Question 1

Telephony-grade sampling with DPCM: A 4 kHz noiseless channel uses one sample every 125 microseconds (i.e., 8000 samples per second) to transmit digital signals. If Differential PCM encodes each sample using a 4-bit relative value, how many bits per…

Accepted Answer

Introduction / Context: In baseband telephony and many voice codecs, the sampling frequency is typically 8000 samples per second because the voice channel bandwidth is about 4 kHz. When using Differential PCM (DPCM), each sample conveys the change relative to a predictor, often allowing fewer bits per sample than standard PCM. Calculating the resulting bit rate requires multiplying samples per second by bits per sample.Given Data / Assumptions:Channel is treated as noiseless for the purpose of bit rate calculation.Sampling interval is 125 microseconds → 8000 samples/s.Each DPCM codeword carries 4 bits per sample.No framing or overhead is considered (payload bit rate only).Concept / Approach: Bit rate for a sampled system using fixed-length codewords is simply: bit_rate = samples_per_second * bits_per_sample. For DPCM here, the predictor reduces the number of bits needed to 4 per sample compared to 8 in basic PCM, halving the payload rate from 64 Kbps to 32 Kbps.Step-by-Step Solution:Compute samples per second: 1 / (125 * 10^-6) = 8000 samples/s.Multiply by bits per sample: 8000 * 4 = 32000 bits/s.Express in kilobits per second: 32000 bps = 32 Kbps.Verification / Alternative check: Compare with standard PCM: 8000 samples/s * 8 bits/sample = 64 Kbps. Using half the bits per sample (4) yields exactly half the rate, confirming 32 Kbps.Why Other Options Are Wrong:64 Kbps: corresponds to 8 bits/sample PCM, not 4-bit DPCM.8 Kbps: would require 1 bit/sample at 8000 samples/s, not the case here.128 Kbps: double the standard PCM rate; not applicable.None of the above: incorrect because 32 Kbps matches the calculation.Common Pitfalls: Misreading ‘‘every 125’’ as milliseconds rather than microseconds; forgetting to convert units or to multiply by the correct samples per second.Final Answer: 32 Kbps.

Question 2

X.121 addressing (NUA): within a Network User Address, which digit positions are reserved for the user’s sub-address digits?

Accepted Answer

Introduction / Context: Public packet-switched data networks that followed CCITT/ITU-T X.25 used X.121 addressing. A Network User Address (NUA) contains a country/network identifier segment and a national terminal number segment, with certain trailing digits sometimes designated for user sub-addressing purposes (for example, to reach a specific application mailbox or port on the destination system).Given Data / Assumptions:We are selecting digit positions used as user sub-address digits.The NUA length may extend up to 14 digits in common practice.Early digits identify the network (DNIC), later digits identify the user endpoint and sub-address.Concept / Approach: In a typical X.121 NUA structure, digits 1–4 are the Data Network Identification Code (DNIC). Subsequent digits form the national terminal number. The last two digits (positions 13–14 when a 14-digit NUA is used) are often treated as user-controlled sub-address digits, enabling further selection at the subscriber’s site.Step-by-Step Solution:Recognize DNIC occupies positions 1–4 → not sub-address.Identify that trailing digits are available for user-level sub-addressing.Select positions 13–14 as the sub-address digit pair.Verification / Alternative check: Vendor documentation and historical NUA plans frequently describe the final two digits as user-selected sub-addresses, distinct from the carrier-assigned network and subscriber portions.Why Other Options Are Wrong:5–7 or 8–12: these typically belong to the national terminal numbering plan, not expressly reserved for sub-addressing.1–4: DNIC (country+network), not user sub-address.None of the above: incorrect because the trailing digits 13–14 serve the role.Common Pitfalls: Assuming a shorter NUA length and forgetting the special role of the final digits; confusing national numbering with sub-address space.Final Answer: 13-14.

Question 3

PC-to-mainframe connectivity: the software that allows a personal computer to behave like a character or block-mode terminal (for example, VT100 or 3270) is called:

