TECHNIQUES OF CIRCUIT ANALYSIS C.T. Pan 1 4.1 Introduction 4.2 The Node-Voltage Method ( Nodal Analysis ) 4.3 The Mesh-Current Method ( Mesh Analysis ) 4.4 Fundamental Loop Analysis 4.5 Fundamental Cutset Analysis C.T. Pan 2
4.1 Introduction A circuit consists of b branches and n nodes. A direct algebraic approach:2b method Example. C.T. Pan 3 4.1 Introduction b=6,n=4 3 independent equations only (n-1) KCL: node A: I 1 +I 4 +I 6 =0 node B:-I 4 +I 2 +I 5 =0 node C: I 3 -I 5 -I 6 =0 node D: I 1 +I 2 +I 3 =0 C.T. Pan 4
4.1 Introduction KVL: loop ABDA:V 4 +V 2 -V 1 =0 loop BCDB:V 5 +V 3 -V 2 =0 loop ACBA:V 6 -V 5 -V 4 =0 C.T. Pan 5 4.1 Introduction Component model (b branches) V 1 = E 1, V 4 = R 4 I 4, V 2 = R 2 I 2, V 5 = R 5 I 5, V 3 = E 3, V 6 = R 6 I 6, C.T. Pan 6
4.1 Introduction Want to find V k, I k, K=1,2,3,,6 There are 2b unknowns. number of KCL eqs:n-1 number of KVL eqs:b-(n-1) number of component model:b The total number of equations. n-1+[b-(n-1)]+b=2b C.T. Pan 7 4.1 Introduction Just enough information to find the solution. Problem:Too many variables!! It is not efficient! Question:Is it possible to find an optimal method with minimum unknowns? C.T. Pan 8
4.1 Introduction Nodal analysis is based on a systematic application of KCL and is a general method. Mesh Analysis is based on a systematic application of KVL and can be used for planar circuits only. C.T. Pan 9 4.1 Introduction Fundamental loop analysis is based on a systematic application of KVL to the fundamental loops. It requires the definition of tree. Fundamental cutset analysis is based on a systematic application of KCL to the fundamental cutsets. It also requires the definition of tree. C.T. Pan 10
4.1 Introduction Node: A point where two or more elements join Path: A trace of adjoining basic elements with no elements included more than once Branch: A path that connects two nodes Essential node: A node where three or more elements join Essential branch: A path that connects two essential nodes without passing through an essential C.T. Pan node 11 4.1 Introduction Loop: A path whose last node is the same as the starting node Mesh: A loop that does not enclose any other loops Planar circuit: A circuit that can be drawn on a plane with no crossing branches C.T. Pan 12
Example 4.1.1 4.1 Introduction The nodes are { a, b, c, d, e, f } The essential nodes are { b, d, e, f } The branches are { R 1, R 2, R 3, R 4, R 5, R 6, V 7, R 8 } The essential branches are { R 4, R 6, V 7, R 8, R 2 R 3, R 1 R 5 } C.T. Pan 13 Example 4.1.1 4.1 Introduction The meshes are m 1 : b c d e b m 2 : a b e f a m 3 : f e d f C.T. Pan 14
4.1 Introduction A graph G is defined as a set: G(B, N, F) B: set of branches N: set of nodes F: a function specifying how a branch is connected. Oriented (directed) graph G, if each branch of G is assigned a direction. ^ C.T. Pan 15 Example 4.1.2 GBN ˆ (,, F) 4.1 Introduction B = { b 1, b 2, b 3, b 4, b 5, b 6, b 7, b 8 } N = { a, b, c, d, e, f } F = { f 1, f 2, f 3, f 4, f 5, f 6, f 7, f 8 } f 1 : a b f 2 : b c C.T. Pan 16
4.1 Introduction For each oriented graph G one can construct a tree T. ^ A tree T of G is a connected graph which (a) contains all the nodes. (b) without forming any loop. ^ The branches which do not belong to tree T are called links. Each tree branch defines a unique cutset (or super node), called fundamental cutset. C.T. Pan 17 4.1 Introduction Each link together with its unique path connecting the two nodes of the link defines a unique loop, called fundamental loop. A network with b branches, n nodes, and l links will satisfy the fundamental theorem of network topology : b = l + n -1 C.T. Pan 18
