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The document discusses matrix analysis of networks, focusing on understanding the structure and properties of interconnected systems. It explains fundamental concepts such as tree and co-tree formations, the use of Kirchhoff's laws in matrix form, and techniques for circuit analysis using mesh and nodal methods. The analysis includes equations, graphs, and examples to illustrate how to apply matrix techniques for solving circuit problems.

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Get Homework/Assignment Done Homeworkping.com Homework Help https://www.homeworkping.com/ Research Paper help https://www.homeworkping.com/ Online Tutoring https://www.homeworkping.com/ click here for freelancing tutoring sites Matrix Analysis of Networks – J. R. Lucas Used to have a compact and neat form of solution. • necessary to know the structure of a network, and Matrix Analysis of Networks – Professor J R Lucas 1 May 2011
• formulate the problem based on the structure Because Large networks are • tedious to analyse using normal equations • easier/more convenient to formulate in matrix form. Matrix Analysis of Networks – Professor J R Lucas 2 May 2011
Topology Deals with structure of an interconnected system Formulates the problem based on non-measurable properties of network. Geometric structure of the interconnection of network elements completely characterises − number of independent loop currents − number of independent node-pair voltages that are necessary to study the network. Matrix Analysis of Networks – Professor J R Lucas 3 May 2011
Figure 1 – Structure of the network 1(a) and (b) have the same structure (or topology). However elements are quite different. Matrix Analysis of Networks – Professor J R Lucas 4 May 2011 R1 L1 C (a) (b)
Graph of Network Figure 2 – Circuit Matrix Analysis of Networks – Professor J R Lucas 5 May 2011 (a) Network (b) Graph of Network
Figure 2 Circuit of figure 2(a) also has same topology. Figure 2(b) shows structure corresponding to all 3 circuits. Does not indicate any of the elements in the networks. Known as the graph of the network – Has all the nodes of the original network In obtaining the graph, − each element of the network is represented by a line − each voltage source by a short-circuit Matrix Analysis of Networks – Professor J R Lucas 6 May 2011
− and each current source by an open circuit. Matrix Analysis of Networks – Professor J R Lucas 7 May 2011
Tree of a Network Which of the diagrams would also represent a normal tree (without leaves) ? and why ? Only first diagram would fully satisfy the requirements. Second diagram has branches closing on itself − only a tree like Nuga might appear to close on itself Third diagram has branches in mid air not joined to main tree. Matrix Analysis of Networks – Professor J R Lucas 8 May 2011
Properties associated with trees. 1. All branches must be part of the tree 2. There cannot be closed loops formed from branches 3. There cannot be branches isolated from the tree Same properties apply in defining a tree of a network − can be many trees associated with a given network. − need not have a trunk coming from the ground and branches coming from the trunk. − a reduced graph of network with some of the links removed so as to leave all the nodes connected together by graph, but not to have any loop left. Matrix Analysis of Networks – Professor J R Lucas 9 May 2011
Possible trees for the Graph When a tree of the network is removed from graph, what remains is called the co-tree of the network. Co-tree is graph of removed links –compliment of the tree. A co-tree may contain closed loops, and disconnected branches. Matrix Analysis of Networks – Professor J R Lucas 10 May 2011 Graph of Network Some of the possible trees
Analysis structure of network A single branch is required to join two nodes. Joining each additional node would require an additional branch. Let b = number of branches in the network n = number of nodes in the network l = number of independent loops Thus number of branches in tree = n – 1 number of links removed = b – (n – 1) = b – n + 1 Matrix Analysis of Networks – Professor J R Lucas 11 May 2011 Node 1 Node 2 Node 1 Node 2 New Node
Formation of Independent Loops If any one of removed links are added to the tree, then a new loop is formed. ∴ number of links removed from graph to form the tree is equal to the number of independent loops. l = b – n + 1 Oriented Graph − Numbered branches with assigned directions to currents. − Voltage considered to increase in direction opposite to flow of current Matrix Analysis of Networks – Professor J R Lucas 12 May 2011 Oriented Graph 1 32 4 6 5
Matrix Analysis of Networks To solve circuit problems, need to write the equations corresponding to • Ohm’s Law, and • Kirchoff’s Current Law • Kirchoff’s Voltage Law Same is true even when there are a large number of branches. − use matrix analysis Matrix Analysis of Networks – Professor J R Lucas 13 May 2011
k -1 -1 +1 +1 +1 0 0 i1 i2 i3 i4 i6 i7 i5 Kirchoff’s current Law in matrix form For any node k − i1 − i2 − i4 + i6 + i7 = 0 or i1 + i2 + i4 = i6 + i7 or i1 + i2 + i4 – i6 – i7 = 0 or +1. i1 + 1. i2 + 0 .i3 + 1.i4 + 0. i5 – 1.i6 – 1.i7 = 0 Last form is preferred for matrix implementation Matrix Analysis of Networks – Professor J R Lucas 14 May 2011
− all currents in network are included in equation with different coefficients. For computer implementation, there must be a unique method (convention) of obtaining the coefficients ajk. Ij – current in jth branch jth branch − directed away from kth node: ajk = +1 − directed towards kth node: ajk = –1 − not incident on the kth node: ajk = 0 Kirchoff’s current law may be written, for the kth node Matrix Analysis of Networks – Professor J R Lucas 15 May 2011
a1k .i1 + a2k . i2 + a3k .i3 + a4k .i4 + ...... ....... ..... a7k .i7 = 0 or ∑ = =⋅ b j jjk ia 1 0 at kth node, for all k Collection of equations, for each node k, would give [ ] )1()1()( 0 ××× =⋅ nb b bn It A In [A]t , row vectors are dependant, since sum is zero. [A]t written with one row less, giving only (n-1) rows. [A]t – node-branch incidence matrix, (n-1)×b. [A] – branch-node incidence matrix, b× (n-1) ajk = +1 if jth current is directed away from the kth node ajk = −1 if jth current is directed towards the kth node Matrix Analysis of Networks – Professor J R Lucas 16 May 2011
