BE Computer Engineering (Pokhara University) Data Communication (PU, CMM 220) Question Paper 2079 Nepal
This is the official BE Computer Engineering (Pokhara University) Data Communication (PU, CMM 220) question paper for 2079, as set in the regular annual examination. It carries 100 full marks and a time allowance of 180 minutes, across 12 questions. On Kekkei you can attempt this Data Communication (PU, CMM 220) past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your BE Computer Engineering (Pokhara University) Data Communication (PU, CMM 220) exam or solving previous years' question papers, this 2079 paper is a great way to practise under real exam conditions.
Section A: Long Answer Questions
Attempt all / any as specified.
(a) Distinguish between analog and digital signals, and explain the relationship between bit rate, baud rate, and bandwidth with the help of suitable examples. (8)
(b) Compare guided and unguided transmission media. With neat diagrams, explain the construction and operating characteristics of optical fiber cable, and state two advantages it has over coaxial cable. (7)
(a) Analog vs Digital Signals; Bit Rate, Baud Rate, Bandwidth (8)
Analog vs Digital Signals
| Feature | Analog Signal | Digital Signal |
|---|---|---|
| Values | Continuous; infinitely many values over a range | Discrete; finite number of levels (e.g. 0 and 1) |
| Waveform | Smooth, continuously varying (e.g. sine wave) | Abrupt, level changes (square-like) |
| Example | Human voice, audio over a telephone line | Data from a computer, square-wave pulses |
| Bandwidth need | Lower for the same information in many cases | Theoretically infinite (sharp edges = many harmonics) |
| Noise effect | Hard to recover original after noise | Easy to regenerate exact levels |
Bit Rate, Baud Rate, Bandwidth
- Bit rate = number of bits transmitted per second (bps).
- Baud rate (signal rate) = number of signal elements (symbols) transmitted per second (baud).
- Relationship: If each signal element carries bits, then
where is the number of distinct signal levels. So baud rate bit rate always.
- Bandwidth is the range of frequencies a medium passes (Hz). The Nyquist limit links bandwidth to signal rate: , so higher bandwidth permits a higher baud rate and hence a higher bit rate.
Examples:
- A system sending 2 bits per symbol (4 levels, QPSK) at 1000 baud gives a bit rate of bps, while the baud rate is only 1000.
- A binary system (, ) at 1000 baud gives 1000 bps — here bit rate = baud rate.
- For a 3 kHz telephone channel with 4 levels, Nyquist gives bps.
(b) Guided vs Unguided Media; Optical Fiber (7)
Guided vs Unguided
| Guided (wired) | Unguided (wireless) | |
|---|---|---|
| Path | Physical conductor/fiber (twisted pair, coax, fiber) | Free space / air |
| Direction | Confined along the cable | Broadcast in all/specific directions |
| Examples | UTP, coaxial cable, optical fiber | Radio, microwave, infrared |
| Security/EMI | More secure, less interference | More prone to interception and interference |
Optical Fiber — Construction
Described structure (centre outwards):
- Core: thin glass/plastic strand of higher refractive index that carries the light.
- Cladding: glass of lower refractive index surrounding the core.
- Buffer coating / jacket: plastic protective outer layers.
Operating principle: Light is launched into the core at an angle greater than the critical angle, so at the core–cladding boundary it undergoes total internal reflection (). The ray bounces along the core to the far end, where a photodetector converts it back to an electrical signal. Modes: single-mode (very thin core, one ray path, long distance) and multimode (wider core, many paths).
Two advantages over coaxial cable
- Much higher bandwidth and far lower attenuation, allowing higher data rates over longer distances without repeaters.
- Immunity to electromagnetic interference (light, not electric current) and greater security from tapping.
(a) Explain the process of Pulse Code Modulation (PCM) with a block diagram, clearly describing the sampling, quantization, and encoding stages. State the Nyquist sampling theorem. (9)
(b) A signal band-limited to 4 kHz is sampled and each sample is encoded using 8 bits. Calculate the minimum sampling rate and the resulting bit rate of the PCM output. (6)
(a) Pulse Code Modulation (PCM) (9)
PCM converts an analog signal into a digital bit stream in three stages.
