BE Computer Engineering (IOE, TU) Data Communication (IOE, CT 604 / ENCT 253) Question Paper 2079 Nepal

(a) Data rate vs. signal rate (baud rate)

• Data rate (bit rate) is the number of bits transmitted per second, measured in bps. It expresses how fast information travels.
• Signal rate (baud rate) is the number of signal elements (symbols) sent per second, measured in baud. It expresses how fast the physical signal changes.

If each signal element carries $n = \log_2 L$ bits (where $L$ is the number of signal levels), the two are related by:

$$N = S \times \log_2 L \quad\Rightarrow\quad S = \frac{N}{\log_2 L}$$

where $N$ = data rate (bps) and $S$ = signal rate (baud). Increasing the number of levels $L$ packs more bits per symbol, so the data rate rises for the same baud rate (better bandwidth efficiency), but the levels become closer together and more vulnerable to noise.

(b) Nyquist bit rate (noiseless channel)

$$C = 2B \log_2 L$$

with $B = 3\text{ kHz} = 3000\text{ Hz}$.

(i) 2 levels: $C = 2 \times 3000 \times \log_2 2 = 6000 \times 1 = \mathbf{6\,kbps}$

(ii) 8 levels: $C = 2 \times 3000 \times \log_2 8 = 6000 \times 3 = \mathbf{18\,kbps}$

(c) Shannon capacity (noisy channel)

First convert SNR from dB:

$$30 = 10\log_{10}(\text{SNR}) \Rightarrow \text{SNR} = 10^{3} = 1000$$ $$C = B\log_2(1 + \text{SNR}) = 3000 \times \log_2(1001)$$ $$\log_2(1001) \approx 9.97 \Rightarrow C \approx 3000 \times 9.97 \approx \mathbf{29.9\,kbps} \;(\approx 30\,kbps)$$

Why practical rate is lower: Shannon gives a theoretical upper bound assuming ideal coding and infinite-complexity error correction. In practice the rate is limited by finite SNR margins, non-ideal modulation/coding, inter-symbol interference, hardware imperfections, additional impairments (distortion, crosstalk) and the need for a safety margin to keep the bit-error rate acceptably low. Hence achievable rates stay below the Shannon limit.

(a) Digital-to-analog modulation techniques

A carrier $A\cos(2\pi f_c t + \phi)$ has three parameters that can be varied by the digital data: amplitude, frequency, phase.

1. ASK (Amplitude Shift Keying): The amplitude of the carrier is switched between two levels while frequency and phase stay fixed. In binary ASK (OOK), bit 1 = carrier present, bit 0 = no carrier.

• Waveform: full-amplitude bursts for 1s, flat (zero) line for 0s.
• Application: optical fibre / infrared (LED on-off) communication.

2. FSK (Frequency Shift Keying): The frequency is switched between $f_1$ (for 1) and $f_2$ (for 0); amplitude and phase constant.

• Waveform: higher-frequency sinusoid for 1, lower-frequency sinusoid for 0, both at constant amplitude.
• Application: low-speed modems, caller-ID, RFID.

3. PSK (Phase Shift Keying): The phase is changed (e.g. $0°$ for 1, $180°$ for 0 in BPSK); amplitude and frequency constant.

• Waveform: a sinusoid whose phase abruptly flips by $180°$ at each $1\to0$ / $0\to1$ transition.
• Application: Wi-Fi, satellite and cellular links, RFID.

(b) Constellation diagrams

• QPSK: 4 points on a circle at phases $45°, 135°, 225°, 315°$, each carrying 2 bits/symbol.
• 8-PSK: 8 points equally spaced ($45°$ apart) on a circle, each carrying 3 bits/symbol.

The number of points $M$ gives $\log_2 M$ bits per symbol, so more points = higher bandwidth efficiency (more bits per baud). However, for a fixed signal power the points lie on the same circle, so as $M$ grows the angular spacing shrinks and the Euclidean distance between adjacent points decreases, making the symbols easier for noise to confuse. Thus 8-PSK is more bandwidth-efficient than QPSK but is more susceptible to noise (higher BER for the same SNR).

(c) Why QAM over pure PSK at high rates

In high-order PSK all points lie on one circle, so packing many points forces them dangerously close together. QAM varies both amplitude and phase, spreading the points across a 2-D grid. This maximises the minimum distance between constellation points for a given average power, giving a better noise margin and lower BER than PSK of the same order — so QAM achieves higher data rates more reliably.

(a) FDM vs TDM vs WDM

WDM is conceptually FDM applied to very high (optical) frequencies on fibre.

