## Tuesday, January 4, 2011

### Fundamental Loop Matrix

Sources:

http://www.transtutors.com/homework-help/Networks+%26+Systems/Network+Graph+Theories/loop-matrix-question.aspx

http://books.google.com/books?id=9c38IdqCbusC&pg=PA1119&lpg=PA1119&dq=define+fundamental+loop+matrix&source=bl&ots=63Jwqa6QEH&sig=Rqp4AG4sbnI5_eeyaVdgp_DK6AE&hl=en&ei=lfoiTaGdNo3KrAft56D-Cw&sa=X&oi=book_result&ct=result&resnum=5&ved=0CCEQ6AEwBA#v=onepage&q=define%20fundamental%20loop%20matrix&f=false

http://www.transtutors.com/homework-help/Networks+%26+Systems/Network+Graph+Theories/loop-matrix-question.aspx

Definition:

Consider a connected graph G with b branches and nt nodes. Select any arbitrary tree. The tree will contain n = nt - 1 tree branches (twigs) and l = (b - n) link branches. Every link defines a fundamental loop of the network. Let us take the example of the graph shown in Fig. a.

Fig.
Let T be a tree of G as shown in Fig. b. The number of fundamental loops of this graph will be (b - n) = 3. the three f-loop l1, l2 and l3 are shown in Fig. c. In order to apply KVL to each fundamental loop, we take reference direction of the loop which coincides with the reference direction of the link defining the loop.
l1 : v1 + v2 + v5 + = 0
l2 : v2 + v3 + v4 = 0
l3 : v1 + v3 + v6 = 0
In matrix form, we can write

Bf vb = 0   (KVL)
Where  is an  matrix called the fundamental loop matrix or tie-set matrix
Bf = [bkj]
Where bkj are the elements of bf the (k,j) element of the matrix is defined as follows:
bkj = 1 when branch bj is in the f-loop lk and their reference directions (orientations) coincide
bkj = -1 when branch bj is in the f-loop lk and has opposite orientation
bkj = 0 when branch is not in the f-loop

Example:

In the graph shown in Fig. a tree consisting of branches 4, 5, 6, 7, 8 is chosen, as shown by heavy lines. Write the fundamental loop matrix of the graph.
SOLUTION
The fundamental loops defined by links {1, 2, 3} and their orientations are shown in Figs, c and d. Consider loop l1. It contains branches {1, 6, 8, 5}. The orientation of loop l1 is given the same orientation as its defining link 1. Therefore the element b11 is written 1. The directions of branches 6, 8 and 5 in l1 are the same as l1. therefore the entries

FIG.

b16,b18  and b15  are each equal to 1. Since branches 2, 3, 4 and 7 are not in loop b13 = 0,b14 = 0,b17 = 0.
Loop l2 contains branches {2, 6, 7}. The orientation of l2 is given the same orientation as its defining link. 2. Therefore the element b22 = 1. The orientations of branches 6 and 7 are opposite to the orientation of l2 consequently, b126 = -1 and b27 = -1 Since branches {1, 3, 4, 5, 8} are not contained in l2, b21 = 0,b23 = 0,b24 = 0,b25 = 0,b28  = 0.
Loop lhas branches (3, 4, 8, 7). The orientation of l3 is the same as that of its defining link 3.Therefore the element b33 = 1 The directions of branches 4 and 7 coincide with the orientation of l3 Hence b34 = 1, b37 = 1. The orientation of branch 8 does not coincide with the orientation of l3. Hence b38 = -1. Since branches {1, 2, 5, 6} are not contained in l3,b31 = 0, b32 = 0, b35 = 0, b36 = 0. Thus, we obtain the following matrix:

#### 1 comment:

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