Block diagram (described):
Analog input → [Low-pass / anti-alias filter] → [Sampler (PAM)] → [Quantizer] → [Encoder] → PCM digital output
1. Sampling
The continuous analog signal is sampled at regular intervals, producing a Pulse Amplitude Modulated (PAM) sequence. To reconstruct the signal without loss, sampling must obey the Nyquist theorem.
2. Quantization
Each sampled amplitude is rounded to the nearest of a finite set of levels ( levels for -bit encoding). The rounding error introduced is called quantization noise.
3. Encoding
Each quantized level is mapped to an -bit binary code word, giving the serial PCM bit stream.
Nyquist Sampling Theorem
A band-limited signal with maximum frequency can be perfectly reconstructed from its samples if it is sampled at a rate . The minimum rate is the Nyquist rate.
(b) Numerical (6)
Given: bandwidth (max frequency) , bits per sample .
Minimum sampling rate (Nyquist rate):
Bit rate of PCM output:
Results: minimum sampling rate = 8 kHz, PCM bit rate = 64 kbps.
(a) Compare circuit switching, datagram packet switching, and virtual-circuit packet switching in terms of connection setup, addressing overhead, delay, and resource utilization. (9)
(b) Explain the working of the Stop-and-Wait ARQ protocol with a timing diagram, and describe how it handles a lost data frame and a lost acknowledgment. (6)
(a) Comparison of Switching Techniques (9)
| Criterion | Circuit Switching | Datagram Packet Switching | Virtual-Circuit Packet Switching |
|---|---|---|---|
| Connection setup | Dedicated path set up before transfer (call setup phase) | No setup; each packet routed independently | Logical path (VC) set up before transfer |
| Addressing overhead | Address used only during setup; none per data unit | Full destination address in every packet | Short VC identifier in each packet (less overhead) |
| Path | Same physical path for whole session | Packets may take different paths | All packets follow the same VC path |
| Delay | Setup delay, then constant low delay | No setup delay; variable per-packet delay | Setup delay, then fairly uniform delay |
| Ordering | In order | May arrive out of order | In order |
| Resource utilization | Poor — bandwidth reserved even when idle | High — bandwidth shared statistically | High — shared, with reservation benefits |
| Example | Traditional telephone network | The Internet (IP) | X.25, Frame Relay, ATM |
(b) Stop-and-Wait ARQ (6)
Working: The sender transmits one frame and then stops, waiting for an acknowledgment (ACK) before sending the next. Frames and ACKs are numbered modulo 2 (0 and 1) to detect duplicates. A timer is started for every transmitted frame.
Timing diagram (described):
Sender Receiver
| --- Frame 0 -----------> | (accept, send ACK1)
| <----------- ACK 1 ------ |
| --- Frame 1 -----------> | (accept, send ACK0)
| <----------- ACK 0 ------ |
Lost data frame: The receiver gets nothing, so it sends no ACK. The sender's timer expires (timeout) and it retransmits the same frame with the same sequence number.
Lost acknowledgment: The frame arrived but its ACK was lost. The sender times out and resends the frame. The receiver sees a frame with a sequence number it has already accepted, discards the duplicate, and re-sends the ACK, keeping both sides synchronized.
Limitation: Only one frame is outstanding at a time, so link utilization is low on long-delay or high-bandwidth links.
(a) Explain how the Cyclic Redundancy Check (CRC) detects errors. For the data word 1101011011 and the generator polynomial G(x) = x^4 + x + 1, compute the CRC and show the transmitted code word using binary division. (10)
(b) State the relationship between Hamming distance and a code's ability to detect and correct errors. How many bit errors can a code with a minimum Hamming distance of 5 detect and correct? (5)
(a) Cyclic Redundancy Check (CRC) (10)
Principle: The sender treats the data word as a binary polynomial. It appends zero bits (where = degree of generator ) and divides this by using modulo-2 (XOR) division. The remainder is the CRC; it replaces the appended zeros to form the transmitted code word, which is exactly divisible by . The receiver divides the received word by ; a non-zero remainder indicates an error.
Given: Data = 1101011011, divisor = 10011 (degree ).
Append 4 zeros: dividend = 1101011011 0000.
