Reduce the following Boolean expressions to the indicated number of literals .

[Faraj]

Reduce the following Boolean expressions to the indicated number of literals :
(a) (x’y’+z)’ + z + xy + wz to three literals
(b) (A’+C)(A’+C’)(A+B+C’D) to four literals

 

[Faraj]

Reduce the following Boolean expressions to the indicated number of literals

Reduce the following Boolean expressions to the indicated number of literals .



Reduce the following Boolean expressions to the indicated number of literals :
(a) (x’y’+z)’ + z + xy + wz to three literals
(b) (A’+C)(A’+C’)(A+B+C’D) to four literals



I have tomorrow exam .

i domnload this subject here to find solve by faster (

 

[Ratchit]

Faraj,

I have tomorrow exam .

Why don't you use a Karnaugh map (K-map) to solve those problems? If we give you the answers, how is that going to help you on the exam? Where is your attempt to solve those problems? How can the first expression be solved for 3 literals when the expression contains 4 literals? Does one of the literals cancel?

Ratch

 

[Faraj]

Faraj,



Why don't you use a Karnaugh map (K-map) to solve those problems? If we give you the answers, how is that going to help you on the exam? Where is your attempt to solve those problems? How can the first expression be solved for 3 literals when the expression contains 4 literals? Does one of the literals cancel?

Ratch

I have conidtion use by algabre

yes . cancel by algabre law

 

[Ratchit]

Faraj,

Where is your attempt to solve those problems?

Ratch

 

[Faraj]

thank you ,
I am finding solve for this problem on this link
https://www.electro-tech-online.com/custompdfs/2011/11/homework20120solution.pdf

thank you Ratch ^_^

 

[Ratchit]

Faraj,

OK, let's do the first expression.

(x’y’+z)’ + z + xy + wz = xz' +yz'    +z+xy+wz ; first term by DeMorgan's theorem
xz'+yz'+z+xy+wz = xz'+yz'+xy+z(1+w) = xz'+yz'+xy+z  ; now only three terms
xz'+yz'+xy+z = x(y+y')z'+(x+x')yz'+xy(z+z')+(x+x')(y+y')z ;expand out the terms
= xyz'+xy'z'+xyz'+x'yz'+xyz+xyz'+xyz+xy'z+x'yz+x'y'z
=x'y'z+x'yz'+x'yz+xy'z'+xy'z+xyz'+xyz ;eliminating duplicate terms
(xy'z'+xy'z+xyz'+xyz)+(x'y'z+x'yz+xy'z+xyz)+(x'yz'+x'yz+xyz'+xyz) ;rearranging terms, we can duplicate terms as many times as we need to do so
(xy'z'+xy'z+xyz'+xyz)=xy'(z'+z)+xy(z'+z)=xy'+xy=x(y+y')=x  ;reducing the first bracketed term
(x'y'z+x'yz+xy'z+xyz) = z  reducing the second term is just as easy
(x'yz'+x'yz+xyz'+xyz) = y and finally the third term is reduced

Therefore, the result is x+y+z. All the terms in the expanded expression were used to calculate the result. You may ask how I knew how to arrange the terms to get a result. The answer is that I used a K-map to see what expanded terms made up the reduced result. There is no systematic way using only algebra to get a result. You have to try and guess if that is all the methodology available. Keep in mind that you do not have to hand in your K-map. I will do the next problem in another post.

Ratch

 

[Ratchit]

Faraj,

OK, now for the second problem.

(A’+C)(A’+C’)(A+B+C’D) = A'B'C'D+A'BC'D'+A'BC'D+A'BCD'+A'BCD ;multiplying out expression
(A'B'C'D+A'BC'D)+(A'BC'D'+A'BC'D+A'BCD'+A'BCD) arrange the term using secret K-map
(A'B'C'D+A'BC'D) = A'C'D(B'+B)=A'C'D ;reduce first term
(A'BC'D'+A'BC'D+A'BCD'+A'BCD) =A'B ;reduce second term

So the result is A'B+A'C'D

Ratch

 

[MrAl]

Hi,

Here's my reduction using only the properties of Boolean Algebra.

I use a different notation to keep it more readable:
1. nx is the same as x' (just use a 'n' before the letter instead of ' after it)
2. n(x) negates everything inside the parens, so n(x)=x', and n(a+b)=na*nb=a'b' also for example.

Using that notation the first problem is reduced as follows:

n(z+nx*ny)+w*z+z+x*y
nz*x+y+w*z+z+x*y
x+y+w*z+z+x*y
x+y+w*z+z
x+y+z


See if you can figure out what properties were used to reduce each line to each successive line. I think this is more clear than my original post.

They use this kind of technique in the linked to paper also. Notice how short and sweet it comes out?

 

[Faraj]

Thank You everybody ^_^