missed pyq across all chapters

Here are the complete, step-by-step mathematical solutions for all 6 missing edge-case puzzles.

I have formatted them exactly using your [!example] callout blocks so you can literally copy and paste them straight into the bottom of their respective Chapter files in your Obsidian vault.

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📋 To add to: 02 Chapter 2 - Digital Codes.md

PYQ Master Puzzle (2022): The 12-Bit Register Decoder Question: The state of a 12-bit register is 100010010111. What is its content if it represents: (i) Three decimal digits in BCD? (ii) Three decimal digits in the excess-3 code? (iii) Three decimal digits in the 8-4-2-1 code? (iv) A binary number?

Solution Methodology: First, divide the 12-bit string into three 4-bit nibbles: 1000, 1001, 0111.

(i) As BCD (Standard 8421): Evaluate each 4-bit block directly into decimal.

• 1000 = 8
• 1001 = 9
• 0111 = 7
• Answer: 897

(ii) As Excess-3 Code: Excess-3 is generated by adding 3 (0011) to the decimal digit. To decode it, we must subtract 3 from each nibble.

• 1000 (8) - 3 = 5
• 1001 (9) - 3 = 6
• 0111 (7) - 3 = 4
• Answer: 564

(iii) As 8, 4, -2, -1 Code: Apply the specific mathematical weights to each bit in the nibble: $(8\times B_3)+(4\times B_2)-(2\times B_1)-(1\times B_0)$.

• 1000 $\to 8(1)+4(0)-2(0)-1(0)$ = 8
• 1001 $\to 8(1)+4(0)-2(0)-1(1)$ = 7
• 0111 $\to 8(0)+4(1)-2(1)-1(1)\to 4-2-1$ = 1
• Answer: 871

(iv) As a pure Binary Number: Evaluate the entire 12-bit string as a single base-2 integer.

• $100010010111_2=(1\times 2^{11})+(1\times 2^7)+(1\times 2^4)+(1\times 2^2)+(1\times 2^1)+(1\times 2^0)$
• $=2048+128+16+4+2+1$
• Answer: 2199

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📋 To add to: 03 Chapter 3 - Boolean Algebra.md

PYQ Master Proof (2018): Duality & Deep Complementation Question: What is duality principle? Find the complement of the following Boolean function and reduce this complement function to a minimum number of literals: $F=[(AB{)}^{\prime }A][(AB{)}^{\prime }B]$

Part A: The Duality Principle The Duality Principle states that every valid Boolean algebraic expression remains valid if we mathematically interchange all AND ($\cdot$) operators with OR ($+$) operators, and simultaneously interchange all 0s with 1s, while keeping the variables themselves completely unchanged (do not complement the literals).

Part B: Simplification and Complementation Before blindly applying De Morgan’s theorem to the whole massive equation, it is mathematically safer to simplify the original function $F$ first. Original Function: $F=[(AB{)}^{\prime }A]\cdot [(AB{)}^{\prime }B]$ Step 1: Apply De Morgan’s to the inner terms. $F=[(A^{\prime }+B^{\prime })A]\cdot [(A^{\prime }+B^{\prime })B]$ Step 2: Apply the Distributive Law. $F=[A{A}^{\prime }+A{B}^{\prime }]\cdot [A^{\prime }B+B{B}^{\prime }]$ Step 3: Apply the Null Law ($X{X}^{\prime }=0$). $F=[0+A{B}^{\prime }]\cdot [A^{\prime }B+0]$ $F=(A{B}^{\prime })\cdot (A^{\prime }B)$ Step 4: Rearrange (Commutative Law). $F=A{A}^{\prime }B{B}^{\prime }$ $F=0$

The Final Complement: Since the original function perfectly reduces to absolute $0$, its complement $F^{\prime }$ is simply: $F^{\prime }=1$ The minimum number of literals is zero (it requires no variables, just a constant HIGH wire).

PYQ Master Problem (2016): POS Canonical Extraction Question: Given the function $f(A,B,C)=(A+B+C^{\prime })(A+B^{\prime }+C^{\prime })(A^{\prime }+B+C^{\prime })(A^{\prime }+B^{\prime }+C^{\prime })$, construct the truth table and express the function in maxterm and minterm forms.

Step 1: Identify the Maxterms The function is given in strict Canonical Product of Sums form. A maxterm evaluates to 0 when its uncomplemented variables are 0 and its complemented variables are 1.

• $(A+B+C^{\prime })\to 001\to {\mathbf{M}}_1$
• $(A+B^{\prime }+C^{\prime })\to 011\to {\mathbf{M}}_3$
• $(A^{\prime }+B+C^{\prime })\to 101\to {\mathbf{M}}_5$
• $(A^{\prime }+B^{\prime }+C^{\prime })\to 111\to {\mathbf{M}}_7$ Maxterm Form: $f(A,B,C)=\Pi M(1,3,5,7)$

Step 2: Identify the Minterms The minterms are simply all the remaining decimal states not covered by the maxterms. Since there are 3 variables, the domain is 0 to 7. Minterm Form: $f(A,B,C)=\Sigma m(0,2,4,6)$

Step 3: Construct the Truth Table Write the 8 states. Place a 0 at rows 1, 3, 5, and 7. Place a 1 everywhere else. | A B C | f(A,B,C) | | :---: | :---: | | 0 0 0 | 1 | | 0 0 1 | 0 | | 0 1 0 | 1 | | 0 1 1 | 0 | | 1 0 0 | 1 | | 1 0 1 | 0 | | 1 1 0 | 1 | | 1 1 1 | 0 |

Class Test 01: SOP vs POS Gate Optimization Question: Given $F=\Pi M(0,2,5,7,8,10,13,15)$ a) Convert into canonical SOP. b) Which implementation (SOP or POS) requires fewer logic gates?