ajk = 0 if jth current is not incident on the kth node Kirchoff’s voltage Law in matrix form ∑ = =⋅ b r rrs vb 1 0 for sth loop, for all s; where brs = −1, 0, or +1 [ ] )1()1()( 0 ××× =⋅ lb b bl Vt B [B]t – mesh-branch incidence matrix, (l×b) [B] – branch-mesh incidence matrix, (b×l) brs = +1if rth current is in same direction as sth loop brs = −1if rth current is in opposite direction to sth loop brs = 0 if rth current is in the not part of the sth loop Ohm’s Law in matrix form Matrix Analysis of Networks – Professor J R Lucas 17 May 2011 0 0 0+1 -1 -1 0 -1 +1 s 0 0 0
for all branches k = 1, 2, .... ... b vk = – egk + Zk igk + Zk ik Either voltage source or current source would normally be used. Conversion with either Thevenin’s Theorem or Norton’s Theorem. Matrix Analysis of Networks – Professor J R Lucas 18 May 2011 egk igk Zkik + igk ik vk Figure - General branch
With a voltage source only vk = – egk + Zk ik for all branches k = 1, 2, ....... b and in matrix form as[ ] bbgbb IZEV +−= With a current source only ik = Yk vk – igk for all branches k = 1, 2, .... ... b and in matrix form as [ ] bbgbb VYII +−= , where [Yb] = [Zb]-1 Matrix Analysis of Networks – Professor J R Lucas 19 May 2011 egk Zk vk ik igk Ykik vk
In Summary From Kirchoff’s Laws [ ] )1()1()( 0 ××× =⋅ nb b bn It A (1) (n-1) independent equations [ ] )1()1()( 0 ××× =⋅ lb b bl Vt B (2) l independent equations and from Ohm’s Law [ ] bbgbb IZEV +−= (3) b independent equations or [ ] bbgbb VYII +−= (3)* b independent equations Thus total number of independent equations is n – 1 + l + b = b + b = 2 b 2b independent equations 2b unknowns (b branch currents and b branch voltages) Matrix Analysis of Networks – Professor J R Lucas 20 May 2011
Can be solved. Not usual to solve for both current and voltage simultaneously. Reductions can be done in two ways. 1) Eliminate voltages and solve for currents − mesh analysis 2) Eliminate currents and solve for voltages. − nodal analysis. Matrix Analysis of Networks – Professor J R Lucas 21 May 2011
Mesh Analysis − Eliminate the branch voltages from the equations. − Reduce remaining currents to a minimum using Kirchoff’s current law. Apply Kirchoff’s voltage law for solution. Define a set of mesh currents, mI . Branch currents bI related to mesh currents mI by an algebraic summation. [ ] mb IBI = (4) Eliminate Vb from the equations, Pre-multiply equation (3) by [B]t . Matrix Analysis of Networks – Professor J R Lucas 22 May 2011
[ ] [ ] [ ] [ ] bb t gb t b t IZBEBVB +−= from equation (2), [B]t Vb = 0. Also [ ] mb IBI = ∴ [ ] [ ] [ ][ ] mb t gb t IBZBEB = [B]t Vb = 0 → sum of voltages around a loop is zero. i.e. [B]t Vb → sum of voltages around a loop. ∴ [B]t Egb → sum of source voltages around a loop. Defined as mesh source voltage vector Egm . i.e. Egm = [B]t Egb ∴ Egm = [ ] [ ][ ] [ ] mmmb t IZIBZB = Matrix Analysis of Networks – Professor J R Lucas 23 May 2011
where [Zm] = [ ] [ ][ ]BZB b t corresponds to l equations [B] also known as the tie-set matrix (as its elements tie the loop together) Unknowns are l values of current Im Original 2b equations and 2b unknowns reduced to l equations and l unknowns. Elements of [Zm] can be obtained either from above mathematics, or by inspection as follows. Simple evaluation of [Zm] and Egm zjj = self impedance of mesh j = sum of all branch impedances in mesh j Matrix Analysis of Networks – Professor J R Lucas 24 May 2011
zjk = mutual impedance between mesh j and mesh k = sum of all branch impedances common to mesh j and mesh k and traversed in mesh direction − sum of all branch impedances common to mesh j and mesh k, and traversed in opposite direction ej = algebraic sum of the branch voltage sources in mesh j in mesh direction. Matrix Analysis of Networks – Professor J R Lucas 25 May 2011
Example 1 Solve the circuit using Mesh matrix analysis. Work from first principles. Solution Number the branches and the loops. Matrix Analysis of Networks – Professor J R Lucas 26 May 2011 j6 Ω E1 100∠00 V j20 Ω -j120 Ω E2 100∠300 V 10 Ω 20 Ω 10 Ω
Write the loop currents in terms of the branch currents. i1 = I1 i2 = – I3 i3 = I1 – I2 i4 = I2 i5 = I2 – I3 i6 = I3 Matrix Analysis of Networks – Professor J R Lucas 27 May 2011 or in matrix form                               − − − =                     3 2 1 6 5 4 3 2 1 100 110 010 011 100 001 I I I i i i i i i I1 I2 I3 i1 i4 i3 i5 i6 i2 E1 100∠00 V j20 Ω j6 Ω –j120 Ω E2 100∠36.870 10 Ω 20 Ω 10 Ω
This gives the Branch-Mesh incidence matrix [B]. Mesh–Branch incidence matrix [B]t can also independently by writing the relation between the mesh direction and the branch direction. [ ]           −− −= 110010 011100 000101 t B Notice that this corresponds to the transpose of the earlier written matrix. Matrix Analysis of Networks – Professor J R Lucas 28 May 2011
Vector of branch source voltages is Branch impedance matrix is [ ]                     − = 600000 0100000 0020000 00012000 0000100 0000020 j j j Zb Egm = [B]t Egb , and [Zm] = [B]t [Zb] [B]                     ∠ ∠           −− −= 0 0 0 0 87.36100 0100 110010 011100 000101 0 0 gmE , Egm =           ∠− ∠ 0 0 87.36100 0 0100 Matrix Analysis of Networks – Professor J R Lucas 29 May 2011                     ∠ ∠ = 0 0 0 0 87.36100 0100 0 0 gbE
[ ]                     − − −                     −           −− −= 100 110 010 011 100 001 600000 0100000 0020000 00012000 0000100 0000020 110010 011100 000101 j j j Zm [ ]                     − − −           −− −= 600 10100 0200 0120120 1000 0020 110010 011100 000101 j jj j Zm =           +− −− − 620100 1012030120 0120100 j jj jj Both Egm and Zm could have been written by inspection. Thus           ∠− ∠ 0 0 87.36100 0 0100 =                     +− −− − 3 2 1 620100 1012030120 0120100 I I I j jj jj Equations may be solved by inversion or otherwise. Matrix Analysis of Networks – Professor J R Lucas 30 May 2011
