Simplifying a Boolean expression for two-level NAND gate circuits
$begingroup$
The expression is: F = (X' + Y' + Z')(Y' + A')
I have no clear idea on how to go about simplifying this with Boolean algebra. After it's simplified, I'll need to implement it only using NAND gates.
My instinct is to start off with using the distributive property? Or would it be easier to use k-maps?
boolean-algebra
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
$endgroup$
add a comment |
$begingroup$
The expression is: F = (X' + Y' + Z')(Y' + A')
I have no clear idea on how to go about simplifying this with Boolean algebra. After it's simplified, I'll need to implement it only using NAND gates.
My instinct is to start off with using the distributive property? Or would it be easier to use k-maps?
boolean-algebra
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
$endgroup$
add a comment |
$begingroup$
The expression is: F = (X' + Y' + Z')(Y' + A')
I have no clear idea on how to go about simplifying this with Boolean algebra. After it's simplified, I'll need to implement it only using NAND gates.
My instinct is to start off with using the distributive property? Or would it be easier to use k-maps?
boolean-algebra
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
$endgroup$
The expression is: F = (X' + Y' + Z')(Y' + A')
I have no clear idea on how to go about simplifying this with Boolean algebra. After it's simplified, I'll need to implement it only using NAND gates.
My instinct is to start off with using the distributive property? Or would it be easier to use k-maps?
boolean-algebra
boolean-algebra
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
asked Oct 13 '15 at 22:28
borderlineNoviceborderlineNovice
117212
117212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, your question is wrong place however your answer is,
$1. quad F = (X' + Y' + Z')(Y' + A')$
$2. quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$
$3. quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$
$4. quad F=( Y'(X'+Z'+1) + A(X'+Z') )$
$5. quad F=( Y'(1) + A(X'+Z') )$
$6. quad F=( Y' + AX' + AZ' )$
Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.
click here to access circuit
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
$endgroup$
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
add a comment |
$begingroup$
You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').
F = AND (X' + Y' + Z', Y' + A')
F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').
F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.
F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')
F3 = NAND (Y, A)
F4 = NAND (F5, F5) where F5 = X' + Y'
F5 = NAND (X, Y)
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all, your question is wrong place however your answer is,
$1. quad F = (X' + Y' + Z')(Y' + A')$
$2. quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$
$3. quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$
$4. quad F=( Y'(X'+Z'+1) + A(X'+Z') )$
$5. quad F=( Y'(1) + A(X'+Z') )$
$6. quad F=( Y' + AX' + AZ' )$
Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.
click here to access circuit
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
$endgroup$
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
add a comment |
$begingroup$
First of all, your question is wrong place however your answer is,
$1. quad F = (X' + Y' + Z')(Y' + A')$
$2. quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$
$3. quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$
$4. quad F=( Y'(X'+Z'+1) + A(X'+Z') )$
$5. quad F=( Y'(1) + A(X'+Z') )$
$6. quad F=( Y' + AX' + AZ' )$
Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.
click here to access circuit
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
$endgroup$
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
add a comment |
$begingroup$
First of all, your question is wrong place however your answer is,
$1. quad F = (X' + Y' + Z')(Y' + A')$
$2. quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$
$3. quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$
$4. quad F=( Y'(X'+Z'+1) + A(X'+Z') )$
$5. quad F=( Y'(1) + A(X'+Z') )$
$6. quad F=( Y' + AX' + AZ' )$
Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.
click here to access circuit
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
$endgroup$
First of all, your question is wrong place however your answer is,
$1. quad F = (X' + Y' + Z')(Y' + A')$
$2. quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$
$3. quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$
$4. quad F=( Y'(X'+Z'+1) + A(X'+Z') )$
$5. quad F=( Y'(1) + A(X'+Z') )$
$6. quad F=( Y' + AX' + AZ' )$
Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.
click here to access circuit
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
edited Mar 31 '16 at 20:21
daOnlyBG
2,31371734
2,31371734
answered Mar 31 '16 at 19:49
mathemamathema
1011
answered Mar 31 '16 at 19:49
mathemamathema
1011
answered Mar 31 '16 at 19:49
mathemamathema
1011
1011
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
add a comment |
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
$begingroup$
You will have to take deMorgan's on 6. to eliminate OR's (+) to get all NANDs.
$endgroup$
– StainlessSteelRat
Mar 31 '16 at 20:42
add a comment |
$begingroup$
You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').
F = AND (X' + Y' + Z', Y' + A')
F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').
F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.
F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')
F3 = NAND (Y, A)
F4 = NAND (F5, F5) where F5 = X' + Y'
F5 = NAND (X, Y)
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
$endgroup$
add a comment |
$begingroup$
You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').
F = AND (X' + Y' + Z', Y' + A')
F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').
F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.
F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')
F3 = NAND (Y, A)
F4 = NAND (F5, F5) where F5 = X' + Y'
F5 = NAND (X, Y)
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
$endgroup$
add a comment |
$begingroup$
You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').
F = AND (X' + Y' + Z', Y' + A')
F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').
F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.
F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')
F3 = NAND (Y, A)
F4 = NAND (F5, F5) where F5 = X' + Y'
F5 = NAND (X, Y)
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
$endgroup$
You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').
F = AND (X' + Y' + Z', Y' + A')
F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').
F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.
F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')
F3 = NAND (Y, A)
F4 = NAND (F5, F5) where F5 = X' + Y'
F5 = NAND (X, Y)
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
answered Jan 27 '17 at 0:59
gnasher729gnasher729
5,9871028
5,9871028
add a comment |
add a comment |
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