Modulo-2 division (XOR at each step):
1101011011 0000 ÷ 10011
11010 ^ 10011 = 01001
bring down → 010011
10011 ^ 10011 = 00000
bring down bits → 000001 ...continuing the standard XOR long division
Performing the full long division gives:
1 1 0 0 0 1 0 1 0 0 (quotient)
10011 ) 1101011011 0000
10011
-----
010011
10011
-----
000001 1011
1 0011
------
1 1100
1 0011
------
1111 0
1001 1
------
110 10
100 11
------
10 010
10 011
------
001 → remainder bits
The remainder (last bits) is 1110 (CRC).
Transmitted code word = data + CRC =
Note: the CRC is the 4-bit remainder of the modulo-2 division; dividing the transmitted word
11010110111110by10011gives remainder 0, confirming a valid code word.
(b) Hamming Distance, Error Detection & Correction (5)
The minimum Hamming distance of a code is the smallest number of bit positions in which any two valid code words differ.
- To detect up to errors: , so .
- To correct up to errors: , so .
For :
Result: A code with minimum Hamming distance 5 can detect up to 4-bit errors and correct up to 2-bit errors.
Section B: Short Answer Questions
Attempt all / any as specified.
Differentiate between Frequency Division Multiplexing (FDM) and Time Division Multiplexing (TDM). With a neat diagram, explain how synchronous TDM works and why guard bands are required in FDM.
FDM vs TDM
| Feature | FDM (Frequency Division Multiplexing) | TDM (Time Division Multiplexing) |
|---|---|---|
| Sharing of | Bandwidth (frequency) | Time |
| Signal type | Analog | Digital (primarily) |
| Each channel gets | A separate frequency band | A separate time slot |
| Separation | Guard bands (frequency gaps) | Guard times / framing |
| Operation | All channels transmit simultaneously on different frequencies | Channels take turns using the full bandwidth |
Synchronous TDM
In synchronous TDM, the time axis is divided into frames, and each frame is divided into slots, one fixed slot per input line. A multiplexer scans the inputs in round-robin order and places each input's data into its dedicated slot in every frame, even if that input has no data (the slot is then left empty). The demultiplexer at the receiver, synchronized by framing bits, distributes slot data back to the correct output lines.
Diagram (described): Inputs A, B, C feed a MUX; output stream is A1 B1 C1 | A2 B2 C2 | A3 B3 C3 | ... where each | marks a frame boundary.
Drawback: slots reserved for idle inputs waste bandwidth — solved by statistical (asynchronous) TDM.
Why guard bands in FDM
Real filters are not ideal, so adjacent frequency channels can overlap. Guard bands are unused frequency gaps placed between channels to prevent inter-channel interference (crosstalk) and to allow practical band-pass filters to separate the channels cleanly.
Explain the three main causes of transmission impairment: attenuation, distortion, and noise. A signal travels through a transmission medium and its power is reduced to one-half. Express this attenuation in decibels (dB).
Three Causes of Transmission Impairment
- Attenuation — Loss of signal energy/strength as it travels through the medium, due to resistance/absorption. It increases with distance and frequency; amplifiers or repeaters are used to compensate. Measured in decibels (dB).
- Distortion — Change in the shape of the signal. In composite signals, different frequency components travel at different speeds and have different delays, so they arrive with shifted phases and the combined waveform is distorted.
- Noise — Unwanted external energy added to the signal: thermal noise (random electron motion), induced noise (from motors/appliances), crosstalk (coupling between wires), and impulse noise (spikes from lightning/switching).
Attenuation in Decibels
Given the power is reduced to one-half: .
Result: The attenuation is approximately −3 dB (a 3 dB loss), i.e. halving the power corresponds to a 3 dB drop.
With the help of waveforms, explain the three basic digital-to-analog modulation techniques: Amplitude Shift Keying (ASK), Frequency Shift Keying (FSK), and Phase Shift Keying (PSK). Why is QPSK preferred over BPSK?
Digital-to-Analog Modulation Techniques
Digital-to-analog conversion modulates one parameter of a carrier according to the data bits.
1. Amplitude Shift Keying (ASK)
The amplitude of the carrier is changed; frequency and phase stay constant. Typically bit 1 = carrier present (high amplitude), bit 0 = no/low carrier.
Waveform (described): for 1 0 1 1 → carrier burst, flat (off), carrier burst, carrier burst. Simple but very sensitive to noise.
