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Design of Two-Level Multiple-Output Networks. Design a network with four inputs and three outputs which realizes the functions F1(A,B,C,D) = S m(11,12,13,14,15) F2(A,B,C,D) = S m(3,7,11,12,13,15) F3(A,B,C,D) = S m(3,7,12,13,14,15). AB. AB. Direct Realization (each function as a min. - PowerPoint PPT Presentation
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Design of Two-Level Multiple-Output Networks
Design a network with four inputs and threeoutputs which realizes the functionsF1(A,B,C,D) = m(11,12,13,14,15)
F2(A,B,C,D) = m(3,7,11,12,13,15)
F3(A,B,C,D) = m(3,7,12,13,14,15)
ABAB
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Direct Realization(each function as a min.sum of prime implicants)
Note: AB is common to F1, F3
Multiple-Output Realization
= AB+ACD
= ABC’+A’CD+ACD
= A’CD+AB
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Example: Using Common Terms in Multiple Output Network Design
f1 = a’bd + abd+ab’c’ + b’cf2 = c + a’bd f3 = bc + ab’c’ + abd
8 gates, 22 gate inputs
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Best Soln.
Soln. Requires extra gate
Another Example
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Soln. with max.number of common terms (8 gates, 26 inputs)
Best Soln.has no common terms(7 gates, 18 inputs)
Another Example
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Code Converters
Code converters – take an input code, translate to its equivalent output code.
Codeconverter
Inputcode
Outputcode
Example: BCD to Excess-3 Code Converter.
Input: BCD digit
Output: Excess-3 digit
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BCD-to-Excess-3 Code Converter Truth table:
BCD Excess-3A B C D W X Y Z
0 0 0 0 0 0 0 1 11 0 0 0 1 0 1 0 02 0 0 1 0 0 1 0 13 0 0 1 1 0 1 1 04 0 1 0 0 0 1 1 15 0 1 0 1 1 0 0 06 0 1 1 0 1 0 0 17 0 1 1 1 1 0 1 08 1 0 0 0 1 0 1 19 1 0 0 1 1 1 0 0
10 1 0 1 0 X X X X11 1 0 1 1 X X X X12 1 1 0 0 X X X X13 1 1 0 1 X X X X14 1 1 1 0 X X X X15 1 1 1 1 X X X X
W =m(5,6,7,8,9)
x = m(1,2,3,4,9)
y = m(0,3,4,7,8)
z = m(0,2,4,6,8)
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AB 01
1
00 CD 00
1
1
01
11
10
x 1
x 1
x x
x x
11 10
0
1
3
2
4
5
7
6
12 8
9
11
10
13
15
14
x =m(1,2,3,4,9)+d(10,11,12,13,14,15)bc’d’+b’d+b’c=bc’d’+b’(c+d)
y =m(0,3,4,7,8)+ d(10,11,12,13,14,15) = c’d’+cd
011
1 00
CD 00
11
01 11
10
x 0x 1x xx x
11 10
0
1
3
2
4
5
7
6
12 8
9
11
10
13
15
14
AB
01
1 1 00 CD 00
1 1
01
11
10
x 1
x
x x
x x
11 10
0
1
3
2
4
5
7
6
12 8
9
11
10
13
15
14
AB
W =m(5,6,7,8,9)+d(10,11,12,13,14,15)= a+bc+bd = a+b(c+d)
z =m(0,2,4,6,8)+ d(10,11,12,13,14,15) = d’
01
1 00 CD 00
1 1
01
11
10
x 1
x
x x
x x
11 10
0
1
3
2
4
5
7
6
12 8
9
11
10
13
15
14
AB
1
Underlinedterms arecommon
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Multi-Output NAND and NOR Networks
Conversion to anetwork of all NAND or allNOR gates can also be done for the case of multiple-output networks, using method of Section 8.6, p. 194
(a) Network of AND and OR gates
(b) NOR network
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Multiplexers
2/1 Multiplexer (MUX)
I0
I1
Z
S
if S = 0, then Z = I0
if S = 1, then Z = I1
S is the Control Input used to select one of the data inputs I0 , I1 and connect it to the output terminal, Z.
