# Control Systems/Signal Flow Diagrams

## §Signal-flow graphsEdit

Signal-flow graphs are another method for visually representing a system. Signal Flow Diagrams are especially useful, because they allow for particular methods of analysis, such as Mason's Gain Formula.

Signal flow diagrams typically use curved lines to represent wires and systems, instead of using lines at right-angles, and boxes, respectively. Every curved line is considered to have a multiplier value, which can be a constant gain value, or an entire transfer function. Signals travel from one end of a line to the other, and lines that are placed in series with one another have their total multiplier values multiplied together (just like in block diagrams).

Signal flow diagrams help us to identify structures called "loops" in a system, which can be analyzed individually to determine the complete response of the system.

An example of a signal flow diagram.

### §Forward PathsEdit

A forward path is a path in the signal flow diagram that connects the input to the output without touching any single node or path more than once. A single system can have multiple forward paths.

### §LoopsEdit

A loop is a structure in a signal flow diagram that leads back to itself. A loop does not contain the beginning and ending points, and the end of the loop is the same node as the beginning of a loop.

Loops are said to touch if they share a node or a line in common.

The Loop gain is the total gain of the loop, as you travel from one point, around the loop, back to the starting point.

### §Delta ValuesEdit

The Delta value of a system, denoted with a Greek Δ is computed as follows:

$\Delta = 1 - A + B - C + D - E + F......+ \infty$

Where:

• A is the sum of all individual loop gains
• B is the sum of the products of all the pairs of non-touching loops
• C is the sum of the products of all the sets of 3 non-touching loops
• D is the sum of the products of all the sets of 4 non-touching loops
• et cetera.

If the given system has no pairs of loops that do not touch, for instance, B and all additional letters after B will be zero.

### §Mason's RuleEdit

Mason's rule is a rule for determining the gain of a system. Mason's rule can be used with block diagrams, but it is most commonly (and most easily) used with signal flow diagrams.

If we have computed our delta values (above), we can then use Mason's Gain Rule to find the complete gain of the system:

[Mason's Rule]

$M = \frac{y_{out}}{y_{in}} = \sum_{k=1}^N \frac{M_k \Delta\ _k}{ \Delta\ }$

Where M is the total gain of the system, represented as the ratio of the output gain (yout) to the input gain (yin) of the system. Mk is the gain of the kth forward path, and Δk is the loop gain of the kth loop.

## §ExamplesEdit

### §Solving a signal-flow graph by systematic reduction : Two interlocking loopsEdit

This example shows how a system of five equations in five unknowns is solved using systematic reduction rules. The independent variable is $x_{in}$. The dependent variables are $x_1$, $x_2$, $x_3$, $x_4$, $x_{out}$. The coefficients are labeled $a, b, c, d, e$.

Here is the starting flowgraph:

\begin{align} x_1 &= x_\mathrm{in}+e x_3 \\ x_2 &= b x_1+a x_4 \\ x_3 &= c x_2 \\ x_4 &= d x_3 \\ x_\mathrm{out} &= x_4 \\ \end{align}

The steps for solving $x_{out}$ follow.

#### §Removing edge c from x2 to x3Edit

\begin{align} x_1 &= x_\mathrm{in}+e x_3 \\ x_2 &= b x_1+a x_4 \\ x_3 &= c x_2 \\ x_3 &= c (b x_1+a x_4) \\ x_3 &= bc x_1+ca x_4 \\ x_4 &= d x_3 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Removing node x2 and its inflowsEdit

$x_2$ has no outflows, and is not a node of interest.

#### §Removing edge e from x3 to x1Edit

\begin{align} x_1 &= x_\mathrm{in}+e x_3 \\ x_1 &= x_\mathrm{in}+e (bc x_1+ca x_4) \\ x_1 &= x_\mathrm{in}+ bce x_1+ ace x_4 \\ x_2 &= b x_1+a x_4 \\ x_3 &= bc x_1+ca x_4 \\ x_4 &= d x_3 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Remove edge d from x3 to x4Edit

\begin{align} x_1 &= x_\mathrm{in}+ace x_4 + bce x_1 \\ x_3 &= bc x_1+ac x_4 \\ x_4 &= d x_3 \\ x_4 &= d (bc x_1+ac x_4) \\ x_4 &= bcd x_1 + acd x_4 \\ x_\mathrm{out} &= x_4 \\ \end{align}

Node $x_3$ has no outflows and is not a node of interest. It is deleted along with its inflows.

