hw19sol

# hw19sol - Purdue University School of Electrical and...

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #19 Solution (73–76) Problem 73 (a) y-parameters of the circuit (i) Labels Y 1 Y 2 Y 3 + V 1 - I 2 I 1 I in (s) A B C D + V AB - + V CD - + V 2 - I 2 I 1 I 2 1 : a g m V 1 n1 n2 (ii) ± I 1 I 0 2 ² = ± ? ? ? ? ²± V 1 V AB ² To ﬁnd the matrix elements for the above y-parameter matrix, we write two node equations denoted as n1 and n2 in the above diagram. node n1: I 1 = V 1 Y 1 + ( V 1 - V AB ) Y 2 I 1 = ( Y 1 + Y 2 ) V 1 + ( - Y 2 ) V AB node n2: I 0 2 = ( V AB - V 1 ) Y 2 + V AB Y 3 I 0 2 = ( - Y 2 ) V 1 + ( Y 2 + Y 3 ) V AB Putting into a matrix form, ± I 1 I 0 2 ² = ± Y 1 + Y 2 - Y 2 - Y 2 Y 2 + Y 3 ²± V 1 V AB ² (73.1) (iii) From ideal transformer voltage and current relationships, we get V AB = - V CD / a and I 0 2 = - aI 00 2 Substituting for V AB and I 00 2 into eq.(73.1) and rearranging ± I 1 - aI 00 2 ² = ± Y 1 + Y 2 - Y 2 - Y 2 Y 2 + Y 3 ²± V 1 - V CD / a ² ± I 1 I 00 2 ² = ± Y 1 + Y 2 Y 2 / a Y 2 / a ( Y 2 + Y 3 ) / a 2 ²± V 1 V CD ² (73.2) 1/ 10

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (iv) Overall y-parameters From the ﬁgure, we ﬁnd V CD = V 2 and I 00 2 = g m V 1 + I 2 As done in the last part, substituting for V CD and I 00 2 into eq.(73.2) and rearranging ± I 1 g m V 1 + I 2 ² = ± Y 1 + Y 2 Y 2 / a Y 2 / a ( Y 2 + Y 3 ) / a 2 ²± V 1 V 2 ² ± I 1 I 2 ² = ± Y 1 + Y 2 Y 2 / a Y 2 / a - g m ( Y 2 + Y 3 ) / a 2 ²± V 1 V 2 ² (b) Step Response Y 1 = s , Y 2 = 4 , Y 3 = 4 s , g m = 2 . 5 [S], a = 2 and port 2 terminals are terminated in a 1 Ω resistor. + V 1 - I 2 I 1 I in (s)=18/(s+6) + V 2 - Y = y 11 y 12 y 21 y 22 R L =1 Ω When 1 Ω resistor connected to the load, I 2 = - V 2 Y L = - V 2 . We can modify the overall y-parameter matrix equation we’ve found in part (a) as following ± I 1 I 2 ² = ± Y 1 + Y 2 Y 2 / a Y 2 / a - g m ( Y 2 + Y 3 ) / a 2 ²± V 1 V 2 ² Overall y-matrix found in part(a) ± I 1 0 ² = ± Y 1 + Y 2 Y 2 / a Y 2 / a - g m ( Y 2 + Y 3 ) / a 2 + 1 ²± V 1 V 2 ² After substitution for I 2 ± I 1 0 ² = ± s + 4 2 - 1 / 2 s + 2 ²± V 1 V 2 ² (73.3) Using MATLAB, we ﬁnd v 1 ( t ) = ³ 8 e - 3 t - 6 te - 3 t - 8 e - 6 t ´ u ( t ) [V] v 2 ( t ) = ³ - e - 3 t + 3 te - 3 t + e - 6 t ´ u ( t ) [V] 2/ 10
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hw19sol - Purdue University School of Electrical and...

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