conversions-cheatsheet

📑 Master Conversions Cheatsheet: Digital Codes & Canonical Forms

Purpose

This cheatsheet serves as a high-yield, quick-reference guide for all digital code conversions, error control systems, and algebraic canonical transformations. It is designed to be fully compatible with Obsidian reading and editing views.

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1. Binary Coded Decimal (8421 BCD) Conversions

1.1 Decimal $\leftrightarrow$ BCD

• Decimal to BCD: Directly replace each individual decimal digit with its equivalent 4-bit binary nibble [3].
• BCD to Decimal: Partition the binary string into 4-bit groups (starting from the LSB) and translate each group into its decimal digit [3].

Decimal $\leftrightarrow$ BCD Conversion

• $185_{10}⟹\underset{1}{\underbrace{0001}}\underset{8}{\underbrace{1000}}\underset{5}{\underbrace{0101}}⟹(000110000101{)}_{\text{BCD}}$ [3]

1.2 BCD Addition Algorithm

1. Add the two BCD values using standard binary addition [4].
2. For each 4-bit sum nibble:
   • If Sum $\le 9$ and no carry is generated: The nibble is valid [4].
   • If Sum $>9$ OR a carry is generated: Add a correction factor of $6_{10}=0110_2$ to that nibble [4]. This skips the 6 invalid states ($1010$ to $1111$) [4, 8]. Propagate any resulting carry to the next higher-order nibble [4].

BCD Addition (Correction Required)

Add $6_{10}$ and $7_{10}$: $\begin{array}{rl}0110& (6_{10})\\ +0111& (7_{10})\\ 1101& (\text{Sum }=13_{10}>9⟹\text{Invalid})\\ +0110& (\text{Correction Factor})\\ 10011& (\text{Carry pushed to next nibble})\end{array}$ Final BCD: $00010011_{\text{BCD}}=13_{10}$ [5].

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2. Excess-3 (XS-3) Code Conversions

2.1 Decimal $\leftrightarrow$ Excess-3

• Decimal to Excess-3: Add $3_{10}$ to the decimal digit $N$, then convert the result into a 4-bit binary representation [17].
• Excess-3 to Decimal: Convert the 4-bit group into decimal and subtract $3$ [13, 17].

Decimal $\leftrightarrow$ Excess-3 Conversion

• $4_{10}⟹4+3=7_{10}⟹0111_{\text{XS-3}}$ [17].

2.2 Excess-3 Addition Algorithm

1. Perform standard binary addition on the two Excess-3 digits. Since both numbers contain an offset of $+3$, the raw sum contains an offset of $+6$ [19].
2. For each sum nibble:
   • If a carry is generated (Sum $>9$): Add $3_{10}=0011_2$ to restore the correct $+3$ offset [19].
   • If NO carry is generated (Sum $\le 9$): Subtract $3_{10}=0011_2$ (or add $1101_2$) [19].

Excess-3 Addition (No Carry)

Add $2$ and $3$ in Excess-3: $\begin{array}{rl}0101& (2_{\text{XS-3}})\\ +0110& (3_{\text{XS-3}})\\ 1011& (\text{Raw Binary Sum }=11_{10}⟹\text{No Carry})\\ -0011& (\text{Subtract 3})\\ 1000& (\text{Corrected Sum }=8_{\text{XS-3}}⟹5_{10})\end{array}$ [20].

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3. Weighted 4-Bit Code Conversions

Weighted codes represent a decimal value $D$ by multiplying each bit $d_i$ by its assigned positional weight $w_i$ [7]: $D={\sum }_{i=0}^3{d}_i{w}_i=d_3{w}_3+d_2{w}_2+d_1{w}_1+d_0{w}_0[7]$

3.1 Major Weighted Codes Reference Table

Decimal Digit	8421 (BCD) [7]	2421 [7]	5211 [7]	84-2-1 [7]
0	0000	0000	0000	0000
1	0001	0001	0001	0111
2	0010	0010	0011	0110
3	0011	0011	0100	0101
4	0100	0100	0101	0100
5	0101	1011	1000	1011
6	0110	1100	1001	1010
7	0111	1101	1011	1001
8	1000	1110	1100	1000
9	1001	1111	1111	1111

