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Karnaugh Maps

Sample Exam K-Map Question (TOTAL MARKS: 25)

The Karnaugh map in Figure Q.2 below represents a logic function with 4 variables DCBA. The original equation has 9 terms in it, as can be seen from the map.

DC\BA 00 01 11 10
00 1 1 1
01 1 1 1 1
11 1 1
10

Figure Q.2

Question 2a (6 marks)

Write down the original Boolean equation that was used to fill the Karnaugh map in Figure Q.2 above.

$\overline{D}.\overline{C}.\overline{B}.\overline{A} +$ $\overline{D}.\overline{C}.\overline{B}.A +$ $\overline{D}.\overline{C}.B.A +$ $\overline{D}.C.\overline{BA}+$ $\overline{D}.C.\overline{B}A+$ $\overline{D}.C.B.A+$ $\overline{D}.C.B.\overline{A}+$ $D.C.B.A+$ $D.C.B.\overline{A}$

[6 Marks]

Q 2(b) Copy the K-Map into your answer book. Draw the most suitable loops on your map. [6 Marks]

Group 1:

DC\BA 00 01 11 10
00 1 1 1
01 1 1 1 1
11 1 1
10

D is 0 and does not change in this group and B is 0 does not change in this group. The equation for this group is $\overline{D}.\overline{B}$.

Group 2:

DC\BA 00 01 11 10
00 1 1 1
01 1 1 1 1
11 1 1
10

C is 1 and does not change in this group and B is 1 does not change in this group. The equation for this group is $C.B$

Group 3

DC\BA 00 01 11 10
00 1 1 1
01 1 1 1 1
11 1 1
10

D is 0 and does not change in this group and A and B are 1 and do not change in this group. The equation for this group is $\overline{D}.A.B$

Q2(c) Write down the minimized equation.

$\overline{D}.\overline{B} + C.B + \overline{D}.A.B$

[6 Marks]

Question 2.D

Use a 3 variable K-map to minimise the following function :-

$F = C.B.A + C.\overline{B}.A + C.\overline{B}.\overline{A} + \overline{C}.B.A + \overline{C}.\overline{B}.A$

Rewrite the function with varaibles in alphabetical order.

$F = A.B.C + A.\overline{B}.C + \overline{A}.\overline{B}.C + A.B.\overline{C} + A.\overline{B}.\overline{C}$

AB/C 0 1
00 $\overline{A}.\overline{B}.C$
01
10 $A.\overline{B}.\overline{C}$ $A.\overline{B}.C $
11 $A.B.\overline{C}$ $A.B.C$
AB/C 0 1
00 1
01
10 1 1
11 1 1

Group 1:

AB/C 0 1
00 1
01
10 1 1
11 1 1

So A is 1 and does not change in this group. The equation for this group is $A$

Group 2:

AB/C 0 1
00 1
01
10 1 1
11 1 1

So A, B are 0 and do not change in this group, and C is 1 and does not change in this group. The equation for this group is $\overline{A}.\overline{B}.C$

So combining the two groups we get the minimized equation as:

$$\boxed{F = A + \overline{A}.\overline{B}.C}$$

Alt 3

Q 2(d) Use a 3 variable K-map to minimise the following function :-

$F = C.B.A + C.\overline{B}.A + C.\overline{B}.\overline{A} + \overline{C}.B.A + \overline{C}.\overline{B}.A$

C/BA 00 01 10 11
0 1 1
1 1 1 1

$F = BA + \overline{B}.A+ C.\overline{B}$

$F = BA + \overline{B}.A+ C.\overline{B}$ $ F = A(B + \overline{B}) + C.\overline{B}$ $ F = A + C.\overline{B}$