1. Rowan Hall 238A beard@rowan.edu http://rowan.jkbeard.com September 5, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.

2. Slide 2 Networks I for M.E. September 5, 2006 Passive Sign Convention Always mark one terminal with a + sign Voltage is positive measured from the other terminal Current is positive going into the terminal marked with the + sign

3. Slide 3 Networks I for M.E. September 5, 2006 Source Sign Convention Voltage sources marked with + and – signs inside a circle or diamond Current sources marked with an arrow inside a square, circle or diamond Positive  Current out of a voltage source  Voltage out of a current source

4. Slide 4 Networks I for M.E. September 5, 2006 General Methodology Write the loop equations  Pick small or simple loops  Make all of them one direction – clockwise or counterclockwise  Make sure that every trace is covered once  Use node voltage equations  Leverage supernodes – treat as single nodes Write the node equations -- Currents are positive into each node Write the Ohm’s Law equations Rearrange the equations to allow vector-matrix notation Check your work Use computer to solve the matrix equation  TI-89 for smaller problems  Matlab or Mathcad for matrices larger than 4 X 4 or so  Note that free linear algebra packages are available for most HLLs

5. Slide 5 Networks I for M.E. September 5, 2006 Approach Used Here for 4.4-7 Use voltage node notation Leverage supernodes Write Ohm’s Law equations first Knowns are  Resistances  Source voltages and currents  Controlled source multipliers Unknowns are  Node voltages  Currents through voltage sources  Voltages across current sources

6. Slide 6 Networks I for M.E. September 5, 2006 Technique for Loop Equations Add voltage drops around the loop The sum of voltage drops around a closed loop must be zero Voltage drop is voltage at the present node minus voltage on the other terminal of the resistor or source Drop is positive through a resistor when the loop goes into the “+” terminal and out the “-” terminal and its current is positive Look at the equation with and without Ohm’s Law  Supernodes will identify themselves  Node voltages instead of currents often gives simpler equations

7. Slide 7 Networks I for M.E. September 5, 2006 Voltage Drop Direction of loop

8. Slide 8 Networks I for M.E. September 5, 2006 Technique for Node Equations Current into a node through each resistor is the voltage on the other side of each resistor divided by the resistance Subtract the voltage at that node times the sum of the reciprocals of all the resistors connected to the node Add currents through sources  Current sources connected to the node  Currents through voltage sources connected to the node Use Ohm’s Law to pose the currents through resistors in terms of the node voltages

9. Slide 9 Networks I for M.E. September 5, 2006 Matrix Notation Is A way of organizing several linear equations. Nothing is changed from the original equations. TI-89, Matlab, and Mathcad can solve the problem for you from there.

10. Slide 10 Networks I for M.E. September 5, 2006 Problem 4.4-7 Waaaay too messy for a quiz!

11. Slide 11 Networks I for M.E. September 5, 2006 Ohm’s Law Use to make node voltages the unknowns

12. Slide 12 Networks I for M.E. September 5, 2006 Loop Equations

13. Slide 13 Networks I for M.E. September 5, 2006 Node Equations

14. Slide 14 Networks I for M.E. September 5, 2006 Rearranging into Matrix Format

15. Slide 15 Networks I for M.E. September 5, 2006 Solution

16. Slide 16 Networks I for M.E. September 5, 2006 Problem 4.4-7 Solution

17. Slide 17 Networks I for M.E. September 5, 2006 Intuitive Approach From Loop 2 plus Loop 3: From Loop 2 The rest is found from Ohm’s Law Good for many homework problems, but not for most real-world problems

18. Slide 18 Networks I for M.E. September 5, 2006 Homework Problem Put problem 4.4-7 in Matlab and solve it with the matrix method.  Reproduce the results given here.  Save your code, print it out, and turn it in with the homework. Use your program to solve problem 4.4-7 with the parameters to the right Why can’t K vv be 1.0? What is special about the circuit that allows simple intuitive solutions?