Accepted Answer

Introduction / Context: Before pervasive web clients and thin-client agents, users commonly accessed host systems through terminal protocols. To connect using a PC, software was needed that replicated the behavior and control sequences of dedicated terminals so the host believed it was talking to the expected device type.Given Data / Assumptions:Goal: make a PC act like a particular terminal model and protocol.Use cases include mainframes, minicomputers, and network devices.Connectivity may occur over serial lines, Telnet, or SSH.Concept / Approach: Terminal emulation software interprets and generates the same control codes, screen layouts, and key mappings as the target terminal (e.g., ANSI/VT100, IBM 3270/5250). It provides the user interface and protocol handling so legacy hosts function correctly without physical terminals.Step-by-Step Solution:Match requirement (PC behaves as a terminal) to the term of art: terminal emulation.Exclude unrelated functions such as dialing, message forums, or signal conversion hardware.Select ‘‘terminal emulation’’ as the correct choice.Verification / Alternative check: Common programs such as PuTTY (VT100/VT220 emulation) or tn3270 clients demonstrate this concept in modern environments.Why Other Options Are Wrong:auto-dialing: call placement feature, not terminal behavior.bulletin board: a hosted service, not emulation software.modem: hardware for modulation/demodulation, not UI/protocol emulation.None of the above: incorrect, because terminal emulation fits precisely.Common Pitfalls: Confusing transport (modems/TCP) with presentation (terminal type and control sequences).Final Answer: terminal emulation.

Question 4

Error detection concepts: which description best defines an error-detecting code in data communications?

Accepted Answer

Introduction / Context: Communication systems add redundancy to detect (and sometimes correct) errors introduced by noise, interference, or synchronization problems. Understanding the general definition of an error-detecting code helps categorize specific techniques such as parity, checksums, and cyclic redundancy checks (CRCs).Given Data / Assumptions:We need a general definition, not a single technique.Examples of techniques include parity (1-bit), checksum (summation), and CRC (polynomial division).One option describes a measurement metric (error rate), not a code.Concept / Approach: An error-detecting code imposes structured redundancy so that only a subset of bit patterns are valid. If transmission errors occur, the received pattern will usually violate the structure, enabling detection. This conceptual description is broader than any single method and therefore is the best definition among the choices presented.Step-by-Step Solution:Identify the option that defines the principle of error detection rather than one instance or a metric.Note that parity (option b) and checksum (option a) are particular examples, not the general definition.Recognize that option d is the bit/block error rate, not a code.Select the general definition (option c).Verification / Alternative check: CRCs, for example, accept only patterns whose polynomial division leaves a specific remainder; any other remainder signals an error, matching the ‘‘rules of construction’’ idea in the correct option.Why Other Options Are Wrong:Option a: describes a checksum method, not the general concept.Option b: describes parity, a single-bit error detector.Option d: defines error rate, not an error-detecting code.Option e: incorrect because a correct general definition is provided.Common Pitfalls: Treating one technique as the definition; confusing performance metrics with coding methods.Final Answer: a code constrained by rules so invalid patterns flag errors (option c).

Question 5

IPv4 basics: what is the standard loopback address used to test the local TCP/IP stack on a host?

Accepted Answer

Introduction / Context: The loopback address range enables a device to send packets to itself for testing without any physical network interface. Pinging the loopback confirms that the local TCP/IP stack is initialized and functional, independent of cabling or external routers.Given Data / Assumptions:We are dealing with IPv4 addressing.Loopback testing uses a reserved address block.Common diagnostic commands such as ping rely on this address.Concept / Approach: The canonical loopback host address is 127.0.0.1, which belongs to the entire 127.0.0.0/8 block reserved for loopback. Any address in that block loops back, but 127.0.0.1 is universally recognized and used in examples and tools.Step-by-Step Solution:Recall that 127.0.0.0/8 is the loopback range.Identify 127.0.0.1 as the standard single-address reference.Select 127.0.0.1 as the correct answer.Verification / Alternative check: Running ‘‘ping 127.0.0.1’’ on most systems tests IP stack operation even with the network adapter disconnected.Why Other Options Are Wrong:255.0.0.0 and 255.255.0.0: netmasks, not loopback host addresses.255.255.255.255.0: not a valid IPv4 address format.None of the above: incorrect because 127.0.0.1 is correct.Common Pitfalls: Confusing netmask notation with host addresses; thinking only 127.0.0.1 loops back (actually the whole 127/8 range does).Final Answer: 127.0.0.1.

Question 6

Web protocols: which well-known TCP/UDP port number is assigned to HTTP (Hypertext Transfer Protocol) for standard web traffic?

Accepted Answer

Introduction / Context: Application protocols are commonly associated with specific port numbers to simplify configuration and interoperability. Knowing the default port for HTTP helps with firewall rules, URL construction, and basic troubleshooting of web services.Given Data / Assumptions:We refer to the default, well-known port (not alternative ports).HTTP is distinct from HTTPS (which uses a different port by default).We assume standard IANA assignments.Concept / Approach: The default port for HTTP is 80/TCP. While HTTP can technically run on other ports, port 80 is the conventional and well-known assignment. (For comparison, HTTPS uses 443/TCP; Telnet uses 23/TCP.)Step-by-Step Solution:Recall the standard port mappings: HTTP → 80.Match to the list of choices and select 80.Note that option 23 corresponds to Telnet, not HTTP.Verification / Alternative check: Accessing ‘‘http://host/’’ implicitly connects to port 80 unless a different port is specified.Why Other Options Are Wrong:99, 86: not the well-known HTTP ports.23: Telnet's port, not HTTP.None of the above: incorrect because 80 is correct.Common Pitfalls: Confusing HTTP (80) with HTTPS (443); assuming the presence of a port in a URL implies HTTPS automatically.Final Answer: 80.

