A counter is constructed with three D flip-flops. The input-output pairs are named (D0, Q0), (D1, Q1), and (D2, Q2), where the subscript 0 denotes the least significant bit. The output sequence is desired to be the Gray-code sequence 000, 001, 011, 010, 110, 111, 101, and 100, repeating periodically. Note that the bits are listed in the Q2 Q1 Q0 format. The combinational logic expression for D1 is
Q̅2 Q0 + Q1 Q̅0
Understanding the D1 Logic for the Gray Code Counter
This problem involves determining the specific logic gate configuration needed for the D input of the second flip-flop (Q1) in a 3-bit Gray code counter. The counter generates a sequence of Gray codes, and we need to find the function that dictates the transition of Q1 to its next state (D1).
Gray Code Sequence and D1 Derivation
The Gray code sequence provided is for bits Q2, Q1, Q0 (MSB to LSB). We need to determine the value of D1 for each state transition. Since D1 is the input to the D flip-flop Q1, D1 must be equal to the next state of Q1 (denoted as Q'1).
| Current State (Q2Q1Q0) | Next State (Q'2Q'1Q'0) | D1 (Next Q1) |
|---|---|---|
| 000 | 001 | 0 |
| 001 | 011 | 1 |
| 011 | 010 | 1 |
| 010 | 110 | 1 |
| 110 | 111 | 1 |
| 111 | 101 | 0 |
| 101 | 100 | 0 |
| 100 | 000 | 0 |
Karnaugh Map for D1 Simplification
We construct a Karnaugh map (K-map) using the current state variables (Q2, Q1, Q0) as inputs and D1 as the output. The map is filled with the D1 values derived from the state transitions.
| Q2 | Q1Q0 | |||
|---|---|---|---|---|
| 00 | 01 | 11 | 10 | |
| 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 |
To find the minimal sum-of-products (SOP) expression, we group the adjacent 1s in the K-map. The groups should be as large as possible, containing powers of 2 (1, 2, 4, 8).
- Group 1: The two 1s in the row where Q2 = 0, specifically in columns 01 and 11. This group covers states where Q2 is 0 and Q0 is 1. The corresponding term is $\overline{Q_2} Q_0$.
- Group 2: The two 1s in the column where Q1Q0 = 10, spanning across rows Q2 = 0 and Q2 = 1. This group covers states where Q1 is 1 and Q0 is 0. The corresponding term is $Q_1 \overline{Q_0}$.
Combining these terms gives the simplified expression for D1:
$$D_1 = \overline{Q_2} Q_0 + Q_1 \overline{Q_0}$$
Matching the Expression to the Option
The derived minimal combinational logic expression for D1 is $\overline{Q_2} Q_0 + Q_1 \overline{Q_0}$. This matches the logic provided in Option 4.
Important Questions from Counter
Which of the following is not capable of storing binary data?
A _________ is a register capable of counting the number of clock pulses arriving at its clock input.
A MOD 2 and a MOD 5 up-counter when cascaded together results in a MOD ______ counter. (in integer)
A 16-bit synchronous binary up-counter is clocked with a frequency fCLK. The two most significant bits are OR-ed together to form an output Y. Measurements show that Y is periodic, and the duration for which Y remains high in each period is 24 ms. The clock frequency fCLK is ______ MHz. (Round off 2 decimal places.)
A modulus 10 counter must have
