Computer Network Questions and Answers

Answer: A

Correct answer: 3

A multiplexer with \(2^n\) data inputs needs \(n\) select lines to address them. Since \(2^3 = 8\), three select lines are required to choose one of the eight inputs.

Answer: C

Correct answer: 3

An encoder performs the reverse of a decoder, turning one active input among \(2^n\) into an \(n\)-bit code. With 8 inputs, \(2^3 = 8\) gives three output lines.

Answer: B

Correct answer: selects one of several inputs and routes it to a single output

A multiplexer, or data selector, has several data inputs, one output, and a set of select lines that decide which input reaches the output. A multiplexer with \( n \) select lines can choose among \( 2^{n} \) inputs, so a 4-to-1 multiplexer needs two select lines and an 8-to-1 needs three. Its inverse is the demultiplexer, described by the second option, which takes one input and routes it to one of several outputs. The two are used together to send many signals over a single shared line and separate them again at the far end - the principle behind time-division multiplexing in communications. A decoder, the third option, converts an \( n \)-bit code into one of \( 2^{n} \) active outputs and is used for address decoding in memory; an encoder does the reverse. Multiplexers can also implement arbitrary logic functions directly, which makes them useful beyond data routing.

Answer: D

Correct answer: XOR

A half adder adds two single bits and produces two outputs. The SUM is generated by an XOR gate and the CARRY by an AND gate: \[ S = A \oplus B, \qquad C = A \cdot B. \] The reason is visible in the arithmetic. 0+0 gives sum 0; 0+1 and 1+0 give sum 1; and 1+1 gives sum 0 with a carry of 1. The sum column is 1 exactly when the inputs differ, which is the definition of XOR, and the carry column is 1 only when both are 1, which is AND. The half adder cannot accept a carry from a previous stage, which is why a full adder is needed for multi-bit addition. A full adder takes three inputs and can be built from two half adders and an OR gate. Chaining full adders gives a ripple-carry adder, simple but slow because each stage waits for the carry from the one below; carry-lookahead adders exist to remove that delay.

Answer: C

Correct answer: 16

Hexadecimal is base 16, using the digits 0 to 9 followed by the letters A to F for the values 10 to 15. It is used because each hexadecimal digit corresponds to exactly four bits, so binary can be written compactly and converted by inspection rather than by arithmetic. The byte 11010110 splits into 1101 and 0110, which are D and 6, giving D6. Octal, base 8, works the same way with three bits per digit, and was common in the era of 12-, 24- and 36-bit machines; hexadecimal took over once 8-, 16- and 32-bit words became standard, since those divide evenly by four. Hexadecimal is written with a prefix to avoid ambiguity: 0x in C and its descendants, or a trailing H in assembly. It appears throughout computing - memory addresses, colour codes such as #FF0000 for red, and MAC addresses are all hexadecimal.

Answer: B

Correct answer: the time between an input change and the corresponding output change

Propagation delay is the time taken for a change at the input to appear at the output. It is measured between the 50 per cent points of the two transitions, and it is usually quoted separately for a low-to-high and a high-to-low transition, since the two are rarely equal. It is the fundamental limit on clock speed. The longest chain of gates between two flip-flops - the critical path - must settle within one clock period, so the maximum frequency is the reciprocal of that total delay plus the flip-flop’s own setup time. It is also why a ripple counter is slower than a synchronous one: each stage waits for the one below it, and the delays accumulate. The fourth option describes fan-out, the number of gate inputs a single output can drive, which is a different specification and is itself limited by current and by the loading capacitance that increases the delay.

Answer: A

Correct answer: successive values differ in exactly one bit

Gray code is an ordering of binary values in which each differs from the next in exactly one bit. For three bits it runs 000, 001, 011, 010, 110, 111, 101, 100. That single-bit property solves a real problem. When several bits change at once in ordinary binary - 0111 to 1000, for instance - they will not switch at precisely the same moment, and a reading taken during the transition can be wildly wrong. With Gray code only one bit ever changes, so the worst error is one count. It is therefore used in shaft encoders and position sensors, where a mechanical brush or optical reader may sample mid-transition. It is also why the axes of a Karnaugh map are labelled in Gray code: adjacent cells then differ in one variable, which is what makes grouping valid. Conversion from binary is simple: the first bit is copied, and each later bit is the XOR of the corresponding binary bit with the one before it.

Answer: C

Correct answer: 255

With \( n \) bits there are \( 2^{n} \) distinct patterns, so eight bits give 256 of them - but they represent the values 0 to 255, since one pattern is used for zero. Hence the maximum is \[ 2^{8} - 1 = 255. \] The off-by-one is the whole point of the question, and 256 is the classic wrong answer: it is the count of values, not the largest of them. That is unsigned. If the eight bits hold a signed value in 2’s complement, the range becomes −128 to +127 - which is where the other two distractors come from. The asymmetry arises because 2’s complement has a single representation of zero, leaving one more pattern for the negatives than for the positives. The pattern 11111111 is therefore 255 when read as unsigned and −1 when read as signed. Nothing in the bits themselves says which is meant; it is the instruction operating on them that decides.

Answer: B

Correct answer: 13

Each binary digit carries a place value that is a power of two, read from the right: \( 2^{0}, 2^{1}, 2^{2}, 2^{3} \), that is 1, 2, 4 and 8. For 1101 the set bits are in the 8, 4 and 1 positions: \[ 1\times8 + 1\times4 + 0\times2 + 1\times1 = 8 + 4 + 1 = 13. \] The reverse conversion divides repeatedly by two and reads the remainders upward: 13 gives remainder 1, then 6 gives 0, then 3 gives 1, then 1 gives 1 - reading up, 1101. It is worth memorising the first few powers of two, since almost every question in this area rests on them: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. Note also that 1101 is D in hexadecimal, since one hexadecimal digit covers exactly four bits - which is why hexadecimal is used as shorthand for binary.

Answer: C

Correct answer: stores one bit of information

A flip-flop is a bistable circuit: it has two stable states and remains in whichever it was last put into, so it stores exactly one bit. It is the basic element of all sequential logic and of static memory. The distinction from combinational logic is the essential one. A combinational circuit’s output depends only on its present inputs; a sequential circuit’s output depends on its inputs and on its stored state, which is what gives it memory. The common types are the SR flip-flop, which sets and resets but has a forbidden input combination; the D flip-flop, which simply captures the value on its data input at the clock edge; the JK, which resolves the SR ambiguity by toggling; and the T, which toggles on each clock pulse. Chained together, flip-flops make registers, shift registers and counters. A latch is the level-triggered relative of the edge-triggered flip-flop.