Question 7

Networking definition: most local area and wide area networks are interconnected with many others to form a single global internetwork. What is the correct term for this worldwide system?

Accepted Answer

Introduction / Context: Today's LANs and WANs do not exist in isolation; they are typically connected through routers and public carriers into a massive, interoperable system using the Internet Protocol suite. The question asks for the formal name of that global internetwork.Given Data / Assumptions:We are naming the global system, not a technology or social trend.The system uses common protocols and routing to interconnect many networks.The proper noun should be selected.Concept / Approach: The correct term is ‘‘The Internet’’—a proper noun referring to the worldwide public internetwork. Other options either are not recognized phrases or describe related but different ideas (e.g., ‘‘Internet technologies’’ refers to tools and standards, not the network itself).Step-by-Step Solution:Identify which option names the global network.Discard options that are descriptions, slogans, or imprecise phrases.Select ‘‘The Internet’’ as the best fit.Verification / Alternative check: Standards organizations (IETF, ICANN) and textbooks universally use ‘‘the Internet’’ to designate the global internetwork of networks.Why Other Options Are Wrong:Internet works: not a standard term.Internet revolution: describes societal/industry change.Internet technologies: refers to protocols and tools, not the network itself.None of the above: incorrect because ‘‘The Internet’’ is correct.Common Pitfalls: Confusing ‘‘internet’’ (generic internetwork) with ‘‘the Internet’’ (the global public one).Final Answer: The Internet.

Question 8

IPv4 subnetting design: for the private Class B network 172.16.0.0, you want each subnet to support approximately 600 usable hosts. Which single subnet mask should you assign?

Accepted Answer

Introduction / Context: When choosing a subnet mask, two requirements must be met: the number of hosts per subnet and the number of subnets available. Here, we focus on host capacity per subnet for 172.16.0.0 (a Class B private block with default /16). We must select a mask that provides at least ~600 usable addresses per subnet.Given Data / Assumptions:Base network: 172.16.0.0/16.Target: ≈600 usable hosts per subnet.Usable hosts per subnet = 2^(host_bits) - 2.Concept / Approach: Borrow s bits from the /16 host space, leaving (16 - s) host bits. The number of usable addresses is 2^(16 - s) - 2. We need this to be ≥ 600. Compute for successive masks until the requirement is satisfied, choosing the smallest mask length that still meets the host count to conserve address space.Step-by-Step Solution:Try /23 (borrow 7 bits): host_bits = 9 → usable = 2^9 - 2 = 510 (insufficient).Try /22 (borrow 6 bits): host_bits = 10 → usable = 2^10 - 2 = 1022 (meets ≥ 600).Corresponding dotted-decimal mask for /22 is 255.255.252.0.Therefore, /22 is the smallest prefix length that satisfies the requirement.Verification / Alternative check: Cross-check neighboring options: /21 (255.255.248.0) yields 2^11 - 2 = 2046 usable (more than enough but wastes addresses). /23 is too small (only 510). Hence /22 is the optimal choice.Why Other Options Are Wrong:255.255.192.0 (/18): far too many hosts; not targeted and wastes space.255.255.224.0 (/19) and 255.255.240.0 (/20): still much larger than required and do not represent the minimal mask meeting 600.255.255.248.0 (/21): meets hosts but is not the minimal that satisfies ≥ 600.Common Pitfalls: Forgetting to subtract 2 addresses for network/broadcast, or misreading the host requirement as ‘‘exactly 600’’ rather than ‘‘at least 600’’.Final Answer: 255.255.252.0.

Question 9

In computer network topology terminology, what do we call the arrangement in which every pair of nodes is connected by a dedicated bi-directional link (i.e., a fully interconnected “full mesh” of links)?