Theorem 4.1 Introduction Given a connected graph G of n nodes and b branches, and a tree T of G There is unique path along the tree between any pair of nodes. There are n-1 tree branches and b-(n-1) links. C.T. Pan 19 4.1 Introduction Every link of T and the unique tree path between its nodes constitute a unique loop, called the fundamental loop associated with the link. Every tree branch of T together with some links defines a unique cut set of G, called the fundamental cut set associated with the tree branch. C.T. Pan 20
Example 4.1.3 4.1 Introduction G (B, N, F), b=8, n=5 B={b 1 b 2 b 3 b 4 b 5 b 6 b 7 b 8 } N={N 1 N 2 N 3 N 4 N 5 } F=[f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 ] T f 1 : N 4 N 1 f 2 : N 1 N 2 f 3 : N 2 N 3 C.T. Pan 21 ^ Example 4.1.4 4.1 Introduction Fundamental loops : T={b 2 b 3 b 5 b 8 } tree branches b 2, b 3, b 5, b 8 no. of tree branches = n-1=4 L={b 1, b 4, b 6, b 7 } links b 1, b 4, b 6, b 7 no. of links = b-(n-1) = 8-(5-1) = 4 C.T. Pan 22
Example 4.1.4 4.1 Introduction Fundamental loops : For l 1 : N 1 N 5 N 4 N 1 For l 4 : N 4 N 5 N 1 N 2 N 3 N 4 For l 6 : N 5 N 1 N 2 N 5 For l 7 : N 5 N 1 N 2 N 3 N 5 C.T. Pan 23 Example 4.1.5 4.1 Introduction Fundamental cutsets: Cut set of b 8 = {b 8, b 1, b 4 } Cut set of b 8 = {b 3, b 4, b 7 } Cut set of b 8 = {b 2, b 6, b 7, b 4 } Cut set of b 8 = {b 5, b 1, b 6, b 7, b 4 } C.T. Pan 24
Example 4.1.5 4.1 Introduction Choose tree branch current direction as reference. Cut only one tree branch. n=5, b=8 T 2 = {b 5, b 6, b 7, b 8 } Fundamental cut sets : Cut set of b 5 = {b 1, b 2, b 5 } Cut set of b 6 = {b 2, b 3, b 6 } Cut set of b 7 = {b 3, b 4, b 7 } Cut set of b 8 = {b 1, b 4, b 8 } C.T. Pan 25 4.1 Introduction Example 4.1.5 Set of links = {b 1, b 2, b 3,b 4 } Fundamental loops : Choose link current as loop current. Contain only one link. Loop of b 1 = N 1 N 5 N 4 N 1 Loop of b 2 = N 2 N 5 N 1 N 2 Loop of b 3 = N 3 N 5 N 2 N 3 Loop of b 4 = N 4 N 5 N 3 N 4 C.T. Pan 26
4.2 The Node-Voltage Method For Linear Resistive Circuits (Example) KCL KVL Component Model (Ohm s s Law) C.T. Pan 27 4.2 The Node-Voltage Method Instead of using branch voltages and branch currents as unknowns, one can choose nodal voltages as unknowns. Only n-1 nodal voltages are required. Each branch voltage can be obtained from nodal voltages and the corresponding branch currents can be calculated by its component model. KVL is automatically satisfied. C.T. Pan 28
4.2 The Node-Voltage Method One need only find the independent KCL equations,and the component model is used to express the branch current in terms of nodal voltages. Ge = I s Any unknown can be calculated once the nodal voltage vector is solved. C.T. Pan 29 4.2 The Node-Voltage Method Case 1. Containing current source only (A) independent current source by inspection (B) with dependent current source express the controlling parameter in terms of nodal voltages C.T. Pan 30
4.2 The Node-Voltage Method Case 2. Containing voltage source (VS) (A) VS contains datum node trivial case (B) VS does not contain datum node by supernode concept (C) for dependent VS express the controlling parameter in terms of nodal voltages C.T. Pan 31 4.2 The Node-Voltage Method Example 4.2.1. (Case 1A) (Basic Case) 1Ω 3Ω 2Ω 4Ω Step 1 Select a node as datum node and choose nodal variables e 1, e 2, e 3 as unknowns C.T. Pan 32