2. Frequency Shift Keying (FSK)
The frequency is changed; amplitude and phase constant. Bit 1 = frequency (higher), bit 0 = frequency (lower).
Waveform (described): for 1 0 1 → fast oscillation, slow oscillation, fast oscillation. More noise-immune than ASK.
3. Phase Shift Keying (PSK)
The phase is changed; amplitude and frequency constant. In BPSK, bit 1 = phase , bit 0 = phase .
Waveform (described): for 1 0 → carrier, then carrier inverted (180° shift) at the bit boundary. Robust against noise; basis of modern modems.
Why QPSK is preferred over BPSK
QPSK uses four phases (), so each symbol carries 2 bits instead of 1.
- For the same baud (symbol) rate, QPSK gives double the bit rate of BPSK.
- Equivalently, QPSK achieves the same bit rate as BPSK using half the bandwidth, improving spectral (bandwidth) efficiency — with essentially the same bit-error-rate performance.
Explain the sliding window concept used in flow control. Differentiate between Go-Back-N ARQ and Selective Repeat ARQ in terms of window size, buffering, and retransmission behavior.
Sliding Window Concept
Flow control with a sliding window lets the sender transmit several frames before needing an acknowledgment, instead of one at a time. Both sender and receiver maintain a window (a range of sequence numbers, numbered modulo ). The sender may send frames while its window is open; as ACKs arrive, the window slides forward, allowing new frames. This keeps the link busy and greatly improves efficiency over Stop-and-Wait.
Go-Back-N ARQ vs Selective Repeat ARQ
| Feature | Go-Back-N ARQ | Selective Repeat ARQ |
|---|---|---|
| Sender window size | Up to | Up to |
| Receiver window size | 1 | (equal to sender) |
| Buffering at receiver | No buffering of out-of-order frames | Buffers out-of-order frames |
| Acknowledgments | Cumulative ACK | Individual / selective ACK |
| Retransmission on error | Resend the erroneous frame and all frames after it | Resend only the specific damaged/lost frame |
| Bandwidth efficiency | Lower (wastes retransmissions) | Higher (minimal retransmission) |
| Complexity | Simpler | More complex (sorting/buffering logic) |
Summary: Go-Back-N is simpler but retransmits more frames on error; Selective Repeat is more efficient on noisy links because it retransmits only the lost frames, at the cost of receiver buffering and a smaller window.
With appropriate signal diagrams, explain the following line-coding schemes: (a) NRZ-L, (b) NRZ-I, (c) Manchester, and (d) Differential Manchester. State one advantage of self-synchronizing schemes.
Line-Coding Schemes
Line coding converts binary data into a digital signal. (Diagrams described per scheme.)
(a) NRZ-L (Non-Return-to-Zero Level)
The voltage level represents the bit: e.g. high (positive) = 0, low (negative) = 1 (or vice versa). The level stays constant for the whole bit.
Signal for 0 1 1 0: high, low, low, high. Problem: long runs of same bit give no transitions → loss of synchronization.
(b) NRZ-I (Non-Return-to-Zero Invert)
The bit is encoded by transition, not level: a 1 causes an inversion at the start of the bit; a 0 means no change. This is differential, so it tolerates polarity reversal.
Signal for 0 1 1 0: stay, invert, invert, stay.
(c) Manchester
Mid-bit transition carries data: e.g. 0 = high-to-low transition, 1 = low-to-high transition (IEEE convention). There is always a transition in the middle of every bit, so it is self-synchronizing, but needs double the bandwidth.
(d) Differential Manchester
Always has a mid-bit transition for timing. The bit value is shown by the presence/absence of a transition at the start of the bit: 0 = transition at the beginning, 1 = no transition at the beginning. It is both self-synchronizing and differential (polarity-independent).
Advantage of Self-Synchronizing Schemes
Schemes like Manchester and Differential Manchester guarantee a transition in every bit period, so the receiver can extract the clock directly from the data stream and stay synchronized even during long runs of 0s or 1s — no separate clock line is needed.
State and explain the Shannon capacity formula and the Nyquist bit-rate formula. A telephone channel has a bandwidth of 3000 Hz and an SNR of 3162 (35 dB). Calculate the maximum theoretical channel capacity using Shannon's formula.