(b) Synchronous TDM calculation

4 channels, each 1 Mbps, frame carries 1 bit per channel ⇒ 4 bits/frame.

(i) Output frame rate: Each channel sends 1 bit per frame at 1 Mbps, so frame rate = channel bit rate = $1\times10^{6}$ frames/s = 1 Mframe/s (i.e. $10^6$ frames per second).

(ii) Frame duration: $T = \dfrac{1}{\text{frame rate}} = \dfrac{1}{10^{6}} = 1\ \mu s$ per frame.

(iii) Output data rate: bits per frame $\times$ frame rate $= 4 \times 10^{6} = \mathbf{4\ Mbps}$ (= sum of the four input rates).

(c) How statistical TDM improves utilisation

In synchronous TDM each source is assigned a fixed slot in every frame, whether or not it has data — idle sources waste their slots. Statistical (asynchronous) TDM allocates slots dynamically, on demand, only to channels that actually have data to send (slots carry an address). This avoids transmitting empty slots, so the link carries useful data more of the time, allowing the combined input rate to exceed the link rate (relying on bursty, non-simultaneous traffic) and giving higher link utilisation.

(a) Comparison of switching techniques

(b) Three phases of a circuit-switched connection

Described as a space/time diagram between caller A, two switches, and callee B:

1. Setup (connection establishment): A sends a request; switches reserve a channel hop-by-hop until B accepts. A dedicated end-to-end circuit is built (shown as the request travelling A→switch→switch→B and an acknowledgement returning).
2. Data transfer: Once the circuit exists, data flows continuously in both directions over the reserved path with no per-packet routing or addressing and minimal delay.
3. Teardown (disconnection): When finished, either end sends a disconnect signal; each switch releases the reserved resources so they can serve other connections.

On the timing diagram the setup occupies an initial round-trip, the long middle band is the data-transfer phase, and a short teardown exchange ends it.

(c) Modern use of virtual-circuit switching

MPLS (Multi-Protocol Label Switching) used in carrier/ISP backbones establishes label-switched paths that behave like virtual circuits, providing traffic engineering and guaranteed QoS for VoIP/VPN traffic. (ATM in older telecom/DSL backbones is another valid example.)

Guided vs unguided media

• Guided (wired) media confine and direct the signal along a physical conductor or waveguide (copper wire or glass fibre). The signal follows the cable; point-to-point.
• Unguided (wireless) media propagate signals through free space (air/vacuum) as electromagnetic waves with no physical conductor — e.g. radio, microwave, infrared. Coverage is broadcast and direction is controlled only by antennas.

Guided media — construction and application

1. Twisted-pair cable: Two insulated copper conductors twisted together; twisting cancels electromagnetic interference and crosstalk. Available as UTP (unshielded) and STP (shielded).

• Application: telephone local loops and Ethernet LAN wiring (Cat 5e/6 cabling).

2. Coaxial cable: A central copper conductor surrounded by an insulating layer, then a braided metallic shield, and an outer plastic jacket. The shield gives higher bandwidth and better noise immunity than twisted pair.

• Application: cable TV distribution and cable-modem broadband; older Ethernet bus networks.

3. Optical fibre: A thin glass/plastic core surrounded by a lower-refractive-index cladding, with a protective buffer/jacket; light is guided by total internal reflection. Immune to EMI, very high bandwidth and low attenuation.

• Application: high-speed long-distance backbone links, FTTH/internet trunks, undersea cables.

CRC computation

Data word: 1101011011 (10 bits)
Generator (divisor): 10011 (5 bits) ⇒ degree 4 ⇒ append 4 zeros.

Augmented dividend: 1101011011 0000

Perform modulo-2 (XOR) division:

1101011011 0000 ÷ 10011

11010 11011 0000
10011
-----
01001 11011 0000
 10011
 -----
 00010 11011 0000     (bring down bits)
   10110 1011 0000   -> leading 1, XOR 10011
   10011
   -----
   00101 1011 0000
     10111 011 0000   -> XOR 10011
     10011
     -----
     00100 011 0000
       10001 1 0000  -> XOR 10011
       10011
       -----
       00010 1 0000
         10100 000   -> XOR 10011
         10011
         -----
         00111 000
           11100 0  -> XOR 10011
           10011
           -----
           01111 0
            1111 0   remainder (4 bits) = 1110

Carrying the division through carefully, the remainder (CRC) = 1110.

Transmitted code word = data + CRC:

$$\texttt{1101011011} \;\Vert\; \texttt{1110} \;=\; \mathbf{1101011011\,1110}$$

Check: dividing 11010110111110 by 10011 gives remainder 0000, confirming the code word is valid.

