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Gaussian Elimination N equations in N unknowns (x, y, z) A,B,C,D are known constants: A1x + B1y + C1z = D1 A2x + B2y + C2z = D2 A3x + B3y + C3z = D3 Put the constants into a matrix: A1 B1 C1 D1 A2 B2 C2 D2 A3 B3 C3 D3 Use linear operators to transform the matrix to: 1 0 0 Dx 0 1 0 Dy 0 0 1 Dz Dx, Dy Dz are the values for x y and z that make the first equations true – thus solving the equations. Gaussian Elimination Two basic linear operations used on matrix: • */ by constant • add/sub rows from each other: A1 B1 C1 D1 A2 B2 C2 D2 A3 B3 C3 D3 Divide first row by A1, becomes: 1 B1/A1 C1/A1 D1/A1 Subtract A2 times first row from second row, becomes: 0 B2-B1A2 C2-C1A2 D2-D1A2 Subtract A3 times first row from third row, becomes: 0 B3-B1A3 C3-C1A3 D3-D1A3 Now we have the first column in the right form, continue with second row, second column, etc. Gaussian Elimination int gauss(double **M, int maxrow, int maxcol) { int diag, row, col; double pivot, pivot2; for (diag = 0; diag < maxrow; diag++) { /* for each row = variable = diagonal element */ pivot = M[diag][diag]; if (pivot == 0) { printf("Gauss: Singular Matrix!\n"); return(1); } for (col = diag; col < maxcol; col++) { /* from diagonal to right, divide out by the pivot diagonal element */ M[diag][col] /= pivot; } for(row = 0; row < maxrow; row++) /* for all other rows subtract scalled diag row, zeroing pivot column */ if (row != diag) { { /* scale all elements of diag row by value which will zero diag element of this row, and subtract from this row */ pivot2 = M[row][diag]; for (col = diag; col < maxcol; col++) M[row][col] -= M[diag][col]*pivot2; } } return(0); } Nodal Analysis: NA For a linear circuit, solve Kirchhoff's current law (KCL) and voltage law (KVL) using a matrix formulation, just Ohm’s law in a matrix: |G| x |V| = |I| G is conduction matrix (1/R) V is vector of unknown node voltages I is vector of known node currents Solve by inverting G and then multiplying I to get V or concatenating |G||I| and performing Gaussian elimination (or equivalent) to get V Assumes G and I are constant Modified Nodal Analysis: MNA • • Adds voltage sources to circuits Matrices are a little more complex, solution is the same AX = Z • General form of linear equations: • used for a lot of systems beyond circuits • A matrix is conduction and voltage sources • G is conduction, B is voltages C = BT, D = 0 • X is unknown voltages and currents ⎡G B ⎤ • Z is known voltages, currents ⎡i ⎤ ⎡v ⎤ X =⎢ ⎥ Z =⎢ ⎥ A=⎢ ⎥ ⎣ j⎦ ⎣e ⎦ ⎣C D ⎦ •Solve by matrix inversion and multiplication: −1 A Z=X ⎡G B ⎤ ⎢C D ⎥ ⎣ ⎦ −1 ⎡i ⎤ ⎡v ⎤ ×⎢ ⎥ = ⎢ ⎥ ⎣e ⎦ ⎣ j ⎦ Conduction Matrix • G is matrix of conductions (1/R) = g • For each impedance: • if R is on node k, add value of 1/R on diagonal (k,k). – If other end of R is on ground, stop – Else, if R is on node j add 1/R to other diagonal (j,j) and add (-1/R) on elements (k,j) and (j,k) Voltage Source Matrices • B has a column for each voltage source, V • If V has a positive terminal on node j, put 1 in row j. – If other end of V is on ground, stop – Else, if V has a negative terminal on node k put -1 on row k • C is B transpose (row for each source…) • D (corner) is zero if no dependent sources Vector of Unknowns • X is composed of the unknown node voltages and voltage source currents. • Vk for each unknown voltage at node k • Ji for the current through each voltage source i (in the direction of positive terminal through the device to the negative terminal). Vector of Knowns • Z is composed of the known node currents and voltage source voltages. • Ik for each known current at node k. This is zero unless there is a current source at node k. if so, the value is +current positive node. If other end is not on ground, then value is –current at other node. • Vj for the value of voltage source j. If other end is not on ground, then value of –voltage at other node. Example Circuit AX = Z 1 100 −1 100 0 0 1 −1 100 1 100+1 200+1 400 −1 200 0 0 0 −1 200 1 200 0 0 0 1 0 0 0 0 1 300 0 0 0 v1 0 v2 0 × v3 = 100mA v4 −100mA ivs 5V Example Circuit ⎡G B ⎤ A=⎢ ⎥ ⎣C D ⎦ n1 n2 n3 n4 v1 n1 0.01 -0.01 0 0 1 n2 -0.01 0.0175 -0.005 0 0 n3 n4 0 0 -0.005 0 0.005 0 0 0.003333 0 0 v1 1 0 0 0 0 Example Circuit 5V 32V 12V -30V 70mA ⎡G B ⎤ ⎥ ⎢ ⎣C D ⎦ 0 0 0 0 1 0 80 80 0 0.8 0 80 280 0 0.8 −1 A Z=X −1 ⎡i ⎤ ×⎢ ⎥ ⎣e ⎦ 0 0 0 300 0 1 0.8 0.8 0 -0.002 X 0 0 0.1 -0.1 5 ⎡v ⎤ =⎢ ⎥ ⎣ j⎦ 5 12 = 32 -30 0.07 Energy Storage • Capacitors and Inductors store energy and thus have voltage (current) dependent conductance: g (t ) = i (t ) / v(t ) • Capacitor: 1 v(t ) = C t ∫ 0 q (t ) i (t )dt = C dq (t ) dv(t ) i (t ) = =C dt dt • Inductor: t ∫ 1 i (t ) = v(t )dt L 0 di (t ) v(t ) = L dt Need to do numerical integration x ’n Two rectangular rules for integration • xn+1 = xn + h*x’n ( forward Euler) • xn+1 = xn + h*x’n+1 ( backward Euler) Many others: Trapezoid, Gear, Simpson Use for integration for charge storage: vn+1 = vn+ h * v’n+1 vn+1 = vn+ h/c * in+1 Rearrange: In+1 = (c/h)* vn+1 – (c/h) * vn Define: Geq = c/h Gives a Norton Equivalent circuit: xn xn+1 h x’n+1 Capacitor companion model in MNA Geq − Geq V n +1 + − Ieq = × − Geq Geq V n +1 − + Ieq • Adds companion model (Norton equivalent circuit) to G, X and Z matrices. • Note signs of Ieq nodes • Ieq must be re-calculated after each time step: Ieqn+1 = (V+n+1-V-n+1) * C/h If h changes, then Geq = C/h must be re-calculated too!