SystemC-AMS(CUDA)

Ruru 2012 8/14

Outline

Introduction

Modeling Formalisms

ELN(Electrical Linear Networks)

Solve Ordinary Differential Equation

Solve Eln-Cluster Workflow

SystemC Kernel Solver

Performance

Introduction

Positioning SystemC AMS Extensions

Introduction

Modeling formalisms and use cases

Modeling Formalisms

Electrical Linear Networks

ELN

Electrical Linear Networks

Setup of the equation system

Kirchhoff’s current law(KCL)

Electrical Linear Networks

ELN elaboration and simulation phases

Problem : Thermo Model

Thermo Model :

RLC model

Solve ELN equation

Purpose

Using CUDA to Speedup

Solve Ordinary Differential Equation

Euler’s Method

Backward Euler Method

SystemC Kernel Solver

Eln-cluster Initialization

Module(r, l, c …) : modules

Node(sca_terminal, sca_node …) : equations

Sparse Matrix

Spcode

Solve Eln-Cluster Workflow

Solve Differential Equation

ana_solvDATA (Sdata, R,

X)

Next Cluster

Next Timestamp

Cluster?

Timestamp?

…

Modify ana_solv to GPU

Every eln-cluster

Every time stamp

Workflow

Copy matrix data to GPU

Execute ana_solv by GPU

Copy result data to host

ana_solvDATA (Sdata, R,

X)

GPU ana_solvDATA (Sdata, R,

X)

Solve Eln-Cluster Workflow

Solve Differential Equation

ana_solvDATA (Sdata, R,

X)

Next Cluster

Next Timestamp

Cluster?

Timestamp?

…

GPU ana_solvDATA (Sdata, R,

X)

Performance

The result is correct

No speed up

200 module of one cluster : 75 ms (CPU)

200 module of one cluster : 165 ms (GPU)

Thanks for Your

Attention!

Euler’s Method

Question : y’ = f(x, y) , x0 < x < b , y(x0) = y0 , h is constant

xi = x0 + i*h ; (i = 0 1 2 ….)

yi = y(xi) ~ Yi (Yi is result)

yn+1 = yn + h*f(yn, xn)

Example : y’ = y , y(0) = 1 , h = 0.5

Solve :

y1 = y(0) + y’(0)*h = 1 + 1*0.5 = 1.5

y2 = …

Yi = exp(x) => Y(0.5) = 1.649…

1.5 vs 1.649

Backward Euler Method

Question : y’ = f(x, y) , x0 < x < b , y(x0) = y0 , h is constant

xi = x0 + i*h ; (i = 0 1 2 ….)

yi = y(xi) ~ Yi (Yi is result)

yn+1 = yn + h*f(yn, xn)

Example : y’ = -y^3 , y(0) = 1 , h = 0.5

Solve :

y1 = y(0) + f(x1, y1)*h = 1 - 0.5*y1^3

First => y1 = y(0) + f(x0, y0)*h = 1-0.5*y0^3 = 0.5

Then recursive…

y1 = 0.9375 => 0.588 => 0.898 … => 0.7709