Nodal Analysis in MAE140 Linear Circuits - Prof. Jorge Cortes, Study notes of Mechanical Engineering

Download Nodal Analysis in MAE140 Linear Circuits - Prof. Jorge Cortes and more Study notes Mechanical Engineering in PDF only on Docsity! MAE140 Linear Circuits 40 Systematic Circuit Analysis (T&R Chap 3) Node-voltage analysis Using the voltages of the each node relative to a ground node, write down a set of consistent linear equations for these voltages Solve this set of equations using, say, Cramer’s Rule Mesh current analysis Using the loop currents in the circuit, write down a set of consistent linear equations for these variables. Solve. This introduces us to procedures for systematically describing circuit variables and solving for them MAE140 Linear Circuits 41 Nodal Analysis Node voltages Pick one node as the ground node Label all other nodes and assign voltages vA, vB, …, vN and currents with each branch i1, …, iM Recognize that the voltage across a branch is the difference between the end node voltages Thus v3=vB-vC with the direction as indicated Write down the KCL relations at each node Write down the branch i-v relations to express branch currents in terms of node voltages Accommodate current sources Obtain a set of linear equations for the node voltages v5 C B A vC vB vA v4 v3 v2 v1 + + + ++ - - - - - MAE140 Linear Circuits 44 Systematic Nodal Analysis Writing node equations by inspection Note that the matrix equation looks just like Gv=i for matrix G and vector v and i G is symmetric (and non-negative definite) Diagonal (i,i) elements: sum of all conductances connected to node i Off-diagonal (i,j) elements: -conductance between nodes i and j Right-hand side: current sources entering node i There is no equation for the ground node – the column sums give the conductance to ground ! ! " # $ $ % & ! ! " # $ $ % & ! ! " # $ $ % & ' = +' +' ''+ 2 2 1 4202 0311 2121 Si Si Si Cv Bv Av GGG GGG GGGG vCvB vA R4 R2R1 R3 iS2 iS1 i4i3 i2 i1 i0 Reference node i5 MAE140 Linear Circuits 45 Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74) Node A: Conductances Source currents entering Node B: Conductances Source currents entering Node C: Conductances Source currents entering iS vBvA vC R/2 2R2R R/2 R MAE140 Linear Circuits 46 Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74) Node A: Conductances G/2B+2GC=2.5G Source currents entering= iS Node B: Conductances G/2A+G/2C+2Gground=3G Source currents entering= 0 Node C: Conductances 2GA+G/2B+Gground=3.5G Source currents entering= 0 iS vBvA vC R/2 2R2R R/2 R ! ! ! " # $ $ $ % & = ! ! ! " # $ $ $ % & ! ! ! " # $ $ $ % & '' '' '' 0 0 5.35.02 5.035.0 25.05.2 S C B A i v v v GGG GGG GGG MAE140 Linear Circuits 49 Nodal Analysis Ex. 3-2 (T&R, 5th ed, p. 72) Solve this using standard linear equation solvers Cramer’s rule Gaussian elimination Matlab + _ vA vB 500Ω1KΩ 2KΩ iS1 iS2 vO ! " # $ % & ' =! " # $ % & ! ! " # $ $ % & ((' ('( '' '' 2 1 33 33 105.2105.0 105.0105.1 S S B A i i v v MAE140 Linear Circuits 50 Nodal Analysis with Voltage Sources Current through voltage source is not computable from voltage across it. We need some tricks! They actually help us simplify things Method 1 – source transformation +_ Rest of the circuitvS RS vA vB Rest of the circuit vA vB RS SR Sv! Then use standard nodal analysis – one less node! MAE140 Linear Circuits 51 Nodal Analysis with Voltage Sources 2 Method 2 – grounding one node Rest of the circuit vA vB +_ vS This removes the vB variable – simpler analysis But can be done once per circuit MAE140 Linear Circuits 54 Nodal Analysis Ex. 3-4 (T&R, 5th ed, p. 76) This is method 1 – transform the voltage sources Applicable since voltage sources appear in series with Resist Now use nodal analysis with one node, A +_ +__ + v0vS1 vS2R3 R2R1 1 1 R Sv 2 2 R SvR1 R2 R3 _ + v0 vA ! ( ) 321 2211 2211321 GGG vGvG v vGvGvGGG SS A SSA ++ + = +=++ MAE140 Linear Circuits 55 Nodal Analysis Ex. 3-5 (T&R, 5th ed, p. 77) vS vBvA vC R/2 2R2R R/2 R+_ Rin iin What is the circuit input resistance viewed through vS? MAE140 Linear Circuits 56 Nodal Analysis Ex. 3-5 (T&R, 5th ed, p. 77) Rewrite in terms of vS, vB, vC This is method 2 Solve vS vBvA vC R/2 2R2R R/2 R+_ Rin iin What is the circuit input resistance viewed through vS? 05.35.02 05.035.0 =+!! =!