A Semantic Model for VHDL-AMS (CHARME '97)
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The document discusses a semantic model for VHDL-AMS, focusing on explaining the behavior of VHDL-AMS processes without detailed syntactic mapping. It presents a process algebraic analysis and an analog solver with examples, highlighting the semantics of assignments and parallel decomposition in the context of VHDL. The aim is to extend existing models and provide sufficient primitives for better understanding of VHDL-AMS behavior.
![Denotational semantics ?
S[a; b] = S[a]; S[b]
P[a; b] = P[a] ∪ S[a]; P[b]
S[while(true)do a end] = { }
P[while(true)do a end] = P[a] ∪ S[a]; P[while(true)do a end]
= µR : Semantics . R = P[a] ∪ S[a]; R
S[a||b] = S[a] ∩ S[b]
P[a||b] = P[a] ∩ P[b]
CHARME ’97, Montr´eal, Canada - October 1997 11](https://image.slidesharecdn.com/talk-charme97-130917145431-phpapp01/85/A-Semantic-Model-for-VHDL-AMS-CHARME-97-12-320.jpg)
 = w0 = w1
∧ wt1 |= p ∧ ∀t ∈ [t0, t1) . wt |= p
(e, w0, t0)P[wait until p](e, w1, t1) = w0 = w1
∧ ∀t ∈ [t0, t1] . wt |= p
Assignment Semantics
(e, w0, t0)S[x ⇐ y after τ](e, w1, t1) = t0 = t1
∧ w1 = relax(e, t0)(force assign(w0))
P[x ⇐ y after τ] = { }
CHARME ’97, Montr´eal, Canada - October 1997 12](https://image.slidesharecdn.com/talk-charme97-130917145431-phpapp01/85/A-Semantic-Model-for-VHDL-AMS-CHARME-97-13-320.jpg)
![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
∆t = 0.02
CHARME ’97, Montr´eal, Canada - October 1997 16](https://image.slidesharecdn.com/talk-charme97-130917145431-phpapp01/85/A-Semantic-Model-for-VHDL-AMS-CHARME-97-17-320.jpg)
![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 ∗ 0 + v′2
2 =
g ∗ s + v2
2
∆t = 0.06
CHARME ’97, Montr´eal, Canada - October 1997 17](https://image.slidesharecdn.com/talk-charme97-130917145431-phpapp01/85/A-Semantic-Model-for-VHDL-AMS-CHARME-97-18-320.jpg)
A Semantic Model for VHDL-AMS (CHARME '97)
- 1. ? A semantic model for VHDL-AMS Natividad Mart´ınez Madrid, Peter T. Breuer, Carlos Delgado Kloos Universidad Carlos III de Madrid <nmadrid,ptb,cdk>@it.uc3m.es CHARME ’97, Montr´eal, Canada - October 1997 1
- 2. Objectives ? • explain behaviour of VHDL-AMS processes – not a detailed syntactic mapping – provide sufficient primitives • present an underlying model that . . . – extends existing model for VHDL CHARME ’97, Montr´eal, Canada - October 1997 2
- 3. Content of the talk ? • Introduction • A process algebraic analysis • Semantics • The analog solver with example CHARME ’97, Montr´eal, Canada - October 1997 3
- 4. Main idea ? VHDL semantics analog extension imperative discrete-event diff. equations real time domain declarative continuous-time ց ւ INTEGRATION CHARME ’97, Montr´eal, Canada - October 1997 4
- 5. Name space division & Properties ? VARIABLES assigned instantaneously not scheduled SIGNALS at least δ-delayed preemptively scheduled QUANTITIES assigned instantaneously governed by diff. eqns. PROCESSES invariant invariant CHARME ’97, Montr´eal, Canada - October 1997 5
- 6. Semantics of assignment ? 00 00 00 0 0000000000000000000000000 11 11 11 1 1111111111111111111111111 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 00000000000000000000000000000000000000000000000000 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 11111111111111111111111111111111111111111111111111 000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000 0000000000000000000000000 111111111111111111111111111111111111111111111111111111111111111111111111111 1111111111111111111111111 1111111111111111111111111 00000000000000000000000000 0000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000 11111111111111111111111111 1111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111111111 00000111110011 "Forced" signal assignment a <= 1 after 4 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 a 000000000000000000000000000000000000000000000000000000000000000000000000000000 111111111111111111111111111111111111111111111111111111111111111111111111111111 . v = x . 00000111110011 "Relaxed" quantity update 0 00 000 0000 1 11 111 1111 0 00 000 0000 1 11 111 1111 000 111 000 111000000000000000000000000000000000000000000000000000000000000000000000000000000 111111111111111111111111111111111111111111111111111111111111111111111111111111x a = v CHARME ’97, Montr´eal, Canada - October 1997 6
- 7. Semantic Domains ? WorldLine = Time → State State = Id → V alue VHDL Statements Semantics = (WorldLine, Time) ↔ (WorldLine, Time) Time = Int VHDL-AMS Statements Semantics = (EqnSet, WorldLine, Time) ↔ (EqnSet, WorldLine, Time) Time = Real CHARME ’97, Montr´eal, Canada - October 1997 7
- 8. Parallel decomposition ? LRM view of basic VHDL compositionality: • processes in parallel (+ kernel) • process body loops continuously • imperative commands in process body run sequentially Our view: • no kernel CHARME ’97, Montr´eal, Canada - October 1997 8
