hw4_sol - EE 101 Handout 12 Professor Abbas El Gamal Feb 11...

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Unformatted text preview: EE 101 Handout # 12 Professor Abbas El Gamal Feb. 11, 2003 Homework #4 Solutions 1. (a) A =    1 1 1- 1 1 1- 1- 1- 1    ˜ A is obtained from A by appending a row which guarantees that one 1 and one- 1 (and only one 1 and only one- 1 ) appear in each column. Therefore, ˜ A =      1 1 1- 1 1 1- 1- 1- 1- 1- 1 1      (b) Since ˜ A has 4 rows, there are 4 nodes altogether, including the reference ground node. And we have 6 columns, that means we have 6 branches. The circuit therefore looks like: v2 b2 i2 v1 b1 i1 i6 v6 b6 b3 v3 i3 b4 v4 i4 b5 i5 v5 2. We choose nodes (3), (6), and (8) to be the reference ground nodes for each of the subcircuit. We then short (3), (6), and (8) together, and remove the labels (6) and (8). We can now write the incidence matrix ˜ A . By inspection, ˜ A =           1- 1 1 1- 1- 1- 1 1 1- 1 1- 1 1- 1- 1 1           1 Note: For an incidence matrix, the rows are not linearly independent. We can always remove one of the rows to create a reduced incidence matrix that has independent rows. Let us eliminate row 3 here, which is what we customarily do since it is our reference ground node. So, the reduced incidence matrix is as follows: A =         1- 1 1 1 1- 1 1- 1- 1 1         Equation for KCL: Ai = 0 where i is i =                i 1 i 2 i 3 i 4 i 5 i 6 i 7 i 8                Equation for KVL: v = A T e , i.e.                v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8                =                1- 1 1 1 1- 1 1- 1- 1 1                        e 1 e 2 e 4 e 5 e 7         Note: e 3 , e 6 , and e 8 do not appear above since they are all the ground nodes. 3. To prove Tellegen’s theorem is remarkably easy. By KCL for the original circuit we have Ai = 0. By KVL for the modified circuit we have ˜ v = A T ˜ e . The important point here is that the matrix A is the same in these two equations because the two circuits have the same topology and the branch and node labels and branch orientations are the same. Therefore, ˜ v T i = ( A T ˜ e ) T i = ˜ eAi = 0 , since Ai = 0. We can also show that v T ˜ i = 0 in a similar way....
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This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.

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hw4_sol - EE 101 Handout 12 Professor Abbas El Gamal Feb 11...

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