2078

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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  1. 110 marksNumericalDesign with state equations and state reduAnswer

    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...

  2. 210 marksNumericalMultiplexersAnswer

    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...
  3. 310 marksNumericalSynchronous CountersAnswer

    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 ...
  4. 45 marksShift registersAnswer

    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...

  5. 55 marksNumericalcomplimentsAnswer

    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:
  6. 65 marksNumericalAddersAnswer

    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

    ABSum (S)Carry (C)
    0000
    0110
    1010
    1101

    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:

    GateOperationOutput
    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]--- C
    

    Total: 5 NAND gates (G1 shared between Sum and Carry paths).

    Summary

    OutputExpressionNAND 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.

  7. 75 marksNumericalproduct of sum simplificationAnswer

    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...

  8. 85 marksNumericalFour variable mapsAnswer

    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...
  9. 95 marksRipple CountersAnswer

    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...

  10. 105 marksNumericalDesign ProcedureAnswer

    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...
  11. 115 marksMagnitude ComparatorAnswer

    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...

  12. 125 marksBinary logicAnswer

    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 LevelPositive LogicNegative Logic
    High (H)10
    Low (L)01

    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:

    FeatureDescription
    Power ConsumptionVery low (static power near zero)
    Noise MarginHigh
    Fan-outHigh
    SpeedModerate to high
    IntegrationVery 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:

    FeatureDetail
    Bit Length8 bits per character
    Total Characters2^8 = 256 possible characters
    Developed byIBM
    Used inIBM 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:

    CharacterEBCDIC Code (Hex)EBCDIC Code (Binary)
    AC11100 0001
    BC21100 0010
    0F01111 0000
    1F11111 0001

    Comparison with ASCII:

    FeatureASCIIEBCDIC
    Bits7/8 bits8 bits
    DeveloperANSIIBM
    UsageUniversal (PCs, Internet)IBM mainframes
    Characters128/256256

    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.