Digital Logic
1. **Problem Statement:**
We will answer each sub-question related to number systems, binary operations, digital logic, and Boolean algebra.
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1.1 **Symbols of the octal number system:**
The octal system uses digits from 0 to 7.
**Answer:** $\{0,1,2,3,4,5,6,7\}$
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1.2 **Decimal value of a binary 1 in the second position to the right of the radix comma:**
The radix comma separates integer and fractional parts.
The first position right of the radix is $2^{-1} = \frac{1}{2}$.
The second position right of the radix is $2^{-2} = \frac{1}{4} = 0.25$.
**Answer:** $0.25$
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1.3 **Biggest symbol in the hexadecimal number system:**
Hexadecimal digits are $0-9$ and $A-F$ where $A=10$ and $F=15$.
The biggest symbol is $F$.
**Answer:** $F$
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1.4 **Two's complement of $10110101_2$:**
Step 1: Invert all bits:
$10110101 \to 01001010$
Step 2: Add 1:
$$01001010 + 1 = 01001011$$
**Answer:** $01001011_2$
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1.5 **Which binary counter uses output from previous flip-flop to clock the next?**
- Synchronous counters use a common clock.
- Asynchronous counters use output from previous flip-flop as clock for next.
**Answer:** Asynchronous binary counter
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1.6 **Number of select lines for a 4:1 multiplexer:**
Number of select lines $= \log_2(4) = 2$
**Answer:** 2
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1.7 **Main difference between full adder and half adder:**
- Half adder adds two bits, outputs sum and carry.
- Full adder adds three bits (including carry-in), outputs sum and carry-out.
**Answer:** Full adder handles carry-in; half adder does not.
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1.8 **Small scale integration (SSI):**
SSI refers to integrated circuits with a small number of gates (typically less than 100).
Used for simple logic functions.
**Answer:** Integration of a few logic gates on a single chip.
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1.9 **Simplify Boolean expressions:**
1.9.1 $(A + \overline{A}) + \overline{B}B$
- $A + \overline{A} = 1$
- $\overline{B}B = 0$
- So expression $= 1 + 0 = 1$
1.9.2 $AC̅ + 1 + A$
- Anything OR 1 is 1
- So expression $= 1$
1.9.3 $BCD + B + BC̅D̅$
- $B + BCD = B$
- $B + BC̅D̅ = B$
- So expression $= B$
1.9.4 $AB + AB̅\overline{A}$
- Note $\overline{A}A = 0$
- So $AB + AB̅\overline{A} = AB + 0 = AB$
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1.10 **Simplify Boolean expressions from Karnaugh maps:**
1.10.1 Map values:
- Rows CD: 00,01,11,10
- Columns AB: 00,01,11,10
- Values:
- Row 00: 1 1 1 1
- Row 01: 1 0 0 1
- Row 11: 1 0 0 1
- Row 10: 1 1 1 1
**Grouping:**
- Entire first and last rows are 1s (rows 00 and 10)
- First and last columns are 1s in all rows
- Group 1: All 1s in rows 00 and 10 (covers all columns)
- Group 2: First column (AB=00) all 1s
- Group 3: Last column (AB=10) all 1s
**Simplified expression:**
- Group 1: $\overline{C}\overline{D} + C\overline{D} = \overline{D}$
- Group 2: $\overline{A}\overline{B}$
- Group 3: $A\overline{B}$
Combine:
$$\overline{D} + \overline{A}\overline{B} + A\overline{B} = \overline{D} + \overline{B}(\overline{A} + A) = \overline{D} + \overline{B}$$
1.10.2 Map values:
- Row 00: 1 0 0 1
- Row 01: 1 0 0 1
- Row 11: 1 1 1 1
- Row 10: 1 0 0 1
**Grouping:**
- Row 11 all 1s
- First and last columns mostly 1s
- Group 1: Row 11 (C D = 11) all columns
- Group 2: First column (AB=00) all rows
- Group 3: Last column (AB=10) all rows
Simplified expression:
- Row 11: $CD$
- First column: $\overline{A}\overline{B}$
- Last column: $A\overline{B}$
Combine:
$$CD + \overline{B}(\overline{A} + A) = CD + \overline{B}$$
1.10.3 Map values:
- Row 00: 1 ∅ ∅ ∅
- Row 01: 1 0 1 1
- Row 11: ∅ 0 ∅ ∅
- Row 10: 1 1 1 1
**Grouping:**
- Row 10 all 1s
- Column 00 mostly 1s
- Group 1: Row 10 (C D = 10) all columns
- Group 2: Column 00 (AB=00) rows 00,01,10
Simplified expression:
- Row 10: $C\overline{D}$
- Column 00: $\overline{A}\overline{B}$
Combine:
$$C\overline{D} + \overline{A}\overline{B}$$
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**Final answers:**
1.1: $\{0,1,2,3,4,5,6,7\}$
1.2: $0.25$
1.3: $F$
1.4: $01001011_2$
1.5: Asynchronous binary counter
1.6: 2
1.7: Full adder handles carry-in; half adder does not
1.8: Integration of a few logic gates on a single chip
1.9.1: 1
1.9.2: 1
1.9.3: $B$
1.9.4: $AB$
1.10.1: $\overline{D} + \overline{B}$
1.10.2: $CD + \overline{B}$
1.10.3: $C\overline{D} + \overline{A}\overline{B}$