4.2 The Node-Voltage Method 1Ω 3Ω 2Ω 4Ω Step 2 Apply KCL to n-1 nodes, use ohm s law to express the branch currents in terms of nodal voltages C.T. Pan node 1: i i 3A = 0 1 2 node 2: + i + i + i = 0 2 3 4 node 3: + i i + 4A = 0 1 4 i k = 0 leaving 33 node 1: node 2: node 3: 4.2 The Node-Voltage Method ( e e ) ( e e ) + 3A = 0 3 1 1 3 1 2 ( e e ) ( e 0) ( e e ) + + = 0 1 2 4 2 1 2 2 3 ( e e ) ( e e ) 3 1 3 2 + + + = 3 4 4A 0 1 1 1+ 1 3 3 e 3A 1 1 1 1 1 1+ + e2 = 0 2 4 4 e 3 4 A 1 1 1 1 + 3 4 3 4 Use ohm s s law C.T. Pan 34 1Ω 3Ω 2Ω 4Ω
4.2 The Node-Voltage Method G G kk G kj =G jk e k I sk e1 Is1 G e = Ge = I = I e I ( ) jk 2 s2 s 3 s 3 : conduction matrix : sum of the conductances connected to node k jk : negative of the sum of conductances directly connecting nodes k and j, k k j : nodal voltage of the kth node : sum of independent current sources entering to node k C.T. Pan 35 4.2 The Node-Voltage Method Example 4.2.2. (Case 1B) 1 4 Ω 1Ω e e 2 1 e3 i x 1 1 5A 2i x 2 Ω 3 Ω C.T. Pan 36
4.2 The Node-Voltage Method Step 1 Same as Example 1 1+ 4 1 4 e1 5A 1 1+ 2+ 3 3 e = 0 2 4 3 3 4 e 3 2i + x Step 2 express -2i x in terms of nodal voltages 2i = 2 x ( e e ) 1 2 1 C.T. Pan 37 4.2 The Node-Voltage Method Step 3 Move the unknowns to the left hand side of the equation 5 1 4 e1 5 1 6 3 e = 0 2 4+ 2 3 2 3+ 4 e 3 0 Ge= I s 1Ω 1 2 Ω 1 4 Ω 1 3 Ω Note : G matrix is no longer symmetric. C.T. Pan 38
4.2 The Node-Voltage Method Example 4.2.3. (Case 2A) (trivial case) 1Ω 3Ω 4Ω Step 1 Select datum node. choose nodal voltages 2Ω 5Ω e 1, e 2, e 3 Step 2 KCL eqs. e 1 =10V, already known Need 2 KCL eqs.. only! C.T. Pan 39 4.2 The Node-Voltage Method node 1 : e = 10V 1 e 10 e 0 e e 1 2 4 e 10 e e e 3 4 5 A 2 2 2 3 node 2 : + + = 4 3 3 2 3 node 3 : + + = 4 A 1Ω 2Ω 3Ω 4Ω 5Ω 1 1 1 1 + + 10 4 2 4 4 e 2 1 1 1 1 e 10 = 3 + 4 + + 3 4 3 4 5 C.T. Pan 40
4.2 The Node-Voltage Method Example 4.2.4. (Case 2B) (supernode( supernode) 1Ω 1 4 Ω 1 2 Ω 1 3 Ω n-1=4-1=3 independent nodal voltages 1 voltage source Only 2 unknowns are needed. Say, choose e a & e b C.T. Pan 41 4.2 The Node-Voltage Method ( e e ) + ( e e ) = A A LL ( ) Node a : 4 1 5 1 a b a c Let the current through the voltage source be i x. ( eb ea) + eb( ) + ix = ( ) ( ) Node b : 2 0 Node c : i + e 3 + e e 4 = 5A x c c a ( e e ) + e ( 2) + e ( 3) + ( e e ) 4= 5 A LL ( 2) b a b c c a C.T. Pan 42
4.2 The Node-Voltage Method Which is the KCL equation written for the supernode including nodes b and c. From the known voltage source, one can get ( ) e = e 5 V LLLLLLLLLLL 3 c b C.T. Pan 43 4.2 The Node-Voltage Method Substituting (3) into (1) & (2) ( ea eb) + ( ea eb + 54 ) = 4 LLLL( 4) ( e e ) + 2e + 3( e 5) + ( e 5 e ) 4= 5 ( 5) b a b b b a From eqs.(4) and (5) 5 5 ea 20 4 = 5 10 eb 15+ 20+ 5 C.T. Pan 44
4.2 The Node-Voltage Method Example 4.2.5 (Case 2C) 1A 3 Ω 1 2v e a e b 2 Ω x e c 1Ω 1 v x 2A 1 4 Ω n-1=4-1=3, 1=3, nodal voltages There is one (dependent) voltage source. Only two unknown are needed. Say,choose e b and e c. C.T. Pan 45 4.2 The Node-Voltage Method For the supernodecontaining nodes a and b: e b (1) + (e b -e c )2 + (e a -e c )3 = 1A + 2A For node c: (1) (e c -e b )2 + e c (4) + (e c -e a )3 = 0.... (2) C.T. Pan 46 46
