Design a digital system whose output is defined as logically low. If the 4 bit inpur binary number is a multiple of 3. Otherwise, the output will be logically high. The output is defined if and only if the input binary number is greater than 2
The correct answer and explanation is :
Design of a Digital System to Detect Multiples of 3 (with Defined Output for Inputs Greater than 2)
Problem Statement
We need to design a digital system that takes a 4-bit binary input and produces a logically low (0) output if the number is a multiple of 3. Otherwise, the output should be logically high (1). However, the output is only valid if the input number is greater than 2.
Step 1: Identifying Valid Inputs
Since we have a 4-bit binary number, the possible range of inputs is (0000_2) (0) to (1111_2) (15). However, the system’s output is defined only for numbers greater than 2, so the valid range is 3 to 15.
Step 2: Identifying Multiples of 3
The multiples of 3 within this range are:
- (3_{10} = 0011_2)
- (6_{10} = 0110_2)
- (9_{10} = 1001_2)
- (12_{10} = 1100_2)
- (15_{10} = 1111_2)
For these inputs, the output should be 0 (logically low). For all other inputs in the range of 3 to 15, the output should be 1 (logically high).
Step 3: Deriving the Logic Expression
Let ( A, B, C, D ) be the input bits, where ( A ) is the MSB and ( D ) is the LSB. We derive the Boolean function that is low (0) for multiples of 3.
Using Sum of Products (SOP) for when the output is 0:
[
F(A, B, C, D) = \sum m(3, 6, 9, 12, 15)
]
The complement of this function will give the required output ( Y ), which is high for non-multiples of 3.
Using a Karnaugh Map (K-map) simplification, the final expression for ( Y ) (Output) will be derived. The simplest logic circuit can be implemented using AND, OR, and NOT gates.
This design helps to determine whether a 4-bit number is a multiple of 3 efficiently using combinational logic.