\begin{align} x_1 &= x_\mathrm{in}+ace x_4 + bce x_1 \\ x_4 &= bcd x_1 + acd x_4 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Removing self-loop at x1Edit

\begin{align} x_1 &= x_\mathrm{in}+ace x_4 + bce x_1 \\ x_1 (1-bce) &= x_\mathrm{in}+ace x_4 \\ x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} x_4 \\ x_4 &= bcd x_1 + acd x_4 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Removing self-loop at x4Edit

\begin{align} x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} x_4 \\ x_4 &= bcd x_1 + acd x_4 \\ x_4 (1-acd) &= bcd x_1 \\ x_4 &= \frac{bcd}{1-acd} x_1 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Remove edge from x4 to x1Edit

\begin{align} x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} x_4 \\ x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} \times \frac{bcd}{1-acd} x_1 \\ x_4 &= \frac{bcd}{1-acd} x_1 \\ x_\mathrm{out} &= x_4 \\ \end{align}

#### §Remove outflow from x4 to xoutEdit

\begin{align} x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} \times \frac{bcd}{1-acd} x_1 \\ x_\mathrm{out} &= \frac{bcd}{1-acd} x_1 \\ \end{align}

$x_4$'s outflow is then eliminated: $x_\mathrm{out}$ is connected directly to $x_1$ using the product of the gains from the two edges replaced.

$x_4$ is not a variable of interest; thus, its node and its inflows are eliminated.

#### §Eliminating self-loop at x1Edit

\begin{align} x_1 &= \frac{1}{1-bce}x_\mathrm{in}+\frac{ace}{1-bce} \times \frac{bcd}{1-acd} x_1 \\ x_1 (1-\frac{ace}{1-bce} \times \frac{bcd}{1-acd}) &= \frac{1}{1-bce}x_\mathrm{in} \\ x_1 &= \frac{1}{(1-bce) \times (1-\frac{ace}{1-bce} \times \frac{bcd}{1-acd}) }x_\mathrm{in} \\ x_\mathrm{out} &= \frac{bcd}{1-acd} x_1 \\ \end{align}

#### §Eliminating outflow from x1, then eliminating x1 and its inflowsEdit

\begin{align} x_1 &= \frac{1}{(1-bce) \times (1-\frac{ace}{1-bce} \times \frac{bcd}{1-acd}) }x_\mathrm{in} \\ x_\mathrm{out} &= \frac{bcd}{1-acd} x_1 \\ x_\mathrm{out} &= \frac{bcd}{1-acd} \times \frac{1}{(1-bce) \times (1-\frac{ace}{1-bce} \times \frac{bcd}{1-acd}) }x_\mathrm{in} \\ \end{align}

$x_1$ is not a variable of interest; $x_1$ and its inflows are eliminated

\begin{align} x_\mathrm{out} &= \frac{bcd}{1-acd} \times \frac{1}{(1-bce) \times (1-\frac{ace}{1-bce} \times \frac{bcd}{1-acd}) }x_\mathrm{in} \\ \end{align}

#### §Simplifying the gain expressionEdit

\begin{align} x_\mathrm{out} &= \frac{-bcd}{bce+acd-1} x_\mathrm{in} \\ \end{align}


### §Solving a signal-flow graph by systematic reduction: Three equations in three unknownsEdit

This example shows how a system of three equations in three unknowns is solved using systematic reduction rules.

The independent variables are $y_1$, $y_2$, $y_3$. The dependent variables are $x_1$, $x_2$, $x_3$. The coefficients are labeled $c_{jk}$. The steps for solving $x_1$ follow:

## §Electrical engineering: Construction of a flow graph for a RC circuitEdit

This illustration shows the physical connections of the circuit. Independent voltage source S is connected in series with a resistor R and capacitor C. The example is developed from the physical circuit equations and solved using signal-flow graph techniques. Polarity is important:

• S is a source with the positive terminal at N1 and the negative terminal at N3
• R is a resistor with the positive terminal at N1 and the negative terminal at N2
• C is a capacitor with the positive terminal at N2 and the negative terminal at N3.

The unknown variable of interest is the voltage across capacitor C.

Approach to the solution:

• Find the set of equations from the physical network. These equations are acausal in nature.
• Branch equations for the capacitor and resistor. The equations will be developed as transfer functions using Laplace transforms.
• Kirchhoff's voltage and current laws
• Build a signal-flow graph from the equations.
• Solve the signal-flow graph.

### §Branch equationsEdit

The branch equations are shown for R and C.

