hw1 solu 355

hw1 solu 355 - Solutions to Problems 2-7 Cross multiplying,...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Problems 2-7 Cross multiplying, (s 6 +7s 5 +3s 4 +2s 3 +s 2 +3)C(s) = (s 5 +2s 4 +4s 3 +s 2 +3)R(s). Taking the inverse Laplace transform assuming zero initial conditions, d 6 c dt 6 + 7 d 5 c dt 5 + 3 d 4 c dt 4 + 2 d 3 c dt 3 + d 2 c dt 2 + 3 c = d 5 r dt 5 + 2 d 4 r dt 4 + 4 d 3 r dt 3 + d 2 r dt 2 + 3 r . 10. s 4 2 s 3 5 s 2 s C The transfer function is ( s ) R ( s ) = g14 g14 g14 g14 1 s 5 3 s 4 2 s 3 4 s 2 5 s g14 g14 g14 g14 g14 2 . Cross multiplying, (s 5 +3s 4 +2s 3 +4s 2 +5s+2)C(s) = (s 4 +2s 3 +5s 2 +s+1)R(s). Taking the inverse Laplace transform assuming zero initial conditions, d 5 c dc 5 + 3 d 4 c dt 4 + 2 d 3 c dt 3 + 4 d 2 c dt 2 + 5 dc dt + 2 c = dr d 4 r dt 4 + 2 d 3 r dt 3 + 5 d 2 r dt 2 + dt + r . Substituting r(t) = t 3 , d 5 c dc 5 + 3 d 4 c dt 4 + 2 d 3 c dt 3 + 4 d 2 c dt 2 + 5 dc dt + 2 c = 18 g71 (t) + (36 + 90 t + 9 t 2 + 3t 3 ) u( t ). 11. Taking the Laplace transform of the differential equation, s 2 X(s)-s+1+2sX(s)-2+3x(s)=R(s). Collecting terms, (s 2 +2s+3)X(s) = R(s)+s+1. R ( s ) s 2 s g14 1 s 2 Solving for X(s), X(s) = g14 2 s g14 3 + 2 s g14 g14 3 . The block diagram is then, 12. Program: 'Factored' Gzpk=zpk([-15 -26 -72],[0 -55 -56 roots([1 5 30])' roots([1 27 52])'],5) 'Polynomial' Gp=tf(Gzpk) Computer response: ans = Factored Zero/pole/gain: 5 (s+15) (s+26) (s+72) ---------------------------------------------------- s (s+55) (s+56) (s+24.91) (s+2.087) (s^2 + 5s + 30) Solutions to Problems 2-11 b. Thevenizing, 1 s V o ( s ) V i ( s ) Using voltage division, V o ( s ) g32 V i ( s ) 2 g32 1 2 s 2 g14 s g14 2 . Thus, 1 2 g14 s g14 1 s 17. a. Writing mesh equations (s+1)I 1 (s) – I 2 (s) = V i (s) -I 1 (s) + (s+2)I 2 (s) = 0 But, I 1 (s) = (s+2)I 2 (s). Substituting this in the first equation yields, (s+1)(s+2)I 2 (s) – I 2 (s) = V i (s) or I 2 (s)/V i (s) = 1/(s 2 + 3s + 1)...
View Full Document

This note was uploaded on 11/07/2011 for the course MEM 355 taught by Professor Ani during the Fall '08 term at UPenn.

Page1 / 9

hw1 solu 355 - Solutions to Problems 2-7 Cross multiplying,...

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