          ∠− ∠           −−−×−×− ×−+−+− ×−+−−+− ∆ =           0 0 2 2 3 2 1 87.36100 0 0100 )120()12030(1001010010120 10100)620(100)620(120 10120)620(12010)620)(12030( 1 jjjjj jjjjj jjjjj I I I           −−          +−−−− −+−+− −+−− ∆ =           6080 0 100 1440012000300010001200 100060020007202400 1200720240022201220 1 3 2 1 jjjj jjj jjjj I I I ∆ = (1220 – j2220)×(–j100) + (720 – j2400) ×(j120) + (–j1200) ×0 = – j122000 – 222000 + j 86400 + 288000 = 66000 – j 35600 = 74989∠-28.34o I1 = (122000 – j 222000 + 0 + j 96000 – 72000)/74989∠-28.34o = (50000 – j 126000)/ 74989∠-28.34o = 135558∠-68.36o /74989∠-28.34o = 1.808∠-40.02o A [Note: Inversion has not been checked so answers may be in error.] Currents I2 and I3 can be similarly determined. The branch currents i1, i2, ..... may then be determined from the matrix equation. [Normally branch 6 would have been marked as part of branch 2] Matrix Analysis of Networks – Professor J R Lucas 31 May 2011
Nodal Analysis − eliminate branch currents from the equations. − Reduce number of remaining voltages to a minimum using Kirchoff’s voltage law. Apply Kirchoff’s current law for solution. Define a set of nodal voltages, NV which are node pair voltages (i.e. voltage across a pair of nodes) Branch voltages bV are related to nodal voltages NV by an algebraic summation. [ ] Nb VAV = (5) [A] too does not have the reference node. Matrix Analysis of Networks – Professor J R Lucas 32 May 2011
Pre-multiply equation (3)* by [A]t . [ ] [ ] [ ] [ ] bb t gb t b t VYAIAIA +−= from equation (1), [A]t Ib = 0 . Substituting from (5) [ ] [ ] [ ][ ] Nb t gb t VAYAIA +−=0 [ ] [ ] [ ][ ] Nb t gb t VAYAIA = IgN = [YN]VN where [ ] gb t gN IAI = , and [ ] [ ] [ ][ ]AYAY b t N = Source nodal current vector IgN and the nodal admittance matrix [YN] could be written by inspection. yii = sum of all branch admittances incident at node i Matrix Analysis of Networks – Professor J R Lucas 33 May 2011
yij = negative of the sum of all branch admittances connecting node i and node j . Reason for negative sign can be understood as follows: ik = yk vk = yk (Vi – Vj) At any node i, injected current Igi = Σ ik = Σ yk (Vi – Vj) ∑∑ ≠ = ≠ = −= N ij j jk N ij j ikgi VyVyI 11 for all j Since Vi is a constant for a given i, ∑∑∑∑ ≠ = ≠ = ≠ = ≠ = −+         =−= N ij j jki N ij j k N ij j jk N ij j kigi VyVyVyyVI 1 ) 111 ( ∑∑ = ≠ = =+= N j jii N ij j jiiiiigi VyVyVyI 11 corresponds to nodal equation As in the case of mesh analysis, Matrix Analysis of Networks – Professor J R Lucas 34 May 2011 vk ji ykik
IgN = [YN]VN is first solved to give VN and the branch voltages and branch currents then obtained using the matrix equations. Matrix Analysis of Networks – Professor J R Lucas 35 May 2011
Example 2 Example 1 has been reformulated as a problem with current sources rather than with voltage sources. [If voltage sources are present, they would first have to be converted to current sources]. Matrix Analysis of Networks – Professor J R Lucas 36 May 2011 5∠-900 A j20 Ω j6 Ω -j120 Ω 8.575∠5.910 A10 Ω 20 Ω 10 Ω i1 i4 i3 i5 i2 V1 V2
Network may also be drawn in terms of admittances. The branch-node incidence matrix [A], branch injected current Igb, and branch admittance matrix may be written, with reference selected as earthed node as follows. Matrix Analysis of Networks – Professor J R Lucas 37 May 2011 ∠      ∠     i1 i4 i3 i5 i2 V1 V2
[ ]                 − − − = 10 11 01 10 01 A , Igb =                 ∠ −∠ 0 0 0 91.5575.8 905 o o , [ ]                 − − = 1.00000 005.0000 00008333.000 0000441.00735.00 000005.0 j j j Yb As in mesh analysis, nodal current injection vector and nodal admittance matrix may be written from first principles. Left as an exercise for you to work out. This is worked by inspection. [ ] [ ]       −++− −++− =      ∠ −∠ = 0441.00735.01.005.005.0 05.005.000833.005.0 , 91.5575.8 905 j jj YI No o gN             −++− −++− =      ∠ −∠ ∴ 2 1 0441.00735.01.005.005.0 05.005.000833.005.0 91.5575.8 905 V V j jj o o       ∠ −∠       ++− −++ ∆ =      ∴ o o jj j V V 91.5575.8 905 05.000833.005.005.0 05.00441.00735.01.005.01 2 1 ∆ = (–j0.05+j0.00833+0.05)(0.05+0.1+0.0735–j0.0441) – 0.052 Matrix Analysis of Networks – Professor J R Lucas 38 May 2011
= (0.05 – j 0.04167)(0.2235 – j 0.0441) – 0.0025 = 0.06509∠-39.81o ×0.2278∠-11.16o – 0.0025 = 0.01483∠-50.97 – 0.0025 = 0.00934 – 0.0025 – j 0.01152 = 0.00684 – j 0.01152 = 0.0134∠-59.30o V1 = (0.2278∠-11.16o ×5∠-90o +0.05×8.575∠5.91o )/0.0134∠-59.3o = (– 0.2205 – j 1.1175 + 0.4265 + j 0.04415) /0.0134∠-59.30o = (0.2060 – j 1.0733)/0.0134∠-59.30o = 1.093∠-79.14o /0.0134∠-59.30o V1 = 81.6∠-19.84o V ∴branch current i1 = 20 68.273.23 20 84.196.81100 j j j o + = −∠− i1 Aj o 09.40809.1165.1384.1 −∠=−= Matrix Analysis of Networks – Professor J R Lucas 39 May 2011
which is the same answer (to calculation accuracy) that was obtained in example 1. Matrix Analysis of Networks – Professor J R Lucas 40 May 2011
Conversion of Ideal sources (a) Ideal Voltage sources No impedance directly in series with voltage source Ideal voltage sources are distributed to branches connected to one of the nodes of original ideal source. Matrix Analysis of Networks – Professor J R Lucas 41 May 2011 E      ≡                            
(b) Ideal Current sources No admittance appears directly in parallel with current source Ideal current source has been distributed around a loop connecting the two points of original source. Matrix Analysis of Networks – Professor J R Lucas 42 May 2011 ≡        
Matrix Analysis of Networks – Professor J R Lucas 43 May 2011
Port Pair of nodes across which a device can be connected. Voltage is measured across the pair of nodes. Current going into one node is the same as the current coming out of the other node in the pair. These pairs are entry (or exit) points of the network. Compare with an Airport or a Sea Port. Entry and exit points to a country. Planes that enter at a given port are the ones that take off from same port. Matrix Analysis of Networks – Professor J R Lucas 44 May 2011
Two-Port Theory Convenient to develop special methods for systematic treatment of networks. Single-port linear active networks − Thevenin’s or Norton’s equivalent circuit. Linear passive networks − Convenient to study behaviour relative to a pair of designated ports. Matrix Analysis of Networks – Professor J R Lucas 45 May 2011           
