A Semantic Model For VHDL-AMS
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A Semantic Model For VHDL-AMS
- 1. ? A semantic model for VHDL-AMS Natividad MartıĢnez Madrid, Peter T. Breuer, Carlos Delgado Kloos Universidad Carlos III de Madrid <nmadrid,ptb,cdk>@it.uc3m.es CHARME ā97, MontreĢal, Canada - October 1997 1
- 2. Objectives ? ā¢ explain behaviour of VHDL-AMS processes ā not a detailed syntactic mapping ā provide sufficient primitives ā¢ present an underlying model that . . . ā extends existing model for VHDL CHARME ā97, MontreĢal, Canada - October 1997 2
- 3. Content of the talk ? ā¢ Introduction ā¢ A process algebraic analysis ā¢ Semantics ā¢ The analog solver with example CHARME ā97, MontreĢal, Canada - October 1997 3
- 4. Main idea ? VHDL semantics analog extension imperative discrete-event diff. equations real time domain declarative continuous-time Ö Ö INTEGRATION CHARME ā97, MontreĢal, Canada - October 1997 4
- 5. Name space division & Properties ? VARIABLES assigned instantaneously not scheduled SIGNALS at least Ī“-delayed preemptively scheduled QUANTITIES assigned instantaneously governed by diff. eqns. PROCESSES invariant invariant CHARME ā97, MontreĢal, Canada - October 1997 5
- 6. Semantics of assignment ? 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 00000000000000000000000000 00000000000000000000000000 00000000000000000000000000 00000000000000000000000000 00000000000000000000000000 11111111111111111111111111 11111111111111111111111111 11111111111111111111111111 11111111111111111111111111 11111111111111111111111111 00000 11111 00 11 "Forced" signal assignment a <= 1 after 4 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 a 00000000000000000000000000 00000000000000000000000000 00000000000000000000000000 11111111111111111111111111 11111111111111111111111111 11111111111111111111111111 . v = x . 00000 11111 00 11 "Relaxed" quantity update 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 00000000000000000000000000 00000000000000000000000000 00000000000000000000000000 11111111111111111111111111 11111111111111111111111111 11111111111111111111111111 x a = v CHARME ā97, MontreĢal, Canada - October 1997 6
- 7. Semantic Domains ? WorldLine = Time ā State State = Id ā V alue VHDL Statements Semantics = (WorldLine, Time) ā (WorldLine, Time) Time = Int VHDL-AMS Statements Semantics = (EqnSet, WorldLine, Time) ā (EqnSet, WorldLine, Time) Time = Real CHARME ā97, MontreĢal, Canada - October 1997 7
- 8. Parallel decomposition ? LRM view of basic VHDL compositionality: ā¢ processes in parallel (+ kernel) ā¢ process body loops continuously ā¢ imperative commands in process body run sequentially Our view: ā¢ no kernel CHARME ā97, MontreĢal, Canada - October 1997 8
- 9. VHDL-AMS: integer time ā real time Analog Solver
- 10. Process algebraic analysis ? 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 0000000000000000000000000000 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 1111111111111111111111111111 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 Ai Aj Ī£j Ī£i Ī£j Pjā Piā Pi Pj Ki Kj Ī£i 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 00000000000000000000000000000 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 11111111111111111111111111111 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111 AS Ai Aj Ī£j Ī£i Ī£j Pjā Piā Pj Ki Kj Ī£i Pi CHARME ā97, MontreĢal, Canada - October 1997 9
- 11. Process algebraic analysis (II) ? 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 11111111111111111111 i Ī£ Ī£j āi Ī£ āj Ī£ Kāi Ī£ Ī£ Q AS Q AS KAj KAi Pi Pj Ai Aj Kāj Pjā Piā CHARME ā97, MontreĢal, Canada - October 1997 10
- 12. Denotational semantics ? S[a; b] = S[a]; S[b] P[a; b] = P[a] āŖ S[a]; P[b] S[while(true)do a end] = { } P[while(true)do a end] = P[a] āŖ S[a]; P[while(true)do a end] = ĀµR : Semantics . R = P[a] āŖ S[a]; R S[a||b] = S[a] ā© S[b] P[a||b] = P[a] ā© P[b] CHARME ā97, MontreĢal, Canada - October 1997 11
- 13. Wait & Assignment semantics ? Wait Semantics (e, w0, t0)S[wait until p](e, w1, t1) = w0 = w1 ā§ wt1 |= p ā§ āt ā [t0, t1) . wt 6|= p (e, w0, t0)P[wait until p](e, w1, t1) = w0 = w1 ā§ āt ā [t0, t1] . wt 6|= p Assignment Semantics (e, w0, t0)S[x ā y after Ļ](e, w1, t1) = t0 = t1 ā§ w1 = relax(e, t0)(force assign(w0)) P[x ā y after Ļ] = { } CHARME ā97, MontreĢal, Canada - October 1997 12
- 14. Wait semantics ? 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 Generalized Wait a x wait until a = 1 on signals -> discrete time 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 wait until x >= 2/3 on quantities -> continuous time CHARME ā97, MontreĢal, Canada - October 1997 13
- 16. Analog solution ? Bouncing ball example ds dt = v dv dt = āg Ā± av2 Local approximation s(t) = 1 v(t) = āgt origin at s = 1 v = 0 t = 0 Linear approximation (s, v) = (s0, v0) + (t ā t0)(v0, āg Ā± av2 0) Recursion (s, v)(t0,s0,v0)(t) ā¼ ļ£± ļ£² ļ£³ (s0, v0) + (t ā t0)(v0, āg Ā± av2 0) t ā [t0, t1) (s, v)(t1,s1,v1)(t) s1 = s0 + v0āt v1 = v0 + (āg Ā± av2 0)āt t1 = t0 + āt CHARME ā97, MontreĢal, Canada - October 1997 15
- 17. Example ? proc [qty s := 1.0, v := 0.0, g := 9.8, a := 0.1 ] ds/dt == v; dv/dt == -g + if v<0 then a else -a end * v * v; begin wait until s < 0; v := -v; s := 0 ; end āt = 0.02 CHARME ā97, MontreĢal, Canada - October 1997 16
- 18. Example Cont. ? proc [qty s := 1.0, v := 0.0, g := 9.8, a := 0.1 ] ds/dt == v; dv/dt == -g + if v<0 then a else -a end * v * v; begin wait until s < 0; v := 1.4142 * (g * s + 0.5 * v * v)**0.5 ; s := 0 ; end g ā 0 + vā²2 2 = g ā s + v2 2 āt = 0.06 CHARME ā97, MontreĢal, Canada - October 1997 17
- 19. Example Cont. (II) ? Height Speed āt = 0.06 CHARME ā97, MontreĢal, Canada - October 1997 18
- 20. Unresolved areas ? ā¢ Implementation of the parallel composition of pro- cesses if quantities are shared between processes ā¢ Errors in the analog solver may require us to rewrite our program ā¢ The proper execution of the analog solver in Ī“ time when the processes are in parallel CHARME ā97, MontreĢal, Canada - October 1997 19
- 21. Conclusion ? ā¢ What? Extension of our VHDL semantics to cover VHDL-AMS ā¢ How? Embedding of the VHDL semantics in a bigger, continuous time domain, which contains an oracular analog solver ā¢ Why? Clarify the draft standard document CHARME ā97, MontreĢal, Canada - October 1997 20