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Reduced Incidence Matrix Let G be a connected digraph with “n” nodes and “b” branches. Let Aa be the Incidence Matrix of G. The (n-1) x b matrix A obtained by deleting any one row of Aa is called a ReducedIncidence Matrix of G. 1 e1 e2 KCL Equations: 2 1 e3 3 3 2 6 1 2 4 5 3 i1 + i2 − i6 = 0 −i1 − i3 + i4 = 0 −i2 + i3 + i5 = 0 4 Ai = 0 ⇒ node no. 1 1 1 2 −1 3 0 Branch no. 2 3 4 5 6 1 0 0 0 −1 0 −1 1 0 0 −1 1 0 1 0 A i1 i 2 i3 0 i4 = 0 i5 0 i6 0 i A is called the reduced Incidence Matrix of the diagraph G relative to datum node 4 . Reduced Incidence Matrix A Let G be a connected digraph with “n” nodes and “b” branches. Pick any node as the datum node and label the remaining nodes arbitrarily from 1 to n-1 . Label the branches arbitrarily from 1 to b. The reduced incidence matrix A of G is an (n-1) x b matrix where each row j corresponds to node j , and each column k, corresponds to branch k, and where the jkth element ajk of 1 , ajk = −1 , 0 , A is constructed as follow: if branch k leaves node j if branch k enters node j if branch k in not connected to node j Note: Each branch k connected to the datum node has only one non-zero entry in the kth column of A ∴ The (n-1) KCL equations derived from Ai = 0 are linearly independent. Theorem Ai = 0 gives the maximum possible number of linearly-independent KCL equations for a connected circuit. Relationship between A and Aa Let Aa be the n x b Incidence matrix of a connected digraph G with “n” nodes and “b” branches. By deleting corresponding to node any m row from Aa, we obtain the reduced incidence matrix A of G relative to the datum node m . e1 − D6 − v6 + 1 i1 v1 + D1 D2 − v3 + e2 D3 i4 2 + v4 v2 − 3 i3 e3 i5 D5 D4 − − 1 + v5 4 2 1 3 3 KVL around closed node sequence: 2 6 + i2 4 5 4 1 3 2 1 2 3 4 2 1 3 4 2 : v2 + v3 − v1 = 0 : − v3 + v5 − v4 = 0 1 : v2 + v5 − v4 − v1 = 0 These 3 KVL equations are not linearly-independent because the 3rd equation can be obtained by adding the first 2 equations: (v2 + v3 − v1 ) + (−v3 + v5 − v4 ) = v2 − v1 + v5 − v4 = 0 1 Choose 4 as datum node for digraph G e1 e2 2 1 e3 3 3 2 6 KCL Equations: 4 5 1 2 3 i1 + i2 − i6 = 0 −i1 − i3 + i4 = 0 −i2 + i3 + i5 = 0 4 Independent KCL Equations Ai = 0 ⇒ Independent KVL Equations node no. 1 1 1 2 −1 3 0 Branch no. 2 3 4 5 6 1 0 0 0 −1 0 −1 1 0 0 −1 1 0 1 0 A i1 i 2 i3 0 0 i = 4 i5 0 i6 0 i v1 = e1 − e2 v1 1 −1 0 v 1 0 −1 v2 = e1 − e3 2 e v3 0 −1 1 1 v3 = −e2 + e3 = e2 ⇒ v4 = e2 v4 0 1 0 e v5 0 0 1 3 v5 = e3 e v6 = −e1 v6 −1 0 0 v A T Note: Since each equation k from the system of KVL equations generated from T v =A e contains the variable vk, which appears in no other equations, this system of “b” KVL equations are guaranteed to be linearly independent. KVL each branch k connected between node m and the datum node 0 has a KVL equation involving only one node-to-datum voltage em: vk = em How to write An Independent System of KCL and KVL Equations Let N be any connected circuit and let the digraph G associated with N contain “n” nodes and “b” branches. Choose an arbitrary datum node and define the associated node-to-datum voltage e , the branch voltage vector v , and the branch current vector i . Then we have the vector following system of independent KCL and KVL equations. (n-1) Independent KCL Equations : Ai = 0 b Independent KVL Equations : T v = A e ik + For each 2-terminal resistor Rk, Rk vk we have 2 unknown variables {vk, ik} _ ik ij + + For each 3-terminal resistor, we have 4 unknown vj vk _ ij + vj _ _ ik variables {vj, vk, ij, ik} For each 2-port resistor, + vk we have 4 unknown _ variables {vj, vk, ij, ik} ∴ There are more unknown variables than equations derived from constitutive relations of the resistors.