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Chapter 2

Digital Logic

TU BIM past exam questions

Past questions and answers

22 questions set from this chapter. Most repeated first.

  • 2023 AD Model · 1 mark

What do you mean by digital computer?

Answer

A digital computer is an electronic machine that processes data represented in discrete form, using the binary digits 0 and 1. It accepts input, processes it according to stored instructions, and produces output, performing arithmetic and logical operations with high speed and accuracy. Personal computers and laptops are examples.

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  • 2023 AD Model · 1 mark

Classify the binary number.

Answer

A binary number is a number written in base 2 using only the digits 0 and 1. It can be classified as:

  • Integer (whole) binary number, such as 101121011_2, with no binary point.
  • Fractional binary number, such as 0.10120.101_2, with only digits after the binary point.
  • Mixed binary number, such as 101.112101.11_2, with digits on both sides of the point.

Binary numbers can also be unsigned (magnitude only) or signed (sign-magnitude, 1's complement or 2's complement).

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  • 2023 AD Model · 1 mark

What do you mean by 2's complement

Answer

The 2's complement of a binary number is obtained by inverting every bit (the 1's complement) and then adding 1 to the result. It is used to represent negative numbers and to perform subtraction by addition.

Example: for 010120101_2 (5), the 1's complement is 10101010, and adding 1 gives 10111011, which represents −5-5 in 4 bits.

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  • 2023 AD Model · 1 mark

NAND gate is universal gate. Justify

Answer

A gate is universal if the NOT, AND and OR operations can all be built from it alone. NAND can do this:

  • NOT: tie both inputs together, A⋅A‾=Aˉ\overline{A\cdot A}=\bar A.
  • AND: a NAND followed by a NAND used as an inverter, A⋅B‾‾=A⋅B\overline{\overline{A\cdot B}}=A\cdot B.
  • OR: invert each input with a NAND, then NAND the results, Aˉ⋅Bˉ‾=A+B\overline{\bar A\cdot\bar B}=A+B (De Morgan's law).

Since every Boolean function can be written with AND, OR and NOT, NAND gates alone can implement any logic circuit.

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  • 2023 AD Model · 1 mark

What is Boolean algebra?

Answer

Boolean algebra is the algebra of logic in which variables take only two values, 0 and 1 (false and true). It uses the operations AND (⋅\cdot), OR (++) and NOT ( ˉ\bar{\ }), and it is the mathematical basis for analysing and designing digital circuits.

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  • 2023 AD Model · 1 mark

What do you mean by SOP?

Answer

SOP (Sum of Products) is a form of Boolean expression in which several product (AND) terms are ORed together. For example, F=AB+AˉC+BCF = AB + \bar A C + BC. When every product term contains all the variables, it is the canonical SOP, written as a sum of minterms, F=∑m(1,3,7)F=\sum m(1,3,7).

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  • 2023 AD Model · 1 mark

Define PLA.

Answer

A PLA (Programmable Logic Array) is a programmable logic device with a programmable AND array followed by a programmable OR array. Any set of Boolean functions in SOP form can be implemented by choosing which inputs form each product term and which product terms feed each output.

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  • 2023 AD Model · 1 mark

What do you mean by flip flop?

Answer

A flip-flop is a basic sequential circuit with two stable states. It stores one bit of information and keeps its output until a clock or trigger pulse changes it. It is the building block of registers, counters and memory. Examples are the SR, JK, D and T flip-flops.

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  • 2023 AD Model · 1 mark

Define registers.

Answer

A register is a group of flip-flops, one per bit, used to store a binary word (an nn-bit register needs nn flip-flops). Registers may be used simply for storage or, with shift capability, for shifting data left or right (shift registers).

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  • 2023 AD Model · 1 mark

What do you mean by ripple counter?

Answer

A ripple counter is an asynchronous counter in which only the first flip-flop receives the external clock. The output of each flip-flop clocks the next one, so the change "ripples" through the stages. An nn-bit ripple counter counts through 2n2^n states.

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  • 2023 AD Model · 3 marks

Describe the octal and hexadecimal numbers.

Answer

Octal number system. It has base 8 and uses the digits 0 to 7. Each position carries a weight that is a power of 8, so (345)8=3×82+4×8+5=22910(345)_8 = 3\times8^2+4\times8+5 = 229_{10}. Since 8=238=2^3, one octal digit equals exactly 3 binary bits, so (101 110)2=(56)8(101\,110)_2=(56)_8.

