1. Start of presentation September 27, 2012 Algebraic Loops and Structural Singularities The sorting algorithm, as it was demonstrated so far, does not always work correctly. All of the examples shown to this point had been chosen carefully to hide these problems. The aim of the lecture is to generalize the algorithms to systems containing algebraic loops and/or singular structures.

2. Start of presentation September 27, 2012 Table of Contents Algebraic loops Structure diagram Structural singularities Differentiation

3. Start of presentation September 27, 2012 Algebraic Loops: An Example Component equations: U 0 = f(t)u 3 = R 3 · i 3 u 1 = R 1 · i 1 u L = L· di L /dt u 2 = R 2 · i 2 Node equations: i 0 = i 1 + i L i 1 = i 2 + i 3 Mesh equations: U 0 = u 1 + u 3 u L = u 1 + u 2 u 3 = u 2 The circuit contains 5 components  We require 10 equations in 10 unknowns

4. Start of presentation September 27, 2012 Horizontal Sorting I U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L i 1 = i 2 + i 3 U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 1. U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L i 1 = i 2 + i 3 U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L i 1 = i 2 + i 3 U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 2. 3. 4. U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L i 1 = i 2 + i 3 U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2

5. Start of presentation September 27, 2012 Horizontal Sorting II U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L i 1 = i 2 + i 3 U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 Of the six equations that are still a- causal (i.e., the equations not containing a red variable), every one contains at least two unknowns. Furthermore, every one of the unknowns shows up in at least two of the a-causal equations.  Such a situation indicates the existence of algebraic loops.

6. Start of presentation September 27, 2012 Algebraic Loops I 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 We choose one unknown from one equation, e.g. variable i 1 from equation 4. We assume this variable to be known and continue as before. 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 1. 2. 3. 4.

7. Start of presentation September 27, 2012 Algebraic Loops II 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 i2i2 i1i1 u1u1 u3u3 i3i3 u2u2 U0U0 4. 1. 2. 3. 5. 6. Algebraic Loops Structure diagram

8. Start of presentation September 27, 2012 Solution of Algebraic Loops I 1. u 1 = R 1 · i 1 2. u 2 = R 2 · i 2 3. u 3 = R 3 · i 3 4. i 1 = i 2 + i 3 5. U 0 = u 1 + u 3 6. u 3 = u 2 1. u 1 = R 1 · i 1 2. i 2 = u 2 / R 2 3. i 3 = u 3 / R 3 4. i 1 = i 2 + i 3 5. u 3 = U 0 - u 1 6. u 2 = u 3  i 1 = i 2 + i 3 = u 2 / R 2 + u 3 / R 3 = u 3 / R 2 + u 3 / R 3 = ((R 2 + R 3 ) / (R 2 · R 3 )) · u 3 = ((R 2 + R 3 ) / (R 2 · R 3 )) · (U 0 - u 1 ) = ((R 2 + R 3 ) / (R 2 · R 3 )) · (U 0 - R 1 · i 1 )  i 1 = R 2 + R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 · U 0 Equation 4. is replaced by the new equation.

9. Start of presentation September 27, 2012 Solution of Algebraic Loops II U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 i 1 = R 2 + R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 · U 0  The algebraic loop has now been solved, and we can continue with the sorting algorithm as before.

10. Start of presentation September 27, 2012 Horizontal Sorting III U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 i 1 = R 2 + R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 · U 0 U 0 = f(t) u 1 = R 1 · i 1 u 2 = R 2 · i 2 u 3 = R 3 · i 3 u L = L· di L /dt i 0 = i 1 + i L U 0 = u 1 + u 3 u 3 = u 2 u L = u 1 + u 2 i 1 = R 2 + R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 · U 0

11. Start of presentation September 27, 2012 Multiple Coupled Algebraic Loops 1. a = b + 1 2. b = 3·f 3. c = b + d 4. d = h 5. e = a 6. f = e + g 7. g = 2·c 8. h = g 4. 6. 1. 2. 3. 5. c d hg b f a e 3. 6. 7. 8. 1. a = b + 1 2. b = 3·f 3. c = b + d 4. d = h 5. e = a 6. f = e + g 7. g = 2·c 8. h = g  1. a = b + 1 2. b = 3·f 3. c = b + d 4. d = h 5. e = a 6. f = e + g 7. g = 2·c 8. h = g  c = b + d = 3·f + h = 3·f + g = 3·f + 2·c f = e + g = a + 2·c = b + 2·c + 1 = 3·f + 2·c + 1 c + 3·f = 0 2·c + 2·f = -1 c = - 0.75 f = + 0.25

