristic equation — Exercise 5.1 and 5.2.