Upon completion of this extended tutorial, students will be able to:
This tutorial focuses on the critical performance issues that arise at the wireless link layer. While Tutorial 2 covered the physical propagation channel, we now examine how those physical phenomena translate into measurable performance metrics—throughput, error rate, latency, and reliability. We explore the fundamental trade‑offs between data rate, coverage, and robustness, and we dissect the protocols (AMC, ARQ, HARQ) that modern systems use to adapt to the time‑varying wireless channel. We also delve into advanced topics such as the diversity–multiplexing trade‑off, which underpins the design of MIMO systems. This extended version includes rich theoretical derivations, practical case studies, and a substantial set of assessment materials.
The probability of bit error for coherent BPSK in AWGN is:
\[ P_b = Q\left(\sqrt{\frac{2E_b}{N_0}}\right) \]
where \( Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty e^{-t^2/2} dt \). For M‑QAM (square constellations), the approximate BER is:
\[ P_b \approx \frac{4}{\log_2 M} \left(1 - \frac{1}{\sqrt{M}}\right) Q\left(\sqrt{\frac{3 \log_2 M}{M-1} \cdot \frac{E_b}{N_0}}\right) \]
In Rayleigh fading (no diversity), the average BER for BPSK is:
\[ \overline{P_b} = \frac{1}{2}\left(1 - \sqrt{\frac{\bar{\gamma}}{1+\bar{\gamma}}}\right) \approx \frac{1}{4\bar{\gamma}} \quad \text{at high SNR} \]
This inverse SNR dependence (instead of exponential) explains why diversity is essential in fading channels.
Forward Error Correction (FEC) provides coding gain, which is the reduction in required \( E_b/N_0 \) to achieve a target BER. For a code rate \( r \), the effective data rate is reduced by \( r \), but the coding gain can be 3–10 dB depending on the code (e.g., Turbo codes, LDPC).
In a time‑varying channel, capacity can be defined in two ways:
For Rayleigh fading with average SNR \( \bar{\gamma} \), the outage probability for rate \( R \) (bps/Hz) is:
\[ P_{\text{out}} = \Pr\left[\log_2(1+\gamma) < R\right]=1 - e^{-\frac{2^R - 1}{\bar{\gamma}}} \]
Two stations A and C are out of range of each other but both can communicate with AP B. A transmits to B; C senses the channel as idle (since it cannot hear A) and also transmits, causing a collision at B. Consequence: reduced throughput and wasted channel time. Mitigation: RTS/CTS handshake (IEEE 802.11).
A station C hears a transmission from A to B, but C's intended receiver D is outside A's range. C refrains from transmitting even though its transmission would not interfere with A↔B. Consequence: under‑utilisation of the channel. Mitigation: RTS/CTS (C can hear RTS but not CTS, so it knows it may transmit).
A nearby transmitter can drown out signals from a distant transmitter, especially in CDMA systems. Mitigation: fast power control to equalise received power at the base station.
For IEEE 802.11 DCF, the saturation throughput S (fraction of time used for successful data transmission) can be approximated by:
\[ S = \frac{P_s P_{tr} E[P]}{(1-P_{tr})\sigma + P_{tr} P_s T_s + P_{tr}(1-P_s) T_c} \]
where \( P_{tr} \) is the probability of transmission, \( P_s \) is the probability of success given transmission, \( E[P] \) is the payload size, \( \sigma \) is the idle slot time, \( T_s \) is the time for a successful transmission, and \( T_c \) is the time for a collision.
AMC selects the MCS (modulation and code rate) based on the current SINR. Each MCS has a target SINR threshold (for a given BLER, e.g., 10%). The system selects the highest MCS that keeps BLER below the target. Advantages: increased average throughput; Challenges: requires accurate channel estimation and feedback (CQI, CSI).
Open‑loop and closed‑loop power control are used to adjust transmit power to achieve a target SNR or to reduce interference. In cellular uplink, tight power control mitigates the near‑far problem and saves battery.
ARQ schemes: Stop‑and‑Wait (low efficiency), Go‑back‑N (moderate), Selective Repeat (best efficiency). Efficiency depends on window size, round‑trip time, and error rate.