4.2 The Node-Voltage Method Express the controlling parameters in terms of nodal voltages. 2v x = 2(e b -e c ) 1 3 Ω 1 2v e a e b 2 Ω x e c Also, for the known voltage source e a = 2v2 x + e b = 2(e b -e c ) + e b = 3e3 b -2e c.(3) 1A 1Ω v x 2A 1 4 Ω C.T. Pan 47 4.2 The Node-Voltage Method Substituting (3) into (1) and (2) yields e b + 2(e b -e c ) + (3e b -2e c -e c )3 = 3A. (4) (e c -e b )2 + 4e c + (e c -3e b +2e c )3 = 0.(5) 12-11 eb 3A e = -11 15 c 0 1 3 Ω 1 2v e a e b 2 Ω x e c 1A 1Ω v x 2A 1 4 Ω C.T. Pan 48
4.2 The Node-Voltage Method node-to-branch incidence matrix A a A B C 2 3 5 4 1 A B b1 b2 1 1 0-1 b3 b4 b5 0 0 0 1 0 1 D C 0 0-1 1 0 D -1 0 0-1 -1 C.T. Pan 49 4.2 The Node-Voltage Method A a ={a ik }, a ik = 1 if branch i-kleaves node i -1 if branch i-kenters node k 0 if branch i-kis not incident with node i and k An oriented graph G can be represented by its incident matrix A a in computer C.T. Pan 50
4.2 The Node-Voltage Method Consider the general case Example 4.2.5 b=5, n=4 node to branch incidence matrix Aa= a b c d b 1 b 2 b 3 b 4 b 5 b 6 1 1 1 0 0 0 0-1 0 1 0 1 0 0-1 0 1-1 -1 0 0-1 -1 0 C.T. Pan 51 4.2 The Node-Voltage Method Reduced node to branch incidence matrix If node d is chosen as datum node C.T. Pan 52
4.2 The Node-Voltage Method Since each branch voltage can be expressed in terms of nodal voltages: V 1 = e a 0, Let V 2 = e a e V b V 3 = e a e c V 4 = e b 0 V 5 = e c 0 V 3 = e b e c e Then =[ V V V V V V ] T b 1 2 3 4 5 6 T Vb =[ e e e ] a b c T = N e (1) C.T. Pan 53 4.2 The Node-Voltage Method KCL, N I b =0.(2) [ ] I = I I I I I I T b 1 2 3 4 5 6 KVL, automatically satisfied Example, loop abd v v v 2 + 4 1 = 0 ( e e ) + ( e 0) ( e 0) = 0 a b b a C.T. Pan 54
4.2 The Node-Voltage Method General computer model ( ) I = I + V V G k sk k sk k G k 1 = R K = 1,2,..., b using matrix notation k [ ]( )...( 3) I = I + G V V b s b s [ 1 2LL ] [ LL ] T where I = I I I s s s sb V = V V V T s s1 s2 sb C.T. Pan 55 4.2 The Node-Voltage Method Multiply both sides of (3) by N 0=NI +N[ G]( V -V )...( 4) s b s substitute (1) into T s [ ] [ ] T [ ] [ ] 0=NI +N GN e-n GV N G N e=n GV -NI, AX=b A=N[G]N s T b=n[g]v -NI s s s s C.T. Pan 56
4.3 The Mesh Current Method (a) Another general method for analyzing planar circuits. (b) A planar circuit is one that can be drawn in a plane with no branches crossing one another;otherwise otherwise, it is nonplanar. C.T. Pan 57 4.3 The Mesh Current Method (c) A mesh in a planar circuit is a loop that does not contain any other loop within it. (d) Instead of choosing branch currents as unknown variables, one can choose mesh currents as unknown variables to reduce the number of equations which must be solved simultaneously. C.T. Pan 58
4.3 The Mesh Current Method Example 4.3.1 : mesh currents i : mesh ABEFA i i i 1 2 3 4 : mesh B C D E B : mesh E D G H E : mesh F E H K F C.T. Pan 59 4.3 The Mesh Current Method (e) If the mesh currents are known, then any branch current can be obtained from mesh currents and the corresponding branch voltage can be obtained from the component model directly. (f) For a planar circuit consists of b branches and n nodes, one can have b-(nb (n-1) l independent meshes, where l is the number of links. (g) KCL is automatically satisfied. C.T. Pan 60
4.3 The Mesh Current Method Mesh Analysis Method Step 1:Assign 1 mesh currents, i 1, i 2, i l as unknowns. Step 2:Apply 2 KVL to each mesh and use Ohm s law to express the voltages in terms of the mesh currents. Step 3:Solve 3 the algebraic equation Ax=b and find the desired answer. C.T. Pan 61 4.3 The Mesh Current Method Examples: Case 1:Planar 1 circuits with voltage sources (A) Only independent voltage sources. : by inspection. (B) With dependent voltages sources. : Need one more step, namely express the controlling parameter in terms of the mesh currents. C.T. Pan 62