Accepted Answer

Introduction / Context:A fundamental topic in computer networking is “topology,” which describes how devices are interconnected. Understanding the difference between ring, star, tree, and mesh topologies helps network engineers reason about redundancy, path diversity, fault tolerance, and cost. This question asks specifically for the topology where each pair of nodes has its own direct, bi-directional link, which is the textbook definition of a full mesh. Given Data / Assumptions: The network links are bi-directional (traffic can flow both ways). Every possible pair of distinct nodes is directly connected—no intermediate hops required. The term sought is the standard, widely accepted name in networking literature. Concept / Approach:In a full mesh with n nodes, each node connects to every other node. The undirected link count is n*(n-1)/2, and the directed (simplex) count is n*(n-1). This provides maximal redundancy and the shortest possible path (one hop) between any two nodes, at the expense of the highest cabling/port cost and complexity. Partial mesh variants reduce the number of links but preserve multiple alternate paths. Step-by-Step Solution: Identify the requirement: direct bi-directional connectivity between every pair. Map to known topologies: ring (each node has two neighbors), star (all via a central hub), tree (hierarchical), mesh (many interlinks). Select the topology that guarantees a link for every pair: full mesh. Verification / Alternative check:Full mesh is commonly used in small backbone clusters, data center interconnects, or overlay networks where high availability is critical. Routing tables in full meshes are trivial for reachability because each destination is one hop away, confirming the definition aligns with the description given. Why Other Options Are Wrong: Ring: only two neighbors per node; not every pair has a dedicated link. Star: all traffic passes through the central node; pairs are not directly linked. Tree: hierarchical branching without all-pairs direct links. None of the above: incorrect because “Mesh” exactly matches. Common Pitfalls:Confusing “mesh” with “partial mesh”; misreading “bi-directional” as two separate simplex links and overlooking the requirement of a dedicated pairwise connection for every node pair; assuming redundancy automatically means mesh when stars can also be made redundant using multiple cores. Final Answer:Mesh

Question 10

In telecommunications terminology, what does the acronym ISDN expand to in full?

Accepted Answer

Introduction / Context:Many networking acronyms compress long technical names into short forms. Correctly expanding them matters in exams and in the field because the wording often hints at the technology’s goals. ISDN is a classic circuit-switched digital access technology that aimed to carry voice, data, and signaling in an integrated manner over the public telephone network. Given Data / Assumptions: The acronym is ISDN and the domain is telecom/access networks. ISDN historically provided Basic Rate Interface (BRI) and Primary Rate Interface (PRI). The question tests expansion accuracy, not deep technical operation. Concept / Approach:ISDN emphasizes “Integrated Services” (voice + data + signaling) delivered over a “Digital Network.” The canonical expansion found in standards and textbooks is “Integrated Services Digital Network.” Any variation that substitutes “Indian,” “Intelligent,” or “Data” in place of “Digital” and “Services” is incorrect for this acronym. Step-by-Step Solution: Recall standard expansion used by ITU-T and vendors: Integrated Services Digital Network. Compare with distractors: they replace or reorder key words. Select the only industry-correct expansion. Verification / Alternative check:Technical references for BRI (2B + D channels) and PRI (e.g., 23B + D for T1, 30B + D for E1) clearly label ISDN as Integrated Services Digital Network, validating the choice beyond rote memorization. Why Other Options Are Wrong: Indian Standard Digital Network: incorrect geography-specific wording; not the standard term. Intelligent Services Digital Network: nonstandard phrasing; “Integrated” is correct. Integrated Services Data Network: wrong substitution; the accepted expansion uses “Digital,” not “Data.” None of the above: incorrect because the proper expansion is provided in option (b). Common Pitfalls:Mixing up “Digital” with “Data,” or assuming “I” is “Indian” due to regional syllabi; forgetting that the purpose of ISDN was to integrate multiple services over one digital access line. Final Answer:Integrated Services Digital Network

Question 11

In data communications troubleshooting, what is the primary function of a data link content monitor (also called a protocol analyzer at the data link layer)?

Accepted Answer

Introduction / Context:When a network misbehaves, engineers often “sniff” frames on the wire. A data link content monitor—often a protocol analyzer focused on Layer 2—captures, decodes, and displays frame contents and timing so that protocol behavior can be verified against specifications. This helps pinpoint errors such as malformed frames, incorrect checksums, duplex mismatches, or misconfigured VLAN tags. Given Data / Assumptions: The monitor can capture and decode data link frames. We seek the principal purpose in a troubleshooting workflow. Higher-layer analyzers exist, but this tool focuses on the data link. Concept / Approach:Protocol analyzers reveal whether devices adhere to standards and whether sequence, timing, or field values are correct. By inspecting actual traffic, engineers can confirm or refute suspected causes. While a monitor can also infer link characteristics, its central value lies in detecting and diagnosing protocol issues through content analysis and validation of behavior. Step-by-Step Solution: Capture frames from the interface under test. Decode headers and payloads to human-readable fields (MAC addresses, EtherType, VLAN IDs, FCS validity). Compare observed behavior with protocol rules to identify violations or anomalies. Correlate timestamps and sequences to detect retransmissions, loops, or duplex errors. Verification / Alternative check:Tools like Wireshark and dedicated taps are industry staples. They highlight bad FCS, malformed LLDP, or STP loops and confirm the diagnosis by showing concrete frame evidence—exactly what a data link content monitor is meant to accomplish. Why Other Options Are Wrong: Determine the type of transmission: ancillary capability; not the primary goal. Determine switching type: a topology question, not what content monitors chiefly do. Determine flow without decoding: omits the key value—content decoding for protocol issues. None: incorrect because detecting protocol problems is the core purpose. Common Pitfalls:Confusing physical-layer testers (which check signal quality) with protocol analyzers; assuming traffic counters alone can diagnose protocol violations without looking into frame fields. Final Answer:To detect problems in protocols

