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EE 529 Circuit and Systems Analysis Lecture 4 EASTERN MEDITERRANEAN UNIVERSITY Matrices of Oriented Graphs THEOREM: In a graph G let the fundamental circuit and cut-set matrices with respect to a tree to be written as (r ) Bf = B and ( ) U ( ) r Af = U A r If the column orderings are identical then A = B T Matrices of Oriented Graphs Consider the following graph e3 e6 e2 e4 e5 e1 1 0 0 1 1 0 e2 A f 0 1 0 1 0 1 e4 0 0 1 0 1 1 e5 v1 e2 e3 e1 v0 e5 e4 v2 e6 e5 e3 e6 e1 e2 e4 1 1 0 1 0 0 e1 Bf 1 0 1 0 1 0 e3 0 1 1 0 0 1 e6 v3 FUNDAMENTAL POSTULATES Now, Let G be a connected graph having e edges and let xT x1 t , x2 t , and xe t y T y1 t , y2 t , ye t be two vectors where xi and yi, i=1,...,e, correspond to the across and through variables associated with the edge i respectively. FUNDAMENTAL POSTULATES 2. POSTULATE Let B be the circuit matrix of the graph G having e edges then we can write the following algebraic equation for the across variables of G Bx = 0 3. POSTULATE Let A be the cut-set matrix of the graph G having e edges then we can write the following algebraic equation for the through variables of G Ay = 0 FUNDAMENTAL POSTULATES 2. POSTULATE is called the circuit equations of electrical system. (is also referred to as Kirchoff’s Voltage Law) 3. POSTULATE is called the cut-set equations of electrical system. (is also referred to as Kirchoff’s Current Law) Fundamental Circuit & Cut-set Equations Consider a graph G and a tree T in G. Let the vectors x and y partitioned as xT xTb xTc y T y Tb y Tc where xb (yb) and xc (yc) correspond to the across (through) variables associated with the branches and chords of the tree T, respectively. Then yb xb B U x 0 c xc Bxb fundamental circuit equation and U A = 0 yc y b = -Ay c fundamental cut-set equation Series & Parallel Edges Definition: Two edges ei and ek are said to be connected in series if they have exactly one common vertex of degree two. v0 ei ek Series & Parallel Edges Definition: Two edges ei and ek are said to be connected in parallel if they are incident at the same pair of vertices vi and vk. vi ek ei vk (n+1) edges connected in series (x1,y1) (x2,y2) (x0,y0) (xn,yn) 1 x0 x 1 x 1 2 0 x3 xn 1 1 1 n x0 xi i 1 1 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 y0 y 1 0 y2 0 0 y3 1 yn y0 y1 y2 yn (n+1) edges connected in parallel (x0,y0) 1 (x1,y1) y0 y 1 y 1 2 0 y3 yn 1 1 1 n y0 yi i 1 (xn,yn) (x2,y2) 1 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 x0 x1 x2 0 x0 x 1 0 x2 0 0 x3 1 xn xn Mathematical Model of a Resistor A a v(t) i(t) B b v(t ) Ri (t ) Mathematical Model of an Independent Voltage Source v(t) A a v(t) Vs Vs i(t) i(t) B b Mathematical Model of an Independent Voltage Source v(t) a A v(t) Is i(t) Is i(t) B b Circuit Analysis A-Branch Voltages Method: Consider the following circuit. 2 k 15 V 30 V 4 k 3 k 1 k 20 V 10 mA Circuit Analysis A-Branch Voltages Method: 1. Draw the circuit graph 2 k 15 V 30 V 4 k 10 mA 3 k 1 k 20 V 2 a 3 b There are: •5 nodes (n) 4 c 1 6 5 d 7 •3 voltage sources (nv) e 8 •8 edges (e) •1 current source (ni) Circuit Analysis A-Branch Voltages Method: 1. Select a proper tree: (n-1=4 branches) Place voltage sources in tree Place current sources in co-tree Complete the tree from the resistors 2 a 3 b 4 c 1 6 5 d 7 e 8 Circuit Analysis A-Branch Voltages Method: 2. Write the fundamental cut-set equations for the tree branches which do not correspond to voltage sources. i2 i 3 1 1 1 1 1 i5 0 i6 i7 2 a 3 b 4 c 1 6 5 d 7 e 8 Circuit Analysis A-Branch Voltages Method: 2. Write the currents in terms of voltages using terminal equations. i2 i3 i5 i6 i7 v2 2k v3 4k v5 1k v6 3k 10mA 2 a 3 b 4 c 1 6 5 d 7 e 8 Circuit Analysis A-Branch Voltages Method: 2. Substitute the currents into fundamental cut-set equation. v3 v5 v6 v2 10m 2k 4k 1k 3k 2 a 3 b 4 c 1 6 5 d 7 e 8 3. v3, v5, and v6 must be expressed in terms of branch voltages using fundamental circuit equations. Circuit Analysis A-Branch Voltages Method: v3 v2 v1 v2 30 v5 v4 v2 1 15 v2 30 v2 15 v6 v2 v1 v8 v2 30 20 v2 50 v v v v 12k 2 3 5 6 10m 2k 4k 1k 3k 6v2 3v3 12v5 4v6 120 6v2 3(v2 30) 12(v2 15) 4(v2 50) 120 6 3 12 4 v2 90 180 200 120 25v2 350 v2 2 a 3 b 4 c 1 6 5 d 7 e 8 350 14 V 25 Find how much power the 10 mA current source delivers to the circuit Circuit Analysis A-Branch Voltages Method: Find how much power the 10 mA current source delivers to the circuit v7 v8 v1 v2 20 30 14 36V P10mA v7i7 36 10m 360 mW 2 a 3 b 4 c 1 6 5 d 7 e 8 Circuit Analysis Example: Consider the following circuit. Find ix in the circuit. 20 V 2 i1 4 4 ix 15 V 5 i2 i3 3 10 V Circuit Analysis Circuit graph and a proper tree 20 V 2 1 2 3 i1 6 4 4 4 ix 15 V 5 7 8 5 i3 i2 3 ix i5 10 V Circuit Analysis Fundamental cut-set equations 1 2 3 i2 i6 i8 i7 i3 i6 i8 6 4 5 7 8 v2 i2 2 v6 i6 4 v8 i8 3 v3 i3 4 v7 i7 5 Circuit Analysis Fundamental cut-set equations 1 2 3 6 4 5 7 8 v6 v8 v7 v2 2 4 3 5 v3 v6 v8 4 4 3 Circuit Analysis Fundamental circuit equations 1 2 3 6 4 5 7 8 v6 v3 v1 v2 v3 v2 20 v8 v4 v3 v1 v2 v5 15 v3 20 v2 10 v3 v2 15 v7 v1 v2 v5 20 v2 10 v2 30 Circuit Analysis v3 v2 20 v3 v2 15 v2 30 v2 60 4 3 5 2 v3 v3 v2 20 v3 v2 15 12 4 4 3 30v2 15v3 15v2 300 20v3 20v2 300 12v2 360 35v3 77v2 960...................(1) 3v3 3v3 3v2 60 4v3 4v2 60 10v3 7v2 120......................(2) v3= 9.5639V v2=-8.1203 V Circuit Analysis v7 v8 v2 30 v3 v2 15 i5 i7 i8 5 3 5 3 i5 3.48A ix 3.48 A