#### §Resistor R (Branch equation $B_R$)Edit

The resistor's branch equation in the time domain is:

$V_R(t) = R I_R(t)$

In the Laplace-transformed signal space:

$V_R(s) = R I_R(s)$

#### §Capacitor C (Branch equation $B_C$)Edit

The capacitor's branch equation in the time domain is:

$V_C(t) = \frac{Q_C(t)}{C} = \frac{1}{C}\int_{t_0}^t I_C(\tau) \mathrm{d}\tau + V_C(t_0)$

Assuming the capacitor is initially discharged, the equation becomes:

$V_C(t) = \frac{Q_C(t)}{C} = \frac{1}{C}\int_{t_0}^t I_C(\tau) \mathrm{d}\tau$

Taking the derivative of this and multiplying by C yields the derivative form:

$I_C(t) = \frac{\mathrm{d}Q(t)}{\mathrm{d}t} = C\frac{\mathrm{d}V_C(t)}{\mathrm{d}t}$

In the Laplace-transformed signal space:

$I_C(s) = V_C(s) sC$

### §Kirchhoff's laws equationsEdit

#### §Kirchhoff's Voltage Law equation $\mathrm{KVL}_1$Edit

This circuit has only one independent loop. Its equation in the time domain is:

$V_R(t) + V_C(t) - V_S (t) =0$

In the Laplace-transformed signal space:

$V_R(s) + V_C(s) - V_S (s) =0$

#### §Kirchhoff's Current Law equations $\mathrm{KCL}_1, {KCL}_2, {KCL}_3$Edit

The circuit has three nodes, thus three Kirchhoff's current equations (expresses here as the currents flowing from the nodes):

\begin{align} I_S(t) + I_R (t) & = 0 & & \mathrm{(KCL_1)} \\ I_C(t) - I_R (t) & = 0 & & \mathrm{(KCL_2)} \\ I_S(t) - I_C (t) & = 0 & & \mathrm{(KCL_3)} \\ \end{align}

In the Laplace-transformed signal space:

\begin{align} I_S(s) + I_R (s) & = 0 & & \mathrm{(KCL_1)} \\ I_C(s) - I_R (s) & = 0 & & \mathrm{(KCL_2)} \\ I_S(s) - I_C (s) & = 0 & & \mathrm{(KCL_3)} \\ \end{align}

A set of independent equations must be chosen. For the current laws, it is necessary to drop one of these equations. In this example, let us choose $\mathrm{KCL}_1, {KCL}_2$.

### §Building the signal-flow graphEdit

We then look at the inventory of equations, and the signals that each equation relates:

Equation Signals
$\mathrm{B_C}$ $\mathrm{V_C, I_C}$
$\mathrm{B_R}$ $\mathrm{V_R, I_R}$
$\mathrm{KVL}_1$ $\mathrm{V_R, V_C, V_S}$
$\mathrm{KCL}_1$ $\mathrm{I_S, I_R}$
$\mathrm{KCL}_2$ $\mathrm{I_R, I_C}$

The next step consists in assigning to each equation a signal that will be represented as a node. Each independent source signal is represented in the signal-flow graph as a source node, therefore no equation is assigned to the independent source $\mathrm{V_S}$. There are many possible valid signal flow graphs from this set of equations. An equation must only be used once, and the variables of interest must be represented.

Equation Signals Assigned signal node
$\mathrm{B_C}$ $\mathrm{V_C, I_C}$ $\mathrm{I_C}$
$\mathrm{B_R}$ $\mathrm{V_R, I_R}$ $\mathrm{V_R}$
$\mathrm{KVL}_1$ $\mathrm{V_R, V_C, V_S}$ $\mathrm{V_C}$
$\mathrm{KCL}_1$ $\mathrm{I_S, I_R}$ $\mathrm{I_S}$
$\mathrm{KCL}_2$ $\mathrm{I_R, I_C}$ $\mathrm{I_R}$

### §The resulting flow graph is then drawnEdit

The next step consists in solving the signal-flow graph.

Using either Mason or systematic reduction, the resulting signal flow graph is:

### §Mechatronics exampleEdit

Angular position servo and signal flow graph. θC = desired angle command, θL = actual load angle, KP = position loop gain, VωC = velocity command, VωM = motor velocity sense voltage, KV = velocity loop gain, VIC = current command, VIM = current sense voltage, KC = current loop gain, VA = power amplifier output voltage, LM = motor inductance, VM = voltage across motor inductance, IM = motor current, RM = motor resistance, RS = current sense resistance, KM = motor torque constant (Nm/amp) , T = torque, M = momment of inertia of all rotating components α = angular acceleration, ω = angular velocity, β = mechanical damping, GM = motor back EMF constant, GT = tachometer conversion gain constant,. There is one forward path (shown in a different color) and six feedback loops. The drive shaft assumed to be stiff enough to not treat as a spring. Constants are shown in black and variables in purple.