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Dr. Elwin Chandra Monie Department of ECE, RMK Engineering College

Dr. Elwin Chandra Monie Department of ECE, RMK Engineering College. 256K Memory Chip. Applications. Syllabus . Anna University syllabus for VL9253 VLSI Signal processing Text

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Dr. Elwin Chandra Monie Department of ECE, RMK Engineering College

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  1. Dr. Elwin Chandra Monie Department of ECE, RMK Engineering College

  2. 256K Memory Chip Dept. of ECE, RMK Engineering College

  3. Applications Dept. of ECE, R M K Engineering College

  4. Syllabus Anna University syllabus for VL9253 VLSI Signal processing Text Keshab K. Parhi, ‘VLSI Digital Signal Processing Systems, Design and implementation’, Wiley India Pvt. Ltd., 2009 Dept. of ECE, RMK Engineering College

  5. Dept. of ECE, R M K Engineering College

  6. Dept. of ECE, R M K Engineering College

  7. Dept. of ECE, R M K Engineering College

  8. Dept. of ECE, R M K Engineering College

  9. A DSP program is often represented using a Data Flow Graph (DFG), which is a directed graph that describes the program • Consider the following IIR filter y[n] = x[n] + a y[n − 1] Dept. of ECE, RMK Engineering College

  10. In the DFG, nodes represent the tasks or computations (Multiplication/Addition) • Each task is associated with its corresponding execution time • The edges represent the communications between the nodes A → B • Associated with each edge is a non-negative number representing the delay • An iteration of the node is the execution of the node, exactly once Dept. of ECE, RMK Engineering College

  11. Each edge describes a precedence constraint between two nodes • The precedence constraint is an intra-iteration constraint if the edge has zero delays (i.e. computations at nodes connecting the edge occur in the same clock cycle) • The precedence constraint is an inter-iteration constraint if the edge has one or more delays (i.e. computations at nodes connecting the edge occur in different clock cycles) A1 → B1 => A2 → B2 => A3 … Dept. of ECE, RMK Engineering College

  12. Critical Path • the path with the longest computation time among all paths that contain zero delays Critical path length is 26 units Critical path: the lower bound on clock period To achieve high-speed, the length of the critical path should be reduced x(n) D D D D 10 10 10 10 10 y(n) 4 4 4 4 14 18 26 22 Dept. of ECE, RMK Engineering College 26

  13. A recursive DFG has one or more loops • A loop bound for the L-th loop is defined as tL / wL • tL is the loop computation time • wL is the number of delays in the loop • Iteration bound T∞ • Iteration bound is the maximum loop bound of all loops in the DFG • The loop that gives the iteration bound is called the critical loop • The iteration bound determines the minimum critical path of a recursive system represented by that DFG structure! • In other words, no matter how you pipeline or retime the DFG, you cannot get a circuit with lower critical path than the iteration bound! Dept. of ECE, RMK Engineering College

  14. Loops Loop 1: ADBA Loop bound = 4/2 Loop 2: AECBA Loop bound = 5/3 Loop 3: AFCB Loop bound = 5/4 Critical Loop Loop 1 Iteration Bound Max{4/2,5/3,5/4} = 4/2 = 2 T∞=2 units of time. (1) A 2D (2) (1) B D D (1) (2) C E D (2) F That is the minimum clock period (max frequency) this circuit can operate at after pipelining and retiming

  15. Let d be the number of delays in DFG. Define K = [1, 2, · · · , d] Form the matrix L(1) as follows max tqdi→ djif at least one path exists L(1)i,j = q -1 if no such path exists where max tqdi→ djis the maximum of the longest computation time between delay element dito delay element dj Dept. of ECE, RMK Engineering College

  16. Compute the successive matrices L(m+1)i,j = max ( -1, L(1)i,k +L(m)k,j ) kS in which Si,j = { k  K |(li,j  -1) & (lk,j  -1)} The iteration bound is computed from L(m)i,i T∞ = max ---------- i,mKm Dept. of ECE, RMK Engineering College

  17. -1 0 0 -1 4 -1 0 -1 L(1) = 5 -1 -1 0 5 -1 -1 -1 L2,1(2) = max ( -1, L(1)2,k +L(1)k,1) k{1,2,3,4} Dept. of ECE, RMK Engineering College

  18. Longest path matrix algorithm-4 L2,1(2) = max( -1, L(1)2,k +L(1)k,1) k{1,2,3,4} = max( -1,0+5) = 5 L2,2(2) = max( -1, L(1)2,k +L(1)k,2) k{1,2,3,4} = max( -1,4+0 ) = 4 L2,3(2) = max( -1, L(1)2,k +L(1)k,3) k{1,2,3,4} = max(-1) = -1 L2,4(2) = max ( -1, L(1)2,4 +L(1)k,4) k{1,2,3,4} = max(-1,0+0) = 0 Dept. of ECE, RMK Engineering College

  19. Longest path matrix algorithm-5 4 -1 0 -1 5 4 -1 0 L(2) = 5 5 -1 -1 -1 5 -1 -1 5 4 -1 0 8 5 4 -1 L(3) = 9 5 5 -1 T∞ = max 4/2, 4/2, 5/3, 5/3, 5/3, 8/4, 8/4, 5/4, 5/4 9 -1 5 -1 = 2 8 5 4 -1 9 8 5 4 L(4) = 10 9 5 5 10 9 -1 5 Dept. of ECE, RMK Engineering College