Hexadecimal number system. It has base 16 and uses the digits 0 to 9 and the letters A to F (A=10, B=11, C=12, D=13, E=14, F=15). Each weight is a power of 16, so (2F)16=2×16+15=4710(2F)_{16}=2\times16+15=47_{10}. Since 16=2416=2^4, one hex digit equals 4 binary bits, so (1011 1100)2=(BC)16(1011\,1100)_2=(BC)_{16}.

Both systems give a compact way to write long binary numbers and are widely used in computing, for example for memory addresses and machine codes.

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  • 2023 AD Model · 3 marks

Discuss the basic theorems and properties of Boolean algebra.

Answer

Basic postulates, theorems and properties of Boolean algebra:

PropertyAND formOR form
IdentityA⋅1=AA\cdot1=AA+0=AA+0=A
Null (dominance)A⋅0=0A\cdot0=0A+1=1A+1=1
IdempotentA⋅A=AA\cdot A=AA+A=AA+A=A
ComplementA⋅Aˉ=0A\cdot\bar A=0A+Aˉ=1A+\bar A=1
InvolutionAˉ‾=A\overline{\bar A}=A
CommutativeAB=BAAB=BAA+B=B+AA+B=B+A
Associative(AB)C=A(BC)(AB)C=A(BC)(A+B)+C=A+(B+C)(A+B)+C=A+(B+C)
DistributiveA(B+C)=AB+ACA(B+C)=AB+ACA+BC=(A+B)(A+C)A+BC=(A+B)(A+C)
AbsorptionA(A+B)=AA(A+B)=AA+AB=AA+AB=A
De MorganAB‾=Aˉ+Bˉ\overline{AB}=\bar A+\bar BA+B‾=Aˉ⋅Bˉ\overline{A+B}=\bar A\cdot\bar B

Other useful results are A+AˉB=A+BA+\bar AB=A+B and the consensus theorem AB+AˉC+BC=AB+AˉCAB+\bar AC+BC=AB+\bar AC.

Duality. If every AND is swapped with OR, and every 0 with 1, in a valid identity, the result is also a valid identity. This is why each law above comes as a pair.

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  • 2023 AD Model · 3 marks

Explain the don't care condition with example.

Answer

A don't-care condition is an input combination that never occurs, or whose output does not matter, so the function value for it may be taken as either 0 or 1. It is marked X (or d) in the truth table and K-map. Each X can be treated as 1 if that gives a larger group, and as 0 otherwise, which gives a simpler expression.

Example. A BCD digit detector gives F=1F=1 for the digits 5 to 9 on inputs ABCDABCD. The input combinations 1010 to 1111 never occur in BCD, so they are don't-cares:

F=∑m(5,6,7,8,9)+d(10,11,12,13,14,15)F=\sum m(5,6,7,8,9)+d(10,11,12,13,14,15)

Using the don't-cares in the K-map, the cells 8 to 15 form one group giving AA, the cells 5, 7, 13, 15 give BDBD, and the cells 6, 7, 14, 15 give BCBC. Therefore

F=A+BD+BCF = A + BD + BC

Without the don't-cares the expression would need more literals, for example ABˉCˉ+AˉBD+AˉBCA\bar B\bar C+\bar A B D+\bar A B C.

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  • 2023 AD Model · 3 marks

Design full adder with truth table.

Answer

A full adder adds three 1-bit inputs, AA, BB and the carry-in CinC_{in}, and gives a sum SS and a carry-out CoutC_{out}.

Truth table

ABCinC_{in}SCoutC_{out}
00000
00110
01010
01101
10010
10101
11001
11111

Expressions (from the K-maps)

S=A⊕B⊕CinS=A\oplus B\oplus C_{in}

Cout=AB+Cin(A⊕B)=AB+BCin+ACinC_{out}=AB+C_{in}(A\oplus B)=AB+BC_{in}+AC_{in}

Circuit. Two half adders and an OR gate: the first half adder adds AA and BB to give S1=A⊕BS_1=A\oplus B and C1=ABC_1=AB. The second adds S1S_1 and CinC_{in} to give SS and C2=Cin(A⊕B)C_2=C_{in}(A\oplus B). Finally Cout=C1+C2C_{out}=C_1+C_2.