Question 12

In cyclic redundancy checking (CRC), what does the term “CRC” refer to within the frame check sequence process?

Accepted Answer

Introduction / Context:Cyclic Redundancy Check (CRC) is a robust error-detection method used in data link protocols like Ethernet and PPP. It involves treating bit strings as polynomials over a binary field, dividing by a generator polynomial, and appending a check value. Understanding which quantity is called the “CRC” helps clarify how frames are validated at the receiver. Given Data / Assumptions: Binary division by a generator polynomial (the divisor) is performed on the transmitted bits. The process yields a quotient and a remainder. The check field appended to the frame is commonly named “CRC.” Concept / Approach:The sender computes the remainder after polynomial division (using modulo-2 arithmetic) and appends this remainder to the frame. The receiver divides the entire received bitstream by the same generator; a zero remainder indicates no detected errors. Therefore, “CRC” colloquially names the remainder field that is transmitted as the check value, not the divisor, dividend, or quotient. Step-by-Step Solution: Represent data bits as a polynomial; append zeros equal to the degree of the generator. Divide by the generator polynomial; compute remainder R. Append R (the CRC value) to the frame; transmit. Receiver divides received bits by the generator; zero remainder → accept, else error. Verification / Alternative check:Ethernet’s 32-bit Frame Check Sequence (FCS) is precisely such a remainder calculated using a standard 32-bit generator polynomial. Packet analyzers report a “Bad FCS” when the computed remainder at the receiver side is nonzero. Why Other Options Are Wrong: The divisor: that is the generator polynomial, not the CRC value. The quotient: not transmitted or named as the CRC. The dividend: refers to the data (with appended zeros) being divided. None: incorrect because the remainder is explicitly the CRC check field. Common Pitfalls:Mixing up the generator (divisor) with the remainder; assuming the quotient is useful for checking—it is not used; forgetting that CRC is an error-detection mechanism, not error correction. Final Answer:The remainder

Question 13

In cyclic redundancy checking, how does the size of the generator polynomial (divisor) compare to the size of the CRC value that is appended?

Accepted Answer

Introduction / Context:CRC design uses a generator polynomial of degree r. The division process yields a remainder of r bits, which is appended to the frame as the CRC field. Knowing the exact relationship between the generator degree and the remainder length is essential for implementing and verifying CRC algorithms correctly. Given Data / Assumptions: Let the generator polynomial have degree r. The appended CRC (remainder) contains r bits. Polynomial representation uses binary coefficients over modulo-2 arithmetic. Concept / Approach:A polynomial of degree r has r+1 coefficients (from x^r down to x^0). Therefore, the generator (divisor) has r+1 bits, while the CRC remainder has r bits. This creates the precise one-bit difference asked in the question: the divisor is one bit longer than the CRC value appended to the frame. Step-by-Step Solution: Let generator degree = r → generator bit length = r + 1. The CRC remainder length = r. Compare sizes: (r + 1) vs r → generator is one bit more than the CRC. Therefore, the correct relationship is “one bit more than.” Verification / Alternative check:Examples: CRC-16 uses a 17-bit generator and a 16-bit remainder; CRC-32 uses a 33-bit generator and a 32-bit remainder. These well-known cases confirm the general rule and the answer. Why Other Options Are Wrong: The same size: contradicted by all standard CRCs. One bit less / Two bits more: do not match the polynomial degree relationship. None: incorrect since a specific relationship exists. Common Pitfalls:Confusing polynomial degree (r) with bit length; forgetting that the highest-order coefficient implies one additional bit in the generator beyond the remainder length. Final Answer:One bit more than

Question 14

In computer architecture and data communications, what is a “bus” most accurately defined as when discussing device interconnection on a system board or backplane?