HARQ combines FEC and ARQ. Types:
HARQ is used in LTE, 5G, and Wi‑Fi (802.11n/ac/ax).
Mobility causes Doppler spread, which reduces coherence time. Link‑layer effects:
Solutions include robust pilot patterns, faster tracking loops, and make‑before‑break handovers.
Link performance is often expressed as the achieved goodput vs. ideal capacity, with a gap due to overhead and retransmissions.
This fundamental trade‑off (Zheng‑Tse) states that in a MIMO system, diversity gain (reliability) and multiplexing gain (rate) cannot both be maximised simultaneously. For an \( N_t \times N_r \) MIMO channel with diversity order \( d \) and multiplexing gain \( r \) (in bps/Hz), the optimal trade‑off curve is given by a piecewise‑linear function. This framework guides the design of space‑time codes (e.g., Alamouti for diversity, V‑BLAST for multiplexing).
LTE uses 15 MCS levels (QPSK, 16‑QAM, 64‑QAM, with code rates 1/3 to 5/6). The eNodeB estimates SINR from CQI reports and selects an MCS to keep BLER below 10%. Throughput can vary from ~1 Mbps to >100 Mbps depending on channel quality and bandwidth.
5G NR uses adaptive HARQ with asynchronous retransmissions. The maximum number of HARQ retransmissions is configurable. For URLLC, retransmissions are minimised to keep latency low; for eMBB, multiple retransmissions improve reliability.
802.11ax uses 1024‑QAM and OFDMA. It implements AMC based on MCS feedback and also uses BSS colouring and spatial reuse to improve performance in dense deployments.
Test your understanding of link characteristics and performance.
Q1. Define the bit error rate (BER) and explain how it depends on SNR for BPSK in AWGN.
BER is the probability that a transmitted bit is incorrectly received. For BPSK in AWGN, BER = Q(√(2E_b/N₀)). It decreases exponentially with SNR.
Q2. Why does the BER in Rayleigh fading decay only as 1/SNR instead of exponentially?
In Rayleigh fading, the SNR is random (exponential distribution). The average BER is obtained by averaging the AWGN BER over the fading distribution. This averaging removes the exponential tail, resulting in a 1/SNR decay at high SNR.
Q3. What is the hidden terminal problem in wireless LANs? Provide a scenario.
Station A and C are both in range of AP B but out of range of each other. A transmits to B; C senses the medium as idle and transmits, causing a collision at B.
Q4. How does the exposed terminal problem reduce network utilisation?
Station C hears A transmitting to B, so C defers even though its transmission to D would not interfere with A↔B. This reduces concurrent transmissions and lowers throughput.
Q5. Define outage capacity for a fading channel.
Outage capacity is the maximum data rate that can be supported with a given outage probability (i.e., the rate is fixed, and the channel cannot support it for a certain fraction of time).
Q6. Write the expression for outage probability in Rayleigh fading for a rate R (bps/Hz) with average SNR γ̄.
Pout = Pr[log₂(1+γ) < R]=1 - exp(-(2^R - 1)/γ̄).
Q7. What is adaptive modulation and coding (AMC)?
AMC is a technique that selects the modulation order and code rate based on the current channel quality (SINR) to maximise throughput while maintaining a target error rate.
Q8. Explain the difference between Chase Combining (CC) and Incremental Redundancy (IR) in HARQ.
CC retransmits the same packet, and the receiver soft‑combines multiple copies for SNR gain. IR retransmits additional parity bits, increasing coding gain; it is more efficient but requires more memory.
Q9. What is the near‑far problem and how is it mitigated?
A nearby transmitter can overpower a distant transmitter, especially in CDMA. Mitigation: fast closed‑loop power control to equalise received power at the base station.
Q10. Define goodput and how it differs from throughput.
Throughput is the raw data rate including all overhead. Goodput is the application‑layer data rate, excluding retransmissions, headers, and overhead. Goodput ≤ throughput.
Q11. How does Doppler shift affect the link layer performance?
Doppler reduces coherence time, making channel estimation difficult and increasing the error rate. It can cause loss of synchronisation and more frequent retransmissions.
Q12. What is the diversity–multiplexing trade‑off?