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  • 2023 AD Model · 3 marks

Differentiate between multiplexers and demultiplexers

Answer

Multiplexer (MUX)Demultiplexer (DEMUX)
Many inputs, one outputOne input, many outputs
Selects one of 2n2^n inputs using nn select linesSends the single input to one of 2n2^n outputs using nn select lines
Also called a data selectorAlso called a data distributor
Has 2n2^n data inputs, nn selects, 1 outputHas 1 data input, nn selects, 2n2^n outputs
Used in data routing, parallel-to-serial conversion, function implementationUsed in serial-to-parallel conversion and data distribution
Example: 4:1 MUXExample: 1:4 DEMUX

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  • 2023 AD Model · 3 marks

Explain D-flip-flop.

Answer

The D (data or delay) flip-flop has a single data input DD and a clock input. It is made from an SR flip-flop by connecting S=DS=D and R=DˉR=\bar D through an inverter, which removes the invalid state. On the active clock edge, the output QQ takes the value of DD.

Characteristic equation: Qnext=DQ_{next}=D.

DQnextQ_{next}Action
00Reset
11Set

The output follows the input one clock period later, so it is used as a delay element, for storing one bit, and in registers and data latches.

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  • 2023 AD Model · 5 marks

Design a half subtractors logic circuit using only NOR gate.

Answer

A half subtractor computes A−BA-B with two outputs, the difference DD and the borrow BoutB_{out}:

D=A⊕B,Bout=AˉBD = A\oplus B,\qquad B_{out}=\bar A B

ABDBoutB_{out}
0000
0111
1010
1100

Implementation with NOR gates only. Five NOR gates are used:

  1. X=A+B‾X=\overline{A+B}
  2. P=A+X‾=Aˉ (A+B)=AˉBP=\overline{A+X}=\bar A\,(A+B)=\bar AB, which is the borrow BoutB_{out}
  3. Q=B+X‾=Bˉ (A+B)=ABˉQ=\overline{B+X}=\bar B\,(A+B)=A\bar B
  4. Y=P+Q‾Y=\overline{P+Q}
  5. D=Y+Y‾=P+Q=AˉB+ABˉ=A⊕BD=\overline{Y+Y}=P+Q=\bar AB+A\bar B=A\oplus B

The last gate has its two inputs tied together and acts as an inverter. Thus D=A⊕BD=A\oplus B and Bout=AˉBB_{out}=\bar AB are both obtained using NOR gates alone.

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  • 2023 AD Model · 5 marks

Convert the following decimal numbers into hexadecimal and octal number.
(a) 305 (b) 225

Answer

(a) 305

Hexadecimal, dividing by 16:

  • 305÷16=19305 \div 16 = 19 remainder 11
  • 19÷16=119 \div 16 = 1 remainder 33
  • 1÷16=01 \div 16 = 0 remainder 11

Reading the remainders upwards, 30510=(131)16305_{10}=(131)_{16}.

Octal, dividing by 8:

  • 305÷8=38305 \div 8 = 38 remainder 11
  • 38÷8=438 \div 8 = 4 remainder 66
  • 4÷8=04 \div 8 = 0 remainder 44

So 30510=(461)8305_{10}=(461)_8.

(b) 225

Hexadecimal:

  • 225÷16=14225 \div 16 = 14 remainder 11
  • 14÷16=014 \div 16 = 0 remainder 1414 (E)

So 22510=(E1)16225_{10}=(E1)_{16}.

Octal:

  • 225÷8=28225 \div 8 = 28 remainder 11
  • 28÷8=328 \div 8 = 3 remainder 44
  • 3÷8=03 \div 8 = 0 remainder 33

So 22510=(341)8225_{10}=(341)_8.

Check: 3×64+4×8+1=2253\times64+4\times8+1=225 and 14×16+1=22514\times16+1=225.

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  • 2023 AD Model · 5 marks

Describe the three-variable K-map with example.

Answer

A K-map (Karnaugh map) is a graphical method of simplifying Boolean functions. A three-variable map has 23=82^3=8 cells, one for each minterm. It is drawn as 2 rows by 4 columns (or 4 by 2). The columns are labelled in Gray code, 00, 01, 11, 10, so that adjacent cells differ in only one variable.

BCBC=00011110
A=0A=0m0m_0m1m_1m3m_3m2m_2
A=1A=1m4m_4m5m_5m7m_7m6m_6

Rules of grouping. Group adjacent 1s in powers of two (1, 2, 4, 8). The groups can wrap around the edges, and may overlap. Use the largest possible groups. A group of 2k2^k cells eliminates kk variables.

