Question

The following is a truth table of a 3-input, 4-output combinational circuit. Using K- maps obtain the expressions for W, X, Y and Z. Implement using a PAL. Inputs Outputs ΙΑ B С W X Y Z 10 0 0 0 1 10 0 10 0 1 1 1 1 1 10 1 0 1 10 1 1 10 1 1 0 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1

Answer 1

Let's first draw the K-map for each output and obtain it's expression.

K-map for W:

Вс ОО 11 10 О | Q О О

From the above K-map expression for W= AC' + BC' + A'B'C

K-map for X:

BC ol 11 10 А 1 o 1 1 1 o 1

From the above K-map expression for X= B'C' + A'B' + BC

K-map for Y:

И Вс А A 1 1 о O о | 1

From the above K-map , the expression for Y= AC' + AB + BC' + A'B'C

=(AC' + BC' + A'B'C) + AB

= W + AB (Since, W = AC' + BC' + A'B'C )

K-map for Z:

BC 30 ol A 10 O 1 1 1 o

From the above K-map ,expression for Z= A'B + C

Now, let's tabulate the PAL programming table:

Product AND inputs autputs turns AB с W 1 0 - 1 2 W - AC'+ BC'+ A'B'c 3 0 O 1 1 t to 0 - X = BC + B'C'+ ÁB b - 6 7 1 - 8 I 1 Y = W + AB - 9 - - 1 T 10 Z: C+ A'B 11 - 1 (2

To explain the tabular form let me take an example Z = C + AB

First we should take the first element of the expression , here it is C. So in this term the missing elements are A and B , so we have to put "hyphen" mark in the place of A and B and as C elemnt is present we have to put 1 in its place . Suppose if it was C' we have to put 0 in its place.Similarly, in the next line the term A'B is done.

With the help of rogramming table we'll now be able to draw the circuit.

The PAL Diagram will be:

А Á B В B' c Ć W w' * M. 1 -X B Y У cho -Z *

So, with the help of tabular form PAL circuit diagram has been done.

Here, we have to take an expression and mark cross marks in its terms.For example in the expression, W = AC' + BC' + A'B'C

In the first line we have to mark the first term i.e, A'C and in next line BC' and in the last line A'B'C. In similar way other 3 expressions are done.