Accepted Answer

Introduction / Context:The term “bus” appears in both computer architecture and local communications contexts. On a motherboard or backplane, a bus is a shared pathway that multiple components use to send or receive signals. Clarity on this definition helps distinguish it from related but different concepts like carriers, collisions, and general networks. Given Data / Assumptions: Scope is device interconnection within a system or local segment. We are not describing radio carriers or abstract network collections. We seek the precise, classical definition used in textbooks and standards. Concept / Approach:A bus comprises conductors (traces or wires) shared by multiple devices for data, address, or control signaling. Examples include PCIe (logically switched fabric but conceptually descended from the bus idea), legacy PCI, and memory buses. The defining feature is shared connectivity through common conductors rather than point-to-point dedicated pairs for each device. Step-by-Step Solution: Identify key elements: shared conductors; multiple attached devices. Exclude concepts that relate to modulation (radio carrier) or contention outcomes (collisions). Choose the option that states “one or more conductors serving as a common connection.” Verification / Alternative check:Classical buses like ISA or S-100 literally exposed parallel signals on a backplane. Even modern serial interconnects keep the “bus” naming convention to indicate a shared medium or common wiring standard, confirming the correctness of the definition provided. Why Other Options Are Wrong: Continuous frequency: describes a carrier wave, not a wiring interconnect. Two stations attempting same channel: describes a collision event, not a bus itself. Collection of networked units: overly broad and better describes a network than a bus. None: incorrect because (a) is the accepted definition. Common Pitfalls:Equating “bus” with “network” generically; assuming collisions define a bus (they can occur on shared buses but are not the definition). Final Answer:One or more conductors that serve as a common connection for a related group of devices

Question 15

A digital link uses Phase Shift Keying (PSK) at a bit rate of 9600 bps with 16 distinct phase levels (16-PSK). What is the line signaling speed, i.e., the modulation rate in baud (symbols per second)?

Accepted Answer

Introduction / Context:Digital modulation conveys bits by changing a signal’s properties (amplitude, frequency, or phase). With M-ary schemes like 16-PSK, each symbol carries log2(M) bits. The modulation rate in baud equals symbols per second, which may be lower than the bit rate when multiple bits are packed into each symbol. Calculating baud from bps is a frequent exam task because it reinforces the distinction between bit rate and symbol rate. Given Data / Assumptions: Bit rate (R_b) = 9600 bits per second. Modulation = 16-PSK (M = 16 distinct phases). Bits per symbol (k) = log2(M) = log2(16) = 4 bits per symbol. No coding or framing overhead is included in the basic calculation. Concept / Approach:Relationship: R_b = (symbols per second) * (bits per symbol). Therefore, symbols per second (baud) = R_b / k. For 16-PSK, k = 4, which reduces the symbol rate compared to binary schemes. This is the standard approach across M-ary modulation families (M-PSK, M-QAM, etc.). Step-by-Step Solution: Compute k = log2(16) = 4 bits/symbol. Compute baud = 9600 bps / 4 bits/symbol = 2400 symbols/second. Express the modulation rate: 2400 baud. Select the closest matching option exactly: 2400 baud. Verification / Alternative check:As a cross-check, if the system were binary PSK (M=2), k=1 and baud would equal 9600. For 4-PSK (QPSK), k=2 → 4800 baud. For 8-PSK, k=3 → 3200 baud. These comparisons align with the 16-PSK result of 2400 baud, confirming correctness. Why Other Options Are Wrong: 1200 baud: would correspond to k = 8 bits/symbol at 9600 bps (not 16-PSK). 4800 baud: would correspond to QPSK (k=2) at 9600 bps. 9600 baud: true only for binary schemes (k=1) at 9600 bps. 600 baud: does not match any common M for 9600 bps without coding/overhead assumptions. Common Pitfalls:Confusing “baud” and “bps”; forgetting to take log2(M); ignoring overhead (which is excluded unless explicitly stated); assuming baud equals bps for all modulations, which is false for M-ary schemes. Final Answer:2400 baud

Question 16

In Microsoft-centric networks, which service dynamically resolves NetBIOS names to IP addresses (as opposed to static files or DNS hostnames)?