Definitions Driving point impedance is defined as ratio of applied voltage (driving point voltage) across a node-pair to the current entering at the same port. [input impedance of network seen from particular port] Driving point impedance at Port 1 = V1/I1 Driving point impedance at Port 2 = V2/I2 Driving point admittance is similarly defined as the ratio of the current entering at a port to the applied voltage across the same node-pair. Driving point admittance at Port 1 = I1/V1 Driving point admittance at Port 2 = I2/V2 Matrix Analysis of Networks – Professor J R Lucas 46 May 2011 
Immittance is sometimes used to represent either an impedance or an admittance Transfer impedance is defined as the ratio of the applied voltage across a node-pair to the current entering at the other port. Transfer impedance = V1/I2 , V2/I1 Transfer admittance is similarly defined as the ratio of the current entering at a port to the voltage appearing across the other node-pair. Transfer admittance = I1/V2 , I2/V1 Matrix Analysis of Networks – Professor J R Lucas 47 May 2011
Transfer Voltage gain (or ratio) is defined as the ratio of the voltage at a node pair to the voltage appearing at the other node-pair. Transfer voltage gain = V1/V2 , V2/V1 Transfer Current gain (or ratio) is similarly defined as the ratio of the current at a port to the current at the other port. Transfer current gain = I1/I2 , I2/I1 Matrix Analysis of Networks – Professor J R Lucas 48 May 2011
Common Two-port parameters External conditions of a two-port network can be completely defined by currents and voltages at the 2 ports. A general two port network can be characterised by four parameters, derived from the network elements. With symmetry, number of parameters will be reduced. (a) Impedance parameters (b) Admittance parameters (c) Transmission Line parameters (d) Hybrid parameters. Matrix Analysis of Networks – Professor J R Lucas 49 May 2011
(a) Impedance Parameters (z-parameters) or Open-circuit parameters V1 = z11 I1 + z12 I2 If I2 = 0, then z11 = V1/I1 If I1 = 0, then z12 = V1/I2 It follows that, Matrix Analysis of Networks – Professor J R Lucas 50 May 2011               
021 1 11 = = II V z , 012 1 12 = = II V z , 021 2 21 = = II V z , 012 2 22 = = II V z . z-parameters correspond to the driving point and transfer impedances at each port with the other port having zero current (i.e. open circuit). → open circuit parameters. Matrix Analysis of Networks – Professor J R Lucas 51 May 2011
Example 3 Find impedance parameters of the two port T – network. With port 2 on open circuit ba ZZ II V z += = = 021 1 11 bZ II V z = = = 021 2 21 similarly with port 1 open, z12 = Zb z22 = Zb + Zc (b) Admittance Parameters (y-parameters) or Short-circuit parameters Matrix Analysis of Networks – Professor J R Lucas 52 May 2011              [ ]       + + =→ cbb bba ZZZ ZZZ Z
y11, y12, y21, y22 defined with either V1 or V2 zero. y-parameters correspond to driving point and transfer admittances at each port with the other port having zero voltage (i.e. short circuit) → short circuit parameters. Matrix Analysis of Networks – Professor J R Lucas 53 May 2011                
Example 4 Find admittance parameters of the 2 port π–network. y11 = 021 1 =VV I = Ya + Yb y21 = 021 2 =VV I = – Yb Matrix Analysis of Networks – Professor J R Lucas 54 May 2011                      →[Y] =       +− −+ cbb bba YYY YYY
(c) Transmission Line Parameters (ABCD-parameters) Parameters can be defined using either port 2 on short circuit or port 2 on open circuit. In case of symmetrical system, parameter A = D. For a reciprocal system, A.D – B.C = 1 Matrix Analysis of Networks – Professor J R Lucas 55 May 2011                            =      2 2 1 1 I V DC BA I V
Example 5 Find ABCD parameters. A = c cb Y YY + , B = cY 1 C = c accbba Y YYYYYY ++ and D = c ca Y YY + [For symmetrical network, Ya = Yb , A = D]. A.D – B.C = c accbba cc ca c cb Y YYYYYY YY YY Y YY ++ ⋅− + ⋅ + 1 = 2 2 )( c accbbaccbacab Y YYYYYYYYYYYYY ++−+++ =1 (d) Hybrid Parameters (h-parameters) Matrix Analysis of Networks – Professor J R Lucas 56 May 2011                       
The hybrid parameter matrix may be written as             =      2 1 2221 1211 2 1 V I hh hh I V h-parameters can be defined as in other examples, and are commonly used in some electronic circuit analysis. Matrix Analysis of Networks – Professor J R Lucas 57 May 2011  
Interconnection of two-port networks (a) Series connection of two-port networks Series properties are applied to each port at port 1, Ir1 = Is1 = I1, and Vr1 + Vs1 = V1 at port 2 Ir2 = Is2 = I2, and Vr2 + Vs2 = V2 Matrix Analysis of Networks – Professor J R Lucas 58 May 2011                                
[Z] = [Zr] + [Zs] Matrix Analysis of Networks – Professor J R Lucas 59 May 2011
(b) Parallel connection of two-port networks at port 1, Ir1 + Is1 = I1, and Vr1 = Vs1 = V1 Matrix Analysis of Networks – Professor J R Lucas 60 May 2011                          
at port 2, Ir2 + Is2 = I2, and Vr2 = Vs2 = V2 [Y] = [Yr] + [Ys] (c) Cascade connection of networks Output of one network becomes input to next. Ir2 = Is1 Vr2 = Vs1             =      2 2 1 1 r r rr rr r r I V DC BA I V ,             =      2 2 1 1 s s ss ss s s I V DC BA I V                   =      2 2 1 1 I V DC BA DC BA I V ss ss rr rr ABCD matrix of component networks Matrix Analysis of Networks – Professor J R Lucas 61 May 2011                                       
A = 022 1 =IV V = 1, = 1 B = 022 1 =VI V = Z, = 0 C = 022 1 =IV I = 0, =Y D = 022 1 =VI I = 1, = 1 In matrix form       DC BA =       10 1 Z , =       1 01 Y Consider example 5 again Matrix Analysis of Networks – Professor J R Lucas 62 May 2011               ≡   
      DC BA =       1 01 aY         10 11 bY       1 01 cY Simplification of matrix product would give the same answer as in example 5. Matrix Analysis of Networks – Professor J R Lucas 63 May 2011

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240164036 ee2092-4-2011-matrix-analysis

  • 1.
    Get Homework/Assignment Done Homeworkping.com HomeworkHelp https://www.homeworkping.com/ Research Paper help https://www.homeworkping.com/ Online Tutoring https://www.homeworkping.com/ click here for freelancing tutoring sites Matrix Analysis of Networks – J. R. Lucas Used to have a compact and neat form of solution. • necessary to know the structure of a network, and Matrix Analysis of Networks – Professor J R Lucas 1 May 2011
  • 2.