Part A: Canonical SOP The minterms are the exact inverse of the given maxterm list for a 4-variable system (0-15).

• Minterms = $1,3,4,6,9,11,12,14$
• Canonical SOP = $\Sigma m(1,3,4,6,9,11,12,14)$

Part B: Gate Optimization Justification To determine which implementation requires fewer gates, we must simplify both using a K-map.

• SOP Simplification: Plotting the 1s yields a group of four ($B^{\prime }D$) and a second group of four ($B{D}^{\prime }$). $F_{SOP}=B^{\prime }D+B{D}^{\prime }=\mathbf{B}\oplus \mathbf{D}$.
• POS Simplification: Plotting the 0s yields $F^{\prime }=B^{\prime }{D}^{\prime }+BD=(B\oplus D{)}^{\prime }$. Complementing to find F gives $F_{POS}=(B+D)(B^{\prime }+D^{\prime })=\mathbf{B}\oplus \mathbf{D}$.
• Conclusion: Both simplified expressions mathematically reduce to the exact same logic (an XOR gate, or 2 ANDs + 1 OR). Therefore, both SOP and POS require the exact same number of logic gates.

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📋 To add to: 04 Chapter 4 - Universal Gates.md

PYQ Master Design (2015): The 2-NOR Gate Limit Puzzle Question: Implement $F=A^{\prime }{B}^{\prime }{C}^{\prime }+A{B}^{\prime }D+A^{\prime }{B}^{\prime }C{D}^{\prime }$, with don't cares $d=ABC+A{B}^{\prime }{D}^{\prime }$, using no more than Two NOR gates. Assume normal and complement inputs are available.

Step 1: Expand into Minterms We must translate the raw algebra into decimal minterms to safely plot a K-map.

• $A^{\prime }{B}^{\prime }{C}^{\prime }\to 000X\to 0,1$
• $A{B}^{\prime }D\to 10X1\to 9,11$
• $A^{\prime }{B}^{\prime }C{D}^{\prime }\to 0010\to 2$
• $F_{1s}=\Sigma m(0,1,2,9,11)$
• $ABC\to 111X\to 14,15$
• $A{B}^{\prime }{D}^{\prime }\to 10X0\to 8,10$
• $Do{n}^{\prime }tCares=\Sigma m(8,10,14,15)$

Step 2: Group the Zeros ($F^{\prime }$) for NOR Logic Because a NOR gate generates inverted OR logic $(X+Y{)}^{\prime }$, we purposefully simplify for $F^{\prime }$ (the zeros on the K-map) to easily map it to the hardware. The zeros are the remaining cells: $3,4,5,6,7,12,13$.

• Grouping $(4,5,6,7)$ with $(12,13,14X,15X)$ creates an 8-cell block: $B$
• Grouping $(3)$ with $(7)$ creates a pair: $A^{\prime }CD$
• Therefore, the simplified complement is: $F^{\prime }=B+A^{\prime }CD$

Step 3: The 2-NOR Implementation Since $F^{\prime }=B+A^{\prime }CD$, double complementing gives us $F$: $F=(B+A^{\prime }CD{)}^{\prime }$

• NOR Gate 1: We need to generate the $A^{\prime }CD$ term using a NOR gate. Since inputs can be pre-inverted, feed $A,C^{\prime },$ and $D^{\prime }$ into the first NOR gate. $Outpu{t}_1=(A+C^{\prime }+D^{\prime }{)}^{\prime }$ Apply De Morgan’s: $Outpu{t}_1=A^{\prime }CD$
• NOR Gate 2: Feed $B$ and $Outpu{t}_1$ into the second NOR gate. $Outpu{t}_2=(B+Outpu{t}_1{)}^{\prime }$ $Outpu{t}_2=(B+A^{\prime }CD{)}^{\prime }$

Proof Complete: We successfully implemented the complex function using exactly two NOR gates!

PYQ Note (2023): Circuit Analysis to Single Gate Note: While the 2023 exam provided a specific visual circuit (Fig 2c) to reduce to a single gate, the universal methodology for this question type is Mano's Intermediate Variable method.

1. Assign variables ($T_1,T_2$) to the outputs of the first level of gates.
2. Cascade the equations forward until you derive the final Boolean equation.
3. Use Boolean algebra to simplify. If a 10-mark question asks you to “reduce to a single gate,” the algebraic mess will always factor down into either an XOR $(A\oplus B=A^{\prime }B+A{B}^{\prime })$ or an XNOR $(A\odot B=AB+A^{\prime }{B}^{\prime })$ equation!