chapter 6 folding - 國立中興大學

VLSI DSP 2010 Y.T. Hwang 6-1

Chapter 6 FoldingChapter 6 Folding


Introduction (1)Introduction (1)

foldingDSP architecture where multiple operations are multiplexed to a single function unit

Trading area for time in a DSP architecture

Reduce the number of function units by a factor of N at the expense of increasing the computing time by a factor of N

N: folding factor

Present a systematic way to derive the folded DSP architecture



Folding exampley(n) = a(n) + b(n) + c(n)

Time multiplexed on a single pipeline adder

An input sample must remains 2 clock cycles



More on foldingMay lead to an architecture using a large number of registers

Design to minimize the number of registers


Folding transformation (1)Folding transformation (1)

PreliminaryConsider a DFG

An edge e connecting nodes U and V with w(e) delays

Executions of the l-th iterations of U and V at time units Nl+u and Nl+v

u and v: folding orders and 0 ≤ u,v ≤ N-1

N: folding factor, the number of operations folded to a single function unit

HU and HV: function units to execute nodes U and V

HU is pipelined by PU stages



Folding an edgehas w(e) delays

l-th iteration of node U is available at time Nl + u + PU

Generated data is used by the (l+w(e))-th iteration of V

The result must be stored for

VU e⎯→⎯

uvPeNwuPNlvewlNVUD UUe

F −+−=++−++=⎯→⎯ )(][]))(([)(

Folding factor = N



Folding setAn order of operations executed by the same hardware

Example: S1 = {A1,Ø ,A2}

⇒ A1: (S1|0), A2: (S1|2)

Biquad filter exampleAddition : 1 u.t. and 1-stage pipelining, PA = 1

Multiplication: 2 u.t. and 2-stage pipelining, PM = 2

Folding factor N = 4

Assume folding set S1 = {4, 2, 3, 1}, S2 = {5, 8, 6, 7}



Biquad filter example (cont.)Node 3 is executed on adder at time instance (4l + 2)



Biquad filter example (cont.)DF(1→8) = 5 : an edge from the adder to the multiplier in the folded DFG with 5 delays

Because node 8 has (S2|1), the folded edge is switched at the input of the multiplier at 4l + 1



Valid foldingmust hold for all edges in the DFG

Can be achieved by retiming

Recall after retiming has a delay

wr(e) = w(e) + r(V) - r(U)

Let denote the number of folded delay by folding the retimed DFG

0)( ≥⎯→⎯ VUD eF

VU e⎯→⎯

0)(' ≥⎯→⎯ VUD eF

⎥⎦

⎥⎢⎣

⎢ ⎯→⎯≤−⇒

⎯→⎯≤−⇒

≥−+−−+⇒≥−+−⇒≥⎯→⎯

N

VUDVrUr

N

VUDVrUr

uvPUrVrewN

uvPeNwVUD

eF

eF

U

Ure

F

)()()(

)()()(

0))()()((

0)( 0)('



Retiming for valid foldingSolve a system of inequalities

First construct a constraint

Use Floyd-Warshallalgorithm to solve the problem



Retiming for valid folding (cont.)

Constraint graph

Solution

r(1) = -1, r(2) = 0

r(3) = -1, r(4) = 0

r(5) = -1, r(6) = -1

r(7) = -2, r(8) = -1

Leads to the DFG in Fig 6.3

Can be achieved equivalently by cut set retiming using C1 and C2



More on foldingThe original DFG and the N-unfolded version of the folded DFG (synthesized with folding factor N) are retimed and/or pipelined versions of each other

An arbitrary DFG can be unfolded by a factor N and then folded again to generate a family of architectures


Register minimization in folding (1)Register minimization in folding (1)

Lifetime analysisTo compute the minimum number of registers required to implement a DSP algorithm in hardware

A data sample (variable) is live from the time it is produced (excluded) through the time it is consumed (included)

A variable after lifetime is called “dead”

The maximum number of live variables at each time unit is the minimum number of registers required to implement the DSP program


Register minimization in folding (2)Register minimization in folding (2)

ExampleAssume 3 variables a, b, c

Life time of variable a: {1,2,3,4}

Life time of variable b: {2,3,4,5,6,7}

Life time of variable c: {5,6,7}

Number of live variables {1,2,2,2,2,2,2}

2 registers are needed to implement the DSP program


Linear lifetime chart (1) Linear lifetime chart (1)

When the iteration period is less than the span of the scheduling, the scheduling overlapsThe number of live variables at time instance n is the sum of the number of live variables at cycles n-kN, ∀k

Non-overlapped Overlapped withSchedule period 6


Linear lifetime chart (2)Linear lifetime chart (2)