- 9. VHDL-AMS: integer time → real time Analog Solver
- 10. Process algebraic analysis ? 000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000 111111111111111111111111111111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111111111 00000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 11111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 Ai Aj Σj Σi Σj Pj’ Pi’ Pi Pj Ki Kj Σi 00000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000 11111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111 1111111111111111111111111111111111111111111111111111111111 11111111111111111111111111111 000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111 AS Ai Aj Σj Σi Σj Pj’ Pi’ Pj Ki Kj Σi Pi CHARME ’97, Montr´eal, Canada - October 1997 9
- 11. Process algebraic analysis (II) ? 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111 000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000 000000000000000000000000000000000000000 111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 111111111111111111111111111111111111111 111111111111111111111111111111111111111 00000000000000000000 0000000000000000000000000000000000000000 0000000000000000000000000000000000000000 0000000000000000000000000000000000000000 00000000000000000000 000000000000000000000000000000000000000000000000000000000000 11111111111111111111 1111111111111111111111111111111111111111 1111111111111111111111111111111111111111 1111111111111111111111111111111111111111 11111111111111111111 111111111111111111111111111111111111111111111111111111111111 00000000000000000000 00000000000000000000 000000000000000000000000000000000000000000000000000000000000 00000000000000000000 0000000000000000000000000000000000000000 0000000000000000000000000000000000000000 0000000000000000000000000000000000000000 11111111111111111111 11111111111111111111 111111111111111111111111111111111111111111111111111111111111 11111111111111111111 1111111111111111111111111111111111111111 1111111111111111111111111111111111111111 1111111111111111111111111111111111111111 iΣ Σj ’iΣ ’jΣ K’i Σ Σ Q AS Q AS KAj KAi Pi Pj Ai Aj K’j Pj’ Pi’ CHARME ’97, Montr´eal, Canada - October 1997 10
- 12. Denotational semantics ? S[a; b] = S[a]; S[b] P[a; b] = P[a] ∪ S[a]; P[b] S[while(true)do a end] = { } P[while(true)do a end] = P[a] ∪ S[a]; P[while(true)do a end] = µR : Semantics . R = P[a] ∪ S[a]; R S[a||b] = S[a] ∩ S[b] P[a||b] = P[a] ∩ P[b] CHARME ’97, Montr´eal, Canada - October 1997 11
- 13. Wait & Assignment semantics ? Wait Semantics (e, w0, t0)S[wait until p](e, w1, t1) = w0 = w1 ∧ wt1 |= p ∧ ∀t ∈ [t0, t1) . wt |= p (e, w0, t0)P[wait until p](e, w1, t1) = w0 = w1 ∧ ∀t ∈ [t0, t1] . wt |= p Assignment Semantics (e, w0, t0)S[x ⇐ y after τ](e, w1, t1) = t0 = t1 ∧ w1 = relax(e, t0)(force assign(w0)) P[x ⇐ y after τ] = { } CHARME ’97, Montr´eal, Canada - October 1997 12
- 14. Wait semantics ? 0 00 00 0 000 0 00 1 11 11 1 111 1 11 00000000000000000000 0000000000000000000000000000000000000000 11111111111111111111 1111111111111111111111111111111111111111 Generalized Wait a x wait until a = 1 on signals -> discrete time 00000000000000000000 0000000000000000000000000000000000000000 11111111111111111111 1111111111111111111111111111111111111111 00 0 0 000 0 000 0 11 1 1 111 1 111 1 00000000000000000000 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 11111111111111111111 0 0 00 0 0 000 0 00 1 1 11 1 1 111 1 11 wait until x >= 2/3 on quantities -> continuous time CHARME ’97, Montr´eal, Canada - October 1997 13
- 16. Analog solution ? Bouncing ball example ds dt = v dv dt = −g ± av2 Local approximation s(t) = 1 v(t) = −gt origin at s = 1 v = 0 t = 0 Linear approximation (s, v) = (s0, v0) + (t − t0)(v0, −g ± av2 0) Recursion (s, v)(t0,s0,v0)(t) ∼ (s0, v0) + (t − t0)(v0, −g ± av2 0) t ∈ [t0, t1) (s, v)(t1,s1,v1)(t) s1 = s0 + v0∆t v1 = v0 + (−g ± av2 0)∆t t1 = t0 + ∆t CHARME ’97, Montr´eal, 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 ∆t = 0.02 CHARME ’97, Montr´eal, 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 ∗ 0 + v′2 2 = g ∗ s + v2 2 ∆t = 0.06 CHARME ’97, Montr´eal, Canada - October 1997 17
- 19. Example Cont. (II) ? Height Speed ∆t = 0.06 CHARME ’97, Montr´eal, Canada - October 1997 18
- 20. Unresolved areas ? • Implementation of the parallel composition of pro- cesses if quantities are shared between processes • Errors in the analog solver may require us to rewrite our program • The proper execution of the analog solver in δ time when the processes are in parallel CHARME ’97, Montr´eal, Canada - October 1997 19
- 21. Conclusion ? • What? Extension of our VHDL semantics to cover VHDL-AMS • How? Embedding of the VHDL semantics in a bigger, continuous time domain, which contains an oracular analog solver • Why? Clarify the draft standard document CHARME ’97, Montr´eal, Canada - October 1997 20