It is the fundamental trade‑off in MIMO systems: increasing spatial multiplexing gain (higher data rate) reduces diversity gain (reliability), and vice versa. The Zheng‑Tse curve characterises the optimal trade‑off.
Q13. In the IEEE 802.11 DCF, what is the significance of the backoff counter?
The backoff counter randomises the access time, reducing collision probability. It is decremented when the channel is idle; when it reaches zero, the station transmits.
Q14. Explain the concept of "coding gain" in the context of FEC.
Coding gain is the reduction in required Eb/N₀ (in dB) to achieve a target BER when using error‑correcting coding, compared to an uncoded system.
Q15. What factors determine the maximum throughput of a CSMA/CA wireless network?
Key factors: number of stations, packet size, data rate, backoff parameters (CWmin, CWmax), and the probability of collisions. Higher load increases collisions, reducing throughput.
Q16. How does power control help in cellular networks?
Power control reduces interference to other users, saves battery, mitigates the near‑far problem, and improves cell capacity.
Q17. Describe the difference between a stop‑and‑wait ARQ and a selective‑repeat ARQ.
Stop‑and‑wait sends one frame at a time and waits for ACK; simple but low efficiency. Selective repeat allows multiple outstanding frames and retransmits only lost frames; high efficiency but complex buffer management.
Q18. What is the purpose of RTS/CTS in IEEE 802.11?
RTS/CTS reserves the medium for a data transmission, mitigating the hidden terminal problem by alerting all stations within range of either the sender or receiver.
Q19. How does the channel coherence time affect the choice of modulation and coding?
If the coherence time is short (high mobility), the channel changes quickly, requiring robust modulation and coding (lower rate). If it is long, higher‑rate MCS can be used.
Q20. Explain the concept of "link budget" margin and how it relates to outage probability.
Link budget margin is an extra power reserve to account for fading and shadowing. A larger margin reduces outage probability but may increase interference or power consumption.
Q21. What is the difference between open‑loop and closed‑loop power control?
Open‑loop uses the receiver to estimate path loss and adjust power without feedback. Closed‑loop uses feedback from the receiver (e.g., TPC commands) to fine‑tune power.
Q22. In the context of HARQ, what is the benefit of soft‑combining?
Soft‑combining combines the received signals (or LLRs) from multiple retransmissions, increasing the effective SNR and improving the chance of successful decoding.
Q23. Why is the diversity–multiplexing trade‑off important for MIMO system design?
It provides a theoretical framework to guide the selection of space‑time coding schemes. Depending on the application (reliability‑critical vs. rate‑critical), one can choose a scheme that operates at a specific point on the trade‑off curve.
Q24. How does packet size affect the throughput of a wireless link? Consider both overhead and error rate.
Larger packets reduce overhead (higher efficiency) but increase the probability of packet error (since longer packets are more likely to have errors). There is an optimal packet size that maximises goodput.
Q25. Explain the concept of "coverage–capacity trade‑off" in wireless networks.
To increase coverage, one can increase transmit power or use lower frequencies, but that may increase interference and reduce capacity. Conversely, to increase capacity, one can use smaller cells (higher density) but that reduces coverage per cell.
Apply your knowledge to these performance analysis problems.
Exercise 1: Calculate the BER for BPSK in AWGN at Eb/N0 = 10 dB. Then, compute the average BER in Rayleigh fading at the same average SNR.
Eb/N0 = 10 dB = 10 (linear).
AWGN: BER = Q(√(2×10)) = Q(4.47) ≈
3.9×10⁻⁶.
Rayleigh: BER ≈ 1/(4×10) = 0.025 (2.5%). Rayleigh fading severely degrades BER
without diversity.
Exercise 2: A wireless link uses 16‑QAM with a code rate 1/2. The symbol rate is 2 Msps. What is the data rate? If the required SNR for 16‑QAM is 15 dB, and the channel provides 12 dB, what is the outage probability?
Bits per symbol = log₂(16) = 4. Data rate = 2e6 × 4 × 0.5 = 4 Mbps.
