Electronic – Simplification of boolean expression

boolean-algebrakarnaugh mapsop

I am asking you for a help with following boolean expression that i can't solve by myself. I can minimize expression in Karnaugh map but i have to use boolean algebra too.

Expression:

!(C + !D) * !(B + D) + !!(C + !D) * !!(B + D)

(!C * D) * (!B * !D) + (C + !D) * (B+D) – De Morgan law
(!C * D) * (!B * !D) equals 0 because of A * 0 = 0
(C + !D) * (B + D) – multiplying everything
BC + B!D + CD + !DD
!DD equals 0
BC + B!D + CD – that what remains

So i'm stucked and i don't know how should i continue.

Expression after simplification should be:
CD + B!D – (SoP – Sum of Products)

Can you help me, please?

Thank you very much for your time.

Best Answer

You got most of the way. Follow what's below:

$$\begin{align*} F &= \overline{(C + \overline{D})}\: \overline{(B + D)} + \overline{\overline{(C + \overline{D})}}\: \overline{\overline{(B + D)}}\tag{0}\\\\ &=\overline{C} \:D\: \overline{B}\: \overline{D} + (C + \overline{D})\: (B + D)\tag{1}\\\\ &=(C + \overline{D})\: (B + D)\tag{2}\\\\ &=B\:C + C\: D + B\:\overline{D} + D\:\overline{D}\tag{3}\\\\ &=B\:C + C\: D + B\:\overline{D}\tag{4}\\\\ &=B\:C\:D + B\:C\:\overline{D} + C\: D + B\:\overline{D}\tag{5}\\\\ &=(B\:C + C)\: D+(B\:C + B)\:\overline{D}\tag{6}\\\\ &=(C\:[B + 1])\: D+(B\:[C + 1])\:\overline{D}\tag{7}\\\\ &=(C\cdot1)\: D+(B\cdot1)\:\overline{D}\tag{8}\\\\ &=C\: D+B\:\overline{D}\tag{9} \end{align*}$$

Here, you can see that I've expanded the BC term in step 5. This is just turning a single case into two cases, which does NOT change the result. I think you can see that it doesn't, by inspection.

Then, in step 6 I organize the summed terms so that I can factor out D and Not-D, to create two somewhat more complex terms. But now, simple inspection tells you that in the first term (based on D) that if C is true that it doesn't matter whether or not B C is true and that if C is false then so is B C. So that can be reduced down to just C. The same idea also applies to the second term (based on Not-D.) I've added steps 7 and 8 to show this transition.

The final result is in 9, now.

Related Solutions

Electronic – How to simplify the actual function using K-maps

Just simply make a truth table that shows values of F for each of the combinations of A, B and C. Then transfer this information to the K-Map format table to do the minimization. enter image description here

Electronic – Karnaugh map subcube with no unchanging variables

Lets say your five variables are A, B, C, D, E. To "draw" a 5 variable K-map, draw two K-Maps and imagine them stacked on top of one-another.

E=0

                !C          C
            ----------  ---------
     AB\CD  00    01    11   10
!A { 00     0000  0001  0011 0010  > !B
!A { 01     0100  0101  0111 0110  } B
A  < 11     1100  1101  1111 1110  } B
A  < 10     1000  1001  1011 1010  > !B
            vvvv  ---------- vvvv         
             !D       D       !D      


E=1
                !C          C
            ----------  ---------
     AB\CD  00    01    11   10
!A { 00     0000  0001  0011 0010  > !B
!A { 01     0100  0101  0111 0110  } B
A  < 11     1100  1101  1111 1110  } B
A  < 10     1000  1001  1011 1010  > !B
            vvvv  ---------- vvvv         
             !D       D       !D

You see how every grid location (which maps to a min-term) differs from all its neighbors (including wrap-around edge-cases) by exactly one bit? That's more-or-less the whole point of a K-Map representation. Now in a 4-variable K-Map you can "circle" contiguous rectilinear shapes where the output is a "1" so long as their "area" is a power of two (i.e. 1, 2, 4, 8, or 16 terms) in order to achieve a "cover" of the K-map. I believe there are 2^16 possible covers of a 4-variable K-map (i.e. 2^(2^4)).

The only difference in the 5-variable case is "adjacent" also include analogous positions in the two 4-variable maps (sub-spaces), and there are more ways to ways to "circle" a number of cells that is a power of two; that is to say there are "three-dimensional" covers possible.

Hopefully it's clear that "the thing that is constant in a straight line" is what that line identifies, and that the intersections of "the things that are constant" is what makes a particular location a minterm. If you "circle" the upper-left four grid locations in both E=0 and E=1 maps, then E is a don't-care. If you circle the quadrant in just the E=0 map, then the term you are circling must have E in the intersection.

Best Answer

Related Solutions

Electronic – How to simplify the actual function using K-maps

Electronic – Karnaugh map subcube with no unchanging variables

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