Unfolding, Look ahead transformation

Unfolding

• Use when T∞ is not an integer and/or computation time of a node in DFG > T∞ .

• T∞ increase to J T∞.• Critical path may or may not change.• Clock period is still limited by critical path.• Throughput rate increase by a factor of J

(i.e. Tclk/J).

Unfolding

1) Assume all node has computing time =0T∞= 5/10 = 0.5Tc = 1 after retiming

0D-C4-0B4

0D-C3-0B3

0D-C2-0B2

0D0DC1C10B1

0D0DC0C00B0

1D-B0-1A4

0D-B4-1A3

0D-B3-1A2

0D1DB2B01A1

0D0DB1B11A0

J=5J=2J=5J=2

⎣(i+w)/J⎦(i+w)%Jwi

0D-A4-0D4

0D-A3-0D3

0D-A2-0D2

0D0DA1A10D1

0D0DA0A00D0

2D-E1-7C4

2D-E0-7C3

1D-E4-7C2

1D4DE3E07C1

1D3DE2E17C0

J=5J=2J=5J=2

⎣(i+w)/J⎦(i+w)%Jwi

1D-D1-2E4

1D-D0-2E3

0D-D4-2E2

0D1DD3D12E1

0D1DD2D02E0

1D-B2-3E4

1D-B1-3E3

1D-B0-3E2

0D2DB4B13E1

0D1DB3B03E0

J=5J=2J=5J=2

⎣(i+w)/J⎦(i+w)%Jwi

Unfolding with J =2T∞= 5/5 = 1Tc = 1 after retimingTs = ½ = 0.5

Unfolding with J =5T∞= 5/2 = 2.5Tc = 5 after retimingTs = 5/5 = 1

2)

(a) What is iteration bound and critical path. Assume addition takes 2 u.t. and multiplication takes 4 u.t.

(b) Retime it to minimize the critical path. (c) Unfold it to achieve the throughput rate = T∞ .

Unfolding

(a) T∞ = 14/4, Tc = 14(b)

Tc =Tclk = Ts= 4 u.t.T∞ = 14/4

(b)

J = 4, Tc =Tclk = 14 u.t.Ts= 14/4T∞ = 4(14/4) = 14

Look Ahead Transformation

1) Consider linear time invariant recursion equation: y(n) = ay(n-1)+bu(n)

The input sequence u(n) has sampling rate of 1 sample/µs. Assume TA = 1µs and TM= 2µs. Design hardware that can execute this algorithm at the giving sample rate.

Unfolding

1D-A0-1C2

0D1DA2A11C1

0D0DA1A01C0

J=3J=2J=3J=2

⎣(i+w)/J⎦(i+w)%Jwi

Unfolding after 2 order look ahead

1D-D2-3G21D2DD1D03G11D1DD0D13G01D-C1-2I21D1DC0C12I10D1DC2C02I01D-B0-1I20D1DB2B01I10D0DB1B11I0J=3J=2J=3J=2

⎣(i+w)/J⎦(i+w)%Jwi

2) Rewriting equationy(n) = 0.6y(n-1) − 0.25y(n-2) + x(n)into a state-space formulation. Then apply 2nd

order look-ahead transform.

For 2nd order IIR filter, we havey(n) = 2rcos θy(n-1)-r2y(n-2)+x(n)

Hence r = 0.5, cos θ = 0.6, and sin θ = 0.8.

where bT = [1/(2 sin θ) 0] , cT = [2r cos 2θ r sin 2θ], and d = 1.Apply look-ahead transformation m−1 times and note that [R(θ)]m = R(mθ), we have

1

0( 1) (( ) ) ( 1) ( ) ( )

(( ) ) ( 1) ( )( ) ( ) ( )

mm k

km

T

n r R m n m r R k x n k

r R m n m ny n n d x n

θ θ

θ

−

=

+ = ⋅ − + + −

= ⋅ − + +

= + ⋅

∑v v b

v uc v

[ ]

1 1 1

2 2 2

11 2

2

( 1) ( )cos sin( ) ( ) ( ) ( )

( 1) ( )sin cos

( )( ) ( ) ( ) ( )

( )T

v n v n br x n r R n x n

v n v n b

v ny n c c d x n n d x n

v n

θ θθ

θ θ+ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤

= ⋅ + = ⋅ +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥+ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤

= + ⋅ = + ⋅⎢ ⎥⎣ ⎦

v b

c v

+ X+ D

X

+X

X

0.53cos159

0.53cos 159

0.53sin 159

−0.53sin 159

v1(n)

v2(n)

y(n)

u1(n)

x(n)

X

+ X+ D 0.48

-0.28

+X

1

D D

v1(n-2)

D D

v2(n-2)

u2(n)

m=3 (original formulation m = 1)