4.3 The Mesh Current Method Case 2:Planar 2 circuits with current sources (A) Independent current source exist only in one mesh. :Trivial case. (B) Independent current source exists between two meshes. :Use supermesh concept (C) With dependent current source :Need one more step, namely express the controlling parameter in terms of the mesh currents. C.T. Pan 63 4.3 The Mesh Current Method Example 4.3.2:Case 1A(Basic case) 1Ω 2Ω 5V 3Ω 4Ω 10V 5Ω b= 7, n= 6 l = 7-(6-1) = 2 Choose i & i mesh currents. 1 2 mesh 1: KVL, i 1+ i 2+ 5V + ( i - i ) 3= 0 1 1 1 2 mesh 2: KVL, ( i - i ) 3 + (-5) + i 4+ 10V+ i 5= 0 2 1 2 2 C.T. Pan 64
4.3 The Mesh Current Method Rearrange the above equations 1+2+3 Ω -3Ω -3Ω 3+4+5 Ω i 1 i 2 = -5V 5-10 [ R ] i @ Ri = VS ik i:mesh current vector V :sourse voltage vector S R:resistance matrix 1Ω 2Ω 5V 3Ω 4Ω 10V 5Ω C.T. Pan 65 4.3 The Mesh Current Method R kk = sum of all the resistances in mesh k R ik = sum of the resistances between meshes i and k and the algebraic sign depends on the relative direction of meshes i and k, plus (minus)sign for same(opposite) ) direction. C.T. Pan 66
4.3 The Mesh Current Method Example 4.3.3:Case 1B 24V I O 10Ω 12Ω 4Ω 24Ω 4I O Step 1. Choose mesh currents i 1, i 2, i 3 Step 2. Same as example 1 10+12-10 -12-10 10+4+24-4 -12-4 4+12 C.T. Pan 67 i 1 i 2 i 3 = 24 0-4I 0 4.3 The Mesh Current Method Step 3. Express I o in terms of mesh currents I o =i 1 -i 2 Substitute into the above mesh equations 10+12-10 -12-10 10+4+24-4 -12+4-4-4 4+12 i 1 i 2 i 3 = 24 0 0 C.T. Pan 68
4.3 The Mesh Current Method i1 = 2.25A i = 0.75A i 2 3 = 1.5A I = i -i = 1.5A O 1 2 24V I O 10Ω 12Ω 4Ω 24Ω 4I O C.T. Pan 69 4.3 The Mesh Current Method Example 4.3.4:Case 2A 10V 4Ω 3Ω 6Ω 5Ω 5A b= 6, n=4 l= 6-(46 (4-1)=3 Step 1. Choose mesh currents i 1, i 2, i 3 Step 2. For case 2A i 3 =-5A, answer C.T. Pan 70
4.3 The Mesh Current Method KVL for mesh 1: ( i - i ) 4 + ( i - i ) 6= 10V 1 2 1 3 KVL for mesh 2: i 3 + ( i - i ) 5 + ( i - i ) 4= 0 2 2 3 2 1 Substitute i 3 1 2 1 = -5A into KVL equations ( i - i ) 4 + ( i + 5) 6= 10V i 3 + ( i + 5) 5 + ( i - i ) 4= 0 2 2 2 1 10V 4Ω 3Ω 6Ω 5Ω 5A 4+6-4 -4 3+5+4 Ω i 1 i 2 10-30 = -25 C.T. Pan 71 4.3 The Mesh Current Method Example 4.3.5:Case 2B (Supermesh( concept) 6Ω 10Ω Step 1. Choose mesh currents 20V v x 2Ω 6A 4Ω i, i 1 2 Step 2. Algebraic solution approach: Assume the voltage across the current source i s v x C.T. Pan 72
4.3 The Mesh Current Method KVL for mesh 1: i 6 + ( i - i ) 2 + (-v ) = 20V 1 1 2 x KVL for mesh 2: v + ( i - i ) 2+ i 10+ i 4= 0 x 2 1 2 2 20V 6Ω v x 2Ω 6A 10Ω 4Ω Add two KVL equations i 6+ i 10+ i 4= 20V...(A) 1 2 2 This is the KVL equation for the supermesh! Need one more equation form the current source: 6A=i - i...(b) 2 1 From (A) and (B), one can get the solution. C.T. Pan 73 4.3 The Mesh Current Method Example 4.3.6:Case 2C Step1: choose i 1, i 2, i 3, i 4 mesh currents. There are two current sources, one is independent and the other is dependent. And the number of unknowns can be reduced by 2, namely only two 2Ω equations are a required. i 1 5A 4Ω 2Ω I 0 + 6Ω i2 3I 0 3 8Ω 4 10V - C.T. Pan 74