+! = CvBGvAGv CGvBGvAGv SvAv SGvCGvBGv SGvCGvBGv 25.35.0 5.05.03 =+− =− R R inR R Sv R CvSv R BvSv ini Sv Cv Sv Bv 872.0 75.11 25.10 25.10 75.11 2/2 25.10 25.6 , 25.10 75.2 == = ! + ! = == MAE140 Linear Circuits 59 Summary of Nodal Analysis 1. Simplify the cct by combining elements in series or parallel 2. Select as reference node the one with most voltage sources connected 3. Label node voltages and supernode voltages – do not label the nodes directly connected to the reference 4. Use KCL to write node equations. Express element currents in terms of node voltages and ICSs 5. Write expressions relating node voltages and IVSs 6. Substitute from Step 5 into equations from Step 4 Write the equations in standard form 7. Solve using Cramer, Gaussian elimination or matlab MAE140 Linear Circuits 60 Solving sets of linear equations Cramer’s Rule Thomas & Rosa Appendix B pp. A-2 to A-11 ! ! ! " # $ $ $ % & '= ! ! ! " # $ $ $ % & ! ! ! " # $ $ $ % & '' '' '' 6 10 4 833 275 325 3 2 1 x x x ( ) ( ) ( ) 5075125250 )3(7)2()2(3)3()3(8)2(5)2()3(875 27 32 )3( 83 32 )5( 83 27 5 833 275 325 =!!= !"!!"!!!"!!"!+!"!!"= ! !! !+ ! !! !! ! ! = !! !! !! =# ( ) ( ) ( ) 100150250200 )3(7)2()2(6)3()3(8)2(10)2()3(874 27 32 6 83 32 )10( 83 27 4 836 2710 324 1 =+!= !"!!"!+!"!!"!+!"!!"= ! !! + ! !! !! ! ! = ! !! !! =# 2 50 1001 1 ==! ! =x MAE140 Linear Circuits 61 Solving sets of linear equations (contd) Notes: This Cramer is not as much fun as Cosmo Kramer in Seinfeld I do not know of any tricks for symmetric matrices 24114250340 210 34 )3( 86 34 )5( 86 210 5 863 2105 345 2 =++!= !! ! !+ ! !! !! = ! !!! ! =" 8424060 107 42 )3( 63 42 )5( 63 107 5 633 1075 425 3 =+!= ! ! !+ ! ! !! ! ! = !! !! ! =" 48.0 50 242 2 ==! ! =x 68.1 50 843 3 ==! ! =x MAE140 Linear Circuits 64 Mesh Current Analysis Dual of Nodal Voltage Analysis with KCL Mesh Current Analysis with KVL Mesh = loop enclosing no elements Restricted to Planar Ccts – no crossovers (unless you are really clever) Key Idea: If element K is contained in both mesh i and mesh j then its current is ik=ii-ij where we have taken the reference directions as appropriate Same old tricks you already know ++ ++ + + + - -- - - - - R1 R2 R3vS1 vS2v4v3 v2v1 v0 iA iB Mesh A: -v0+v1+v3=0 Mesh B: -v3+v2+v4=0 (R1+R3)iA-R3iB=vS1 -R3iA+(R2+R3)iB=-vS2 v1=R1iA v0=vS1 v2=R2iB v4=vS2 v3=R3(iA-iB) ! " # $ % & ! " # $ % & ! " # $ % & ' = +' '+ 2 1 323 331 Sv Sv Bi Ai RRR RRR MAE140 Linear Circuits 65 Mesh Analysis by inspection Matrix of Resistances R Diagonal ii elements: sum of resistances around loop Off-diagonal ij elements: - resistance shared by loops i and j Vector of currents i As defined by you on your mesh diagram Voltage source vector vS Sum of voltage sources assisting the current in your mesh If this is hard to fathom, go back to the basic KVL to sort these directions out SvRi = - + + - vS1 vS2R1 R3R2 R4 iA iB iC + - v0 MAE140 Linear Circuits 66 Mesh Analysis by inspection Matrix of Resistances R Diagonal ii elements: sum of resistances around loop Off-diagonal ij elements: - resistance shared by loops i and j Vector of currents i As defined by you on your mesh diagram Voltage source vector vS Sum of voltage sources assisting the current in your mesh If this is hard to fathom, go back to the basic KVL to sort these directions out SvRi = - + + - vS1 vS2R1 R3R2 R4 iA iB iC + - v0 ! ! ! " # $ $ $ % & ' ' = ! ! ! " # $ $ $ % & ! ! ! " # $ $ $ % & +'' '+ '+ 1 2 2 3232 343 221 0 0 S S S C B A v v v i i i RRRR RRR RRR MAE140 Linear Circuits 69 Mesh Analysis with ICSs – Method 3 Supermeshes – easier than supernodes Current source in more than one mesh and/or not in parallel with a resistance 1. Create a supermesh by eliminating the whole branch involved 2. Resolve the individual currents last R4 iS1 iS2 R3 R2R1 vO + - iA iC iB Supermesh Excluded branch 0)(342)(1 =!+++! AiCiRCiRBiRAiBiR 1SiAi = 2SiCiBi =! MAE140 Linear Circuits 70 Summary of Mesh Analysis 1. Check if cct is planar or transformable to planar 2. Identify meshes, mesh currents & supermeshes 3. Simplify the cct where possible by combining elements in series or parallel 4. Write KVL for each mesh 5. Include expressions for ICSs 6. Solve for the mesh currents MAE140 Linear Circuits 71 Linearity & Superposition Linear cct – modeled by linear elements and independent sources Linear functions Homogeneity: f(Kx)=Kf(x) Additivity: f(x+y)=f(x)+f(y) Superposition –follows from linearity/additivity Linear cct response to multiple sources is the sum of the responses to each source 1. “Turn off” all independent sources except one and compute cct variables 2. Repeat for each independent source in turn 3. Total value of all cct variables is the sum of the values from all the individual sources