Block diagram algebra: Question bank — Physics deep dive

Before we start, one vocabulary anchor so nothing below uses an undefined word:

$R$ — the reference input: the signal we feed into the loop (the "command", e.g. "go to this altitude"). It is the arrow entering the very first summer.
$E$ — the error signal: the summer's output, $E = R - B$ , i.e. how far the command is from what's fed back. Think "the gap the controller must close."
$B$ — the fed-back signal: what the feedback block returns to the summer, $B = H Y$ .
$Y$ — the output: the signal leaving the forward block, $Y = GE$ . It's what we actually measure or move.
$G$ , $H$ are blocks: boxes that multiply whatever signal enters them by a function of $s$ . Think "a slide that scales your speed."
A summer is a circle that adds or subtracts the arrows entering it.
A take-off point is a dot that copies one signal to several wires (copying changes nothing).
Loop gain $= G H$ : the total factor a signal picks up going once around the loop and back.

The picture below fixes those four symbols to real wires — refer back to it whenever a derivation names $R$ , $E$ , $B$ , or $Y$ .

To see why the numerator never changes when feedback turns positive, look at the two summers side by side.

True or false — justify

Series blocks

G_{1}

then

G_{2}

give the same output if you swap their order to

G_{2}

then

G_{1}

True — for scalar transfer functions multiplication commutes,

G_{1} G_{2} = G_{2} G_{1}

, so the cascade output is identical. (This fails only for matrix/MIMO blocks, which don't commute.)

Parallel blocks

G_{1}, G_{2}

combine to

G_{1} + G_{2}

regardless of the summer's signs

False — the sign is inherited from the summer. If the lower branch enters with a minus, you get

G_{1} - G_{2}

; the plus is not automatic.

For negative feedback the closed-loop transfer function is

\frac{G}{1 + G H}

True — from

E = R - H Y

and

Y = GE

you get

Y (1 + G H) = GR

. The

+

1 + G H

is the fingerprint of the negative summer.

Positive feedback just flips the numerator sign, giving

\frac{- G}{1 + G H}

False — it flips the denominator sign:

\frac{G}{1 - G H}

. The summer now adds the fed-back signal, so

E = R + H Y

, changing the denominator, not the numerator (see the s02 derivation above).

H = 1

(unity feedback) the loop gain equals the forward gain

True — loop gain is

G H = G \cdot 1 = G

, so the round-trip factor is exactly the forward block. Closed-loop TF becomes

\frac{G}{1 + G}

You can move a summing junction downstream past a block

G

with no correction, since summers only add

False — everything the summer added now gets multiplied by

G

. To keep the arithmetic, each moved branch must be pre-multiplied by

G

G (A \pm B) = G A \pm GB

(see figure s03).

The "

1

" in

1 + G H

has no physical meaning; it's just algebra

False — it is the direct passthrough: the output reaching the input once without any feedback trip. The

G H

is the trip; the

1

is "you, going straight through, one time."

Making

H

larger always makes the closed-loop gain larger

False — larger

H

increases

G H

, which sits in the denominator

1 + G H

, so more feedback shrinks the closed-loop gain. That shrinking is the whole point of feedback (tame a huge

G

A take-off point can be slid across a block freely because copying a signal changes nothing

False — copying doesn't change the signal, but where you copy does. Past the block the tap reads

G A

instead of

A

, so you must insert

\frac{1}{G}

to recover the original value (see figure s04).

Spot the error

"Inner and outer loops? I'll reduce the outer loop first since it's the main control loop." — find the flaw

The outer feedback formula needs its forward path to be a single block, but that path still hides the un-reduced inner loop. You must work inside-out: collapse the inner loop to one block first (figure s05).

"Cascade forward path

G_{1}, G_{2}

with feedback

H

: closed-loop

= \frac{G _{1} G _{2}}{1 + G _{2} H}

." — spot it

The loop gain is the full round trip

G_{1} G_{2} H

, not

G_{2} H

. Correct denominator is

1 + G_{1} G_{2} H

; both forward blocks are inside the loop.

"Unity feedback means

H = 1

, so the closed-loop TF is just

\frac{G}{2}

." — spot it

You cannot substitute

G = 1

;

G

is a function of

s

, not the number 1. Unity feedback sets

H = 1

giving

\frac{G}{1 + G}

, which is not

\frac{G}{2}

unless

G

happened to equal 1.

"Moving a take-off downstream past

G

, I multiply the branch by

G

to match." — spot it

Backwards. Downstream the tap already reads

G A

(too big), so you divide by

G

, i.e. insert

\frac{1}{G}

. Multiplying by

G

is the rule for moving a summer past

G

, not a take-off.

