FOUNDATION · CHAPTER 4 OF 8

Block diagrams and signal-flow graphs are equations

A block diagram is a set of linear equations, one per block and summing point; a signal-flow graph is the same set written as a branch table. This chapter derives each reduction rule as algebra and checks Mason's formula against Cramer's rule. You need it to reduce a diagram without guessing.

Updated 13 Oct 2026

What you will be able to do

You will read a block diagram as a list of linear equations and reduce it by solving them. You will derive each rule for moving a take-off point or a summing point as one line of algebra. You will write a signal-flow graph as a branch table. You will check Mason's formula against Cramer's rule and see why products of non-touching loops appear.

Before you start

Chapter 1, What one feedback loop does
Chapter 2, The Laplace transform as a working tool
Solving linear equations by substitution
Writing linear equations in matrix form
A 3 by 3 determinant as a sum of products (one entry from each row and each column) and by cofactor expansion
Cramer's rule

A block diagram is a list of equations

Recap from Chapter 2: with the input applied at $t = 0$ and zero initial conditions, a block obeys $Y (s) = G (s) U (s)$ , with $G (s)$ the transform of its impulse response. So every block gives one equation. Every summing point gives one more: its output is the sum of its inputs, each with its sign. A take-off point gives none; it only reuses a signal. Substituting the block equations into the summing-point equations leaves one equation per summing point. Eliminating the internal signals is reduction: blocks in series multiply, blocks in parallel add with their signs, and a negative-feedback loop closes to $G / (1 + G H)$ (Chapter 1).

EQUATIONS OF A DIAGRAM

Y_{i} = G_{i} U_{i} X_{k} = j \sum \pm S_{j}

Holds for linear time-invariant blocks, zero initial conditions and no loading (a block's equation does not change with what its output drives). Y_i, U_i: output and input of block i; G_i(s): its transfer function, in units of Y_i over U_i; X_k: output of summing point k; S_j: its input signals, each with its sign. All are functions of s.

A move of a take-off point or summing point is allowed when every signal the rest of the diagram uses keeps its value; signals inside the moved part may change. Take the take-off point first. A block gives $Y = G U$ . A branch that leaves after the block carries $Y$ . Drawn from before the block instead, the branch must still carry $Y$ , so it needs a copy of the block.

MOVING A TAKE-OFF POINT PAST A BLOCK

Y = G U ⟺ U = \frac{1}{G} Y

After to before: the branch needs a block G, since Y = G U. Before to after: it needs a block 1/G, since U = Y/G, which needs G(s) not identically zero. U and Y are the signals before and after the block.

MOVING A SUMMING POINT PAST A BLOCK

Y Y = G A + B = G (A + \frac{B}{G}) = G (A + B) = G A + GB

First line: B enters after the block; the same Y results if B enters before it through a block 1/G (needs G not identically zero). Second line: B enters before the block; the same Y results if it enters after through a block G. A, B: inputs; Y: output.

WATCH OUT FOR

Moving a take-off point past a summing point as if nothing changed. Before it the branch carries A; after it, A + B. Test: write the branch's equation before and after. If they differ, pass the branch through a new summing point that subtracts B.

WORKED EXAMPLE

A diagram has these equations. Summing point 1 forms the error: $E = R - C$ . Block 1: $W = G_{1} E$ with $G_{1} = 10$ . Summing point 2: $V = W - H_{1} C$ , where the feedback block $H_{1} = 0.2$ takes $C$ and its output $H_{1} C$ is already written in. Block 2: $C = G_{2} V$ with $G_{2} = 5$ . All gains are constants without units. Find $T = C / R$ .

Solution

List the equations: $E = R - C$ , $W = 10 E$ , $V = W - 0.2 C$ , $C = 5 V$ . The feedback block's equation is already inside the third, so these are four equations for the four signals $E$ , $W$ , $V$ , $C$ , with $R$ as the input.
Eliminate $W$ by substitution: $V = 10 E - 0.2 C$ . Substituting one block's equation into the next is the series rule.
Eliminate $V$ : $C = 5 (10 E - 0.2 C) = 50 E - C$ . So $2 C = 50 E$ and $C = 25 E$ . The inner loop closes to $5/2 = 2.5$ because $1 + G_{2} H_{1} = 2$ , and in series with 10 that gives 25.
Eliminate $E$ : $C = 25 (R - C)$ , so $26 C = 25 R$ .
Divide by $R$ . This is Chapter 1's $G / (1 + G H)$ with $G = 25$ and $H = 1$ : for any finite positive $G$ , $G / (1 + G)$ is below 1, and it approaches 1 as $G$ grows.