4.3 The Mesh Current Method Step2: We can write KVL equations for mesh 4 and the supermesh as indicated in Fig. mesh 4: (i( 4 - i 3 )*8 + i 4 *2 = -10 V (A) supermesh: i 2 *6 + i 1 *2 + i 3 *4 + (i( 3 -i 4 )*8 = 0 (B) 2Ω i 1 5A 4Ω 2Ω I 0 6Ω i 3I i 2 0 3 8Ω i 4 C.T. Pan 75 + 10V - 4.3 The Mesh Current Method For two current sources, we can get two equations to eliminate two unknowns 5A = i 2 -i 1 => i 2 =i 1 +5 (C) 3I 0 = i 2 -i 3 = i 1 +5-i 3 => i 3 = i 1 +5-3I 0 (D) C.T. Pan 76
4.3 The Mesh Current Method For this case with dependent source, we need one more step, namely to express the controlling parameter I o in terms of mesh currents. I 0 = -i 4 (E) Substitute (C), (D), and (E) into (A) and (B) : [i 4 (i 1 + 5 + 3i 4 )]*8 + i 4 *2 = -10 (i 1 + 5)*6 + i 1 *2 + (i 1 + 5 + 3i 4 )*4 + [(i 1 + 5 + 3i 4 ) - i 4 ]*8=0 C.T. Pan 77 4.3 The Mesh Current Method Equivalently -8-14 i1 = 30 28 36 i -90 4 i = -7.5 A i i i 1 2 3 4 = -2.5 A = 3.93 A = 2.143 A C.T. Pan 78
4.3 The Mesh Current Method (i) General Case mesh1 mesh2 mesh3 outer mesh b1 b2 b3 b4 b5 b6-1 1 0 1 0 0 0-1 1 0-1 0 0 0 0-1 1 1 1 0-1 0 0-1 C.T. Pan 79 4.3 The Mesh Current Method M ik = 1 (-1)( if branch k is included in mesh i and the branch current direction is in the same (opposite) direction as the mesh current. M ik = 0 if branch k is not included in mesh i. Reduced mesh to branch matrix M M = -1 1 0 1 0 0 0-1 1 0-1 0 0 0 0-1 1 1 C.T. Pan 80
4.3 The Mesh Current Method KCL is automatically satisfied KVL MV B = 0 (A) V B = [ v 1 v 2 v 3 v 4 v 5 v 6 ] T v1 v 2-1 1 0 1 0 0 0 3 0-1 1 0-1 0 v = 0 v 4 0 0 0-1 1 1 0 v 5 v 6 C.T. Pan 81 4.3 The Mesh Current Method -v 1 + v 2 + v 4 = 0 : KVL for mesh 1 -v 2 + v 3 v 5 = 0 : KVL for mesh 2 -v 4 + v 5 + v 6 = 0 : KVL for mesh 3 C.T. Pan 82
4.3 The Mesh Current Method The branch currents can be expressed in terms of mesh currents I B = M T J (B) I B [ i 1 i 2 i 3 i 4 i 5 i 6 ] T J = [ j 1 j 2 j 3 ] T C.T. Pan 83 4.3 The Mesh Current Method i 1 = -j 1 i 4 = j 1 -j 3 i 2 = j 1 -j 2 i 5 = -j 2 +j 3 i 3 = j 2 i 6 = j 3 C.T. Pan 84
4.3 The Mesh Current Method A i k + v k - B v k = E SK + (i( k - I sk ) R K C.T. Pan 85 4.3 The Mesh Current Method In matrix form V B = [R]I B + E S -[R]I S (C) [R] = diag [R 1 R 2 R b ] branch resistance matrix E S = [E S1 E S2 E Sb ] T branch voltage source vector I S = [I S1 I S2 I Sb ] T branch current source vector C.T. Pan 86
4.3 The Mesh Current Method Multiply M on both sides of (C) and use relation (A) to get 0 = MV B = M[R] I B + M E S M[R]I S (D) Substitute (B) into (D) M[R]M T J = M[R]I S ME S AX = b C.T. Pan 87 4.3 The Mesh Current Method (j) Nodal Versus Mesh Analysis Both methods provide a systematic way of analyzing a complex network. Which method should be chosen for a particular problem? The choice is based on two considerations. (1)First consideration: choose the method which contains minimum number of unknowns that requires simultaneous solution. C.T. Pan 88
4.3 The Mesh Current Method (1) First consideration: choose the method which contains minimum number of unknowns that requires simultaneous solution. For a circuit contains b branches, n nodes, among them there are m v voltage sources and m i current sources, including dependent source, then the number of unknowns to be solved simultaneously is n-l-m v : for nodal analysis b-(n-1) 1)-m i : for mesh analysis C.T. Pan 89 4.3 The Mesh Current Method (2)Second consideration: Depends on the solution required, if nodal voltages are required, it may be expedient to use nodal analysis. If branch or mesh currents are required it may be better to use mesh analysis. C.T. Pan 90