Required SNR = 15 dB =
31.62 linear. Provided SNR = 12 dB = 15.85 linear. Outage occurs when SNR < 31.62. For Rayleigh
with average 15.85, Pout = 1 - exp(-31.62/15.85) = 1 - exp(-2) = 1 - 0.135 =
0.865 (86.5%).
Exercise 3: In an 802.11 WLAN, there are 10 stations contending for the channel. The slot time is 20 µs, successful transmission time is 1 ms, collision time is 500 µs. Use the simplified Bianchi model to estimate throughput for a payload of 1000 bytes (8000 bits) at 100 Mbps. Assume Ptr = 0.2 and Ps = 0.8.
Idle time = (1-Ptr)×σ = 0.8×20e-6 = 16
µs.
Ptr×Ps×Ts = 0.2×0.8×1e-3 = 160
µs.
Ptr×(1-Ps)×Tc = 0.2×0.2×0.5e-3 = 20 µs.
Total average
time = 16+160+20 = 196 µs.
Payload bits = 8000 bits (plus overhead? The formula uses E[P] as
bits, but we need to account for PHY overhead. Assuming E[P] = 8000 bits, S = 8000 / 196e-6 =
40.8 Mbps. But this is ideal; actual is lower due to backoff and other overhead.
Exercise 4: A HARQ system uses Chase Combining. The initial transmission has SNR of 5 dB. After one retransmission, the combined SNR is 8 dB. What is the effective SNR gain? If the required SNR for correct decoding is 10 dB, how many retransmissions are needed?
SNR gain = 8 - 5 = 3 dB (linear factor of 2). Each retransmission doubles the SNR (in linear). Required 10 dB = 10 linear; 5 dB = 3.16 linear. Need factor 10/3.16 = 3.16. Each retransmission factor = 2. So need log₂(3.16) = 1.66 → 2 retransmissions (total 3 transmissions).
Exercise 5: A mobile station moves at 50 km/h in a 2.4 GHz system. Calculate the maximum Doppler shift. If the coherence time is 0.423/fd, how does this affect the link adaptation time?
v = 13.89 m/s. λ = 0.125 m. fd = 13.89/0.125 = 111.1 Hz. Tc = 0.423/111.1 = 3.8 ms. Link adaptation must be faster than 3.8 ms to track the channel; otherwise, outdated CSI leads to errors.
Exercise 6: A system uses adaptive modulation with three modes: QPSK (required SNR 8 dB), 16‑QAM (15 dB), 64‑QAM (22 dB). If the SNR varies uniformly between 5 dB and 25 dB, what is the average throughput (symbol rate = 1 Msps)?
Uniform SNR from 5 to 25 dB. Modes thresholds: 8, 15, 22 dB. Region 1: <8 dB → no transmission (or fallback). Region 2: 8-15 dB → QPSK (2 bits/sym). Region 3: 15-22 dB → 16-QAM (4 bits/sym). Region 4: 22-25 dB → 64-QAM (6 bits/sym). Average bits/sym=(2×7/20 + 4×7/20 + 6×3/20)=(14+28+18)/20=60/20=3 bits/sym. Throughput=1e6×3=3 Mbps.
Exercise 7: Compare the throughput of a stop‑and‑wait ARQ vs. selective‑repeat ARQ for a link with frame error rate 0.1, window size 7, and propagation delay 20 ms, data rate 1 Mbps, frame size 1000 bits.
Transmission time Tf = 1000/1e6 = 1 ms. Round‑trip time = 2×20 + 1 = 41 ms (assuming
ACK negligible). Stop‑and‑wait efficiency = (1 - P)/(1 + 2Tprop/Tf) =
0.9/(1 + 40) = 0.9/41 = 0.022 → 22 kbps.
Selective repeat efficiency = (1 - P) = 0.9 (if
window size large enough to keep pipe full). Throughput = 0.9×1 Mbps = 900 kbps.
Exercise 8: A 2×2 MIMO system uses Alamouti coding (diversity scheme). The SNR per branch is 10 dB. What is the diversity order? What would be the diversity order if the system used V‑BLAST (spatial multiplexing)?
Alamouti provides full diversity order of Nt×Nr = 4 (in independent Rayleigh fading). V‑BLAST (uncoded) provides diversity order of Nr = 2 (since multiplexing trades diversity). With ML decoding, V‑BLAST can achieve diversity Nr - Nt + 1, but with linear receivers it's lower.
