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Unformatted text preview: EE 101 Handout # 25 Professor Abbas El Gamal March 11, 2003 Homework #7 Solutions 1. (a) The first thing we do is change the circuit to its equivalent phasor representation. To find the equivalennt impedance place a test source V x in parallel with the VCCS and measure the current I x flowing through it. KCL results in: I x = g V + V x R j/wC You can find the voltage V by voltage divider: V = V x j wRC j Plug the second equation into the first: I x = V x jg ( wRC j ) + wC ( RwC j ) The equivalent impedance is defined as Z = V x / I x : Z = wRC j jg + wC We want to seperate the real and imaginary part. A common technique is to multiply the equation by ( wC + jg ) / ( wC + jg ). Note that ( wc + jg ) is the conjugate of the denominator. After some manipulation: Z = w 2 C 2 R + g g 2 + w 2 C 2 + j wC ( Rg 1) g 2 + w 2 C 2 (b) Looking at the imaginary part it is easy to see that the impedance will be positive only when R > 1 /g , i.e. , R > 100kΩ, independent of the values of ω and C . 2. (a) After transforming each element into its appropriate phasor, we can write the node voltage equations for each node as follows: n 1 : n 1 = 2 n 2 : n 2 n 1 1 + n 1 1 2 j = 0 n 3 : n 3 n 4 2 = 2 V = 2 n 2 n 4 : n 4 n 3 2 + n 4 1 2 j + n 4 j +1 = 0 1 Rearranging the equations so that all of the variables ( n 1 , n 2 , n 3 , and n 4 ) are on the left side, the matrix equation is then: 1 1 1 + j 2 2 . 5 . 5 . 5 1 + 1 . 5 j E 1 E 2 E 3 V out = 2 (b) To get v out ( t ), solve for the matrix from part (a) to get V out = . 8 . 8 j . Thus, v out ( t ) = 0 . 8 √ 2 cos (2 t 135 ◦ ) = 1 . 13 cos (2 t 135 ◦ ). 3. First, let’s write what we know about the experiments in terms of phasors. Experiment V I P avg 1 j 10 1 . 414 e j 45 ◦ — 2 — j 2 10 3 e j 45 ◦ 1 2.5 4 1 j . 1 + 0 . 1 — (a) Let’s look at the first experiment. The phasor of the voltage is 10 e jπ/ 2 and the phasor of the current is √ 2 e jπ/ 4 , therefore the impedance should be 5 √ 2 e jπ/ 4 = 5 5 j ....
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