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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?

0

1

$begingroup$

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

share|cite|improve this question

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 | 

0

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 29 at 15:39

Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης

290111

$endgroup$

share|cite|improve this question

asked Mar 29 at 15:39

Κωνσταντίνος ΚορναράκηςΚωνσταντίνος Κορναράκης

290111

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

1

$begingroup$

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'

share|cite|improve this answer

answered Mar 30 at 13:48

Axel KemperAxel Kemper

3,51611418

$endgroup$

add a comment | 

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1

$begingroup$

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'

share|cite|improve this answer

answered Mar 30 at 13:48

Axel KemperAxel Kemper

3,51611418

$endgroup$

add a comment | 

1

$begingroup$

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'

share|cite|improve this answer

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'

share|cite|improve this answer

answered Mar 30 at 13:48

Axel KemperAxel Kemper

3,51611418

$endgroup$

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

x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'

share|cite|improve this answer

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