2079

CSC116 · TU past paper

Digital Logic 2079 question paper

The complete TU 2079 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 T flip flops.[10]

    Critical issue: The actual state diagram image is NOT provided in the question. No states, transitions, inputs, or outputs can be read from the prompt. Missing data: - Number of states - State labels and encoding - State transitions (pre...

  2. 210 marksNumericalMultiplexersAnswer

    Implement $F = \Sigma(1,4,5,7)$ using a. Multiplexer b. Decoder c. PLA [10]

    • Function: $F(A,B,C) = \Sigma(1,4,5,7)$ - 3 variables: A (MSB), B, C (LSB) - Minterms where F = 1: 1, 4, 5, 7 A B C Minterm F --------------------- 0 0 0 m0 0 0 0 1 m1 1 0 1 0 m2 0 0 1 1 m3 0 1 0 0 m4 1 1 0 1 m5 1 1 1 0 m6 0 1 1 1 m7 1 ...
  3. 310 marksNumericalSynchronous CountersAnswer

    Differentiate between combinational circuit and sequential circuit. Design mod-10 synchronous counter using J-K flip-flop. Show necessary truth tables and K-maps.[10]

    • Counter type: Mod-10 (decade / BCD) synchronous counter - Flip-flop type: J-K flip-flop - Count sequence: $0 \to 9$ then back to $0$ - Number of flip-flops: $2^4 = 16 \ge 10 \Rightarrow 4$ flip-flops - States $1010$ to $1111$ (10-15) a...
  4. 45 marksShift registersAnswer

    Provide one example where shift left operator can be used. Explain serial-in-serial-out register. [5]

    --- The shift left operation is commonly used for fast multiplication by powers of 2. Example: Let the binary number stored in a register be: After applying one shift left operation: After applying two shift lefts: Observation: Operation...

  5. 55 marksNumericalcomplimentsAnswer

    Carry out the following tasks:a. Perform 2's complement subtraction of 010110 - 100101b. Represent decimal number 10 in its BCD form [5]

    • Part a: Subtraction $010110 - 100101$ (both 6-bit binary numbers) - Part b: Represent decimal 10 in BCD --- - Minuend $A = 010110$ - Subtrahend $B = 100101$ 1's complement (invert bits): $$100101 \rightarrow 011010$$ Add 1: $$011010 + ...
  6. 65 marksNumericalAddersAnswer

    Derive the Boolean expression for sum and carry of half adder. Draw its combinational circuit. Implement it using only NOR gates. [5]

    Half Adder: Derivation, Circuit, and NOR Implementation

    STEP 1 - Given Data

    • Inputs: two single-bit binary values, $A$ and $B$
    • Outputs: Sum $S$, Carry $C$
    • Constraint: implement using only NOR gates

    All required information is present in the question.


    STEP 2 - Solution

    Truth Table

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

    Boolean Expressions

    Sum: $S = 1$ for the 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 both inputs are 1:

    $$C = A \cdot B$$

    Combinational Circuit (XOR + AND)

    A ──┬───────────┐
        │          [XOR]───── S = A ⊕ B
    B ──┼───────────┘
        │
        ├───────────┐
        │          [AND]───── C = A·B
    B ──┘───────────┘
    

    NOR-Only Implementation

    Basic NOR identities:

    • $\bar{X} = X \text{ NOR } X$
    • $X + Y = (X \text{ NOR } Y) \text{ NOR } (X \text{ NOR } Y)$
    • $X \cdot Y = (X \text{ NOR } X) \text{ NOR } (Y \text{ NOR } Y)$

    Carry $C = A \cdot B$. Using AND-from-NOR:

    $$C = (A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B) = \overline{\bar A + \bar B} = A\cdot B$$

    Sum $S = A \oplus B$. The standard 5-gate NOR realization of XOR:

    GateInputsOutput
    G1$A$ NOR $B$$\overline{A+B}$
    G2$A$ NOR G1$\overline{A + \overline{A+B}} = \bar A(A+B) = \bar A B$
    G3$B$ NOR G1$\overline{B + \overline{A+B}} = \bar B(A+B) = A\bar B$
    G4G2 NOR G3$\overline{\bar A B + A\bar B} = \overline{A\oplus B}$
    G5G4 NOR G4$A \oplus B = S$

    Verification of G4/G5: $\text{G4} = \overline{S}$, so $\text{G5}=\overline{\overline S}=S$. Correct.

    Circuit diagram (NOR gates only):

    A ─┬───────────[G1]────┬───────[G2]───┐
       │      B ────┘       │  A ───┘      │
       │                    └──[G3]────┐   ├─[G4]──[G5]── S
       │              B ───────────────┘   │  (G4 fed twice)
       │
       ├─[Ga: A NOR A = A']──┐
       │                     ├─[Gc]── C
       └─────────────────────┘
           B ─[Gb: B NOR B = B']──┘
    

    Gate count:

    • Sum: 5 NOR gates (G1-G5)
    • Carry: 3 NOR gates (Ga = $\bar A$, Gb = $\bar B$, Gc = final NOR)
    • G1 is shared, but counting independently: Sum = 5, Carry = 3.