Answer: $T = C / R = 25/26 \approx 0.96$ (dimensionless)

A signal-flow graph is the same equations

A signal-flow graph writes the same equations without blocks. Each signal is a node. Each block becomes a directed branch, with the block's gain, from its input node to its output node. A summing point needs no symbol: a node's value is the sum of what arrives on its incoming branches, and a minus sign goes into the branch gain. A take-off needs none either: any number of branches may leave a node, each carrying its full value. The first example becomes the branches (R, E, 1), (C, E, $- 1$ ), (E, W, 10), (W, V, 1), (C, V, $- 0.2$ ) and (V, C, 5), written as (from, to, gain).

NODE EQUATION

x_{j} = i \sum t_{ij} x_{i}

Holds at every node except the source (the input node, which has no incoming branch), under the same conditions as before. The sum runs over the branches entering node j. x_j: value of node j (a function of s); t_ij: gain of the branch from node i to node j, in units of x_j over x_i, zero if there is none; t_jj is a self-loop.

Take a graph with an input $R$ and three nodes $x_{1}$ , $x_{2}$ , $x_{3}$ . The table lists every branch as (from node, to node, gain). The gains $a, b, c, d$ are dimensionless constants or functions of $s$ .

From node	To node	Gain
R	x1	1
x2	x1	a
x1	x2	b
x2	x3	c
x3	x3	d

NODE EQUATIONS AS A MATRIX EQUATION

x_{1} x_{2} x_{3} = R + a x_{2} = b x_{1} = c x_{2} + d x_{3} ⟹ 1 - b 0 - a 1 - c 00 1 - d x_{1} x_{2} x_{3} = R 00

Each equation is read from the table; the unknowns move to the left and R stays on the right. The entry 1 - d comes from x3 - d x3.

WATCH OUT FOR

Reading an arrow backwards. Node 2 sends values to nodes 1 and 3, but its own equation holds only the branches that enter it. Test: x2 = b x1 has one term because exactly one arrow ends at node 2.

Mason's formula from Cramer's rule

Cramer's rule gives any unknown as a ratio of two determinants, and the denominator is the determinant of the matrix above. Mason's formula is that ratio read off the graph, so you never write the matrix. Three words first. A loop is a closed path along the arrows that passes through no node twice. Two loops touch if they share a node. A forward path runs along the arrows from the input to the output, through no node twice.

THE DETERMINANT OF THE EXAMPLE

Δ = det 1 - b 0 - a 1 - c 00 1 - d = (1 - ab) (1 - d) = 1 - (ab + d) + ab d

The third column has one nonzero entry, 1 - d, so expanding along it gives (1 - d) times the 2 by 2 determinant 1 - ab. The graph has two loops: x1 to x2 and back, with gain ab, and the self-loop at x3, with gain d. They share no node, so they do not touch.

Read the expansion term by term. The 1 is the empty product, $- ab$ and $- d$ are single loops, and $+ ab d$ is a product of two loops. Why only non-touching ones? Row $i$ is node $i$ 's equation and column $j$ holds the coefficients of $x_{j}$ . So the entry in row $i$ , column $j \neq = i$ is $- t_{j i}$ , the branch from $j$ into $i$ , and a diagonal entry is $1 - t_{ii}$ . Each term of a determinant uses one entry from every row and every column. Follow one term:

Say it uses the entry in row $i$ , column $j \neq = i$ : a branch from $j$ into $i$ .
Row $j$ needs an entry too. Column $j$ is taken, so that entry lies off the diagonal, in some column $k$ : a branch from $k$ into $j$ .
Repeat. Each step reaches a new node or column $i$ , the only free column among the nodes visited. The nodes are finite, so the chain closes at $i$ : its branches form a loop.
An entry with no branch is 0, so a nonzero term holds only real loops. Loops in one term use different rows and columns, so they share no node.
That each loop enters with a minus sign, and a pair of loops with a plus, is the general rule, stated here without proof; the expansion above shows it.

THE OUTPUT BY CRAMER'S RULE

x_{3} = \frac{det 1 - b 0 - a 1 - c R 0 0}{Δ} = \frac{R b c}{Δ}

Replace the third column, that of x3, by the right-hand side and divide by Δ. Expanding the numerator along that column leaves R times the 2 by 2 determinant of rows 2 and 3, columns 1 and 2, which is bc (position 1,3 carries a plus sign). Needs Δ ≠ 0. Also bc is the gain of the only forward path: 1 · b · c.