Accepted Answer

Introduction / Context:Legacy Windows networking used NetBIOS names for discovery and access. Mapping those names to IP addresses could be done dynamically or statically. Knowing which mechanism provides dynamic NetBIOS-to-IP resolution is important for troubleshooting name conflicts, browsing issues, and legacy application connectivity in mixed environments. Given Data / Assumptions: The focus is NetBIOS name resolution, not general DNS hostname resolution. Dynamic rather than static mapping is required. Windows environments include several related services and files. Concept / Approach:Windows Internet Name Service (WINS) is the dynamic NetBIOS name server. Clients register their NetBIOS names and IP addresses with WINS and query it to resolve others. This differs from DNS (resolves hostnames, not NetBIOS names) and from LMHOSTS (a static text file that maps NetBIOS names to IP addresses). DHCP assigns IP configuration but does not resolve NetBIOS names. Step-by-Step Solution: Identify the namespace: NetBIOS names. Identify the requirement: dynamic resolution service. Select WINS as the dedicated dynamic NetBIOS name service. Verification / Alternative check:Administrative consoles show WINS databases with registered names and lifetimes; network traces capture NetBIOS Name Service queries to WINS servers, validating that WINS performs dynamic resolution in this context. Why Other Options Are Wrong: DNS: primarily resolves DNS hostnames (FQDNs) via UDP/TCP 53; not specifically NetBIOS names. DHCP: provides IP configuration (address, mask, gateway, DNS/WINS options) but not name resolution. LMHOSTS: static file on clients; no dynamic registration. None: incorrect because WINS exists precisely for this task. Common Pitfalls:Assuming DNS entirely replaces NetBIOS resolution in all legacy scenarios; forgetting that LMHOSTS entries are static and must be manually maintained. Final Answer:WINS

Question 17

Which portion of the radio spectrum is used for standard AM (Amplitude Modulation) broadcast transmission?

Accepted Answer

Introduction / Context:Radio services are categorized by frequency bands, each with distinctive propagation and usage patterns. Standard AM broadcast radio (the familiar medium-wave band) occupies a well-defined slice of spectrum. Knowing these band limits is foundational for communications engineering and exam questions on spectrum allocation. Given Data / Assumptions: The question concerns standard AM broadcast, not shortwave or FM broadcast. We use conventional ITU band naming (VLF, LF, MF, HF, VHF, etc.). Approximate decimal ranges are acceptable for classification. Concept / Approach:AM broadcast (medium-wave) lies in the Medium Frequency (MF) band. Typical channel plans place AM broadcast from roughly 530 kHz to 1700 kHz in many regions, which is squarely within the MF range of 300 kHz to 3 MHz. Other bands (HF and VHF) are used for shortwave broadcasting and FM/TV services respectively, not standard medium-wave AM. Step-by-Step Solution: Identify target service: standard AM broadcast (medium-wave). Match to spectrum band: Medium Frequency (MF). MF band numerical limits: 300 kHz to 3 MHz. Select the option that states MF: 300 kHz to 3 MHz. Verification / Alternative check:Regulatory charts and textbooks list AM broadcast allocations inside MF, while HF (3–30 MHz) covers shortwave services and VHF (30–300 MHz) includes FM broadcast near 88–108 MHz. This confirms the MF choice is correct for standard AM broadcast. Why Other Options Are Wrong: VLF (3–30 kHz): used for specialized long-range/underwater comms, not AM broadcast. HF (3–30 MHz): supports shortwave broadcasting, not standard medium-wave AM. VHF (30–300 MHz): home to FM broadcast and TV, not AM medium-wave. None: incorrect because the MF range option is correct. Common Pitfalls:Confusing AM “medium-wave” with shortwave (HF) AM; assuming “AM” generically means any amplitude-modulated service, whereas the question points to standard broadcast AM in the medium-wave band. Final Answer:Medium Frequency: 300 kHz to 3 MHz

Question 18

Videotex, an early interactive information service for households and kiosks, combined which foundational technologies to let users select on-screen content on demand?

Accepted Answer

Introduction / Context:Before the modern World Wide Web, several systems delivered interactive pages to users over telephone or cable lines. Videotex was among the most notable, enabling users to browse information, shop, and message using a TV or dedicated terminal. Understanding what technologies Videotex brought together gives historical perspective on today’s web and smart TV platforms. Given Data / Assumptions: Videotex presented on-screen pages and menus. Users navigated interactively, selecting content to view. The service worked over telecommunications networks to remote databases. Concept / Approach:Videotex blended: (1) television display hardware for rendering pages on consumer screens; (2) communications over the public network for two-way interactivity; and (3) computer technology to store, generate, and serve the information. This trio together enabled an early form of on-demand, menu-driven digital content long before broadband and HTML browsers became ubiquitous. Step-by-Step Solution: Identify display component: television used to render visuals to users at home. Identify connectivity: telecommunication links (often dial-up) to reach content servers. Identify processing: computers generated and managed the interactive pages and services. Conclude that Videotex is a fusion of all three categories. Verification / Alternative check:Well-known deployments such as Minitel (France) and Prestel (UK) illustrate this mix: TV/terminal displays, modems/lines for connectivity, and central computers offering directories, banking, and messaging—validating “All of the above.” Why Other Options Are Wrong: Television alone: lacks interactivity and remote data. Communication alone: needs a display and computing to present content. Computer technology alone: without home displays and networks, there is no consumer service. None: incorrect because Videotex inherently combined all three. Common Pitfalls:Confusing Videotex with Teletext (a one-way broadcast overlay carried in TV signals). Videotex was two-way and interactive, whereas Teletext was not. Final Answer:All of the above

Question 19

In classic IEEE 802.3 (Ethernet) design fundamentals, which capability is NOT provided by the basic MAC/framing system (assume no higher-layer recovery or 802.1Q enhancements)?

