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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online