    Summary

    OutputExpressionNOR gates
    Sum$S = A \oplus B$5
    Carry$C = A \cdot B$3

    A wiring trap worth knowing: one tempting XOR construction does not work. Taking:

    • G4 = G1 NOR G2 = $\overline{\overline{A+B} + \bar A}$. Evaluate: $= (A+B)\cdot A = A$, not $A\bar B$.
    • G5 = G1 NOR G3 = $(A+B)\cdot B = B$, not $\bar A B$.
    • G6 = G4 NOR G5 = $A$ NOR $B = \overline{A+B}$, which is NOT $A\oplus B$.

    So those gate assignments do not yield the XOR. The correct minimal XOR uses 5 NOR gates with the wiring shown above, giving $S = A\oplus B$ and $C = A\cdot B$.

  7. 75 marksNumericalBoolean FunctionsAnswer

    Express F′=(x+y+z)′F' = (x + y + z)'F′=(x+y+z)′ as a sum of min-terms. [5]

    • Function: $F' = (x + y + z)'$ - Variables: $x, y, z$ (3 variables, so $2^3 = 8$ possible combinations) Evaluate $F = (x + y + z)$ and its complement $F' = (x+y+z)'$ for all 8 rows. The sum $(x+y+z) = 0$ only when all variables are 0; o...
  8. 85 marksNumericalFour variable mapsAnswer

    Minimize the following Boolean function using K-map $F(A, B, C, D) = \Sigma m(0,1,2,5,7,8,9,10,13,15)$. [5]

    Minterms with value 1: 0, 1, 2, 5, 7, 8, 9, 10, 13, 15 Rows = AB, Columns = CD (Gray code order): Verify 1s: m0✓ m1✓ m2✓ m5✓ m7✓ m8✓ m9✓ m10✓ m13✓ m15✓ (10 ones) Group 1 (Quad): m{0,1,8,9} - AB = 00 and 10 (A varies), B = 0; CD = 00 and ...

  9. 95 marksRipple CountersAnswer

    What are the practical implications of down counter? Explain BCD ripple counter. [5]

    --- A down counter is a counter that counts in decreasing order (e.g., from 15 to 0 for a 4-bit counter). Its practical implications include: 1. Timer/Countdown Applications: Used in digital timers, microwave ovens, and bomb-timer circui...

  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 even number. [5]

    • Number of inputs: 3 (let them be $A$ = MSB, $B$, $C$ = LSB) - Number of outputs: 1 (call it $F$) - Condition: $F = 1$ when the binary value of inputs is an even number No missing data. The problem is fully specified. --- Three input bi...
  11. 115 marksRead-only-MemoryAnswer

    Differentiate between ROM and PLA. Explain different types of ROM. [5]

    --- Feature ROM (Read Only Memory) PLA (Programmable Logic Array) --------- Full Form Read Only Memory Programmable Logic Array Structure Fixed AND array + Programmable OR array Programmable AND array + Programmable OR array AND Plane Fi...

  12. 125 marksIntegrated CircuitsAnswer

    Write short notes on (Any two): a. Positive Logic b. I2L c. ASCII [5]

    Short Notes (Any Two): a. Positive Logic and c. ASCII


    a. Positive Logic

    Positive Logic is a convention used in digital systems to assign binary logic values (0 and 1) to the two voltage levels present in a circuit.

    Definition:

    In positive logic, the higher voltage level is assigned the logic value 1 (HIGH) and the lower voltage level is assigned the logic value 0 (LOW).

    Voltage LevelLogic Value
    High Voltage (e.g., +5V)Logic 1
    Low Voltage (e.g., 0V)Logic 0

    Key Points:

    • It is the most commonly used logic convention in digital circuits and computers.
    • Logic gates (AND, OR, NOT, etc.) are designed and analyzed based on this convention.
    • The opposite convention is called Negative Logic, where the lower voltage level represents logic 1 and the higher voltage level represents logic 0.
    • The same physical circuit can perform different logical operations depending on whether positive or negative logic convention is applied.

    Example:

    • In a TTL (Transistor-Transistor Logic) circuit, +5V = Logic 1 and 0V = Logic 0 under positive logic convention.

    c. ASCII

    ASCII stands for American Standard Code for Information Interchange.

    Definition:

    ASCII is a standard character encoding scheme used to represent text and control characters in computers and communication devices using binary numbers.

    Key Features (from notes):

    • ASCII uses a 7-bit code, consisting of:
      • 3 zone bits (higher order bits)
      • 4 numeric bits (lower order bits)
    • Total possible combinations: 2^7 = 128 characters

    Character Representation:

    CategoryNumber of Characters
    Alphabets (uppercase A-Z and lowercase a-z)62
    Numerals (0-9)10
    Special Characters66

    Example:

    CharacterASCII (Decimal)ASCII (Binary)
    A65100 0001
    a97110 0001
    048011 0000

    Additional Notes:

    • ASCII is considered an improvement over the BCD code as it can represent a much wider range of characters.
    • It is widely used in data communication, keyboards, and text files.
    • An extended version (ASCII-8) uses 8 bits, allowing 256 characters, which includes additional special and graphical symbols.
    • Another extended code, EBCDIC (Extended Binary Coded Decimal Interchange Code), is used mainly in IBM mainframe computers.