Marking note: award full marks for correct method (append 4 zeros, modulo-2 division) and a 4-bit remainder appended to form the 14-bit code word; the key result is CRC = 1110.

(a) Hamming distance relationships

Let $d_{min}$ be the minimum Hamming distance of the code.

• To detect up to $s$ bit errors: $d_{min} \ge s + 1$.
• To correct up to $t$ bit errors: $d_{min} \ge 2t + 1$.

So a larger minimum distance allows more errors to be detected/corrected.

(b) Hamming code for data 1001101 (even parity)

For 7 data bits we need $r$ redundant bits where $2^r \ge m + r + 1$. With $m = 7$, $r = 4$ ($2^4 = 16 \ge 7 + 4 + 1 = 12$). Total length = 11 bits.

Redundant (parity) bits go in positions that are powers of two: 1, 2, 4, 8. Data bits fill the rest. Number positions left→right as 1…11. Data 1001101 = $d_1 d_2 d_3 d_4 d_5 d_6 d_7$ = 1 0 0 1 1 0 1.

Each parity bit covers positions whose binary index has that bit set (even parity ⇒ make the covered set sum even):

• P1 covers 1,3,5,7,9,11 → data 1,0,1,1,1 → ones = 4 (even) ⇒ P1 = 0
• P2 covers 2,3,6,7,10,11 → data 1,0,1,0,1 → ones = 3 (odd) ⇒ P2 = 1
• P4 covers 4,5,6,7,12.. → 5,6,7 = 0,0,1 → ones = 1 (odd) ⇒ P4 = 1
• P8 covers 8,9,10,11 → 9,10,11 = 1,0,1 → ones = 2 (even) ⇒ P8 = 0

Transmitted 11-bit code word (pos 1→11):

$$\underbrace{0}_{P1}\ \underbrace{1}_{P2}\ 1\ \underbrace{1}_{P4}\ 0\ 0\ 1\ \underbrace{0}_{P8}\ 1\ 0\ 1 \;=\; \mathbf{01110010101}$$

Given

Bit rate $R = 1\text{ Mbps} = 10^{6}$ bps, propagation delay $T_p = 270$ ms (one-way), frame size $L = 1000$ bits.

Transmission time: $T_t = \dfrac{L}{R} = \dfrac{1000}{10^{6}} = 1\text{ ms}$.

$$a = \frac{T_p}{T_t} = \frac{270}{1} = 270$$

(a) Stop-and-Wait ARQ

Efficiency:

$$\eta = \frac{T_t}{T_t + 2T_p} = \frac{1}{1 + 2a} = \frac{1}{1 + 2(270)} = \frac{1}{541} \approx 0.00185 \approx \mathbf{0.18\%}$$

Throughput $\approx 0.00185 \times 1\text{ Mbps} \approx 1.85$ kbps — extremely poor.

(b) Go-Back-N ARQ, window $N = 7$

$$\eta = \frac{N}{1 + 2a} = \frac{7}{1 + 2(270)} = \frac{7}{541} \approx 0.01294 \approx \mathbf{1.29\%}$$

Throughput $\approx 12.9$ kbps — 7× better than Stop-and-Wait, but still very low.

Comment

The link has a huge bandwidth–delay product ($a = 270$), so Stop-and-Wait wastes almost the entire round-trip waiting for ACKs. Go-Back-N with a larger window is more suitable, but even $N = 7$ is far too small here; for full utilisation we need $N \ge 1 + 2a = 541$, which requires a much larger sequence-number field (sliding-window / pipelining). Hence a sliding-window protocol with a large window (ideally Selective-Repeat) is the right choice for high-delay satellite links.

Pulse Code Modulation (PCM)

PCM converts an analog signal into a digital bit stream in three main steps. Block diagram (in words):

$$\text{Analog in} \rightarrow \boxed{\text{Sampler (PAM)}} \rightarrow \boxed{\text{Quantizer}} \rightarrow \boxed{\text{Encoder}} \rightarrow \text{PCM bit stream}$$

1. Sampling: The continuous analog signal is sampled at regular intervals (rate $f_s$), producing a PAM (pulse-amplitude-modulated) sequence of discrete-time samples.

2. Quantization: Each sample amplitude is approximated to the nearest of a finite set of levels (e.g. $L = 2^n$ levels). This introduces quantization noise/error.

3. Encoding: Each quantized level is mapped to an $n$-bit binary code word, giving the digital PCM output. (At the receiver: decode → reconstruct → low-pass filter to recover the analog signal.)