4.4 Fundamental Loop Analysis Two other variations besides nodal and mesh analysis are introduced, namely the fundamental loop analysis and the fundamental cutset analyses. They are useful for understanding how to write the state equation of a circuit. C.T. Pan 91 4.4 Fundamental Loop Analysis Theorem Given an oriented graph G of n nodes and b branches, and a tree T of G, then 1. Between any pair of nodes there is a unique path along the tree. 2. There are n-1 n 1 tree branches and b-n+1 b links. C.T. Pan 92
4.4 Fundamental Loop Analysis Theorem 3. Every link of T and the corresponding unique tree path between its two terminals constitute a unique loop, called the fundamental loop associated with this link. 4. Every tree branch of T together with some links defines a unique cutset of G, called the fundamental cutset associated with this tree branch. C.T. Pan 93 4.4 Fundamental Loop Analysis G Given the oriented graph G of n (= 4) nodes and b (= 6) branches. C.T. Pan 94
4.4 Fundamental Loop Analysis I 2 Step 1 I 1 I 4 J 1 J 2 ` I 5 J 3 I 6 I 3 Pick an arbitrary tree T. Number the links from 1 to l and then tree branches from l +1 to b. C.T. Pan 95 4.4 Fundamental Loop Analysis Step 2 Choose link currents J1, J2, J3 as unknowns and adopt each link current direction as the reference of the corresponding fundamental loop current. Write down KVL equations for the l fundamental loops. C.T. Pan 96
4.4 Fundamental Loop Analysis Loop1 Loop2 Loop3 V1 V 2 1 0 0-1 -1 0 0 3 0 1 0-1 0-1 V = 0 V 4 0 0 1 0 1-1 0 V 5 V 6 branch voltage vector branch current vector [ ] [ ] V @ V V V V V V T B 1 2 3 4 5 6 I @ I I I I I I T B 1 2 3 4 5 6 In matrix form L V B = 0... ( A) C.T. Pan 97 4.4 Fundamental Loop Analysis Step 3 KCL If every branch current is formed by linear combination of loop currents, then KCL will be satisfied automatically. I = J 1 1 I = J 2 2 I = J 3 3 I = -J -J 4 1 2 I = -J+J 5 1 3 I = -J -J 6 2 3 [ ] In matrix form ( ) T I B = L J... B J = J J J Link current vector T 1 2 3 C.T. Pan 98
4.4 Fundamental Loop Analysis Step 4 Component model branch k k k sk C.T. Pan ( ) LL [ ]( ) ( ) V = I - I R+ V, k = 1, 2,, b k k sk k sk or V = R I - I + V... C B B s s K sk 4.4 Fundamental Loop Analysis C.T. Pan Multiply (C) by L matrix and substitute (A) and (B) [ ] [ ] [ ] T [ ] L V = L RI - L R I+ L V B B s s 0 = L R LJ - L RI + L V T [ ] [ ] Z J @ e L R LJ = L R I - L V s s s s ( )... D [R] :diagonal matrix Z :l l matrix, called loop impedance matrix e s :equivalent loop voltage source vector s
4.4 Fundamental Loop Analysis Step 5 Solve equation (D) and calculate I B and V B. Note that 1. Link currents are used as the unknowns. 2. The number of links is the same as that of the mesh currents. Because they are all optimal methods in terms of using minimum number of unknown current variables. C.T. Pan 101 4.4 Fundamental Loop Analysis 3. The resulting circuit loop equations depend on the choice of a tree. 4. Compared with mesh analysis, the fundament loop analysis is more general, i.e. not restricted to planar circuits, but requires defining a tree. C.T. Pan 102
4.5 Fundamental Cutset Analysis Use the same example as an demonstration. a 1 4 5 2 + e 1 - b + e 3 - + e 2-6 3 c Given the oriented graph G. Step 1 Pick a tree T. Number the links from 1 to l and then tree branches from l +1 to b. d C.T. Pan 103 4.5 Fundamental Cutset Analysis Step 2 Choose the tree branch voltages e 1, e 2, e 3 as unknowns and adopt the tree branch current as the reference direction of the associated fundamental cutset.. Write down KCL equations for the n-1 n fundamental cutsets. Cutset 1 for branch 4 Cutset 2 for branch 5 Cutset 3 for branch 6 i4 + i1 + i2= 0 i5 + i1 - i3= 0 i6 + i2 + i3= 0 C.T. Pan 104