12. Start of presentation September 27, 2012 Structural Singularities: An Example The mixed rotational and translational system exhibits three bodies: the inertiae J 1 and J 2 as well as the mass m. Therefore, we would expect the system to be of 6 th order. 3 bodies  6 differential equations + 3 algebraic equations (D’Alembert) 3 frictions  3 algebraic equations (friction forces) 2 springs  2 algebraic equations (spring forces) 1 gear  2 algebraic equations (transmission)  16 equations 16 unknowns

13. Start of presentation September 27, 2012 Modeling of the Gear We cut through the gear. The effect of the other body is replaced by a cutting force.  The torque  is proportional to the cutting force F, and the displacement x is proportional to the angle .  = r · F x = r · 

14. Start of presentation September 27, 2012 Cutting the System  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G x = r ·  2  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x  16 equations 16 unknowns

15. Start of presentation September 27, 2012 Horizontal Sorting I  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G x = r ·  2  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x This equation cannot be used since it contains no unknown. Idea: If an equation holds true for all times, then every derivative of that equation holds true as well.  Replace the unusable equation by its time derivative.

16. Start of presentation September 27, 2012 Differentiation I  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G v = r ·  2  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x Unfortunately, the equation is still not usable, because it still does not contain any unknown.  Differentiate the unusable equation once more.

17. Start of presentation September 27, 2012 Differentiation II  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt The equation has now become usable, since both of the variables contained in it are unknowns. The two derivatives had been red until now, because they appeared only once in the equation system. As they now appear twice, they need to be reset to black.

18. Start of presentation September 27, 2012 Horizontal Sorting II  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt 

19. Start of presentation September 27, 2012 Horizontal Sorting III  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt There still remain 6 equations in 6 unknowns. Every one of these equations contains at least two unknowns. Every one of the unknowns appears at least in two of the remaining equations.  We are again confronted with at least one algebraic loop.

20. Start of presentation September 27, 2012 Algebraic Loop  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt Choice  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt 

21. Start of presentation September 27, 2012 Horizontal Sorting IV  (t) =  T1 +  B1 +  B3  B1 =  T2 +  k1 +  G F G = F I + F k2 + F B2 + m · g  T1 = J 1 · d1d1 dt d1d1 =  1  T2 = J 2 · d2d2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt  T1  =  (t) -  B1 -  B3  T2  =  B1 -  k1 -  G F G = F I + F k2 + F B2 + m · g =  T1 / J 1 d1d1 dt d1d1 =  1 =  T2 / J 2 d  2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt 

22. Start of presentation September 27, 2012 Solution of the Algebraic Loop I d  2 dt =  T2 / J 2 = (  B1 -  k1 -  G ) / J 2 = (  B1 -  k1 ) / J 2 -  G /J 2 = (  B1 -  k1 ) / J 2 - (r /J 2 ) · F G  T1  =  (t) -  B1 -  B3  T2  =  B1 -  k1 -  G F G = F I + F k2 + F B2 + m · g =  T1 / J 1 d1d1 dt d1d1 =  1 =  T2 / J 2 d  2 dt d2d2 =  2 F I = m· dv dt dx dt = v  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt = (  B1 -  k1 ) / J 2 - (r /J 2 ) · (F I + F k2 + F B2 + m·g) = (  B1 -  k1 ) / J 2 - (r /J 2 ) · (F k2 + F B2 + m·g) - (r /J 2 ) · F I = (  B1 -  k1 ) / J 2 - (r /J 2 ) · (F k2 + F B2 + m·g) - (m·r /J 2 ) · dv/dt = (  B1 -  k1 ) / J 2 - (r /J 2 ) · (F k2 + F B2 + m·g) - (m·r 2 /J 2 ) · d  2 /dt

23. Start of presentation September 27, 2012 Solution of the Algebraic Loop II  T1  =  (t) -  B1 -  B3  T2  =  B1 -  k1 -  G F G = F I + F k2 + F B2 + m · g  G = r · F G  B1 = B 1 · (  1 –  2 )  B3 = B 3 ·  1 F B2 = B 2 · v  k1 = k 1 ·  2 F k2 = k 2 · x dv dt = r · d2d2 dt =  T1 / J 1 d1d1 dt d1d1 =  1 d2d2 dt =  2 F I = m· dv dt dx dt = v d  2 dt =  B1 -  k1 – r · (F k2 + F B2 ) – m·g·r J 2 + m·r 2

24. Start of presentation September 27, 2012 Comments The problem of the structural singularity occurred, because the mass m and the inertia J 2 cannot be moved independently of each other. For this reason, it had to be possible to describe the system by only 4 differential equations. The solution approach presented here does not exploit that simplification directly. A better approach shall be explained in due course.

25. Start of presentation September 27, 2012 References Cellier, F.E. and H. Elmqvist (1993), “Automated formula manipulation supports object-oriented continuous-system modeling,” IEEE Control Systems, 13(2), pp. 28-38.Automated formula manipulation supports object-oriented continuous-system modeling