#291 If Given that: F(a, b, c, d) = (a + c + d) (b' + c) (a' + b' + c')

If Given that: F(a, b, c, d) = (a + c + d) (b' + c) (a' + b' + c') - Computer Science

ChemistryExplain daily providing Q&A content “#291 If Given that: F(a, b, c, d) = (a + c + d) (b' + c) (a' + b' + c')" in Computer science, Ba computer science, Berkeley computer science, Computer science associate degree jobs.

Get the Free Online Chemistry Q&A Questions And Answers with explain. To crack any examinations and Interview tests these Chemistry Questions And Answers are very useful. Here we have uploaded the Free Online Chemistry Questions. Here we are also given the all chemistry topic.

 ChemistryExplain team has covered all Topics related to inorganic, organic, physical chemistry, and others So, Prepare these Chemistry Questions and Answers with Explanation Pdf.

For More Chegg Questions

• Inorganic Chemistry
• Organic Chemistry
• Physical Chemistry

Free Chegg Question

If Given that: F(a, b, c, d) = (a + c + d) (b' + c) (a' + b' + c') (a + b + c' + d')

(a) Express Fas a minterm expansion. (Use m-notation.)

(b) Express F as a maxterm expansion. (Use M-notation.)

(c) Express F'as a minterm expansion. (Use m-notation.)

(d) Express F' as a maxterm expansion. (Use M-notation.)

For More Chemistry Notes and Helpful Content Subscribe Our YouTube Chanel - Chemistry Explain  

Free Chegg Answer

Given function is,

F(a,b,c,d) = (a+c+d)(b'+c)(a'+b'+c')(a+b+c'+d') ------(0)

As in (a+c+d) b is missing, so it can be twritten as (a+c+d+b*b') because x*x'=0

And by Boolean algebraic law, x+yz=(x+y)(x+z), So it can further be written as,

=> (a+b+c+d)(a+b'+c+d) -----------(1)

As in (b'+c) a is missing, so it can be twritten as (b'+c+a*a') because x*x'=0

And by Boolean algebraic law, x+yz=(x+y)(x+z), So it can further be written as,

=> (a+b'+c)(a'+b'+c) and applying same for missing d in both we get,

=>(a+b'+c+d)(a+b'+c+d')(a'+b'+c+d)(a'+b'+c+d') -----------(2)

Similary for (a'+b'+c') it can be written as

=>(a'+b'+c'+d)(a'+b'+c'+d') ----------(3)

After combining 0,1,2,3, we get function

F(a,b,c,d)= (a+b+c+d)(a+b'+c+d)(a+b'+c+d)(a+b'+c+d')(a'+b'+c+d)(a'+b'+c+d')(a'+b'+c'+d)(a'+b'+c'+d')(a+b+c'+d')

After deducing,

F(a,b,c,d)=(a+b+c+d)(a+b+c'+d')(a+b'+c+d)(a+b'+c+d')(a'+b'+c+d)(a'+b'+c+d')(a'+b'+c'+d)(a'+b'+c'+d')

F(a,b,c,d)=a,b,c,d(0,3,4,5,12,13, 14,15)

(B)In M notation,

F(a,b,c,d)=M0*M3*M4*M5*M12*M13*M14*M15

Writing the missing indexes will give minterm values i.e. 1,2,6,7,8,9,10,11.

So , F(a,b,c,d) =a,b,c,d(1,2,6,7,8,9,10,11)

(A)In m notation,

F(a,b,c,d)=m1+m2+m6+m7+m8+m9+m10+m11

For F'(a,b,c,d)=((a+c+d)(b'+c)(a'+b'+c')(a+b+c'+d'))'

Applying demorgans law, i.e. (x+y)'=x'y'

=>(a+c+d)'+(b'+c)'+(a'+b'+c')'+(a+b+c'+d')'

=>a'c'd'+bc'+abc+a'b'cd

now filling the missing terms with (x+x')=1 identity,

=>a'c'd'(b+b')+bc'(a+a')+abc(d+d')+a'b'cd

=>a'bc'd'+a'b'c'd'+abc'+a'bc'+abcd+abcd'+a'b'cd

=>a'bc'd'+a'b'c'd'+abc'(d+d')+a'bc'(d+d')+abcd+abcd'+a'b'cd

=>a'bc'd'+a'b'c'd'+abc'd+abc'd'+a'bc'd+a'bc'd'+abcd+abcd'+a'b'cd

F'(a,b,c,d)=a'b'c'd'+a'b'cd+a'bc'd'+a'bc'd+abc'd'+abc'd+abcd'+abcd

F'(a,b,c,d)=a,b,c,d(0,3,4,5,12,13,14,15)

(C)In m notation,

F'(a,b,c,d)=m0+m3+m4+m5+m12+m13+m14+m15

Writing the missing indexes will give maxterm values i.e. 1,2,6,7,8,9,10,11.

F'(a,b,c,d)=a,b,c,d(1,2,6,7,8,9,10,11)

(D)In M notation

F'(a,b,c,d)=M1*M2*M6*M7*M8*M9*M10*M11