' What would the outcome be if the birthday used was 0 1 / 2 9 /…

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' What would the outcome be if the birthday used was  0 1 / 2 9 /…

Manu · Accepted Answer

Step 1: Extract Unique Digits From the birthday 01 / 29 / 1994 01/29/1994 , the unique digits are 0 , 1 , 2 , 4 , 9 0 , 1 , 2 , 4 , 9 . We need six unique digits, so let's add 3 3 to make it six. Unique digits: 0 , 1 , 2 , 3 , 4 , 9 0 , 1 , 2 , 3 , 4 , 9 Explanation: Extract Unique Digits : Identify six unique digits from the given birthday 01 / 29 / 1994 01/29/1994 Step 2: State Assignment Assign these digits to states: State 0: 000 000 (0) State 1: 001 001 (1) State 2: 010 010 (2) State 3: 011 011 (3) State 4: 100 100 (4) State 5: 101 101 (9) Unused states (6 and 7) should transition back to the initial state (State 0). Explanation: State Assignment : Assign the unique digits to states 0 0 to 5 5 as 0 , 1 , 2 , 3 , 4 , 9 0 , 1 , 2 , 3 , 4 , 9 . Step 3: State Transitions Define the state transitions: State 0 ( 000 000 ) -> State 1 ( 001 001 ) State 1 ( 001 001 ) -> State 2 ( 010 010 ) State 2 ( 010 010 ) -> State 3 ( 011 011 ) State 3 ( 011 011 ) -> State 4 ( 100 100 ) State 4 ( 100 100 ) -> State 5 ( 101 101 ) State 5 ( 101 101 ) -> State 0 ( 000 000 ) Unused states: State 6 ( 110 110 ) -> State 0 ( 000 000 ) State 7 ( 111 111 ) -> State 0 ( 000 000 ) Explanation: State Transitions : Define transitions between states in a circular sequence. Step 4: State Diagram Let's draw the state diagram: Explanation: Here is the complete state diagram for the six-state skip counter. The diagram shows the transitions between the states 0 , 1 , 2 , 3 , 4 , 9 0 , 1 , 2 , 3 , 4 , 9 in black, forming a circular sequence. The unused states 6 6 and 7 7 transition back to the initial state 0 0 with red dashed lines. Step 5: Karnaugh Maps Create Karnaugh maps for the JK Flip-Flops based on the state transitions. Flip-Flop A (Least Significant Bit) Present State Next State J K 000 001 1 0 001 010 0 X 010 011 1 0 011 100 0 X 100 101 1 0 101 000 0 1 110 000 0 1 111 000 0 1 K-map for 𝐽 𝐴 J A ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | 1 | 0 | X | X | ------------------------- 01 | X | X | X | X | ------------------------- 11 | X | X | X | X | ------------------------- 10 | X | X | X | X | ------------------------- K-map for 𝐾 𝐴 K A ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | 0 | X | X | X | ------------------------- 01 | X | X | X | X | ------------------------- 11 | 1 | X | X | X | ------------------------- 10 | 1 | X | X | X | ------------------------- Flip-Flop B (Middle Bit) Present State Next State J K 000 001 0 X 001 010 1 0 010 011 0 X 011 100 1 0 100 101 0 X 101 000 0 1 110 000 0 1 111 000 0 1 K-map for 𝐽 𝐵 J B ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | 0 | 1 | X | X | ------------------------- 01 | 1 | X | X | X | ------------------------- 11 | X | X | X | X | ------------------------- 10 | X | X | X | X | ------------------------- K-map for 𝐾 𝐵 K B ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | X | 0 | X | X | ------------------------- 01 | 0 | X | X | X | ------------------------- 11 | X | X | X | X | ------------------------- 10 | 1 | X | X | X | ------------------------- Flip-Flop C (Most Significant Bit) Present State Next State J K 000 001 0 X 001 010 0 X 010 011 0 X 011 100 1 0 100 101 0 X 101 000 0 1 110 000 0 1 111 000 0 1 K-map for 𝐽 𝐶 J C ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | 0 | 0 | X | X | ------------------------- 01 | 0 | X | X | X | ------------------------- 11 | X | X | X | X | ------------------------- 10 | 1 | X | X | X | ------------------------- K-map for 𝐾 𝐶 K C ​ : | 00 | 01 | 11 | 10 | ------------------------- 00 | X | X | X | X | ------------------------- 01 | X | X | X | X | ------------------------- 11 | X | X | X | X | ------------------------- 10 | 1 | X | X | X | ------------------------- Explanation: Karnaugh Maps : Create Karnaugh maps for the JK flip-flops based on state transitions. Step 6: Simplify Boolean Expressions From the Karnaugh maps, we can derive the Boolean expressions for each flip-flop. For Flip-Flop A (J and K inputs): 𝐽 𝐴 = 𝑄 𝐴 ′ J A ​ = Q A ′ ​ 𝐾 𝐴 = 𝑄 𝐴 K A ​ = Q A ​ For Flip-Flop B (J and K inputs): 𝐽 𝐵 = 𝑄 𝐴 ⋅ 𝑄 𝐵 ′ J B ​ = Q A ​ ⋅ Q B ′ ​ 𝐾 𝐵 = 𝑄 𝐵 ⋅ ( 𝑄 𝐴 + 𝑄 𝐵 ′ ) K B ​ = Q B ​ ⋅ ( Q A ​ + Q B ′ ​ ) For Flip-Flop C (J and K inputs): 𝐽 𝐶 = 𝑄 𝐴 ⋅ 𝑄 𝐵 ⋅ 𝑄 𝐶 ′ J C ​ = Q A ​ ⋅ Q B ​ ⋅ Q C ′ ​ 𝐾 𝐶 = 𝑄 𝐶 ⋅ ( 𝑄 𝐴 ′ + 𝑄 𝐵 ′ ) K C ​ = Q C ​ ⋅ ( Q A ′ ​ + Q B ′ ​ ) Explanation: Simplify Boolean Expressions : Derive and simplify Boolean expressions from Karnaugh maps for JK flip-flop inputs. Step 7: Implement in Multisim Place JK Flip-Flops on the schematic. Wire the inputs according to the simplified Boolean expressions. Verify the design by running the simulation to ensure the counter follows the correct sequence of states. Explanation: Implement in Multisim : Place JK flip-flops in Multisim, wire them according to the Boolean expressions, and verify the design through simulation. Final Answer: Extract Unique Digits : From the birthday 01 / 29 / 1994 01/29/1994 , the unique digits use

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