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The solutions to problem 3 of the electrical engineering 571 course, focusing on circuit analysis and control. It includes the analogous circuits, state variables, and eigenvalues and eigenvectors calculations for various problems.

Typology: Assignments

Pre 2010

1 / 3

Download EE571 - Solutions to Homework #3: Circuit Analysis and Control and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! EE571 Solution to HW#3 0. Have you picked a lab partner to work with you? Your first lab is coming soon! 1a) The analogous circuit for the first problem is: Original Circuit Analogous Circuit J1 J2 F1 K w=input Torque F2 Na: 1 θ θ 1 2 J1=1/3 1/F1=1 w 2:1 Na:1 1/F2=1 J2=1/2 1/K=1/4 where F1=F2=1 Nms/rad, K= 4 Nm/rad, J1= 1/3 Nms2,J2= 1/2 Nms2 and Na=2 The admittance seen to the left of the transformer is Y=J2s+K/s. Therefore, this admittance reflected across the transformer is: Yeq=Y/Na2=(J2s+K/s)/Na2. This reflected admittance would come from a capacitor of J2/Na2 in parallel with an inductor with a value of Na2/K Thus, the equivalent circuit is given by: Equivalent Circuit Circuit with State Variables Replaced J1=1/3 1/F1=1 w 1/F2=1 J2/Na =1/8 Na /K=1 2 2 Vc2 - + Vc1 - + iL 1/F1=1 w 1/F2=1 Vc2 Vc1 iL + - + - The state variable of the analogous electrical circuit is x=[Vc1 Vc2 iL ]T. Thus: & & & & / ( ) / ( ) / ( ) ( ) ( ) ( ) ( ) ( ) ( x V V i C i C i L v i i v i i i i i i i i v v c c L c c L c c L c V c V c i c w c V c V c i c w L V L c c L c c L c = = = = + + + + + + + 1 2 1 1 2 2 1 2 1 1 1 1 2 1 2 2 1 1 1 3 8 1 3 8 1 1 2 1 2 1 V L i L w c c L c c L c c L c L v v v v i w v v i w v v i w 2 3 2 1 0 1 8 1 1 1 0 1 0 0 0 1 2 1 2 1 2+ + = − + + + − − + + + + ) ( ) ( ) ( ) or &x w= − − − + 6 3 0 8 8 8 0 1 0 3 0 0 1b) The analogous circuit for problem 1b) is: Original Circuit Input Flow Change q1 R1 Tank 1 with Capacitance C1 Tank 2 with Capacitance C2 Change in Fluid Level qin h2 Change in Fluid Level h1 R2 Flow Resistance (both R1 and R2 are connected between tank 1 and tank2 ) qout q2 R2 Analogous Circuit C1 w =qin Req = R1||R2 h2 - + h1 - + q1+q2 C2 R2 qout The state variable of the analogous electrical circuit is x=[hc h2 ]T. Thus: ++− ++− = ++ ++ = = = )0)/1/1(/(/1 )//(/1 )(/1 )(/1 )(/1 )(/1 22112 2211 2222 1111 22 11 2 1 21 21 wRRhRhC wRhRhC iiiC iiiC iC iC h h x eq eq wchchc wchchc c c & & & or w C x CRCR RR CR CRCR x eq eq eq eqeq + + − − = 0 /1 1 11 1 222 2 2 11& where Req= R1R2/(R1+R2) 2a) i) First let’s find the eigenvalues of A: 310)3)(1(34 21 12 21 12 21 2 ==∴=−−=+−= −− −− =−⇒ = sandsssss s s AsIA Next, let’s find the eigenvectors of A: [ ] − = =∴= −− −− = −− −− =− 1 1 0 11 11 211 121 21 11 1 21 11 2 1 11 p p P p p p p PAIs [ ] = =∴= − − = −− −− =− 1 1 0 11 11 231 123 22 12 2 22 12 2 1 22 p p P p p p p PAIs ii) First let’s find the eigenvalues of A: 460)4)(6(122 212 2 212 20 21 2 −==∴=+−=−+−= −− − =−⇒ = sandsssss s s AsIA Next, let’s find the eigenvectors of A: [ ] = =∴= − − = −− − =− 3 1 0 412 26 2612 26 21 11 1 21 11 2 1 11 p p P p p p p PAIs [ ] − = =∴= −− −− = −−− −− =− 2 1 0 612 24 2412 24 22 12 2 22 12 2 1 22 p p P p p p p PAIs iii) First let’s find the eigenvalues of A: