# Processing a Linear Graph Into a Normal Tree

### Overview

In order to use a linear graph to find the equations which describe a system, it first must be processed into a normal tree (also known as a system graph tree). The normal tree can be used to determine the order of a system, and to write a system of equations that describe the system. This article assumes a basic understanding of how to draw the linear graph of a system, and knowledge of through and across variables.

### Normal Tree Components

There are two components that make up a normal tree, twigs (also known as branches) and links. Links are represented as hashed lines, and twigs are represented as normal lines.

The equations that result from the normal tree are sorted according to two types of variables, primary and secondary. Primary variables are represented by the across variable of twigs and the through variable of links. Secondary variables are represented by the through variable of twigs and the across variable of links.

### Sample Problem

Draw the normal tree for the following system, and determine the order of the system.

The first step of drawing a normal tree is creating a linear graph for the system. The creation of linear graphs is covered in a separate article.

A correctly drawn normal tree is composed of as many twigs as there are non-ground nodes in the linear graph. These twigs should never form a complete loop.

### Steps for drawing links and twigs

Include all across variable sources as twigs

Include as many as possible A-Type elements as twigs

Include as many as possible D-Type elements as twigs

Include as many as possible T-type elements as twigs

In the case of our example problem

### Step 1

There is one across (voltage) variable source in the example problem, \(V_S\). This must be represented as a twig

### Step 2

There are two A-type elements (\(C_1\) and \(C_2\)) in our example problem. Both can be included as twigs without creating a loop in the normal tree.

### Step 3

There are two D-Type elements (\(R_1\) and \(R_2\)) in the example problem, however if either is added as a twig, a loop will be formed.

### Step 4

There are no T-Type elements (Inductors) in this problem.

Hence, there are 3 twigs in the system (\(V_S\), \(C_1\), and \(C_2\)). The remaining segments of the linear graph (\(R_1\) and \(R_2\)) must be links.

Each A-Type twig and each T-Type link represents an independent energy storage element, and contributes to the order of the system.

There are two A-Type twigs in the system (\(C_1\) and \(C_2\)) and no T-Type links in the system. This system is of order N=2.

If you would like to know how to represent this system as a set of equations, look into how to develop elemental and constraint equations from a normal tree.

### Other Notes

It is possible that when adding twigs to the normal tree, there will be several possible options: ie \(C_1\) or \(C_2\) could be added as twigs without forming a loop but not both. In this case, both choices are viable and will result in the same system of equations.

It is worth noting that when a problem has multiple A-type elements in direct series or multiple T-type elements in direct parallel, these elements must be combined into a single equivalent element to obtain the correct order of the system.

This process was conducted for an electrical problem, however the process of drawing normal trees also applies to the mechanical (translational/rotational), thermal, and fluid domains.