Exercise 9: A link has a packet error rate (PER) of 0.2. If the maximum number of HARQ retransmissions is 3, what is the residual packet error probability?
Each transmission has PER = 0.2. With 3 retransmissions (total 4 attempts), the residual error = (0.2)⁴ = 0.0016 = 0.16%.
Exercise 10: In a wireless network, the CCA (Clear Channel Assessment) threshold is -82 dBm. The received signal from a desired transmitter is -75 dBm. An interfering transmitter is also present at -80 dBm. Will the station consider the channel busy? What if the interferer is at -85 dBm?
CCA threshold = -82 dBm. If interference is -80 dBm (above threshold), the station senses the channel as busy (carrier sense). If interference is -85 dBm (below threshold), the station may sense idle, leading to potential collisions if the interferer is actually active (if it can be heard above the noise floor). This demonstrates the hidden terminal problem.
Exercise 11: A 5G system uses a link adaptation algorithm that measures SINR every 1 ms. The channel coherence time is 5 ms. Is this measurement rate adequate? What if the mobility is 100 km/h at 28 GHz?
Coherence time = 5 ms. Measurement every 1 ms is faster than coherence, so it can track the channel well. At 100 km/h, fd = v/λ. λ at 28 GHz = 0.0107 m. v=27.78 m/s. fd = 27.78/0.0107 = 2596 Hz. Tc = 0.423/2596 = 0.16 ms. The measurement period (1 ms) is longer than coherence, causing outdated CSI. Need faster measurement or lower mobility.
Exercise 12: Design an AMC table with 4 MCS levels for a system that must keep BLER < 10%. The SINR thresholds are: MCS0 (QPSK, R=1/3) threshold 0 dB, MCS1 (QPSK, R=1/2) threshold 5 dB, MCS2 (16-QAM, R=1/2) threshold 10 dB, MCS3 (64-QAM, R=3/4) threshold 15 dB. If the measured SINR is 12 dB, which MCS should be selected? What is the resulting data rate if the symbol rate is 1 Msps?
SINR = 12 dB. Select highest MCS with threshold ≤ 12 dB: MCS2 (10 dB) is suitable; MCS3 (15 dB) would cause high BLER. MCS2: 16-QAM (4 bits/sym) × 1/2 = 2 bits/sym. Data rate = 1e6 × 2 = 2 Mbps.
Independent research and analytical problems.
HW1. Derive the BER expression for QPSK in AWGN. How does it compare to BPSK in terms of required Eb/N0 for the same BER?
QPSK can be seen as two orthogonal BPSK channels; the symbol error rate is approximately 2Q(√(Es/N0)) and the bit error rate is about Q(√(2Eb/N0)) for Gray mapping, which is the same as BPSK. Thus, QPSK achieves the same BER as BPSK with the same Eb/N0, but uses twice the bandwidth efficiency.
HW2. Analyse the performance of CSMA/CA with RTS/CTS versus without RTS/CTS in a network with hidden terminals. Derive the throughput gain.
Without RTS/CTS, hidden terminals cause collisions that waste airtime. With RTS/CTS, the channel is reserved, reducing collisions but adding overhead (RTS, CTS frames). The gain depends on the packet size: for large packets, RTS/CTS is beneficial; for small packets, the overhead may outweigh the benefit. The throughput with RTS/CTS is higher when the probability of hidden terminal collisions is significant.
HW3. Derive the outage probability for a MIMO system with two transmit and one receive antenna (Alamouti) in Rayleigh fading. Compare with SISO.
For Alamouti (2×1), the effective SNR is the sum of two independent exponential SNRs (chi‑square with 4 degrees of freedom). The outage probability is Pout = 1 - e^{-γth/γ̄} - (γth/γ̄) e^{-γth/γ̄}. This is lower than SISO for the same average SNR due to diversity.
HW4. Research and compare the HARQ schemes used in LTE (HARQ with 8 processes) and 5G NR (adaptive HARQ with flexible timing). Discuss the impact on latency.
