Respuesta :

This question is incomplete, the complete question is;

Your boss asks you to design a Boolean circuit that verifies whether a given integer 0 < x < 16 is divisible by 5.

Every such number is represented in binary using four bits, say b₃b₂b₁b₀, and so your Boolean circuit will have four inputs. For instance, the number 13 is written in binary as 1101 and so to test its divisibility by 5 a user would feed the values b₀ = 1, b₁ = 0, b₂ = 1, and b₃ = 1 into the inputs of your circuit.

The Boolean circuit will have a single output, which should deliver the value 1 if the iput values represent a number that is divisible by 5 and 0 otherwise.

a) write down the truth table of the Boolean function F(b₀,b₁,b₂b₃) that implements this "divisible by 5" operation

b) construct a Boolean expression in disjunctive normal form that implements the Boolean function yo wrote down in a)

Answer:

Given that;

integer range = 0≤ x ≤ 16

within 4bits, we can represnt each number

(0,5,10,15)

a)  

Truth table for function that implements  divisible by 5

Integer    B3    B2    B1    B0     Y

0              0       0      0      0      1

1              0       0      0       1      0

2              0       0      1        0     0

3              0       0      1        1      0

4              0       1       0       0     0

5              0       1       0       1      1

6              0       1       1        0     0

7              0       1       1        1       0

8              1       0      0        0      0

9              1       0      0        1       0

10             1       0      1         0      1

11              1       0      1          1      0

12             1        1      0         0     0

13             1         1     0          1      0

14             1         1       1         0     0

15            1         1        1        1       1

b)

Boolean expression that implements the Boolean function from a)

from the truth table;

Boolean expression Y is;

Y = b⁻₃b⁻₂b⁻₁b⁻₀ / y₁   +   b⁻₃b₂b⁻₁b₀ / y₂   +   b₃b⁻₂b₁b⁻₀ / y₃   +   b₃b₂b₁b₀ / y₄

Answer:

  circuit is in the second attachment

Step-by-step explanation:

Attached is a truth table for the desired circuit, where b3–b0 are the input bits, MSB–LSB. We notice that the output is true whenever b3=b1 and b2=b0. This can be written in DNF as ...

  [tex]Y=b_3'b_2'b_1'b_0'+b_3b_2'b_1b_0'+b_3'b_2b_1'b_0+b_3b_2b_1b_0[/tex]

More compactly, it can be written in terms of the exclusive-nor function as ...

  [tex]y=(b_3\odot b_1)\wedge(b_2\odot b_0)[/tex]

A circuit diagram showing this circuit is the second attachment. (A–D are the input bits, in order MSB–LSB (or its reverse)).

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