    • formulate theproblem based on the structure Because Large networks are • tedious to analyse using normal equations • easier/more convenient to formulate in matrix form. Matrix Analysis of Networks – Professor J R Lucas 2 May 2011
  • 3.
    Topology Deals with structureof an interconnected system Formulates the problem based on non-measurable properties of network. Geometric structure of the interconnection of network elements completely characterises − number of independent loop currents − number of independent node-pair voltages that are necessary to study the network. Matrix Analysis of Networks – Professor J R Lucas 3 May 2011
  • 4.
    Figure 1 –Structure of the network 1(a) and (b) have the same structure (or topology). However elements are quite different. Matrix Analysis of Networks – Professor J R Lucas 4 May 2011 R1 L1 C (a) (b)
  • 5.
    Graph of Network Figure2 – Circuit Matrix Analysis of Networks – Professor J R Lucas 5 May 2011 (a) Network (b) Graph of Network
  • 6.
    Figure 2 Circuit offigure 2(a) also has same topology. Figure 2(b) shows structure corresponding to all 3 circuits. Does not indicate any of the elements in the networks. Known as the graph of the network – Has all the nodes of the original network In obtaining the graph, − each element of the network is represented by a line − each voltage source by a short-circuit Matrix Analysis of Networks – Professor J R Lucas 6 May 2011
  • 7.
    − and eachcurrent source by an open circuit. Matrix Analysis of Networks – Professor J R Lucas 7 May 2011
  • 8.
    Tree of aNetwork Which of the diagrams would also represent a normal tree (without leaves) ? and why ? Only first diagram would fully satisfy the requirements. Second diagram has branches closing on itself − only a tree like Nuga might appear to close on itself Third diagram has branches in mid air not joined to main tree. Matrix Analysis of Networks – Professor J R Lucas 8 May 2011
  • 9.
    Properties associated withtrees. 1. All branches must be part of the tree 2. There cannot be closed loops formed from branches 3. There cannot be branches isolated from the tree Same properties apply in defining a tree of a network − can be many trees associated with a given network. − need not have a trunk coming from the ground and branches coming from the trunk. − a reduced graph of network with some of the links removed so as to leave all the nodes connected together by graph, but not to have any loop left. Matrix Analysis of Networks – Professor J R Lucas 9 May 2011
  • 10.
    Possible trees forthe Graph When a tree of the network is removed from graph, what remains is called the co-tree of the network. Co-tree is graph of removed links –compliment of the tree. A co-tree may contain closed loops, and disconnected branches. Matrix Analysis of Networks – Professor J R Lucas 10 May 2011 Graph of Network Some of the possible trees
  • 11.
    Analysis structure ofnetwork A single branch is required to join two nodes. Joining each additional node would require an additional branch. Let b = number of branches in the network n = number of nodes in the network l = number of independent loops Thus number of branches in tree = n – 1 number of links removed = b – (n – 1) = b – n + 1 Matrix Analysis of Networks – Professor J R Lucas 11 May 2011 Node 1 Node 2 Node 1 Node 2 New Node
  • 12.
    Formation of IndependentLoops If any one of removed links are added to the tree, then a new loop is formed. ∴ number of links removed from graph to form the tree is equal to the number of independent loops. l = b – n + 1 Oriented Graph − Numbered branches with assigned directions to currents. − Voltage considered to increase in direction opposite to flow of current Matrix Analysis of Networks – Professor J R Lucas 12 May 2011 Oriented Graph 1 32 4 6 5
  • 13.
    Matrix Analysis ofNetworks To solve circuit problems, need to write the equations corresponding to • Ohm’s Law, and • Kirchoff’s Current Law • Kirchoff’s Voltage Law Same is true even when there are a large number of branches. − use matrix analysis Matrix Analysis of Networks – Professor J R Lucas 13 May 2011
  • 14.
    k -1 -1 +1 +1 +1 0 0 i1 i2 i3 i4 i6 i7 i5 Kirchoff’s current Lawin matrix form For any node k − i1 − i2 − i4 + i6 + i7 = 0 or i1 + i2 + i4 = i6 + i7 or i1 + i2 + i4 – i6 – i7 = 0 or +1. i1 + 1. i2 + 0 .i3 + 1.i4 + 0. i5 – 1.i6 – 1.i7 = 0 Last form is preferred for matrix implementation Matrix Analysis of Networks – Professor J R Lucas 14 May 2011
  • 15.
    − all currentsin network are included in equation with different coefficients. For computer implementation, there must be a unique method (convention) of obtaining the coefficients ajk. Ij – current in jth branch jth branch − directed away from kth node: ajk = +1 − directed towards kth node: ajk = –1 − not incident on the kth node: ajk = 0 Kirchoff’s current law may be written, for the kth node Matrix Analysis of Networks – Professor J R Lucas 15 May 2011
  • 16.
    a1k .i1 +a2k . i2 + a3k .i3 + a4k .i4 + ...... ....... ..... a7k .i7 = 0 or ∑ = =⋅ b j jjk ia 1 0 at kth node, for all k Collection of equations, for each node k, would give [ ] )1()1()( 0 ××× =⋅ nb b bn It A In [A]t , row vectors are dependant, since sum is zero. [A]t written with one row less, giving only (n-1) rows. [A]t – node-branch incidence matrix, (n-1)×b. [A] – branch-node incidence matrix, b× (n-1) ajk = +1 if jth current is directed away from the kth node ajk = −1 if jth current is directed towards the kth node Matrix Analysis of Networks – Professor J R Lucas 16 May 2011
  • 17.
    ajk = 0if jth current is not incident on the kth node Kirchoff’s voltage Law in matrix form ∑ = =⋅ b r rrs vb 1 0 for sth loop, for all s; where brs = −1, 0, or +1 [ ] )1()1()( 0 ××× =⋅ lb b bl Vt B [B]t – mesh-branch incidence matrix, (l×b) [B] – branch-mesh incidence matrix, (b×l) brs = +1if rth current is in same direction as sth loop brs = −1if rth current is in opposite direction to sth loop brs = 0 if rth current is in the not part of the sth loop Ohm’s Law in matrix form Matrix Analysis of Networks – Professor J R Lucas 17 May 2011 0 0 0+1 -1 -1 0 -1 +1 s 0 0 0
  • 18.
    for all branchesk = 1, 2, .... ... b vk = – egk + Zk igk + Zk ik Either voltage source or current source would normally be used. Conversion with either Thevenin’s Theorem or Norton’s Theorem. Matrix Analysis of Networks – Professor J R Lucas 18 May 2011 egk igk Zkik + igk ik vk Figure - General branch
  • 19.
