P ROGRAMMABLE L OGIC D EVICES, T HRESHOLD L OGIC Unit-4.
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Transcript of P ROGRAMMABLE L OGIC D EVICES, T HRESHOLD L OGIC Unit-4.
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PROGRAMMABLE LOGIC DEVICES, THRESHOLD LOGIC
Unit-4
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TOPICS:
Basic PLD’s- ROM, PROM, PLA and PLD Realization of Switching Functions using
PLD’s Threshold Logic: Capabilities of Threshold
gate Synthesis of threshold functions Multi gate Synthesis
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INTRODUCTION
A programmable Logic Device is an integrated
circuit with internal logic gates that are
connected through electronic fuses.
Programming the device involves the blowing of
fuses along the paths that must be disconnected
so as to obtain a particular configuration.
The word programming here refers to a
hardware procedure that specifies the internal
configuration of the device.
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INTRODUCTION
The gates in a PLD are divided into an AND
array and an OR array that are connected
together to provide an AND-OR sum of
product implementation.
The initial state of a PLD has all the fuses
intact.
Programming the device involves the blowing
of internal fuses to achieve a desired logic
function.
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ROM VS. PLA/PAL
(a) Programmable read-only memory (PROM)
InputsFixed
AND array(decoder)
ProgrammableOR array
OutputsProgrammableConnections
(b) Programmable array logic (PAL) device
Inputs ProgrammableAND array
FixedOR array
OutputsProgrammableConnections
(c) Programmable logic array (PLA) device
Inputs ProgrammableOR array
OutputsProgrammableConnections
ProgrammableConnections
ProgrammableAND array
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READ-ONLY MEMORYROM
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ROM
A ‘B’C’D’ A ‘B’C’D A ‘B’CD’A ‘B’CD A ‘BC’D’ A ‘BC’D A ‘BCD’ A ‘ BCD A B’C’D’ A B’C’D A B’CD’ A B’CD
A B C’D’A B C’D A B C D’ A B C D
F 1
F 3
F 2
A
B
C
D
S2
S1
S0
S3
0 1 2 3 4 5 6 7 8 9
10 1 1 12 13 14 15
4:16 dec
Enb
• Decoder Produces minterms
• ORs Produce SOP’s
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ROM
D7D6D5D4D3D2D1D0
A2
A1A0
A
B
C
F0F1F2F3
X XX
XX
X
XX
XX
• ROM A decoder A set of programmable
OR’s
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EXAMPLEFind a ROM-based circuit implementation for:f(a,b,c) = a’b’ + abcg(a,b,c) = a’b’c’ + ab + bch(a,b,c) = a’b’ + c
Solution:Express f(), g(), and h() in m()
format (use truth tables)Program the ROM based on the 3
m()’s
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EXAMPLE There are 3 inputs and 3 outputs, thus we need a
8x3 ROM block. f = m(0, 1, 7) g = m(0, 3, 6, 7) h = m(0, 1, 3, 5, 7)
3-to-8decoder
01234567
a
b
c
f g h
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ROM AS A MEMORY Read Only Memories (ROM) or Programmable
Read Only Memories (PROM) have: N input lines, M output lines, and
2N decoded minterms. Can be viewed as a memory with the inputs
as addresses of data.
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(MEMORIES)Volatile:
Random Access Memory (RAM): SRAM "static" DRAM "dynamic"
Non-Volatile: Read Only Memory (ROM):
Mask ROM "mask programmable" EPROM "electrically programmable" EEPROM “electrically erasable electrically
programmable" FLASH memory - similar to EEPROM with
programmer integrated on chip
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ROM AS MEMORY
0 1 1 0 1
1 0 0 0 0
2 1 0 0 1
3 0 0 1 0
4 0 0 0 0
5 1 0 0 0
6 0 0 1 1
7 0 1 0 0
Address
3 4
8x4 ROM
D0D1D2D3D4D5D6D7
A2
A1A0
A2
A1
A0
F3F2F1F0
X XX
XX
X
XX
XX
Example: For input (A2,A1,A0) = 011, output is (F0,F1,F2,F3 ) = 0010.•What are functions F3, F2 , F1 and F0 in terms of (A2, A1, A0)?
A[2:0] F[3:0]
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PROGRAMMABLE LOGICPAL, PLA
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16
PLASProgrammable Logic Array
Pre-fabricated building block of many AND/OR gates (or NOR, NAND) "Personalized" by making/ breaking connections among the gates.
General purpose logic building blocks.
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17
PLA
Inputs
Dense array of AND gates Product
terms
Dense array of OR gates
Outputs
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18
PLA
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19
PLA
• A 3×2 PLA with 4 product terms.