LTE uses synchronous HARQ with fixed timing (8 processes for FDD), providing predictable latency (~8 ms RTT). 5G NR uses asynchronous HARQ with flexible scheduling, allowing faster retransmissions (e.g., 1‑2 ms RTT) for URLLC, but requires more dynamic scheduling and control signalling.
HW5. Explain the concept of "fairness" in wireless MAC protocols. How does the IEEE 802.11 binary exponential backoff provide fairness?
Fairness means each station gets an equal share of the channel. The BEB mechanism gives stations that experience collisions a longer backoff, which reduces their transmission probability, allowing other stations to access the medium. However, this can be unfair in some scenarios (e.g., stations with different packet sizes).
HW6. In a time‑varying channel, what is the difference between ergodic capacity and outage capacity? When would you choose one over the other?
Ergodic capacity is the average rate over all fading states, achievable with long coding and interleaving. Outage capacity is a fixed rate with a guaranteed outage probability. For delay‑tolerant applications (e.g., file download), ergodic capacity is suitable. For real‑time applications (e.g., voice), outage capacity is more appropriate to ensure a minimum rate with high probability.
HW7. Analyse the effect of packet aggregation on MAC efficiency. How does it improve throughput, and what are the drawbacks?
Aggregation combines multiple small packets into a larger frame, reducing overhead (headers, ACKs) and improving efficiency. Drawbacks: increased frame error rate (larger frame is more likely to have errors), increased latency for the last packet, and more complex buffer management.
HW8. Derive the link budget for a satellite link and calculate the fade margin required for 99.9% availability given rain attenuation statistics.
Link budget includes EIRP, path loss, receiver G/T, and required C/N. Rain attenuation is a log‑normal distribution; the fade margin is set to the attenuation value exceeded for 0.1% of the time (for 99.9% availability). This is obtained from ITU‑R rain maps and the specific attenuation model.
HW9. Compare the performance of equal‑gain combining (EGC) and maximal‑ratio combining (MRC) for a 2‑branch diversity system in Rayleigh fading. Which provides better diversity gain and why?
MRC weights each branch by its SNR and co‑phases, providing the maximum possible SNR (sum of branch SNRs). EGC equally weights all branches, resulting in a slightly lower SNR. MRC provides about 1‑2 dB gain over EGC for 2 branches, depending on the correlation.
HW10. Research the concept of "link adaptation" in 5G NR, including the use of CQI, PMI, and RI. How do these feedback mechanisms enable efficient adaptation?
CQI (Channel Quality Indicator) reports the recommended MCS. PMI (Precoding Matrix Indicator) suggests a precoding matrix for beamforming. RI (Rank Indicator) recommends the number of spatial layers. The gNB uses these to select MCS, precoding, and rank, adapting to the channel conditions in near real‑time.
HW11. Analyse the trade‑offs between using a higher code rate and a lower modulation order for a given target data rate and SNR.
For a fixed data rate, one can use high modulation (e.g., 64‑QAM) with low code rate or low modulation (QPSK) with high code rate. The best choice depends on the SNR: at high SNR, 64‑QAM with a moderate code rate is efficient; at low SNR, QPSK with a low code rate is more robust. The optimal pair is given by the capacity of the AWGN channel with modulation constraints.
HW12. Write a critical evaluation of the statement: "With advanced FEC and HARQ, wireless links can achieve near‑Shannon capacity." Discuss the limitations and practical constraints.
Modern codes (LDPC, Turbo) can approach Shannon capacity within 1‑2 dB. However, practical systems have limitations: finite block length, channel estimation errors, hardware impairments (phase noise, non‑linearity), and MAC overhead. Additionally, the variable nature of fading and interference means that average throughput is often much lower than peak capacity. HARQ helps but adds latency. Near‑capacity operation is possible only in ideal conditions with sufficient SNR and low mobility.
This extended tutorial covered the critical performance issues of wireless links, including:
These principles are essential for understanding how modern systems like Wi‑Fi, LTE, and 5G achieve high performance despite the challenging wireless environment. The next tutorial will apply these concepts to the specific architecture and operation of IEEE 802.11 Wireless LANs.
© COMP347 – Unit 7: Wireless and Mobile Networks (Extended Tutorial 3)