    With a voltagesource only vk = – egk + Zk ik for all branches k = 1, 2, ....... b and in matrix form as[ ] bbgbb IZEV +−= With a current source only ik = Yk vk – igk for all branches k = 1, 2, .... ... b and in matrix form as [ ] bbgbb VYII +−= , where [Yb] = [Zb]-1 Matrix Analysis of Networks – Professor J R Lucas 19 May 2011 egk Zk vk ik igk Ykik vk
  • 20.
    In Summary From Kirchoff’sLaws [ ] )1()1()( 0 ××× =⋅ nb b bn It A (1) (n-1) independent equations [ ] )1()1()( 0 ××× =⋅ lb b bl Vt B (2) l independent equations and from Ohm’s Law [ ] bbgbb IZEV +−= (3) b independent equations or [ ] bbgbb VYII +−= (3)* b independent equations Thus total number of independent equations is n – 1 + l + b = b + b = 2 b 2b independent equations 2b unknowns (b branch currents and b branch voltages) Matrix Analysis of Networks – Professor J R Lucas 20 May 2011
  • 21.
    Can be solved. Notusual to solve for both current and voltage simultaneously. Reductions can be done in two ways. 1) Eliminate voltages and solve for currents − mesh analysis 2) Eliminate currents and solve for voltages. − nodal analysis. Matrix Analysis of Networks – Professor J R Lucas 21 May 2011
  • 22.
    Mesh Analysis − Eliminatethe branch voltages from the equations. − Reduce remaining currents to a minimum using Kirchoff’s current law. Apply Kirchoff’s voltage law for solution. Define a set of mesh currents, mI . Branch currents bI related to mesh currents mI by an algebraic summation. [ ] mb IBI = (4) Eliminate Vb from the equations, Pre-multiply equation (3) by [B]t . Matrix Analysis of Networks – Professor J R Lucas 22 May 2011
  • 23.
    [ ] [] [ ] [ ] bb t gb t b t IZBEBVB +−= from equation (2), [B]t Vb = 0. Also [ ] mb IBI = ∴ [ ] [ ] [ ][ ] mb t gb t IBZBEB = [B]t Vb = 0 → sum of voltages around a loop is zero. i.e. [B]t Vb → sum of voltages around a loop. ∴ [B]t Egb → sum of source voltages around a loop. Defined as mesh source voltage vector Egm . i.e. Egm = [B]t Egb ∴ Egm = [ ] [ ][ ] [ ] mmmb t IZIBZB = Matrix Analysis of Networks – Professor J R Lucas 23 May 2011
  • 24.
    where [Zm] =[ ] [ ][ ]BZB b t corresponds to l equations [B] also known as the tie-set matrix (as its elements tie the loop together) Unknowns are l values of current Im Original 2b equations and 2b unknowns reduced to l equations and l unknowns. Elements of [Zm] can be obtained either from above mathematics, or by inspection as follows. Simple evaluation of [Zm] and Egm zjj = self impedance of mesh j = sum of all branch impedances in mesh j Matrix Analysis of Networks – Professor J R Lucas 24 May 2011
  • 25.
    zjk = mutualimpedance between mesh j and mesh k = sum of all branch impedances common to mesh j and mesh k and traversed in mesh direction − sum of all branch impedances common to mesh j and mesh k, and traversed in opposite direction ej = algebraic sum of the branch voltage sources in mesh j in mesh direction. Matrix Analysis of Networks – Professor J R Lucas 25 May 2011
  • 26.
    Example 1 Solve thecircuit using Mesh matrix analysis. Work from first principles. Solution Number the branches and the loops. Matrix Analysis of Networks – Professor J R Lucas 26 May 2011 j6 Ω E1 100∠00 V j20 Ω -j120 Ω E2 100∠300 V 10 Ω 20 Ω 10 Ω
  • 27.
    Write the loopcurrents in terms of the branch currents. i1 = I1 i2 = – I3 i3 = I1 – I2 i4 = I2 i5 = I2 – I3 i6 = I3 Matrix Analysis of Networks – Professor J R Lucas 27 May 2011 or in matrix form                               − − − =                     3 2 1 6 5 4 3 2 1 100 110 010 011 100 001 I I I i i i i i i I1 I2 I3 i1 i4 i3 i5 i6 i2 E1 100∠00 V j20 Ω j6 Ω –j120 Ω E2 100∠36.870 10 Ω 20 Ω 10 Ω
  • 28.
    This gives theBranch-Mesh incidence matrix [B]. Mesh–Branch incidence matrix [B]t can also independently by writing the relation between the mesh direction and the branch direction. [ ]           −− −= 110010 011100 000101 t B Notice that this corresponds to the transpose of the earlier written matrix. Matrix Analysis of Networks – Professor J R Lucas 28 May 2011
  • 29.
    Vector of branchsource voltages is Branch impedance matrix is [ ]                     − = 600000 0100000 0020000 00012000 0000100 0000020 j j j Zb Egm = [B]t Egb , and [Zm] = [B]t [Zb] [B]                     ∠ ∠           −− −= 0 0 0 0 87.36100 0100 110010 011100 000101 0 0 gmE , Egm =           ∠− ∠ 0 0 87.36100 0 0100 Matrix Analysis of Networks – Professor J R Lucas 29 May 2011                     ∠ ∠ = 0 0 0 0 87.36100 0100 0 0 gbE
  • 30.
    [ ]                     − − −                     −           −− −= 100 110 010 011 100 001 600000 0100000 0020000 00012000 0000100 0000020 110010 011100 000101 j j j Zm [ ]                     − − −           −− −= 600 10100 0200 0120120 1000 0020 110010 011100 000101 j jj j Zm =          +− −− − 620100 1012030120 0120100 j jj jj Both Egm and Zm could have been written by inspection. Thus           ∠− ∠ 0 0 87.36100 0 0100 =                     +− −− − 3 2 1 620100 1012030120 0120100 I I I j jj jj Equations may be solved by inversion or otherwise. Matrix Analysis of Networks – Professor J R Lucas 30 May 2011
  • 31.
              ∠− ∠           −−−×−×− ×−+−+− ×−+−−+− ∆ =           0 0 2 2 3 2 1 87.36100 0 0100 )120()12030(1001010010120 10100)620(100)620(120 10120)620(12010)620)(12030( 1 jjjjj jjjjj jjjjj I I I           −−          +−−−− −+−+− −+−− ∆ =           6080 0 100 1440012000300010001200 100060020007202400 1200720240022201220 1 3 2 1 jjjj jjj jjjj I I I ∆ = (1220– j2220)×(–j100) + (720 – j2400) ×(j120) + (–j1200) ×0 = – j122000 – 222000 + j 86400 + 288000 = 66000 – j 35600 = 74989∠-28.34o I1 = (122000 – j 222000 + 0 + j 96000 – 72000)/74989∠-28.34o = (50000 – j 126000)/ 74989∠-28.34o = 135558∠-68.36o /74989∠-28.34o = 1.808∠-40.02o A [Note: Inversion has not been checked so answers may be in error.] Currents I2 and I3 can be similarly determined. The branch currents i1, i2, ..... may then be determined from the matrix equation. [Normally branch 6 would have been marked as part of branch 2] Matrix Analysis of Networks – Professor J R Lucas 31 May 2011
  • 32.