Matrix transpose example

Assume row-wise access

Input time: Tinput

Zero latency output time: Tzlout

Tdiff = Tzlout – Tinput

Required latency Tlat = magnitude of the most negative value of Tdiff

Toutput = Tzlout + Tlat

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ifc

heb

gda

ihg

fed

cba


Linear lifetime chart (3)Linear lifetime chart (3)

Matrix transpose example (cont.)Assume iteration period of the DSP program is N = 9


Circular lifetime chartCircular lifetime chart

Circular lifetime chartPoint i represents the time partition i and all time instances {(Nl+i)}

linear circular


Data allocation (1)Data allocation (1)

Forward backward register allocationTo achieve minimum number of registersDetermine how variables are assigned to registers in the “allocation table”

Step 1: determine the minimum number of registers using lifetime analysis

Step 2: Input each variable at the time step corresponding to the beginning of its lifetimeIf multiple variables are input in a given cycle, they are allocated to multiple registers according to lifetime in a descending order



Forward allocationIf register i holds the variable in the current cycle, then register i+1 holds the same variable in the next cycleIf the register i+1 is not available, then the variable is allocated to the first available forward register

Step 3:Each register is allocated in a forward manner until it is dead or reaches the last register

Step 4:In periodic scheduling, the allocation of current iteration also repeats itself in subsequent iterationsIf Rj is occupied by a variable in cycle l, hash the position for Rj at time unit l+N



Step 5:For a variable that reaches the last register and is not yet dead, allocate it in backward manner

If multiple registers available, choose the one with least but sufficient number of forward registers capable of completing the allocation

After a variable has been allocated backward, allocate it in a forward manner until it is dead or again reaches the last register

Step 6:Repeat step 4 and 5 as required until the allocation is complete



3X3 matrix transpose example with N = 9

After steps 1~4 completion

hashing



Another example

Linear lifetime chart

Step 1~4 completion



architecture design after register allocation



architecture design after register allocation


Register minimization in foldingRegister minimization in folding

GoalTo synthesize control circuits in folded architectures with minimum number of registers

ProceduresPerform retiming for folding

Write folding equations

Use the folding equations to construct a lifetime table

Draw the lifetime chart and determine the required number of registers

Perform forward-backward register allocation

Draw the folded architecture that uses the minimum number of registers


BiquadBiquad filter example (1)filter example (1)

Biquad filter

Original bi-quadFilter design

Design afterretiming



Design without register minimizationTotal of 6 external and 3 internal pipelining registers

Folding equations Folded architecture



Construct a lifetime tableEach a node with lifetime (Tinput→Toutput) corresponds to an entry in the lifetime table

Tinput : u (folding order) + PU (# of pipelining stages of the function unit)

Toutput : u+ PU +maxV{DF(U→V)

Foe node 1, folding order is 3, adder’s PU is 1⇒ Tinput = 3+1=4

Toutput = u+ PU +maxV{DF(U→V) = 3+1+max{1,0,2,3,5}=9



Construct a lifetime table and lifetime chartAssume N (iteration period) is 4

Minimum number of registers required is 2



Allocation tableOnly variables n1, n7 and n8

with non-zero duration are shown

Variable n1 is output in cycles 4,5,6,8,9, only the latest cycle 9 is shown in the table



Folded design with 2 registersEdge (1→2) has DF(1→2)= 1 delay

after 1 delay the variable n1 is located in R1

An edge from R1 to adder switched at 4l+1 because the node 2 has folding order 1



Folded design with 2 registers (cont.)Edge (1→7) has DF(1→7)= 3 delays

after 3 delays the variable n1 is located in R2

An edge from R2 to multiplier switched at 4l+2 because the node 7 has folding order 2


IIR filter example (1)IIR filter example (1)

IIR filter before retimingy(n) = ay(n-3) + by(n-5) + x(n)

Folding factor = 2

Folding set: ADD S1 = {1,2}, MPY S2 = {4,3}

Retiming solution

r(1) = 0, r(2) = 0, r(3) = -2, r(4) = -1



IIR filter after retiming

Folding equations for the retimed DFG

DF(1→2)= 2(0) − 1 + 1 − 0 = 0

DF(2→3)= 2(3) − 1 + 1 − 1 = 5

DF(2→4)= 2(2) − 1 + 0 − 1 = 2

DF(3→1)= 2(2) − 2 + 0 − 1 = 1

DF(4→1)= 2(1) − 2 + 0 − 0 = 0

Lifetime table



Lifetime chartA total of 3 registers is needed



Allocation table and folded design3 registers (minimized) v.s. 6 registers (unminimized)

chapter 6 folding - 國立中興大學

Documents