  20. Data Independence Graph x’=x y x0 x1 x2 x3 x4 x5 b b’=b y’= y+bx 0 0 0 0 0 x b2 2 0 b1 1 0 0 b0 y1 y2 y3 y4 y5 y0 1 2 3 4 5 6 y(n)= b0 x(n) + b1 x(n-1) + b2 x(n-2) Dept. of ECE, RMK Engineering College

  21. Pipelining in FIR filters • Reduce the critical path • Increase the clock speed or sample speed • Reduce power consumption • Introduce pipelining latches along the data path Dept. of ECE, RMK Engineering College

  22. Pipelining in FIR filters Critical path : TM+2TA => TM+TA Dept. of ECE, RMK Engineering College

  23. General Method of Pipelining • Pipelining latches can only be placed across any feed-forward cutset of the graph without affecting of the structure • Cutset: A cutset is a set of edges of a graph such that if these edges are removed from the graph, the graph becomes disjoint. • Feed-forward cutset: A cutset is called a feed-forward cutset if the data move in the forward direction on all the edges of the cutset Limitations of Pipelining • Increase in Latency : The difference in the availability of the first output • Increase in the number of latches Dept. of ECE, RMK Engineering College

  24. General Method of Pipelining Critical path: 4 Feed forward cutset Not Correct ! Critical Path: 2 Dept. of ECE, RMK Engineering College

  25. Transposition Theorem x(n) c b a Z-1 Z-1 y(n) Reverse the direction of all edges in a given SFG and interchanging the input and output ports preserve the functionality of the system Critical Path : TM+2TA => TM+TA Dept. of ECE, RMK Engineering College

  26. Fine-Grain Pipelining Multiplier with processing time of 10 is split into two units with processing times 6 and 4 Critical path: 12 => 6 Dept. of ECE, RMK Engineering College

  27. Parallel processing FIR Filters y(n)= ax(n)+bx(n-1)+cx(n-2) y(3k) = ax(3k)+bx(3k-1)+cx(3k-2) y(3k+1)= ax(3k+1)+bx(3k)+cx(3k-1) y(3k+2)= ax(3k+2)+bx(3k+1)+cx(3k) Sample speed is increased since multiple samples are processed at the same time. Clock speed remains the same Dept. of ECE, RMK Engineering College

  28. Parallel processing FIR Filters Iteration Time= 1/3 (TM+2TA ) Used 3 sets of resources for 3-parallel system Dept. of ECE, RMK Engineering College

  29. Pipelining for Low Power Ccharge V0 Propagation delay = --------------- k(V0- Vt)2 Power consumption = Ctotal V02 f For M Level pipelining Ccharge is reduced by 1/M Keeping f same reduce V0 by β V0 whereβ 0 to 1 Ppip = Ctotalβ2 V02 f = β2 Pseq Ccharge/M β V0 Propagation delaypip = -------------------- k(βV0- Vt)2 If the clock period is kept the same Ccharge V0Ccharge/M β V0 ------------ = ------------------- k(V0- Vt)2 k(βV0- Vt)2 (βV0- Vt)2 = β (V0- Vt)2 Solve for β Dept. of ECE, RMK Engineering College

  30. Example on Pipelining Consider an original 3-tap FIR filter and its fine-grain pipeline. Assume TM=10 ut, TA=2 ut, Vt=0.6V, Vo=5V, and CM=5CA.In fine-grain pipeline filter, the multiplier is broken into 2 parts, m1 and m2 with computation time of 6 u.t. and 4 u.t. respectively, with capacitance 3 times and 2 times that of an adder, respectively. (a) What is the supply voltage of the pipelined filter if the clock period remains unchanged? (b) What is the power consumption of the pipelined filter as a percentage of the original filter? Dept. of ECE, RMK Engineering College

  31. Solution Solution: Original : C charge = CM + CA = 6 CA Pipelining : C charge = 3 C A (5 β - 0.6)2 = β(5 - 0.6)2 β = 0.6033 or 0.0239 ( not valid) Vpip = 3.0165V0 Ppip = 0.364 Pseq Dept. of ECE, RMK Engineering College

  32. Parallel System for Low power Power consumption : Ppar= (L Ctotal) (βV0)2 f / L = P seq for L- Parallel System Propagation delay: CchargeV0 Cchargeβ V0 Tseq = --------------- Tpar= ---------------- k(V0- Vt)2 k(βV0- Vt)2 L Tseq = Tpar β(V0- Vt)2 = L (βV0- Vt)2 Solve for β Dept. of ECE, RMK Engineering College

  33. Example on Parallel system Consider a 4-tap FIR filter shown in Fig. 3.18(a) and its 2-parallel version in 3.18(b). The two architectures are operated at the sample period 9 u.t. Assume TM=8, TA=1, Vt=0.45V, Vo=3.3V, CM=8CA (a) What is the supply voltage of the 2-parallel filter? (b) What is the power consumption of the 2- parallel filter as a percentage of the original filter? Dept. of ECE, RMK Engineering College

  34. Solution Ccharge = CM + CA 2- parallel: Ccharge = CM + 2CA = 10CA 9 (β3.3 - 0.45)2 = 5 β (3.3 - 0.45)2 β = 0.6585 or 0.0282 (not valid) Vpar = 2.1743 Vo Ppar = 0.4341 P Dept. of ECE, RMK Engineering College

  35. Problems & Assignments • Prob. 2.7.1 (a) • Prob. 2.7.4 Assignment • Design a Low pass filter with sample rate of 48KHz and order 40 with cut off frequency of 10KHz. Write VHDL/Verilog code and simulate Hint: Use Matlab to find the coefficients and test the filter functionality by testing the impulse response 2) Implement a 4-tap filter in direct form and in transpose form. Introduce pipelining and compare the performance Dept. of ECE, RMK Engineering College

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