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1 HARDWARE / SOFTWARE PARTITIONING Devang Sachdev Lizheng Zhang.

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Presentation on theme: "1 HARDWARE / SOFTWARE PARTITIONING Devang Sachdev Lizheng Zhang."— Presentation transcript:

1 1 HARDWARE / SOFTWARE PARTITIONING Devang Sachdev Lizheng Zhang

2 HW SW ECE 734 Project Presentation2 Motivation Hardware – Software Codesign (HSC) –Most efficient implementation of a system –Unified H-S integration –Higher confidence in the systems functionality –Lower costs and smaller development cycles Hardware – Software Partitioning –Definition: The process of deciding, for each subsystem, whether the required functionality is more advantageously implemented in hardware or software –Fundamental phase of HSC

3 HW SW ECE 734 Project Presentation3 Basics Goal: To achieve a partition that will give us the required performance within the overall system requirements (in performance, area, power, cost etc.) Approach –Software – oriented partitioning: Start with all functionality in software and move portions into hardware which are time-critical and can not be allocated to software –Hardware – oriented partitioning: Start with all functionality in hardware and move portions into software implementation

4 HW SW ECE 734 Project Presentation4 Design Methodology MATLAB C languageC + VHDLVHDL Functional verification SW HW

5 SW ECE 734 Project Presentation5 Target Architecture Processor FPGA Local Memory Shared Memory Global Memory

6 HW SW ECE 734 Project Presentation6 Test Application Discrete Cosine Transform (DCT) –Computationally intensive block in JPEG –Equation

7 HW SW ECE 734 Project Presentation7 Design Issue – Truncation Error Decide bits needed to represent data –Cosine coefficients [-0.5,+0.5] s.xxxxx00000000 –Data path length [-4096,+4095] sxxx_xxxx_xxxx0.00000 JPEG files compression

8 HW SW ECE 734 Project Presentation8 Cosine Coefficients 243 5678 1

9 HW SW ECE 734 Project Presentation9 Computation Data Length 11 79810 131214

10 HW SW ECE 734 Project Presentation10 DCT Partitioning for v=1:8 for u=1:8 sum2(u,v)=0; for n=1:8 sum1(n,u) = 0; for m=1:8 x1 = f(m,n)*dctcos(m,u); sum1(n,u)= sum1(n,u)+ x1; end y1 = sum1(n,u)*dctcos(n,v); sum2(u,v)= sum2(u,v)+y1; end

11 HW SW ECE 734 Project Presentation11 Thank You


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