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20
DESIGN FOR PLA:EXAMPLE
Implement the following functions using PLAF0 = A + B' C'F1 = A C' + A BF2 = B' C' + A BF3 = B' C + A
Personality Matrix
1 = asserted in term0 = negated in term- = does not participate
Input Side:
1 = term connected to output0 = no connection to output
Output Side:Outputs Inputs Product
t erm
Reuse of
t erms
A 1 - 1 - 1
B 1 0 - 0 -
C - 1 0 0 -
F 0 0 0 0 1 1
F 1 1 0 1 0 0
F 2 1 0 0 1 0
F 3 0 1 0 0 1
A B B C A C B C A
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EXAMPLE: CONTINUED
F0 = A + B' C'F1 = A C' + A BF2 = B' C' + A BF3 = B' C + A
Personality Matrix
Outputs Inputs Product t erm
Reuse of
t erms
A 1 - 1 - 1
B 1 0 - 0 -
C - 1 0 0 -
F 0 0 0 0 1 1
F 1 1 0 1 0 0
F 2 1 0 0 1 0
F 3 0 1 0 0 1
A B B C A C B C A
A B C
F0 F1 F2 F3
AB
B’C
AC’
B’C’
A
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CONSTANTS
Sometimes a PLA output must be programmed to be a constant 1 or a constant 0. P1 is always 1 because
its product line is connected to no inputs and is therefore always pulled HIGH;
this constant-1 term drives the O1 output.
No product term drives the O2 output, which is therefore always 0.
Another method of obtaining a constant-0 output is shown for O3.
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PALS Programmable Array Logic
a fixed OR array.
Inputs
Dense array of AND gates Product
terms
Dense array of OR gates
Outputs
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PALinputs
1st output section
2nd output section
3rd output section
4th output section
Only functions withat most four products can be implemented
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PAL
W = ABC + CDX = ABC + ACD + ACD + BCD Y = ACD + ACD + ABD
x
x
x
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END of Unit-4 (Part-1)
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THRESHOLD
LOGICSHOLD LOGICUNIT-4(PART-2)
(STLD AUTONOMOUS)
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Logic design of sw functions constructed of electronic gates different type of switching element : threshold element.
Threshold element can be considered as a generalization of the conventional gates.
A large class of sw fn. can be realized by a single threshold element through lots of combinations of weights and threshold.
Threshold function : A fn f(x1, x2, ••• , xn) is defined as a threshold fn iff there exists a set of weights {w1, w2, ••• , wn} and a threshold T such that
f(x1, x2, ••• , xn) = 1, iff
f(x1, x2, ••• , xn) = 0, iff
,where T and weights wi’s are real number, xi’s are binary inputs, and each weight wi is associated
with a particular input var xi.
A straightforward approach to identify a threshold fn is to derive from the truth table, a set of 2n linear simultaneous inequalities and to solve them.
w1 ••• wi ••• wn
Tx1 ••• xi ••• xn
f
<Threshold element>
n
iii Txw
1
,
n
iii Txw
1
.
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Eg) f(x1, x2, x3) = (0,1,3)
0 T w3 T w2 < T w2+w3 T
w1 < T w1+w3 < T w1+w2 < T
w1+w2+w3 < T
11010000
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
Inequality( )fx1 x2 x3
3
1iii Txw
w3>w2
w3>w1
w3>w1+w3 w1<0
To find more effective methods, we will study the properties of a Threshold fn.
Capabilities and limitation (A two input NOR gate)
-1
-1-½x1
x2
f 0 0 1 0 1 0 1 0 0 1 1 0
x1 x2 f<NOR gate>
Since a NOR gate is strongly logically complete, any combinational sw. fn can be realized by threshold elements alone. Because of huge possible combinations of weights and threshold, lots of sw. fn can be realized by only one threshold element threshold fn
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Is every sw. fn a threshold fn? No.
Problem 1: Given a sw. fn f(x1, x2, ••• , xn), determine whether or not it is a threshold fn. and if it is, find weights and a threshold.
Problem 2: If the given sw. fn is not a threshold fn, how many threshold elements are required to realized the given sw. fn.
Properties: A fn f(x1, x2, ••• , xn) is positive in a var xi iff
f(x1, x2, ••• , xi-1, 1, xi+1, ••• , xn) f(x1, x2, ••• , xi-1, 0, xi+1, ••• , xn)
A fn f(x1, x2, ••• , xn) is negative in a var xi iff
f(x1, x2, ••• , xi-1, 0, xi+1, ••• , xn) < f(x1, x2, ••• , xi-1, 1, xi+1, ••• , xn)
A fn which is positive or negative in every var is called unate.
positive unate : positive in every var negative unate : negative in every var
A completely specified fn is unate iff any minimal S. of P. representation of the fn uses either literal xi or literal xi’ but not both for 1 i n.