    Nodal Analysis − eliminatebranch currents from the equations. − Reduce number of remaining voltages to a minimum using Kirchoff’s voltage law. Apply Kirchoff’s current law for solution. Define a set of nodal voltages, NV which are node pair voltages (i.e. voltage across a pair of nodes) Branch voltages bV are related to nodal voltages NV by an algebraic summation. [ ] Nb VAV = (5) [A] too does not have the reference node. Matrix Analysis of Networks – Professor J R Lucas 32 May 2011
  • 33.
    Pre-multiply equation (3)*by [A]t . [ ] [ ] [ ] [ ] bb t gb t b t VYAIAIA +−= from equation (1), [A]t Ib = 0 . Substituting from (5) [ ] [ ] [ ][ ] Nb t gb t VAYAIA +−=0 [ ] [ ] [ ][ ] Nb t gb t VAYAIA = IgN = [YN]VN where [ ] gb t gN IAI = , and [ ] [ ] [ ][ ]AYAY b t N = Source nodal current vector IgN and the nodal admittance matrix [YN] could be written by inspection. yii = sum of all branch admittances incident at node i Matrix Analysis of Networks – Professor J R Lucas 33 May 2011
  • 34.
    yij = negativeof the sum of all branch admittances connecting node i and node j . Reason for negative sign can be understood as follows: ik = yk vk = yk (Vi – Vj) At any node i, injected current Igi = Σ ik = Σ yk (Vi – Vj) ∑∑ ≠ = ≠ = −= N ij j jk N ij j ikgi VyVyI 11 for all j Since Vi is a constant for a given i, ∑∑∑∑ ≠ = ≠ = ≠ = ≠ = −+         =−= N ij j jki N ij j k N ij j jk N ij j kigi VyVyVyyVI 1 ) 111 ( ∑∑ = ≠ = =+= N j jii N ij j jiiiiigi VyVyVyI 11 corresponds to nodal equation As in the case of mesh analysis, Matrix Analysis of Networks – Professor J R Lucas 34 May 2011 vk ji ykik
  • 35.
    IgN = [YN]VN isfirst solved to give VN and the branch voltages and branch currents then obtained using the matrix equations. Matrix Analysis of Networks – Professor J R Lucas 35 May 2011
  • 36.
    Example 2 Example 1has been reformulated as a problem with current sources rather than with voltage sources. [If voltage sources are present, they would first have to be converted to current sources]. Matrix Analysis of Networks – Professor J R Lucas 36 May 2011 5∠-900 A j20 Ω j6 Ω -j120 Ω 8.575∠5.910 A10 Ω 20 Ω 10 Ω i1 i4 i3 i5 i2 V1 V2
  • 37.
    Network may alsobe drawn in terms of admittances. The branch-node incidence matrix [A], branch injected current Igb, and branch admittance matrix may be written, with reference selected as earthed node as follows. Matrix Analysis of Networks – Professor J R Lucas 37 May 2011 ∠      ∠     i1 i4 i3 i5 i2 V1 V2
  • 38.
    [ ]                 − − − = 10 11 01 10 01 A , Igb=                 ∠ −∠ 0 0 0 91.5575.8 905 o o , [ ]                 − − = 1.00000 005.0000 00008333.000 0000441.00735.00 000005.0 j j j Yb As in mesh analysis, nodal current injection vector and nodal admittance matrix may be written from first principles. Left as an exercise for you to work out. This is worked by inspection. [ ] [ ]       −++− −++− =      ∠ −∠ = 0441.00735.01.005.005.0 05.005.000833.005.0 , 91.5575.8 905 j jj YI No o gN             −++− −++− =      ∠ −∠ ∴ 2 1 0441.00735.01.005.005.0 05.005.000833.005.0 91.5575.8 905 V V j jj o o       ∠ −∠       ++− −++ ∆ =      ∴ o o jj j V V 91.5575.8 905 05.000833.005.005.0 05.00441.00735.01.005.01 2 1 ∆ = (–j0.05+j0.00833+0.05)(0.05+0.1+0.0735–j0.0441) – 0.052 Matrix Analysis of Networks – Professor J R Lucas 38 May 2011
  • 39.
    = (0.05 –j 0.04167)(0.2235 – j 0.0441) – 0.0025 = 0.06509∠-39.81o ×0.2278∠-11.16o – 0.0025 = 0.01483∠-50.97 – 0.0025 = 0.00934 – 0.0025 – j 0.01152 = 0.00684 – j 0.01152 = 0.0134∠-59.30o V1 = (0.2278∠-11.16o ×5∠-90o +0.05×8.575∠5.91o )/0.0134∠-59.3o = (– 0.2205 – j 1.1175 + 0.4265 + j 0.04415) /0.0134∠-59.30o = (0.2060 – j 1.0733)/0.0134∠-59.30o = 1.093∠-79.14o /0.0134∠-59.30o V1 = 81.6∠-19.84o V ∴branch current i1 = 20 68.273.23 20 84.196.81100 j j j o + = −∠− i1 Aj o 09.40809.1165.1384.1 −∠=−= Matrix Analysis of Networks – Professor J R Lucas 39 May 2011
  • 40.
    which is thesame answer (to calculation accuracy) that was obtained in example 1. Matrix Analysis of Networks – Professor J R Lucas 40 May 2011
  • 41.
    Conversion of Idealsources (a) Ideal Voltage sources No impedance directly in series with voltage source Ideal voltage sources are distributed to branches connected to one of the nodes of original ideal source. Matrix Analysis of Networks – Professor J R Lucas 41 May 2011 E      ≡                            
  • 42.
    (b) Ideal Currentsources No admittance appears directly in parallel with current source Ideal current source has been distributed around a loop connecting the two points of original source. Matrix Analysis of Networks – Professor J R Lucas 42 May 2011 ≡        
  • 43.
    Matrix Analysis ofNetworks – Professor J R Lucas 43 May 2011
  • 44.
    Port Pair of nodesacross which a device can be connected. Voltage is measured across the pair of nodes. Current going into one node is the same as the current coming out of the other node in the pair. These pairs are entry (or exit) points of the network. Compare with an Airport or a Sea Port. Entry and exit points to a country. Planes that enter at a given port are the ones that take off from same port. Matrix Analysis of Networks – Professor J R Lucas 44 May 2011
  • 45.
    Two-Port Theory Convenient todevelop special methods for systematic treatment of networks. Single-port linear active networks − Thevenin’s or Norton’s equivalent circuit. Linear passive networks − Convenient to study behaviour relative to a pair of designated ports. Matrix Analysis of Networks – Professor J R Lucas 45 May 2011           
  • 46.
