Problem finding SOP form of f + gConvert boolean expression into SOP and POSMinimize SOP and POS algebraically?Is canonical SOP/POS form for a boolean expression unique?How many minimized forms can a boolean expression have with 4 variables?Karnaugh Map minimal SOP formIs the term 'Sum of min-terms' same as 'Standard SOP form' in Boolean Algebra?Simplify sop expression using Boolean algebraSOP expression simplifyCan there be prime implicants or essential prime implicants of SOP form?
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Problem finding SOP form of f + g
Convert boolean expression into SOP and POSMinimize SOP and POS algebraically?Is canonical SOP/POS form for a boolean expression unique?How many minimized forms can a boolean expression have with 4 variables?Karnaugh Map minimal SOP formIs the term 'Sum of min-terms' same as 'Standard SOP form' in Boolean Algebra?Simplify sop expression using Boolean algebraSOP expression simplifyCan there be prime implicants or essential prime implicants of SOP form?
$begingroup$
I have a project and I am asked to find the sop form of f+g and find its cost and then compare it to the cost if I implement f and g separately.
I am trying to find SOP form of f+g and I am stack because f and g has nothing in common based on the form of and g that I've found:
f = x2'x5 + x1'x2'x4' + x2x4x5'
g = x1'x2' + x2x3' + x1x2x5
And based on those f+g = x2'x5 + x1'x2'x4' + x2x4x5' + x1'x2' + x2x3' + x1x2x5
Here are my karnaugh maps, f above, g below.
My question is: Should I keep what I have done? Or should I find new forms for f and g so that I can get a form for f + g that makes more sense?
Any tip or answer or any suggestion for my question is welcome. Thanks in advance.
boolean-algebra
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
$endgroup$
add a comment |
$begingroup$
I have a project and I am asked to find the sop form of f+g and find its cost and then compare it to the cost if I implement f and g separately.
I am trying to find SOP form of f+g and I am stack because f and g has nothing in common based on the form of and g that I've found:
f = x2'x5 + x1'x2'x4' + x2x4x5'
g = x1'x2' + x2x3' + x1x2x5
And based on those f+g = x2'x5 + x1'x2'x4' + x2x4x5' + x1'x2' + x2x3' + x1x2x5
Here are my karnaugh maps, f above, g below.
My question is: Should I keep what I have done? Or should I find new forms for f and g so that I can get a form for f + g that makes more sense?
Any tip or answer or any suggestion for my question is welcome. Thanks in advance.
boolean-algebra
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
$endgroup$
1
$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29
add a comment |
$begingroup$
I have a project and I am asked to find the sop form of f+g and find its cost and then compare it to the cost if I implement f and g separately.
I am trying to find SOP form of f+g and I am stack because f and g has nothing in common based on the form of and g that I've found:
f = x2'x5 + x1'x2'x4' + x2x4x5'
g = x1'x2' + x2x3' + x1x2x5
And based on those f+g = x2'x5 + x1'x2'x4' + x2x4x5' + x1'x2' + x2x3' + x1x2x5
Here are my karnaugh maps, f above, g below.
My question is: Should I keep what I have done? Or should I find new forms for f and g so that I can get a form for f + g that makes more sense?
Any tip or answer or any suggestion for my question is welcome. Thanks in advance.
boolean-algebra
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
$endgroup$
I have a project and I am asked to find the sop form of f+g and find its cost and then compare it to the cost if I implement f and g separately.
I am trying to find SOP form of f+g and I am stack because f and g has nothing in common based on the form of and g that I've found:
f = x2'x5 + x1'x2'x4' + x2x4x5'
g = x1'x2' + x2x3' + x1x2x5
And based on those f+g = x2'x5 + x1'x2'x4' + x2x4x5' + x1'x2' + x2x3' + x1x2x5
Here are my karnaugh maps, f above, g below.
My question is: Should I keep what I have done? Or should I find new forms for f and g so that I can get a form for f + g that makes more sense?
Any tip or answer or any suggestion for my question is welcome. Thanks in advance.
boolean-algebra
boolean-algebra
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
asked Mar 29 at 15:39
Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης
290111
290111
1
$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29
add a comment |
1
$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29
1
1
$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following three-valued $0, 1, X$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
The following three-valued $0, 1, X$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
$endgroup$
add a comment |
$begingroup$
The following three-valued $0, 1, X$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
$endgroup$
add a comment |
$begingroup$
The following three-valued $0, 1, X$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
$endgroup$
The following three-valued $0, 1, X$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
answered Mar 30 at 13:48
Axel KemperAxel Kemper
3,51611418
3,51611418
add a comment |
add a comment |
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$begingroup$
Your expressions for f and g are correct. Note that you took x5 to be the least-significant and x1 to be the most-significant input for your term numbering. The terms are often numbered in the opposite order.
$endgroup$
– Axel Kemper
Mar 29 at 21:44
$begingroup$
@AxelKemper Thanks for your help ... i am not really sure how you end up with that $f+g$ though.
$endgroup$
– Κωνσταντίνος Κορναράκης
Mar 30 at 9:29