Control Input
Data
Inputs
Z = I0S’ + I1S
I0
S’
I1S
Z
Logic Diagram
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Higher Order Muxes
I0
I1 ZI2
I3
4/1 Mux
if S = “00”, then Z = I0
if S = “01”, then Z = I1
if S = “10”, then Z = I2
if S = “11”, then Z = I3
I0
I1
Z
S[2:0]
I2
I3
8/1 Mux
3
I4
I5I6
I7
Z = I0 S1’ S0’ + I1 S1’ S0 + I2 S1 S0’ + I3 S1 S0
S1 S0
Note: S[2:0] means the three control inputs S2 S1 S0 <--lsb
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Logic Diagram for 8-to-1 Multiplexer
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To build a 2/1 mux for 4-bit wide busses, need four 1-bit 2/1 muxes.
I0
I1
Z
S
A[3:0]
B[3:0]D[3:0]
I0I1 Z
S
I0I1
ZS
I0I1
Z
S
I0I1
ZS
A0B0
A1B1
A2B2
A3B3
Z0
Z1
Z2
Z3
S
Muxes are often used to select groups of bits arranged in busses.
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A 4-to-1 MUX can realize any 3-variable function with no added Logic gates.
Example: Realize F(A,B,C)= A’B’ + AC
Soln.: Expanding F so that all terms include both control inputs, A and B, yieldsF = A’B’ + AC(B’+B) = A’B’1 + AB’C + ABCThe general eqn. For 4-to-1 MUX isF = A’B’I0 + A’BI1+ AB’I2+ ABI3Comparing eqns. We see they will be identical ifI0 =1, I1 =0, I2 =C, I3 =C
Function Realization using MUX
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Example: Consider the function Z(A,B,C,D) = m(0,1,3,6,7,8,11,12,14) and implement it
using an 8-to-1MUXSoln.: Assume that A,B,C are applied to control inputs s2,s1,s0 resp.Then all 16 possible minterms that can be generated by the MUX can be represented in a table.
000 001 010 011 100 101 110 111
I0 I1 I2 I3 I4 I5 I6 I7
D’ 0 2 4 6 8 10 12 14
D 1 3 5 7 9 11 13 15
1 D 0 1 D’ D D’ D’
Working out the bottom row tells us what to apply to I0, I1, 12, etc.How to work it out? e.g. I0 col D’+D=1, I1 col only minterm with D appears,(apply D), I2 col no minterms used (apply 0), I3 col D’+D=1, I4 col. only minterm with D’ appears(apply D’).
Function Realization using MUX
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Example: Continued
The function is then realized with an 8-to-1 MUXwith ABC applied to the control inputs and the values found in the table to I0, I1, I2, etc.
S2 S1 S0
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Example: Continued Alternate Solution
Represent all of the minterms of Z on a k-map. Then obtaingroupings corresponding to [I0 = 000 with both D’ and D][I1=001 with both D’ and D] etc. Each of the two-minterm groupingscan be thought of as a 1-variable k-map in D only and simplified.
One-variable k-maps
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Example: Continued Yet anotherAlternate Solution
This time assume that A,B,D are applied to control inputs S2,S1,S0 resp.
000 001 010 011 100 101 110 111
I0 I1 I2 I3 I4 I5 I6 I7
C’ 0 1 4 5 8 9 12 13
C 2 3 6 7 10 11 14 15
C’ 1 C C C’ C 1 0
D
C’
C’
1
0
C
C
1
C
S0S1S2
Mux Realization with Control Inputs A, B, and D
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Example: Continued Yet anotherAlternate Solution usingexternal gates
00 01 10 11
I0 I1 I2 I3
00 0 4 8 12
01 1 5 9 13
11 3 7 11 15
CD
10 2 6 10 14
Table method in this case isequivalent to four 2-variable K-maps
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Decoders
This decoder generates all minterms of of the three input variables
Exactly one of the output lines will be 1 for each combination of the input variables.
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Decoders
Note: Active low outputs
7442 4-to-10 decoder
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f1 = (m1’m2’ m4’)’
= m1 + m2 + m4
f2 = (m4’m7’ m9’)’
=m4 + m7 + m9
Realization of Multiple-Output Network using Decoder
Can realize a function by ORing together selected minterm outputs.In this case outputs are active low, so NAND gates are used (effectively ORing together these outputs).