Transcript VLSI Signal Processing

VLSI Signal Processing
Dr. Elwin Chandra Monie
Department of ECE, RMK Engineering College
256K MEMORY CHIP
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APPLICATIONS
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Dept. of ECE, R M K Engineering College
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
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Need for VLSI DSP System
Processors for DSP system
• General Purpose
Microprocessors/Microcontrollers
• General Purpose DSPs
• Custom Processors in VLSI- FPGA, ASIC
Real time throughput
• Sampling rates from 20KHz to 500 MHz
• Present sample is to be processed before the
arrival of the next sample; if not buffered
• Processing rate upto 100 GOPs/sec is required
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Need for VLSI DSP system ….
Data Driven property
• Systems are synchronized by data and not by
clock
• Asynchronous operation possible
Reduced size
• For portable and mobile applications
• High density circuits available - 90MnTr/cm2
• Increases according to Moore’s Law
• Submicron fabrication technology feasible
0.07µm
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Typical DSP Algorithms
Filtering
• FIR, IIR filters
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•
y(n) = ∑kak y(n-k) + ∑kbk x(n-k)
With (Recursive) and without feedback
Convolution and Correlation
y(n) = ∑ x(k) h (n-k)
y(n) = ∑ a(k) x (n+k) n= 1 to ∞
Non-terminating programs – Execute the same
code repetitively
• Adaptive Filters –LMS Algorithm
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Typical DSP Algorithms …
Transforms
• FFT, DCT, DWT
• FFT : X(k) = ∑n x(n) e -j2πkn/N
components
Real and imaginary
Decomposition
• SVD, LU Matrix factorization, QR decomposition
Operations involved
• Arithmetic – Multiplication, Addition
• MAC operation
• Logic – Shifting, barrel shifiting – Delay
• Dot Product/ Matrix-Vector operations
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Data Flow Graph
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]
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Data Flow Graph ….


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

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
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Data Flow Graph ….



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 …
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Data Flow Graph ….

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
10
D
D
10
10
10
10
D
y(n)
26
4
4
22
18
4
14
4
12
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Loop Bound
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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!
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Example of Iteration Bound
(1)

A
Loops

2D
B

D
(1)
C

(2)
(1)
D

Loop bound = 5/4
Critical Loop


Loop bound = 5/3
Loop 3: AFCB

E
(2)

D
Loop bound = 4/2
Loop 2: AECBA

(2)
F
Loop 1: ADBA
Loop 1
Iteration Bound


Max{4/2,5/3,5/4} = 4/2 = 2
T∞=2 units of time.
That is the minimum clock period (max frequency) this
circuit can operate at after pipelining and retiming
Longest path matrix algorithm-1
Let d be the number of delays in DFG. Define
K = [1, 2, · · · , d]
(1)
Form the matrix L as follows
(1)
L
i,j
=
max
tq
q
-1
d i → dj
if at least one path exists
if no such path exists
d →d
where max tq i j is the maximum of the longest
computation time between delay element di to delay
element dj
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Longest path matrix algorithm-2
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
T∞ = max
i,mK
L(m)i,i
---------m
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Longest path matrix algorithm-3
L(1) =
-1 0 0
4 -1 0
5 -1 -1
5 -1 -1
-1
-1
0
-1
L2,1(2) = max ( -1, L(1)2,k + L(1)k,1)
k{1,2,3,4}
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LONGEST PATH MATRIX ALGORITHM-4
L2,1
(2)
= max( -1, L
(1)
(1)
2,k + L
k,1)
k{1,2,3,4}
= max( -1,0+5) = 5
(2)
(1)
(1)
L2,2 = max( -1, L 2,k + L k,2)
k{1,2,3,4}
= max( -1,4+0 ) = 4
(2)
(1)
(1)
L2,3 = max( -1, L 2,k + L k,3)
k{1,2,3,4}
= max(-1) = -1
(2)
(1)
(1)
L2,4 = max ( -1, L 2,4 + L k,4)
k{1,2,3,4}
= max(-1,0+0) = 0
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LONGEST PATH MATRIX ALGORITHM-5
(2)
L
=
L(3) =
4 -1 0 -1
5 4 -1 0
5 5 -1 -1
-1 5 -1 -1
5 4
8 5
9 5
9 -1
-1 0
4 -1
5 -1
5 -1
T∞ = max
4/2, 4/2, 5/3, 5/3, 5/3, 8/4, 8/4, 5/4, 5/4
= 2
L(4)
8
9
= 10
10
5 4 -1
8 5 4
9 5 5
9 -1 5
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DATA INDEPENDENCE GRAPH
x’=x
y
x0
0
2
1
x1
0
x2
0
x3
0
x4
x5
b
b’=b
0
x
y’= y+bx
b2
0
b1
0
0
b0
y1
y0
1
2
y2
3
y3
4
y4
5
y5
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y(n)= b0 x(n) + b1 x(n-1) + b2 x(n-2)
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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
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PIPELINING IN FIR FILTERS
Critical path : TM+2TA => TM+TA
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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 feedforward 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

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GENERAL METHOD OF PIPELINING
Critical path: 4
Feed forward
cutset
Not Correct !
Critical Path: 2
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TRANSPOSITION THEOREM
x(n)
c
b
Z-1
Reverse the direction of all edges in a given SFG and
interchanging the input and output ports preserve the
functionality of the system
a
Z-1
y(n)
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Critical Path : TM+2TA => TM+TA
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FINE-GRAIN PIPELINING
Multiplier with processing time of 10 is split into two units
with processing times 6 and 4
Critical path: 12 => 6
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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
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PARALLEL PROCESSING FIR FILTERS
Iteration Time=
1/3 (TM+2TA )
Used 3 sets of resources
for 3-parallel system
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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 where β 0 to 1
Ppip = Ctotal β2 V02 f = β2 Pseq
Ccharge/M β V0
Propagation delaypip =
-------------------k(βV0- Vt)2
If the clock period is kept the same
Ccharge V0
-----------k(V0- Vt)2
=
(βV0- Vt)2
=
Ccharge/M β V0
------------------k(βV0- Vt)2
β (V0- Vt)2
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Solve for β
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EXAMPLE ON PIPELINING
Consider an original 3-tap FIR filter and its finegrain 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?
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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
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PARALLEL SYSTEM FOR LOW POWER
Power consumption :
Ppar = (L Ctotal) (β V0)2 f / L = P seq
for L- Parallel System
Propagation delay:
Ccharge V0
Tseq = --------------k(V0- Vt)2
L Tseq = Tpar
Tpar =
Ccharge β V0
---------------k(βV0- Vt)2
β(V0- Vt)2 = L (βV0- Vt)2
Solve for β
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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 2parallel filter as a percentage of the original
filter?
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SOLUTION
Ccharge = CM + CA
2- parallel: Ccharge = CM + 2CA = 10CA
9 (β 3.3 - 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
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PROBLEMS & ASSIGNMENTS
1) Prob. 2.7.1 (a)
2) Prob. 2.7.4
Assignment
1) 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
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