Sampling rate for 4 kHz voice

By the Nyquist sampling theorem, a signal band-limited to $f_{max}$ must be sampled at

$$f_s \ge 2 f_{max}$$

to allow perfect reconstruction without aliasing. For $f_{max} = 4\text{ kHz}$:

$$f_s = 2 \times 4\text{ kHz} = \mathbf{8\,kHz} \;(8000\text{ samples/s}).$$

Justification: Sampling below $2f_{max}$ causes overlapping spectra (aliasing) and irrecoverable distortion; at $f_s = 2f_{max}$ the original signal can be exactly reconstructed by an ideal low-pass filter. This is why standard telephony uses 8 kHz sampling (with 8-bit encoding ⇒ 64 kbps PCM).

Line-coding schemes

Assume the data sequence; high/low refer to voltage levels. (Waveforms described per bit interval.)

1. NRZ-L (Non-Return-to-Zero – Level): The voltage level itself represents the bit — e.g. high = 0, low = 1 (or vice-versa). No transition within a bit; a long run of identical bits keeps a constant level. Simple but no built-in clocking.

2. NRZ-I (Non-Return-to-Zero – Invert): Bit value is encoded by transition vs no transition at the start of the bit: a 1 causes an inversion of the level, a 0 keeps the previous level. Differential, so it is immune to polarity reversal, but long runs of 0s still give no transitions.

3. Manchester: Each bit has a transition in the middle of the bit period: e.g. low→high = 1, high→low = 0. The mid-bit transition carries both data and timing.

4. Differential Manchester: There is always a mid-bit transition (used for clocking); the data is encoded by the presence/absence of a transition at the start of the bit — 0 = transition at start, 1 = no transition at start (differential, polarity-independent).

Self-synchronization

Manchester and Differential Manchester are self-synchronizing, because their guaranteed mid-bit transition provides a clock edge in every bit, letting the receiver recover timing regardless of the data. NRZ-L and NRZ-I are not self-synchronizing — long runs of the same bit produce no transitions and the receiver can lose bit timing.

Transmission impairments

1. Attenuation: Loss of signal energy/power as it travels through the medium, because the medium has resistance and the energy converts to heat. Stronger over longer distances; compensated by amplifiers/repeaters. Measured in dB.

2. Distortion: A change in the shape of the signal. In a composite signal, different frequency components travel at slightly different speeds and arrive with different delays, so the recombined waveform is distorted relative to the original.

3. Noise: Unwanted random energy added to the signal — e.g. thermal noise (random electron motion), induced noise (from motors/appliances), crosstalk (coupling from another wire) and impulse noise (spikes from lightning/switching). Noise corrupts the received signal and raises the bit-error rate.

Attenuation calculation

$$\text{Attenuation (dB)} = 10\log_{10}\frac{P_{out}}{P_{in}} = 10\log_{10}\frac{2.5}{10} = 10\log_{10}(0.25)$$ $$= 10 \times (-0.602) \approx \mathbf{-6.02\ dB}$$

The negative sign indicates a loss of about 6 dB (the power has dropped to one-quarter, and each halving of power is $\approx -3$ dB, so two halvings give $\approx -6$ dB).

Sliding-window flow control

Sliding-window flow control lets the sender transmit several frames before needing an acknowledgement, up to a limit set by the window size, instead of one-at-a-time. As ACKs arrive, the window slides forward to allow new frames. This keeps the pipe full and greatly improves utilisation on links with significant propagation delay. Sequence numbers are taken modulo $2^m$ for an $m$-bit field.

Sender window vs receiver window

• Sender window: the set/range of sequence numbers of frames the sender may transmit without waiting for an ACK (outstanding, unacknowledged frames). It shrinks as frames are sent and grows (slides) as ACKs return.
• Receiver window: the range of sequence numbers the receiver is prepared to accept. It defines which incoming frames are valid and slides as frames are correctly received and delivered.

Maximum window size for an $m$-bit sequence number

• Go-Back-N ARQ: $W_{max} = 2^{m} - 1$ (receiver window = 1).
• Selective-Repeat ARQ: $W_{max} = 2^{m-1}$ (sender and receiver windows equal, each half the sequence-number space, to avoid ambiguity).

Spread spectrum

Spread spectrum is a technique in which the signal is deliberately spread over a bandwidth much wider than the minimum needed to carry the information, using a code (spreading sequence) independent of the data. This provides resistance to interference and jamming, low probability of intercept, privacy, and the ability to share a band among many users (CDMA).

FHSS vs DSSS