3.2 Conversion Example: $(605{)}_{10}$ to 5211 Weighted Code

To convert, translate each decimal digit independently using the weighted positional rules:

1. Digit 6: $5(1)+2(0)+1(0)+1(1)=6⟹1001_2$ [11].
2. Digit 0: $5(0)+2(0)+1(0)+1(0)=0⟹0000_2$ [11].
3. Digit 5: $5(1)+2(0)+1(0)+1(0)=5⟹1000_2$ [11].

• Result: $(605{)}_{10}=(100100001000{)}_{5211}$ [11, 12].

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4. Reflected (Gray) Code Conversions

4.1 Binary $\to$ Gray

1. Keep the Most Significant Bit (MSB) identical [31]: $G_{MSB}=B_{MSB}$
2. XOR each subsequent binary bit with the binary bit immediately to its left [31]: $G_i=B_i\oplus B_{i+1}(\text{for }i=N-2,\dots ,0)$

Binary:  B3 -------> B2 -------> B1 -------> B0
         |           |           |           |
         |           XOR         XOR         XOR
         v           v           v           v
Gray:    G3          G2          G1          G0

Binary to Gray Conversion (1011 to Gray)

• $G_3=B_3=1$
• $G_2=B_3\oplus B_2=1\oplus 0=1$
• $G_1=B_2\oplus B_1=0\oplus 1=1$
• $G_0=B_1\oplus B_0=1\oplus 1=0$
• Result: $1110_{\text{Gray}}$ [32].

4.2 Gray $\to$ Binary

1. Keep the MSB identical [32]: $B_{MSB}=G_{MSB}$
2. XOR the previously calculated binary bit with the next incoming Gray code bit [32]: $B_i=B_{i+1}\oplus G_i(\text{for }i=N-2,\dots ,0)$

Gray to Binary Conversion (1110 to Binary)

• $B_3=G_3=1$
• $B_2=B_3\oplus G_2=1\oplus 1=0$
• $B_1=B_2\oplus G_1=0\oplus 1=1$
• $B_0=B_1\oplus G_0=1\oplus 0=1$
• Result: $1011_2$ [33].

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7. Standard $\to$ Canonical Boolean Conversions

• Canonical Form: Every term contains all domain variables exactly once [67].
• SOP Canonical Expansion Rule: Multiply each non-canonical product term by $(x+x^{\prime })=1$ for every missing variable $x$ [74].
• POS Canonical Expansion Rule: Add $(x\cdot x^{\prime })=0$ to each non-canonical sum term for every missing variable $x$, then expand via distribution: $A+BC=(A+B)(A+C)$ [74].

SOP & POS Canonical Conversion

Convert $F(A,B,C,D)=(A+B^{\prime }+C)(B+D^{\prime })$ into Maxterms ($\Pi M$) and Minterms ($\Sigma m$):

1. Term 1 $(A+B^{\prime }+C)$ is missing $D$: $(A+B^{\prime }+C+D{D}^{\prime })=(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })$ $\text{Binary: }0100_2=M_4,0101_2=M_5[75]$
2. Term 2 $(B+D^{\prime })$ is missing $A$ and $C$: Expand $B+D^{\prime }+A{A}^{\prime }+C{C}^{\prime }$ using distribution into 4 maxterms:
   • $(A+B+C+D^{\prime })\to 0001_2=M_1$
   • $(A+B+C^{\prime }+D^{\prime })\to 0011_2=M_3$
   • $(A^{\prime }+B+C+D^{\prime })\to 1001_2=M_9$
   • $(A^{\prime }+B+C^{\prime }+D^{\prime })\to 1011_2=M_{11}[75]$

• Product of Maxterms: $F(A,B,C,D)=\prod M(1,3,4,5,9,11)$ [76].
• Sum of Minterms: Minterm indices are the remaining states (0 to 15): $F(A,B,C,D)=\sum m(0,2,6,7,8,10,12,13,14,15)[76]$