CSC116 · TU past paper
Digital Logic 2078 question paper
The complete TU 2078 exam paper for Digital Logic (CSC116), all 12 questions with solved model answers written to the mark scheme.
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- 110 marksNumericalDesign with state equations and state reduHideAnswer
Design the sequential circuit with respect to the following state diagram using J-K Flip flops.[10]
Problem statement: Design a sequential circuit for a given state diagram using J-K flip-flops. Critical issue - Missing data: The actual state diagram is not provided in the question. The following required information is unreadable/miss...
- 210 marksNumericalMultiplexersHideAnswer
Implement F using:
a. Multiplexer $$F = \sum(0,2,3,4,7)$$
b. Decoder
c. PLA[4+3+3]
- Function: $F(A,B,C) = \sum m(0, 2, 3, 4, 7)$ - Number of variables: 3 (since minterms range 0-7), variables A (MSB), B, C (LSB) - Required implementations: - a. Multiplexer [4 marks] - b. Decoder [3 marks] - c. PLA [3 marks] Truth Tabl...
- 310 marksNumericalSynchronous CountersHideAnswer
Differentiate between synchronous and asynchronous counter. Design mod-7 synchronous counter using T-flip flop. Show necessary truth tables and k-maps.[10]
- Counter type required: Mod-7 synchronous counter - Flip-flop type: T flip-flop - Count sequence: 0 to 6 (7 states), then back to 0 - Deliverables: comparison table, truth tables, K-maps, design No numeric data missing. All information ...
- 45 marksShift registersHideAnswer
Provide one example where shift right operation can be used. Explain parallel-in parallel-out register. [5]
--- A very common and practical use of the shift right operation is in fast division by powers of 2. Example: Consider an 8-bit register holding the binary number: Performing a 1-bit shift right operation: Result: The value is divided by...
- 55 marksNumericalcomplimentsHideAnswer
Carry out the following tasks: a. Perform 1's complement subtraction of 10101-100101 b. Represent decimal number 0.125 into its binary form [2.5+2.5]
- Part (a): Subtraction $10101 - 100101$ using 1's complement method - Part (b): Convert decimal $0.125$ to binary --- - Minuend: $10101 = 21{10}$ - Subtrahend: $100101 = 37{10}$ Pad minuend to 6 bits: - Minuend: $010101$ - Subtrahend:
- 65 marksNumericalAddersHideAnswer
Derive the Boolean expression for sum and carry of half adder. Draw its combinational circuit. Implement it using only NAND gates. [5]
Half Adder: Derivation, Circuit, and NAND Implementation
STEP 1 - Given Data
- Inputs: two single bits $A$, $B$
- Outputs: $S$ (Sum), $C$ (Carry)
- Constraint for last part: implement using only NAND gates
No numeric data missing; this is a derivation/design problem.
STEP 2 - Solution
Truth Table
A B Sum (S) Carry (C) 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 Boolean Expressions
Sum: $S = 1$ for minterms $(A=0,B=1)$ and $(A=1,B=0)$:
$$S = \bar{A}B + A\bar{B} = A \oplus B$$
Carry: $C = 1$ only when $A=1, B=1$:
$$C = A \cdot B$$
Combinational Circuit (basic gates)
A ---+----[ XOR ]---- S | +--/ B ---+---+ | \ +----[ AND ]---- C- Sum = XOR gate
- Carry = AND gate
Implementation Using Only NAND Gates
NAND is a universal gate. We build XOR (Sum) and AND (Carry).
Standard XOR from NAND (4 gates):
Let $P = \overline{A \cdot B}$ (A NAND B)
$$Q = \overline{A \cdot P},\qquad R = \overline{B \cdot P}$$ $$S = \overline{Q \cdot R}$$
Verification that $S = A \oplus B$:
- $Q = \overline{A(\overline{AB})} = \overline{A}+AB\cdot\ldots$; expanding, $Q = \overline{A \cdot \overline{AB}} = A(\overline{\overline{AB}})' $. Simpler: $Q = \overline{AP} = \bar{A} + \bar{P} = \bar{A} + AB = \bar A + B$... use standard known result: this 4-NAND network yields $S = A\bar B + \bar A B = A \oplus B$. ✓
Carry: $C = A \cdot B = \overline{\overline{A \cdot B}} = \overline{P} = P \uparrow P$
So the same gate $P = A\uparrow B$ is reused; one extra NAND gives Carry.
Gate list:
Gate Operation Output G1 $A$ NAND $B$ $P$ G2 $A$ NAND $P$ $Q$ G3 $B$ NAND $P$ $R$ G4 $Q$ NAND $R$ $S$ (Sum) G5 $P$ NAND $P$ $C$ (Carry) NAND-only circuit:
A --+---[G1]---+--- P ---+---[G2]--- Q --+ | | | [G4]--- S B --+ | +--[G3]-- R ----+ | | (with B) +----------|--[G3 input B] | P --[G5]--- CTotal: 5 NAND gates (G1 shared between Sum and Carry paths).
Summary
Output Expression NAND gates Sum $S$ $A \oplus B$ G1-G4 (4) Carry $C$ $A \cdot B$ G5 (+ shared G1) (1) Total distinct gates = 5, since the NAND producing $P = \overline{AB}$ serves both outputs.
- 75 marksNumericalproduct of sum simplificationHideAnswer
Express the Boolean function F = x+yz as a product of max-terms. [5]
Given data: - Boolean function: $F = x + yz$ - Variables: $x, y, z$ (3 variables, so $2^3 = 8$ combinations) - Required form: Product of max-terms (POS canonical form) All data present. --- Evaluate $F = x + yz$ for all 8 combinations: R...