Eg) f(x1, x2, x3) = (0,1,3) = x1’x2’+x1’x3’(unate), x1’x2+x1x2’(not unate)
11
110
10110100 x1x2 x3
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Theorem: all threshold functions are unate.
Proof: let f(x1, x2, ••• , xn) be a non unate fn. Suppose it is a threshold fn then there should be some xi such that f is neither positive or negative in xi. Since f ins non unate, there
exist some input combination, say (a1, a2, ••• , an) such that
f(a1, a2, ••• , ai-1, 0, ai+1, ••• , an) = 1
f(a1, a2, ••• , ai-1, 1, ai+1, ••• , an) = 0
and some input combination say (b1, b2, ••• , bn) such that
f(b1, b2, ••• , bi-1, 0, bi+1, ••• , bn) = 0
f(b1, b2, ••• , bi-1, 1, bi+1, ••• , bn) = 1
Wi<0
n
ikkikk
n
ikk
biikk
n
kkk
n
ikkkk
n
ikk
biikk
n
kkk
TwbwbwbwTbw
TbwbwbwTbw
i
i
,1,1
1
1
,1,1
0
1 Wi>0
n
ikkikk
n
ikk
aiikk
n
kkk
n
ikkkk
n
ikk
aiikk
n
kkk
TwawawawTaw
TawawawTaw
i
i
,1,1
1
1
,1,1
0
1
contradiction
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Theorem: Not all unate function are a threshold fn.
Proof) We need to give one example, consider f(x1, x2, x3, x4) = (3,7,11,12,13,14,15) = x1x2+x3x4 unate fn
f(1,1,0,0)=1 w1+w2 T
f(1,0,1,0)=0 w1+w3 < T
f(0,0,1,1)=1 w3+w4 T
f(0,1,0,1)=0 w2+w4 < T
contradiction!! The given fn is not a threshold fn.
Consider that fn f(x1, x2, ••• , xn) is realized by V1= {w1, w2, ••• , wn; T} whose inputs are x1, x2, ••• , xn. If one of input, say xj, is complemented, the function can be realizable by a single threshold element having a weight-threshold vector.
V2= {w1, w2, •••, -wj, ••• , wn; T-wj} whose inputs are x1, x2, •••, -xj, ••• , xn.
01
00
10110100 x1x2 x3x4
1
1
1
1111
10
11
w1+w2>w1+w3 w2>w3
w3+w4>w2+w4 w3>w2
w1 ••• wj ••• wn
Tx1 ••• xj ••• xn
fw1 ••• -wj ••• wn
T-wj
x1 ••• xj ••• xn
g
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We have to prove that f and g are identical fn.
n
jiijiijj
n
kkk
n
jiijiijj
n
iii
gwTxwxwfTaw
gwTxwxwfTxw
,11
,11
0')(0
1')(1
Case 1: xj=0 (xj’=1)
left side (f) right side (g)
n
jiiiijj fTxwwx
,1
10
n
jiijiijj gwTxwxw
,1
1)(1
n
jiiiijj fTxwwx
,1
00
n
jiijiijj gwTxwxw
,1
0)(1
Case 1: xj=1 (xj’=0)
left side (f) right side (g)
n
jiiiijj fTxwwx
,1
11
n
jii
n
jiiiijjii gTxwwwTxw
,1 ,1
1
n
jiiiijj fTxwwx
,1
01
n
jiiii
n
jiijjii gTxwwwTxw
,1,1
0
Therefore, f and g are identical.
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If a function is realizable by a single threshold element (i.e. threshold function) then by an appropriate selection of complemented and uncomplemented input variables, it is realizable by an threshold element whose weight have any desired sign distribution.
Let a threshold fn f(x1, x2, ••• , xn) be positive in var xi and the weight-threshold vector
V= {w1, w2, ••• , wn; T}
,since f is positive in var xi, there exists a set of values
a1, a2, ••• , ai-1, ai+1, ••• , an for inputs x1, x2, ••• , xi-1, xi+1, ••• , xn such that
f(a1, a2, ••• , ai-1, 1, ai+1, ••• , an) = 1 and f(a1, a2, ••• , ai-1, 0, ai+1, ••• , an) = 0
Hence,
w1a1+w2a2+ ••• + wi-1ai-1+wi+wi+1ai+1+ ••• +wnan T and w1a1+w2a2+ ••• + wi-1ai-1 +wi+1ai+1+ ••• +wnan < T
®wi>0
The weight wi associated with a threshold function which is positive in var xi, is positive.
ÞThe weights associated with a threshold function which is a positive in all its vars, are all positive.