4.5 Fundamental Cutset Analysis In matrix form i1 i 2 1 1 0 1 0 0 0 3 1 0-1 0 1 0 i = 0 i 4 0 1 1 0 0 1 0 i 5 i 6 B ( ) C I = 0... A C.T. Pan 105 4.5 Fundamental Cutset Analysis Step 3 V = C B KVL Every branch voltage can be expressed in terms of the tree branch voltages. In matrix form T e T e = e e e ( )... B 1 2 3 Tree branch voltage vector C.T. Pan 106
4.5 Fundamental Cutset Analysis Step 4 Component model branch k ( ) I = G V - V + I, k = 1, 2, 3, L, b k k k sk sk In matrix form [ ]( ) ( ) I = G V - V + I... C B B s s C.T. Pan 107 4.5 Fundamental Cutset Analysis From (A), (B) and (C) [ ] [ ] [ ] Te [ ] C I = C GV - C GV+ C I B B s s 0 = C GC - C GV+ C I T [ ] e [ ] Y e@ J CGC = C GV- C I s s C.T. Pan 108 s s ( )... D Y :cutsetadmittance matrix [G] :diagonal conductance matrix J s :cutsetcurrent source vector s
4.5 Fundamental Cutset Analysis Step 5 Solve equation (D) and calculate I B and V B. Note that 1. Tree branch voltages are used as unknowns. 2. The number of tree branches is the same as that of the nodal analysis. Because they are all optimal methods in terms of using minimum number of unknown voltages. C.T. Pan 109 4.5 Fundamental Cutset Analysis 3. The resulting cutset equations depend on the choice of a tree.. 4. Compared with nodal analysis, the fundament loop method requires defining a tree. C.T. Pan 110
Summary Table 1 Comparison of Nodal and Fundamental Cutset Analyses Methods KCL KVL Component Model Equation to be Solved Nodal Analysis Fundamental Cutset Analysis NI = 0 CI = 0 VB B = T Ne C.T. Pan 111 V B B = T Cet I = I + [ G]( V V ) I = I + [ G]( V V ) B S B S Ax= b T A= NGN [ ] b= NGV [ ] NI x= e e : node voltage vector e t : tree branch voltage vector S S B S B S Ax= b T A= CGC [ ] b= CGV [ ] CI x= e t S S Summary Table 2 Comparison of Mesh and Fundamental Loop Analyses Methods KCL KVL Component Model Equation to be Solved Nodal Analysis I B = T M Jm B = B B MV 0 Fundamental Cutset Analysis T LJ C.T. Pan 112 I = LV = 0 V = E + [ R]( I I ) V = V + [ R]( I I ) B S B S B S B S Ax= b Ax= b A= MRM [ ] b= MRI [ ] ME x= J J m : mesh current vector J : link current vector m S T S A= LRL [ ] b= LRI [ ] LV x= J S T S
Summary An interesting and practical circuit [Problem 4.16] R R R R v v 1 2 v3 vn v o C.T. Pan 113 Then n k = 1 Summary By using node-voltage method: Choose node b as the datum node. vo v1 vo v2 vo v3 vo vn + + +... + = 0 R R R R nv = ( v + v + v +... + v ) o n 1 vo = v 1 2 3 k C.T. Pan 114 n
Summary It is an averaging circuit used to find the average value of n voltages. Note that there is a common node and all resisters are equal, called sharing bus method. C.T. Pan 115 Summary Objective 1 : Understand and be able to use the node-voltage method to solve a circuit. Objective 2 : Understand and be able to use the mesh-current method to solve a circuit. Objective 3 : Be able to decide whether the node- voltage method or the mesh current method is preferred approach to solving a particular circuit. C.T. Pan 116
Summary Objective 4 : Understand how a circuit is stored in a computer and some basic graph theory. Objective 5 : Understand the basic principle of fundamental loop analysis. Objective 6 : Understand the basic principle of fundamental cutset analysis. C.T. Pan 117 Problem : 4.9 4.17 4.24 4.31 4.39 4.43 Due within one week. Summary C.T. Pan 118