MASON'S GAIN FORMULA

T = \frac{x _{out}}{R} = \frac{\sum _{k} P _{k} Δ _{k}}{Δ}, Δ = 1 - \sum L_{i} + \sum L_{i} L_{j} - \sum L_{i} L_{j} L_{k} + \dots

Holds for a linear graph with Δ ≠ 0. P_k: gain of forward path k. L_i: gain of loop i, self-loops included. Second sum: pairs of loops with no common node; third: triples, no two touching. Δ_k: Δ with every loop that touches path k removed (here Δ_1 = 1). Matches the Cramer result above.

WORKED EXAMPLE

In the three-node graph take $a = 0.1$ , $b = 4$ , $c = 2$ , $d = 0.5$ , all dimensionless. Find $x_{3} / R$ by Cramer's rule and by Mason's formula, and compare.

Solution

Find the loops. $ab = 0.4$ passes through $x_{1}$ and $x_{2}$ ; $d = 0.5$ is the self-loop at $x_{3}$ . They share no node, so they are non-touching.
Cramer, using the results above: $Δ = (1 - ab) (1 - d) = 0.6 \times 0.5 = 0.30$ and the numerator is $R b c = 8 R$ . So $x_{3} / R = 8/0.30 \approx 26.7$ .
Mason, denominator: $Δ = 1 - (0.4 + 0.5) + 0.4 \times 0.5 = 0.1 + 0.2 = 0.30$ . The same number. Without the cross term 0.2 you would get 0.1 and a wrong answer of 80.
Mason, numerator: one forward path, $P_{1} = 1 \times 4 \times 2 = 8$ . It passes through $x_{1}$ , $x_{2}$ and $x_{3}$ , so it touches both loops and $Δ_{1} = 1$ . Then $T = 8/0.30 \approx 26.7$ .

Answer: $x_{3} / R = 8/0.30 \approx 26.7$ (dimensionless), the same by both methods

Check yourself

CHECK YOURSELF 1

A block has output $Y = G U$ . A branch is taken off at $Y$ , after the block. It is moved to a take-off at $U$ , before the block, while still carrying $Y$ . What must the branch contain?

(A)
A block 1/G
(B)
A block G
(C)
A block $G^{2}$
(D)
A block $- G$

Answer and solution

Answer: (B)

The branch must still carry $Y$ .
From $U$ , the signal $Y$ is obtained as $Y = G U$ .
So the branch needs a copy of the block G.

CHECK YOURSELF 2

A block with $G = 4$ has input $A$ . A summing point after it adds an input $B$ , so $Y = 4 A + B$ . The input $B$ is moved to a summing point before the block, and $Y$ must not change. What gain must the branch that carries $B$ contain?

Answer and solution

Answer: $0.25$ (accepted range $\pm 0.001$ )

Write $Y = 4 A + B = 4 (A + B /4)$ .
So $B$ must enter the new summing point through a gain $1/ G = 1/4$ .
The gain is 0.25.

CHECK YOURSELF 3One or more options are correct

Which statements about a signal-flow graph of a linear system with zero initial conditions are correct?

(A)
The value of a node other than the input node is the sum, over the branches entering it, of each branch gain times the value of the node the branch starts from.
(B)
A node with two branches leaving it splits its value between them.
(C)
A minus sign at a summing point of a block diagram appears as a negative branch gain.
(D)
Two loops that share a node are non-touching.

Answer and solution

Answer: (A), (C)

A is the node equation, so it is correct.
B is wrong: every branch leaving a node carries the node's full value.
C is correct: the sign is absorbed into the branch gain, for example $- H$ .
D is wrong: loops that share a node touch.

CHECK YOURSELF 4

A graph has one forward path of gain 6. It has two loops, of gains $- 0.5$ and $- 1$ . The two loops share no node with each other, and each shares a node with the forward path. Find $T$ by Mason's formula.

Answer and solution

Answer: $2$ (accepted range $\pm 0.01$ )

The loops are non-touching, so $Δ = 1 - (- 0.5 - 1) + (- 0.5) (- 1) = 1 + 1.5 + 0.5 = 3$ .
The path touches both loops, so $Δ_{1} = 1$ .
$T = 6 \times 1/3 = 2$ .

Used in these GATE topics

Modelling, block diagrams and signal flow graphs