KNEC Data Communication and Networking Past Paper with Answers — June/July 2023

June/July 2023 — Data Communication and Networking

Paper: Kenya National Examinations Council (KNEC), Diploma in Electrical and Electronic Engineering (Telecommunication option), Module III — Data Communication and Networking, June/July 2023. Paper code: 2602/305.

This paper has eight questions in two sections. Candidates answer five questions, choosing at least two from each section. Each question carries 20 marks.

These are independently prepared revision explanations based on the supplied paper. They are not an official KNEC marking scheme.

Section A: Data Communication

Answer at least two questions from this section.

Question 1

1(a) Define each of the following with respect to error control: (i) redundancy; (ii) code distance; (iii) code weight. (3 marks)

Answer

• Redundancy: Extra bits added to the original data so that errors can be detected or corrected.
• Code distance: The number of bit positions in which two codewords differ. The minimum distance of a code is the smallest such distance among its valid codewords.
• Code weight: The number of 1s in a binary codeword.

1(b) With the aid of a time flow diagram, describe the stop-and-wait ARQ error-control technique in data communication. (7 marks)

Answer

The sender transmits one numbered frame and starts a timer. The receiver checks the frame for errors. If it is correct, the receiver sends an acknowledgement; only then does the sender transmit the next frame. If the frame or acknowledgement is lost or corrupted, the timer expires and the sender retransmits the same frame. Alternating sequence numbers (0 and 1) let the receiver identify and discard a duplicate while acknowledging it again.

Sender                                      Receiver
  |                                             |
  |---------------- Frame 0 ------------------>|
  |                                    check frame
  ||
  |<--------------- ACK 0 ---------------------|
  |                                             |

If Frame 0 is lost:
Sender -------- Frame 0 -----X
        (timer expires)
Sender -------- Frame 0 ------------------------> Receiver
        <---------------------- ACK 1 (duplicate-safe)

1(c) Table 1 shows codewords of 7 bits each to be transmitted using the longitudinal redundancy check (LRC). Using even parity, determine (i) the frame-check sequence and (ii) the transmitted codeword. (6 marks)

Answer

Table 1 — supplied 7-bit codewords

Codeword	Bits
1	1 0 0 1 0 1 0
2	0 1 1 1 1 0 0
3	1 1 1 0 0 0 1
4	1 1 1 1 0 1 0

For even parity, set a parity bit to 1 when its column contains an odd number of 1s, and to 0 when it contains an even number.

Append the LRC row to the four data rows to form the transmitted block:

1(d) An optical-fibre cable with core and cladding refractive indices of 1.62 and 1.52 respectively is used to transmit data. Launching takes place in air. Determine (i) the numerical aperture and (ii) the critical angle. (4 marks)

Answer

The critical angle is measured from the normal at the core–cladding boundary.

Question 2

2(a) Differentiate between half-duplex and full-duplex data communication modes. (2 marks)

Answer

In half-duplex, both ends can transmit, but only one end transmits at a time. In full-duplex, both ends can transmit simultaneously.

2(b) With the aid of a block diagram, describe the components of a data communication system. (6 marks)

Answer

Sender / source → Transmitter / encoder → Communication channel → Receiver / decoder → Destination
                                      ↑
                                 Noise may enter

The main components are the sender (originates the data), the message (the information), a transmitter (converts the message into signals), the transmission medium (carries the signals), a receiver (recovers the message), the destination, and agreed protocols that govern the exchange.

2(c) Explain each of the following data-transmission impairments: (i) thermal noise; (ii) jitter. (4 marks)

Answer

• Thermal noise is random electrical noise caused by the thermal motion of charge carriers in a component or transmission medium. It adds unwanted energy to the signal and reduces the signal-to-noise ratio.
• Jitter is unwanted variation in the timing of signal transitions or received bits. It can cause the receiver to sample a bit at the wrong time.

2(d) A channel transmits data asynchronously using bits of 20 ms duration. The receiver clock is operating at 48 Hz. Determine (i) the receiver clock period; (ii) the slip in the receiver clock on each successive bit; and (iii) the number of slips for the receiver to slide beyond half of the received bit period. (8 marks)

Answer

The receiver clock is slower than the incoming bit timing, so the slip accumulates. Half a bit period is 10 ms; 12 slips reach exactly 10 ms, so the drift is beyond half a bit after the next slip.

Question 3

3(a) Differentiate between switching and routing as used in data networks. (2 marks)

Answer

Switching forwards frames between devices within a local network, commonly using MAC addresses. Routing forwards packets between networks using network-layer addresses and routing information.

3(b) Describe the phases in the circuit-switching technique in data communication. (5 marks)

Answer

1. Connection setup: a dedicated end-to-end path is established and resources are reserved.
2. Data transfer: the two endpoints exchange data over the established path.
3. Connection release: either endpoint terminates the call; the path and reserved resources are released.