Accepted Answer

Introduction: Ethernet's original IEEE 802.3 design focused on a simple, fast link-layer method for sending frames on a shared medium. Understanding exactly what classic Ethernet does—and does not—do helps separate link-layer responsibilities from those of upper layers like TCP/IP. This question asks which specific capability is not supplied by basic Ethernet (without VLAN extensions or higher-layer recovery).Given Data / Assumptions:Context is legacy/basic IEEE 802.3 operation (no 802.1Q VLANs, no link-layer ARQ).Half-duplex shared media historically used CSMA/CD; full-duplex switching removed collisions but kept framing rules.Upper layers (e.g., TCP) may provide reliability independent of Ethernet.Concept / Approach: At Layer 2, Ethernet offers addressing (MAC), framing, and an integrity check (FCS). Historical shared-media networks used CSMA/CD for media access control. If a frame is corrupted or lost, Ethernet does not automatically retransmit it by itself; the receiver simply discards a bad frame (FCS failure). Reliable delivery—automatic retransmission, ordering, flow/congestion control—belongs to higher layers (e.g., TCP) or to special link types that implement ARQ (not standard Ethernet).Step-by-Step Solution:Identify normal Ethernet features: MAC addressing, framing, FCS.Recall media access on classic shared Ethernet: CSMA/CD.Check reliability: no built-in ARQ or link-layer retransmission in basic 802.3.Conclude the missing feature is automatic retransmission at the link itself.Verification / Alternative check: Packet captures show that receivers drop frames with bad FCS; no Ethernet "NACK" exists in standard 802.3. Recovery—if any—arises from upper-layer protocols (e.g., TCP sequence/ACK).Why Other Options Are Wrong:CSMA/CD is the classic Ethernet access method for half-duplex media.MAC addressing is core to Ethernet frame delivery.FCS/CRC is explicitly defined for frame integrity checking.None of the above is incorrect because one option is indeed not provided.Common Pitfalls: Assuming link-layer reliability exists because Wi-Fi uses ARQ; conflating switched full-duplex (no collisions) with guaranteed delivery; confusing VLANs (802.1Q) with basic 802.3.Final Answer: Automatic retransmission of a corrupted or lost frame by the link itself.

Question 20

At a digital receiver in data communications, what is the process called that converts a modulated analog carrier back into a baseband digital bitstream so it can be processed by the computer?

Accepted Answer

Introduction: In a typical digital communication system, the transmitter modulates a digital bitstream onto an analog carrier for efficient transmission over band-limited channels (radio, coax, fiber optics via electrical/optical carriers). At the receiver, the inverse operation extracts the original baseband data so a computer can process it. This question asks for the precise term describing that inverse operation at the receiver.Given Data / Assumptions:The incoming signal is an analog, modulated carrier (e.g., FSK, PSK, QAM).The receiver must recover the original digital symbols/bits.Focus is on the carrier removal and symbol decision process, not merely timing recovery.Concept / Approach: Demodulation is the process that converts a modulated analog signal back into its baseband representation. For digital systems, demodulation maps received waveforms to symbol decisions, applies filtering/equalization as needed, recovers carrier phase/frequency, and outputs the bitstream. Modulation is the complementary transmitter operation; synchronizing/clock recovery is a supporting function that helps demodulation but is not a full replacement. Line coding (e.g., NRZ, Manchester) shapes baseband digital signals on a physical medium—again distinct from removing a carrier.Step-by-Step Solution:Identify the received waveform type: analog carrier holding digital information.Determine the inverse operation needed: remove carrier and map to bits.Name that operation: demodulation.Distinguish supporting functions (synchronization, line coding) from demodulation itself.Verification / Alternative check: In modems and radios, the digital signal processor chain includes carrier recovery, matched filtering, symbol timing, decision, and decoding—collectively framed under demodulation to deliver a digital bitstream to the host system.Why Other Options Are Wrong:Modulation: transmitter-side process.Synchronizing: assists by aligning timing/phase but does not by itself produce the bitstream.Line coding: baseband symbol mapping (no carrier) used on some wired links.None of the above: incorrect because demodulation is the standard term.Common Pitfalls: Equating “analog-to-digital conversion” (ADC) with demodulation; ADC digitizes the waveform, while demodulation interprets it to recover the bits. Many receivers do both in sequence (ADC then digital demodulation), but the conceptual term here is demodulation.Final Answer: Demodulation.

Networking Questions

Questions

Practice smarter