Example. F(A,B,C)=∑m(1,3,5,7)F(A,B,C)=\sum m(1,3,5,7).

00011110
A=0A=00110
A=1A=10110

The four 1s form one group (columns 01 and 11). In this group AA and BB change and C=1C=1 stays constant, so

F=CF=C

Another example is F=∑m(0,2,4,6)F=\sum m(0,2,4,6), which gives F=CˉF=\bar C.

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  • 2023 AD Model · 5 marks

Explain the triggering of flip-flop with example.

Answer

Triggering is the way a clock signal makes a flip-flop change state. There are two main types.

1. Level triggering. The flip-flop responds while the clock stays at a particular level (high or low). During that time the output follows changes in the input. This can cause race-around or multiple changes within one pulse, as in a simple JK latch with J=K=1J=K=1.

2. Edge triggering. The flip-flop changes state only at the instant of a clock transition, so the input is sampled once per pulse.

  • Positive (rising) edge triggered: changes on the 0 to 1 transition.
  • Negative (falling) edge triggered: changes on the 1 to 0 transition. In symbols this is shown by a bubble at the clock input.

Example. A positive edge triggered D flip-flop with D=1D=1 sets Q=1Q=1 only at the rising edge. If DD changes to 0 while the clock is high, QQ stays 1 until the next rising edge, when it takes the new value. Master-slave flip-flops also avoid race-around.

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  • 2023 AD Model · 10 marks

What do you mean by ripple counter? Design the 4-bit ripple counter with timing diagram.

Answer

Ripple counter. A ripple counter is an asynchronous counter in which the clock pulse is applied only to the first flip-flop. The output of each flip-flop drives the clock of the next, so the change ripples through the stages. An nn-bit counter has 2n2^n states, and the propagation delays add up, which limits the speed.

4-bit binary (up) ripple counter.

Use four JK flip-flops with J=K=1J=K=1 (or T flip-flops with T=1T=1), so each toggles when its clock input falls (negative-edge triggered).

  • The external clock goes to FF0FF_0, which gives Q0Q_0 (LSB).
  • Q0Q_0 clocks FF1FF_1, Q1Q_1 clocks FF2FF_2, and Q2Q_2 clocks FF3FF_3, which gives Q3Q_3 (MSB).

Each stage divides the frequency by 2, so Q0=f/2Q_0=f/2, Q1=f/4Q_1=f/4, Q2=f/8Q_2=f/8 and Q3=f/16Q_3=f/16. It is a MOD-16 counter counting from 0000 to 1111, then returning to 0000.

Count sequence and timing.

Clock pulseQ3Q2Q1Q0Q_3Q_2Q_1Q_0
00000
10001
20010
30011
40100
50101
60110
70111
81000
91001
101010
111011
121100
131101
141110
151111
160000

Timing diagram (each column is one clock pulse, 1 means high):

CLK : _|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|_|‾|
Q0  : 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1   toggles on every clock fall
Q1  : 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1   toggles when Q0 falls
Q2  : 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1   toggles when Q1 falls
Q3  : 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1   toggles when Q2 falls

Each waveform has half the frequency of the one before it. A down counter is obtained by clocking each stage from the Qˉ\bar Q output of the previous stage.

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  • 2023 AD Model · 10 marks

Differentiate between sequential and combinational logic. Explain the steps to design of combinational logic.

Answer

Difference between sequential and combinational logic

Combinational logicSequential logic
Output depends only on the present inputsOutput depends on the present inputs and the past state
No memory elementHas memory (flip-flops)
Usually no clock neededUsually clocked
Faster and simplerSlower and more complex
Examples: adder, MUX, decoder, encoderExamples: registers, counters, flip-flops

Steps to design a combinational circuit

  1. Specification. State the problem and decide the number of inputs and outputs.
  2. Assign symbols to the input and output variables.
  3. Truth table. Write the output for every combination of inputs.
  4. Boolean expression. Write each output as a sum of minterms (or product of maxterms) from the truth table.
  5. Simplify using Boolean algebra or a K-map (using don't-cares where available).
  6. Draw the logic diagram with AND, OR, NOT gates, or NAND/NOR only if required.
  7. Verify the circuit by checking it against the truth table.

Example. For a half adder, the inputs are A,BA,B and the outputs are S=A⊕BS=A\oplus B and C=ABC=AB, which are implemented with an XOR gate and an AND gate.

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