1. Start of Presentation September 27, 2012 1 st Homework Problem We wish to analyze the following electrical circuit.

2. Start of Presentation September 27, 2012 Problems I 1.Write down all of the equations that you need to describe this circuit mathematically. You may use either the technique involving potentials or that involving mesh equations. 2.Sort the equation system horizontally. 3.Sort the horizontally sorted equation system vertically. 4.Convert the horizontally and vertically sorted equation system to state-space form.

3. Start of Presentation September 27, 2012 Component equations: Mesh equations: Node equations: u 0 u 4 v 1 v 2 v L i 0 i 1 i 2 i L i C i 3 u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 13 equations in 13 unknowns  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt i 4 = 4 · v 3

4. Start of Presentation September 27, 2012 Problems II 1.Write down all of the equations that you need to describe this circuit mathematically. You may use either the technique involving potentials or that involving mesh equations. 2.Sort the equation system horizontally. 3.Sort the horizontally sorted equation system vertically. 4.Convert the horizontally and vertically sorted equation system to state-space form.

5. Start of Presentation September 27, 2012 Horizontal Sorting I  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

6. Start of Presentation September 27, 2012 Horizontal Sorting II  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

7. Start of Presentation September 27, 2012 Horizontal Sorting III  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

8. Start of Presentation September 27, 2012 Horizontal Sorting IV  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

9. Start of Presentation September 27, 2012 Horizontal Sorting V  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

10. Start of Presentation September 27, 2012 Horizontal Sorting VI  u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

11. Start of Presentation September 27, 2012 Horizontal Sorting VII  u 0 = f 1 (t) i 1 = v 1 / R 1 v 2 = R 2 · i 2 i 3 = v 3 / R 3 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) v 1 = R 1 · i 1 v 2 = R 2 · i 2 v 3 = R 3 · i 3 v L = L · di L dt i C = C · dv 3 dt u 0 = v 1 + v 3 v L = v 1 + v 2 v 2 = u 4 + v 3 i 0 = i 1 + i L i 1 = i 2 + i 3 + i C i 4 = i L + i 2 i 4 = 4 · v 3

12. Start of Presentation September 27, 2012 Problems III 1.Write down all of the equations that you need to describe this circuit mathematically. You may use either the technique involving potentials or that involving mesh equations. 2.Sort the equation system horizontally. 3.Sort the horizontally sorted equation system vertically. 4.Convert the horizontally and vertically sorted equation system to state-space form.

13. Start of Presentation September 27, 2012 Vertical Sorting I  u 0 = f 1 (t) i 1 = v 1 / R 1 v 2 = R 2 · i 2 i 3 = v 3 / R 3 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3

14. Start of Presentation September 27, 2012 Vertical Sorting II  u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3

15. Start of Presentation September 27, 2012 Vertical Sorting III  u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3

16. Start of Presentation September 27, 2012 Vertical Sorting IV  u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3

17. Start of Presentation September 27, 2012 Vertical Sorting V  u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3

18. Start of Presentation September 27, 2012 Problems IV 1.Write down all of the equations that you need to describe this circuit mathematically. You may use either the technique involving potentials or that involving mesh equations. 2.Sort the equation system horizontally. 3.Sort the horizontally sorted equation system vertically. 4.Convert the horizontally and vertically sorted equation system to state-space form.

19. Start of Presentation September 27, 2012 State-space Description I u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 = v L / L di L dt = (v 1 + v 2 ) / L = v 1 / L + v 2 / L = (u 0  v 3 ) / L + (R 2 / L) · i 2 = u 0 / L  v 3 / L + (R 2 / L)· (i 4  i L ) = u 0 / L  v 3 / L + (4R 2 / L) · v 3  (R 2 / L)· i L =  (R 2 /L)· i L + (4R 2 / L  1 / L)· v 3 + (1/L)· u 0

20. Start of Presentation September 27, 2012 State-space Description II u 0 = f 1 (t) i 3 = v 3 / R 3 i 1 = v 1 / R 1 v 2 = R 2 · i 2 = v L / L di L dt = i C / C dv 3 dt v 1 = u 0  v 3 v L = v 1 + v 2 u 4 = v 2  v 3 i 0 = i 1 + i L i C = i 1  i 2  i 3 i 2 = i 4  i L i 4 = 4 · v 3 = i C / C dv 3 dt = (i 1  i 2  i 3 ) / C = v 1 / (R 1 · C)  ( i 4  i L ) /C  v 3 / (R 3 · C) = (u 0  v 3 ) / (R 1 · C)  (4 /C) · v 3 + (1/C) · i L  v 3 / (R 3 · C) = (1/C) · i L  (1/(R 1 · C) + 1/(R 3 · C) + 4 /C) · v 3 +  /(R 1 · C)· u 0

21. Start of Presentation September 27, 2012 State-space Description III dv 3 dt di L dt =  (R 2 /L)· i L + (4R 2 / L  1 / L)· v 3 + (1/L)· u 0 = (1/C) · i L  (1/(R 1 · C) + 1/(R 3 · C) + 4 /C) · v 3 +  /(R 1 · C)· u 0 ·· u 0 + L 1 R 1 · C 1 iLiL v3v3 di L dt dv 3 dt = L R2R2  1 C  L 4R 2 L 1   R 1 · C 1 R 3 · C 1 + +  4 C

22. Start of Presentation September 27, 2012 State-space Description IV iLiL v3v3 ·v3v3 = 0 1 ·· u 0 + L 1 R 1 · C 1 iLiL v3v3 di L dt dv 3 dt = L R2R2  1 C  L 4R 2 L 1   R 1 · C 1 R 3 · C 1 + +  4 C

23. Start of Presentation September 27, 2012 Problems V 5.Program the state-space model in Matlab. 6.Simulate the circuit across 50  sec of simulated time, und plot the variable v 3 as a function of time. You may assume that the initial Voltage across the capacitance and the initial current through the inductivity are equal to 0.0.

24. Start of Presentation September 27, 2012