Shannon Capacity Formula
For a noisy channel, the maximum theoretical data rate is
where = bandwidth (Hz) and = signal-to-noise power ratio (not in dB). It gives an upper bound independent of the number of signal levels and accounts for noise.
Nyquist Bit-Rate Formula
For a noiseless channel,
where = number of discrete signal levels. It shows how many levels are needed but ignores noise; in practice the Shannon limit caps how large can usefully be.
Numerical — Shannon Capacity
Given: Hz, (≈ 35 dB).
Result: The maximum theoretical capacity is approximately 34.9 kbps (≈ 35 kbps). (A handy check: at 35 dB SNR, kbps.)
Explain the principle of error detection using a simple parity check and a two-dimensional parity check. Show with an example how a two-dimensional parity check can both detect and correct a single-bit error.
Simple Parity Check
A single parity bit is appended so the total number of 1s is even (even parity) or odd (odd parity). The receiver recomputes parity; a mismatch signals an error. It detects any odd number of bit errors but misses all even-number errors, and cannot correct errors.
Two-Dimensional (2-D) Parity Check
Data is arranged as a table of rows. A parity bit is computed for each row and for each column, giving a row-parity column and a column-parity row. This detects more errors and can pinpoint and correct a single-bit error by intersection.
Example (even parity)
Data block (3 rows × 4 bits) with row parity (P_row) and column parity (P_col):
| b1 | b2 | b3 | b4 | P_row | |
|---|---|---|---|---|---|
| Row1 | 1 | 0 | 1 | 1 | 1 |
| Row2 | 0 | 1 | 1 | 0 | 0 |
| Row3 | 1 | 1 | 0 | 1 | 1 |
| P_col | 0 | 0 | 0 | 0 | 0 |
Now suppose bit at Row2, b3 flips during transmission (1 → 0). At the receiver:
- Row2 parity fails (its 1-count parity is now wrong).
- Column b3 parity fails.
The error lies at the intersection of the failing row and failing column = Row2, b3. The receiver flips that bit back to correct it.
Capability: A single-bit error causes exactly one row and one column parity to fail, uniquely locating and correcting it; two errors in the same row or column can be detected but not always corrected.
Write short notes on any TWO of the following: (a) Spread spectrum techniques (FHSS and DSSS) (b) Wavelength Division Multiplexing (WDM) (c) Unguided media: radio, microwave, and infrared waves
Short Notes (any TWO)
(a) Spread Spectrum Techniques (FHSS & DSSS)
Spread spectrum spreads a signal over a bandwidth much larger than required, improving resistance to jamming, interception, and interference.
- FHSS (Frequency Hopping Spread Spectrum): The carrier hops among many frequencies in a pseudo-random sequence known to both transmitter and receiver. A different frequency is used for each short time interval, so interference on one frequency affects only a fraction of the transmission. Used in Bluetooth.
- DSSS (Direct Sequence Spread Spectrum): Each data bit is XORed with a high-rate pseudo-random chip code (spreading code), multiplying the signal bandwidth by the spreading factor. The receiver uses the same code to de-spread. It offers a processing gain that rejects narrowband noise. Used in early Wi-Fi (802.11b) and CDMA.
(b) Wavelength Division Multiplexing (WDM)
WDM is FDM applied to optical fiber: several optical signals at different wavelengths (colours) of light are combined onto one fiber. A multiplexer (prism/grating-based combiner) merges the wavelengths at the sending end and a demultiplexer separates them at the receiving end. It hugely increases fiber capacity. DWDM (Dense WDM) packs many closely-spaced wavelengths for very high-capacity long-haul links.
(c) Unguided Media: Radio, Microwave, Infrared
- Radio waves (3 kHz–1 GHz): Omnidirectional, can penetrate walls, propagate long distances (including via the ionosphere). Used for AM/FM radio, TV, cordless phones.
- Microwaves (1–300 GHz): Line-of-sight, highly directional (parabolic dishes); need tower-to-tower alignment. Used in terrestrial point-to-point links, cellular, and satellite communication.
- Infrared (300 GHz–400 THz): Very high frequency, short range, cannot pass through walls (good for confined, secure communication). Used in TV remotes and short-range device links.
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