3(c) (i) Draw a labelled block diagram of the Bisynchronous (BISYNC) frame format. (ii) State how transparency is achieved in the frame in part (i). (5 marks)

Answer

SYN | SYN | SOH | Header | STX | Text / data | ETB or ETX | BCC
sync         header start   start of text     block / text    block-check
characters   information    data              terminator      character

SYN characters establish synchronisation. SOH starts the header, STX starts the text, ETB ends a block or ETX ends the text, and BCC is the block-check character. Transparency is provided by inserting an escape character (DLE) before a control character that occurs in the data; a literal DLE in the payload is escaped as well. The receiver removes the inserted escape character so the original data is recovered.

3(d) Eight data terminals, each transmitting at 6,600 bps, are to be multiplexed using time-division multiplexing (TDM). (i) Determine the total capacity required for synchronous TDM. (ii) If the link utilization is 0.6, using statistical TDM determine (I) the data rate that can be supported and (II) the number of terminals that can be supported. (8 marks)

Answer

At 0.6 activity per terminal, the average data rate generated by the eight terminals is:

Using the 52.8 kbps line capacity from part (i), each statistically multiplexed terminal contributes an average of 0.6 × 6,600 bps:

This uses the stated utilization as the average fraction of time each terminal is active and ignores framing overhead.

Question 4

4(a) Distinguish between data rate and baud rate with respect to data transmission. (2 marks)

Answer

Data rate is the number of bits transmitted per second (bit/s). Baud rate is the number of signalling elements transmitted per second (symbols/s). With M-ary signalling, one symbol may represent more than one bit.

4(b) State three merits of scrambling in data encoding. (3 marks)

Answer

• It prevents long runs of identical bits, helping the receiver maintain clock synchronisation.
• It reduces a direct-current (DC) component and improves the signal spectrum for transmission.
• It creates sufficient transitions for reliable timing recovery without increasing the data rate.

4(c) The data sequence 1011101001 is to be encoded. Draw the resultant waveforms for each of the following encoding schemes: (i) unipolar NRZ; (ii) bipolar AMI; (iii) polar NRZ. (9 marks)

Answer

Use a positive level for binary 1 in unipolar NRZ and NRZ-L, and alternate the AMI mark polarity beginning with +V. A zero level is shown as 0.

Bit position	1	2	3	4	5	6	7	8	9	10
Data bit	1	0	1	1	1	0	1	0	0	1
Unipolar NRZ	+V	0	+V	+V	+V	0	+V	0	0	+V
Bipolar AMI	+V	0	−V	+V	−V	0	+V	0	0	−V
Polar NRZ-L	+V	−V	+V	+V	+V	−V	+V	−V	−V	+V

Polar NRZ is interpreted here as NRZ-L with 1 = +V and 0 = −V.

4(d) A modem transmits 60,671 bits per second using signalling elements chosen from an 8,192-point constellation. Determine (i) the modulation scheme used; (ii) the number of bits that may be encoded in each signalling element; and (iii) the modulation rate of the modem. (6 marks)

Answer

The source specifies the constellation size but not its geometry. This is an 8,192-ary modulation alphabet (8,192-QAM if the constellation is QAM).

Section B: Networking

Answer at least two questions from this section.

Question 5

5(a) (i) Describe voice over internet protocol (VoIP). (ii) State three merits of using the VoIP phone service. (5 marks)

Answer

VoIP carries voice as packetised digital data over an Internet Protocol network instead of using a dedicated analogue telephone circuit. Its merits include lower long-distance costs, easy integration of voice with other data services, and flexible use across compatible devices and locations.

5(b) Outline the steps in the carrier-sense multiple-access with collision-detection (CSMA/CD) medium-access control protocol. (5 marks)

Answer

1. A station senses the shared medium. If it is busy, it waits; if it is idle, it begins transmitting.
2. While transmitting, the station continues monitoring the medium for a collision.
3. If no collision occurs, the transmission completes successfully.
4. If a collision is detected, the station stops transmitting and sends a jam signal so all stations detect it.
5. Each colliding station waits for a random interval selected by the binary exponential back-off procedure, then senses the medium and retries.

5(c) Table 2 shows devices owned by an organization that are to be connected to an ISDN network. Determine the required number of basic-rate interfaces. (7 marks)

Answer

Table 2 — device bit rates

Device	Bit rate
PABX	384 kbps
Computer	564 kbps
Security camera	800 kbps

An ISDN basic-rate interface has two 64 kbps B-channels for user data. The 16 kbps D-channel is for signalling, so the payload capacity per interface is 128 kbps.

5(d) State three merits of ISDN networks. (3 marks)

Answer

• They carry voice, data and other services over a digital network using a common access interface.
• Digital transmission provides reliable, consistent call quality.
• They support faster call setup and multiple channels over one subscriber connection.

Question 6

6(a) State three functions of the asynchronous transfer mode (ATM) physical layer. (3 marks)

Answer

• It transmits and receives the physical bit stream over the selected medium.
• It provides line coding, timing and synchronisation with the physical transmission system.
• Its transmission-convergence functions adapt the cell stream to the physical frame and identify cell boundaries; header-error-control generation/checking is also performed at this layer.