    Definitions Driving point impedanceis defined as ratio of applied voltage (driving point voltage) across a node-pair to the current entering at the same port. [input impedance of network seen from particular port] Driving point impedance at Port 1 = V1/I1 Driving point impedance at Port 2 = V2/I2 Driving point admittance is similarly defined as the ratio of the current entering at a port to the applied voltage across the same node-pair. Driving point admittance at Port 1 = I1/V1 Driving point admittance at Port 2 = I2/V2 Matrix Analysis of Networks – Professor J R Lucas 46 May 2011 
  • 47.
    Immittance is sometimesused to represent either an impedance or an admittance Transfer impedance is defined as the ratio of the applied voltage across a node-pair to the current entering at the other port. Transfer impedance = V1/I2 , V2/I1 Transfer admittance is similarly defined as the ratio of the current entering at a port to the voltage appearing across the other node-pair. Transfer admittance = I1/V2 , I2/V1 Matrix Analysis of Networks – Professor J R Lucas 47 May 2011
  • 48.
    Transfer Voltage gain(or ratio) is defined as the ratio of the voltage at a node pair to the voltage appearing at the other node-pair. Transfer voltage gain = V1/V2 , V2/V1 Transfer Current gain (or ratio) is similarly defined as the ratio of the current at a port to the current at the other port. Transfer current gain = I1/I2 , I2/I1 Matrix Analysis of Networks – Professor J R Lucas 48 May 2011
  • 49.
    Common Two-port parameters Externalconditions of a two-port network can be completely defined by currents and voltages at the 2 ports. A general two port network can be characterised by four parameters, derived from the network elements. With symmetry, number of parameters will be reduced. (a) Impedance parameters (b) Admittance parameters (c) Transmission Line parameters (d) Hybrid parameters. Matrix Analysis of Networks – Professor J R Lucas 49 May 2011
  • 50.
    (a) Impedance Parameters(z-parameters) or Open-circuit parameters V1 = z11 I1 + z12 I2 If I2 = 0, then z11 = V1/I1 If I1 = 0, then z12 = V1/I2 It follows that, Matrix Analysis of Networks – Professor J R Lucas 50 May 2011               
  • 51.
    021 1 11 = = II V z , 012 1 12 = = II V z , 021 2 21 = = II V z , 012 2 22 = = II V z . z-parameterscorrespond to the driving point and transfer impedances at each port with the other port having zero current (i.e. open circuit). → open circuit parameters. Matrix Analysis of Networks – Professor J R Lucas 51 May 2011
  • 52.
    Example 3 Find impedanceparameters of the two port T – network. With port 2 on open circuit ba ZZ II V z += = = 021 1 11 bZ II V z = = = 021 2 21 similarly with port 1 open, z12 = Zb z22 = Zb + Zc (b) Admittance Parameters (y-parameters) or Short-circuit parameters Matrix Analysis of Networks – Professor J R Lucas 52 May 2011              [ ]       + + =→ cbb bba ZZZ ZZZ Z
  • 53.
    y11, y12, y21, y22defined with either V1 or V2 zero. y-parameters correspond to driving point and transfer admittances at each port with the other port having zero voltage (i.e. short circuit) → short circuit parameters. Matrix Analysis of Networks – Professor J R Lucas 53 May 2011                
  • 54.
    Example 4 Find admittanceparameters of the 2 port π–network. y11 = 021 1 =VV I = Ya + Yb y21 = 021 2 =VV I = – Yb Matrix Analysis of Networks – Professor J R Lucas 54 May 2011                      →[Y] =       +− −+ cbb bba YYY YYY
  • 55.
    (c) Transmission LineParameters (ABCD-parameters) Parameters can be defined using either port 2 on short circuit or port 2 on open circuit. In case of symmetrical system, parameter A = D. For a reciprocal system, A.D – B.C = 1 Matrix Analysis of Networks – Professor J R Lucas 55 May 2011                            =      2 2 1 1 I V DC BA I V
  • 56.
    Example 5 Find ABCDparameters. A = c cb Y YY + , B = cY 1 C = c accbba Y YYYYYY ++ and D = c ca Y YY + [For symmetrical network, Ya = Yb , A = D]. A.D – B.C = c accbba cc ca c cb Y YYYYYY YY YY Y YY ++ ⋅− + ⋅ + 1 = 2 2 )( c accbbaccbacab Y YYYYYYYYYYYYY ++−+++ =1 (d) Hybrid Parameters (h-parameters) Matrix Analysis of Networks – Professor J R Lucas 56 May 2011                       
  • 57.
    The hybrid parametermatrix may be written as             =      2 1 2221 1211 2 1 V I hh hh I V h-parameters can be defined as in other examples, and are commonly used in some electronic circuit analysis. Matrix Analysis of Networks – Professor J R Lucas 57 May 2011  
  • 58.
    Interconnection of two-portnetworks (a) Series connection of two-port networks Series properties are applied to each port at port 1, Ir1 = Is1 = I1, and Vr1 + Vs1 = V1 at port 2 Ir2 = Is2 = I2, and Vr2 + Vs2 = V2 Matrix Analysis of Networks – Professor J R Lucas 58 May 2011                                
  • 59.
    [Z] = [Zr]+ [Zs] Matrix Analysis of Networks – Professor J R Lucas 59 May 2011
  • 60.
    (b) Parallel connectionof two-port networks at port 1, Ir1 + Is1 = I1, and Vr1 = Vs1 = V1 Matrix Analysis of Networks – Professor J R Lucas 60 May 2011                          
  • 61.
    at port 2,Ir2 + Is2 = I2, and Vr2 = Vs2 = V2 [Y] = [Yr] + [Ys] (c) Cascade connection of networks Output of one network becomes input to next. Ir2 = Is1 Vr2 = Vs1             =      2 2 1 1 r r rr rr r r I V DC BA I V ,             =      2 2 1 1 s s ss ss s s I V DC BA I V                   =      2 2 1 1 I V DC BA DC BA I V ss ss rr rr ABCD matrix of component networks Matrix Analysis of Networks – Professor J R Lucas 61 May 2011                                       
  • 62.
    A = 022 1 =IV V =1, = 1 B = 022 1 =VI V = Z, = 0 C = 022 1 =IV I = 0, =Y D = 022 1 =VI I = 1, = 1 In matrix form       DC BA =       10 1 Z , =       1 01 Y Consider example 5 again Matrix Analysis of Networks – Professor J R Lucas 62 May 2011               ≡   
  • 63.
          DC BA =       1 01 aY         10 11 bY       1 01 cY Simplificationof matrix product would give the same answer as in example 5. Matrix Analysis of Networks – Professor J R Lucas 63 May 2011