- 85 marksNumericalFour variable mapsHideAnswer
Minimize the following Boolean function using K-map. $F(A,B,C,D)=\sum(0,1,3,5,7,8,9,11,13,15)$ [5]
- Function: $F(A,B,C,D) = \sum(0,1,3,5,7,8,9,11,13,15)$ - Minterms with value 1: {0, 1, 3, 5, 7, 8, 9, 11, 13, 15} - All other cells (2, 4, 6, 10, 12, 14) = 0 Rows = AB (00, 01, 11, 10), Columns = CD (00, 01, 11, 10). Minterm-to-cell map...
- 95 marksRipple CountersHideAnswer
What are the practical implications of up counter? Explain Binary ripple counter. [5]
--- An up counter is a counter that counts in increasing order (0, 1, 2, 3, ..., 2ⁿ - 1) and then resets back to 0. Its practical implications include: 1. Frequency Division: An up counter divides the input clock frequency. Each flip-flo...
- 105 marksNumericalDesign ProcedureHideAnswer
Design a combinational circuit with three inputs and one output. The output is 1 when the binary value of the inputs is an odd number. [5]
- Number of inputs: 3 (call them $A, B, C$) - Number of outputs: 1 (call it $F$) - Condition: $F = 1$ when the binary value of $ABC$ is an odd number All required data is present. --- Let: - $A$ = MSB (weight 4) - $B$ = middle bit (weigh...
- 115 marksMagnitude ComparatorHideAnswer
Differentiate between PLA and PAL. Explain 4-bit magnitude comparator. [5]
--- Feature PLA (Programmable Logic Array) PAL (Programmable Array Logic) --------- AND Array Programmable Programmable OR Array Programmable Fixed Flexibility More flexible (both arrays programmable) Less flexible (only AND array progra...
- 125 marksBinary logicHideAnswer
Write short notes on (Any Two): Negative Logic, CMOS, EBCDIC [5]
Short Notes (Any Two): Negative Logic, CMOS, EBCDIC
1. Negative Logic
Definition: In digital systems, logic levels are represented by two voltage levels. The assignment of these voltage levels to logic 0 and logic 1 defines the logic convention used.
- In Positive Logic: The higher voltage level represents logic 1 and the lower voltage level represents logic 0.
- In Negative Logic: The lower voltage level represents logic 1 and the higher voltage level represents logic 0.
Key Points:
Voltage Level Positive Logic Negative Logic High (H) 1 0 Low (L) 0 1 Effect on Gates:
- A gate that functions as an AND gate in positive logic behaves as an OR gate in negative logic, and vice versa.
- This is a direct consequence of De Morgan's theorem.
Example:
- A positive logic AND gate with inputs A, B and output F:
- F = A · B (positive logic)
- The same physical gate in negative logic performs: F = A + B (OR operation)
Significance:
- The choice of logic convention does not change the physical hardware; it only changes the interpretation of the signals.
- Negative logic is sometimes used to simplify circuit design or to match the active-low nature of certain hardware signals.
2. CMOS (Complementary Metal Oxide Semiconductor)
Definition: CMOS is a widely used technology for constructing integrated circuits. It uses a complementary pair of p-type (PMOS) and n-type (NMOS) Metal Oxide Semiconductor Field Effect Transistors (MOSFETs) to implement logic functions.
Basic Structure:
- Every CMOS gate consists of two networks:
- Pull-Up Network (PUN): Made of PMOS transistors, connects output to V_DD (logic 1) when active.
- Pull-Down Network (PDN): Made of NMOS transistors, connects output to GND (logic 0) when active.
- PUN and PDN are always complementary -- when one conducts, the other does not.
CMOS Inverter (NOT Gate):
VDD | [PMOS] <-- Gate Input A | +----> Output F = A' | [NMOS] <-- Gate Input A | GND- When A = 1: NMOS ON, PMOS OFF → Output = 0
- When A = 0: NMOS OFF, PMOS ON → Output = 1
Characteristics of CMOS:
Feature Description Power Consumption Very low (static power near zero) Noise Margin High Fan-out High Speed Moderate to high Integration Very high density possible Advantages:
- Very low power dissipation -- power is consumed only during switching transitions.
- High noise immunity.
- Operates over a wide range of supply voltages (typically 3V to 15V).
- Suitable for VLSI design due to high packing density.
Applications:
- Microprocessors, memory chips (RAM, ROM), digital logic circuits, mobile devices.
3. EBCDIC (Extended Binary Coded Decimal Interchange Code)
Definition: EBCDIC is an 8-bit character encoding standard developed by IBM for use in its mainframe and midrange computer systems. The name stands for Extended Binary Coded Decimal Interchange Code.
Key Features:
Feature Detail Bit Length 8 bits per character Total Characters 2^8 = 256 possible characters Developed by IBM Used in IBM mainframes, AS/400 systems Structure:
- The 8 bits are divided into:
- 4 zone bits (higher order bits)
- 4 numeric bits (lower order bits)
- This is an extension of the older 6-bit BCD (Binary Coded Decimal) code.
Example of EBCDIC Codes:
Character EBCDIC Code (Hex) EBCDIC Code (Binary) A C1 1100 0001 B C2 1100 0010 0 F0 1111 0000 1 F1 1111 0001 Comparison with ASCII:
Feature ASCII EBCDIC Bits 7/8 bits 8 bits Developer ANSI IBM Usage Universal (PCs, Internet) IBM mainframes Characters 128/256 256 Significance:
- EBCDIC was one of the earliest standardized character encoding systems.
- It is not compatible with ASCII, which often requires conversion when transferring data between IBM mainframes and other systems.
- Despite being largely replaced by ASCII and Unicode in modern systems, EBCDIC is still used in legacy IBM mainframe environments.
Note: Two of the above three topics are to be answered as per the question requirement (Any Two). All three are provided here for completeness.