"Parallel branches both start from the same wire, so I don't need a take-off point." — spot it

You do — a single wire cannot legally fan out into two blocks without a take-off dot marking the copy. It's the primitive that licenses one signal feeding two places.

"Loop gain

G H

can't affect stability; only the numerator poles matter." — spot it

Stability is set by the denominator

1 + G H = 0

(the characteristic equation), which is built entirely from the loop gain. The numerator does not set the poles.

Why questions

Why do blocks multiply their inputs rather than add them?

Because an LTI system acts by convolution in time, and the Laplace transform turns convolution into multiplication in

s

. So "pass through a block" literally is "multiply by

G (s)

Why must nested loops be reduced from the inside out?

Every feedback reduction requires its forward path to already be a single transfer function. An outer loop wrapping an un-reduced inner loop violates that, so you collapse innermost first.

Why does feedback "divide by one-plus-loop-gain" instead of subtracting the loop gain?

Because the fed-back signal is proportional to

Y

itself:

Y = G (R - H Y) = GR - G H Y

. Moving the

- G H Y

across gives

Y + G H Y = GR

, then factoring

Y (1 + G H) = GR

, so

\frac{Y}{R} = \frac{G}{1 + G H}

. The

(1 + G H)

is a collect-and-factor of the

Y

terms — never a subtraction of gains.

Why is the same

\frac{G}{1 + G H}

formula the heart of Op-amp gain?

An op-amp has enormous, poorly-controlled open-loop gain

G

; wrapping feedback

H

divides it down to a stable, resistor-set closed-loop gain

\frac{G}{1 + G H} \approx \frac{1}{H}

when

G H ≫ 1

. Same block algebra, different hardware.

Why can we treat a whole feedback loop as one equivalent block once reduced?

Because block algebra never changes the equation the diagram encodes — it only rewrites the picture. The reduced single block has an identical input-output relation

Y / R

Why do we sometimes prefer Mason's gain formula over these moves?

Because equivalence moves get error-prone in tangled multi-loop diagrams; Mason's formula reads the answer straight off the graph's paths and loops without sliding anything.

Edge cases

What is the closed-loop TF if the loop gain

G H \to \infty

(huge open-loop gain)?

\frac{G}{1 + G H} \to \frac{1}{H}

— the output becomes governed by the feedback path alone, independent of

G

. This is exactly why op-amps with big

G

give clean,

H

-controlled gain.

What happens to

\frac{G}{1 - G H}

(positive feedback) when

G H \to 1

The denominator

\to 0

, so the closed-loop gain blows up — the system is on the edge of instability/oscillation. This is the resonance/runaway boundary of positive feedback.

What is the equivalent block of a parallel pair when

G_{2} = - G_{1}

into an adding summer?

G_{1} + G_{2} = G_{1} - G_{1} = 0

— the branches cancel and the equivalent block is zero; no signal reaches the output. A degenerate but perfectly legal result.

If a feedback block is

H = 0

, what does the loop reduce to?

\frac{G}{1 + G \cdot 0} = \frac{G}{1} = G

— no feedback at all, so you recover the open-loop forward block, as expected when the return path is cut.

What is the series-equivalent block if one block in the chain is

G = 1

(a pure wire)?

It contributes a factor of 1:

G_{1} \cdot 1 \cdot G_{2} = G_{1} G_{2}

. A unity block is an identity element — it passes the signal through unchanged.

Can you move a summing junction past a take-off point with no correction?

No — you must preserve every signal value at both. Such rearrangements need compensating blocks or duplicated summers; always re-derive by insisting each wire keeps its exact value, then test with a dummy input.

Connections

Transfer functions — the $G (s)$ , $H (s)$ objects that live inside each block; every trap above is really a statement about how these multiply and combine.
Laplace transform — the reason "pass through a block" means "multiply", since it turns time-domain convolution into $s$ -domain multiplication (answers the first Why question).
Feedback control loops — the engineering context for Rule 3; the $R$ / $E$ / $B$ / $Y$ derivation on this page is exactly the loop those systems close.
Signal flow graphs & Mason's gain formula — the move-free alternative that sidesteps the take-off/summer sliding traps in the Spot-the-error section.
Op-amp gain — the hardware where the $G H \to \infty \Rightarrow 1/ H$ edge case actually gets used to build stable amplifiers.
Stability & characteristic equation — explains why the denominator $1 + G H = 0$ (not the numerator) sets the poles, justifying the last Spot-the-error item.