6(b) With the aid of diagrams, describe each of the following LAN topologies: (i) bus topology; (ii) ring topology. (8 marks)

Answer

Bus topology: all stations connect to one shared backbone. Terminators at both ends prevent signal reflections. A fault in the backbone can affect the whole segment.

Terminator ───────┬─────────┬─────────┬────── Terminator
                  │         │         │
                 PC1       PC2       PC3

Ring topology: each station connects to two neighbours, forming a closed loop. Data passes from station to station around the ring; a break can interrupt communication unless a protected or dual ring is used.

          PC1 ─────→ PC2
           ↑           ↓
          PC4 ←───── PC3

6(c) Table 3 shows devices used in computer networks and the corresponding layer of the OSI model on which they operate. Complete the table. (4 marks)

Answer

Table 3 — network devices and OSI layers

Network device	OSI layer
Bridge	Layer 2 — Data Link
Hub	Layer 1 — Physical
Gateway	Often Layer 7 — Application; a gateway may translate protocols across multiple layers
Router	Layer 3 — Network

6(d) Draw the X.25 data-link-layer frame. (5 marks)

Answer

┌───────────┬─────────┬─────────┬─────────────────┬─────────┬───────────┐
│ Flag      │ Address │ Control │ Information     │ FCS     │ Flag      │
│ 01111110  │ 8 bits  │ 8/16 bit│ variable length │ CRC     │ 01111110  │
└───────────┴─────────┴─────────┴─────────────────┴─────────┴───────────┘

The FCS is typically a 16-bit CRC. The information field is optional in supervisory and unnumbered frames. X.25 LAPB uses HDLC-style bit stuffing: a 0 is inserted after five consecutive 1s in the frame body so the flag pattern is not mistaken for data.

Question 7

7(a) Describe each of the following cellular technologies: (i) frequency-division multiple access (FDMA); (ii) time-division multiple access (TDMA). (4 marks)

Answer

• FDMA divides the available spectrum into frequency channels and assigns a separate channel to each user during a call.
• TDMA assigns users different time slots on a shared frequency channel; each user transmits in its allocated slot.

7(b) State three features of wide-area networks that distinguish them from local-area networks. (3 marks)

Answer

• A WAN covers a much larger geographic area, often connecting sites in different towns, countries or continents.
• It commonly uses carrier-provided links and public or leased telecommunications infrastructure.
• It usually has greater propagation delay and more varied link technologies than a LAN.

7(c) Draw a labelled diagram of a structured-cabling system from the work area up to the ISP. Indicate the PCs, patch panel, switch and routers. (7 marks)

Answer

WORK AREA                         TELECOMMUNICATIONS ROOM                   WAN
PCs → patch cords → wall outlets → horizontal cabling → patch panel → switch → router / firewall → ISP router → ISP

The patch panel terminates and organises horizontal cable runs. Patch cords connect the patch panel to the switch; the router or firewall connects the LAN to the ISP-facing link.

7(d) Table 4 shows modulation schemes used in wireless networks and their data rates. Complete the table. (6 marks)

Answer

The source table does not print rate units; the listed wireless rates are interpreted as Mbps. For M-ary modulation, each signalling element carries log₂(M) bits. The given rates correspond to a 54 Mbaud symbol rate.

Table 4 — completed modulation values

Modulation scheme	Code bits per sub-carrier	Data rate	Baud rate
BPSK	1	54 Mbps	54 Mbaud
QPSK	2	108 Mbps	54 Mbaud
16-QAM	4	216 Mbps	54 Mbaud

Question 8

8(a) Identify three types of keys used in an encryption scheme. (3 marks)

Answer

• Secret (symmetric) key: the same shared key is used to encrypt and decrypt.
• Public key: the openly distributed key in an asymmetric key pair.
• Private key: the confidential key paired with the public key.

8(b) With the aid of a block diagram, describe the digital-signature technique in relation to network security. (6 marks)

Answer

SENDER
Message → hash function → message digest → sign with sender's private key → digital signature
                 Message + digital signature ───────────────→ Receiver

RECEIVER
Received message → hash function → digest A
Digital signature → verify with sender's public key → digest B
Compare digest A with digest B: match = valid signature and unchanged message

The signature provides message integrity and helps authenticate the sender. Because only the sender should control the private key, it also supports non-repudiation. The message itself is not made confidential by signing; encryption is a separate operation.

8(c) Describe each of the following routing techniques: (i) static routing; (ii) dynamic routing. (4 marks)

Answer

• Static routing uses routes entered manually by an administrator. Routes remain fixed until changed and do not automatically respond to a link failure.
• Dynamic routing uses routing protocols to exchange network information and update routes when topology or link conditions change.

8(d) An organization with an IP address 196.22.50.0 has a subnet mask of 255.255.255.248. Determine (i) the class of the network; (ii) the number of subnets; and (iii) the host addresses of the first subnet. (7 marks)

Answer

The first subnet using modern CIDR rules is 196.22.50.0/29:

Some older classful exercises reserve subnet zero and the all-ones subnet; under that convention there are 30 subnets and the first non-zero subnet has usable hosts 196.22.50.9–196.22.50.14. The convention used by an examiner should be stated.