If a function f(x1, x2, ••• , xn) is threshold fn with V1= {w1, w2, ••• , wn; T}, is the complement f ’(x1, x2, ••• , xn) a threshold fn? Then what is weigh-threshold vector, V2?
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We assume that the weighted sum is not equal to T for any input combination
wixi>T f=1wixi<T f=0 multiply by (-1) to each side.
(-wi)xi<-T f=1 (f ’=0) (-wi)xi>-T f=0 (f ’=1)
V2= {-w1, -w2, ••• , -wn; -T}
Linear separability
If we use the n-cube representation for n-var threshold functions and regard the vertices as points in n-D space,
w1x1+w2x2+ ••• +wnxn = T
Correspond to an (n-1)-D hyperspace which cuts through the n-cube.
Now, if w1x1+w2x2+ ••• +wnxn T then f=1, if w1x1+w2x2+ ••• +wnxn < T then f=0, the hyper plane separates the true vertices from the false vertices Threshold function = linearly separable function.
If n=2, w1x1+w2x2=T
x1
x2
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If n=3, w1x1+w2x2 +w3x3=T Plane equation
x2
x3
x1
True vertices
x2
x3
x1
Linear equation
False vertices
XOR function
0110
0 0 0 1 1 0 1 1
fx1 x2
OR function
0111
0 0 0 1 1 0 1 1
fx1 x2
x1
x2
x1
x2
(1,0) (1,0)(0,0)
(1,1)(1,1) (0,1)
(0,0)
(0,1)
separableNot separable XOR isn’t threshold function We can separate between
true and false vertices
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Our objective is to present a procedure which will determine whether a given sw fn is a threshold fn, and if it is, will provide the value of weights and threshold
1. Test the given fn for unateness. convert to a positive unate fn
2. Try to find only positive weights for a positive unate fn using minimal true vertices and maximal false vertices.
3. Convert to original inputs and adjust weights and threshold.
Ex) f(x1, x2, x3, x4) = (0,1,4,5,6,7,8,9,11,12,13,14,15)
01
00
10110100 x1x2 x3x4
1111
1111
11
111
10
11
f = x2+x3’+x1x4
We can say
positive unate in x1, x2, x4 negative unate in x3
Let g(y1, y2, y3, y4) = f(x1, x2, x3’, x4) then
g(y1, y2, y3, y4) = y2+y3+y1y4
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Minimal true vertices of a positive unate fn: any true vertex, say (a1, a2, ••• , an) such that there is no other true vertex, say (b1, b2, ••• , bn) such that
(b1, b2, ••• , bn) < (a1, a2, ••• , an)
Maximal false vertices of a positive unate fn: any false vertex, say (a1, a2, ••• , an) such that there is no other false vertex, say (b1, b2, ••• , bn) such that
(b1, b2, ••• , bn) > (a1, a2, ••• , an)
01
00
10110100 y1y2 y3y4
111
11
1111
1111
10
11
false vertices
0000 0001
1000
true vertices
0010 0011 0100 0101 0111 1001 1010
1011 1100
1101 1110 1111
Tmaximal false vertices minimal
true vertices
f 0 wixi < T
f 1 wixi T
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w4 w1
w3 w2 w1+w4
T
Conti. example) w4<w3 w4<w2
w4<w1+w4 w1<w3 w1<w2
If we choose, w3=4, w4=2, w1=2, and w2=4
Ex) f(x1, x2, x3, x4) = (0,1,3,4,5,6,7,12,13)
01
00
10110100 x1x2 x3x4
111
111
1
11
10
11
f = x1’x2+x1’x3’+x1’x4+x2x3’
positive unate in x2, x4 negative unate in x1, x3
Let g(y1, y2, y3, y4) = f(x1’, x2, x3’, x4) then
g(y1, y2, y3, y4) = y1y2+y1y3+y1y4+y2y3
2 4 -4 2
3 3-4 fy1 y2 y3 y4
x1 x2 x3 x4
2 4 4 2
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01
00
10110100 y1y2 y3y4
11
1
111
111
10
11
false vertices
1000 0101 0100 0011 0010 0001
0000
true vertices
0110 0111 1001 1010 1011 1100 1101
1110 1111
T
w1 w2+w4
w3+w4
w2+w3 w1+w4 w1+w3
w1+w2
T
w1<w2+w3 w2+w4<w2+w3 w2<w1 w2+w4<w1+w3
w4 <w1 w4<w2
w3 <w1
w3+w4<w1+w2
w1>w2,w3,w4
w4<w3
If we choose, w1=6, w2=5, w3=4, w4=2, and T=8
6 5 4 2
8 -2 fy1 y2 y3 y4
x1 x2 x3 x4
-6 5 -4 2
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