1. Linear Path Loss

a) Define linear path loss in a wireless communication system.

Linear path loss

Linear path loss is the ratio of the transmitted power to the received power in a wireless communication system, expressed as a simple number without units. It shows how much the signal strength weakens as it travels from the sender (transmitter) to the receiver. For example, if the transmitter sends 10 units of power and the receiver gets 1 unit, the linear path loss is 10 (meaning the signal lost 90% of its power).

b) Explain why engineers often prefer to express path loss in decibels (dB) instead of linear scale when analyzing wireless links.

Engineers prefer decibels (dB) because large numbers in linear scale (like millions or billions) become easier to handle as smaller numbers in dB. For example, a linear loss of 1,000,000 is just 60 dB. dB also makes addition and subtraction simple for combining gains and losses, instead of multiplication in linear scale. This helps in quick calculations for complex systems like antennas and cables.

c) A base station transmits a signal at 25 W and a mobile device receives it with power 5 × 10⁻⁷ W.

i) Calculate the linear path loss between the transmitter and receiver.

Formula: Linear Path Loss (L) = P_t / P_r

• Definitions of variables:
• P_t: Transmitted power (in Watts) = 25 W
• P_r: Received power (in Watts) = 5 × 10⁻⁷ W

ii) Convert the path loss you obtained into decibels (dB).

Formula: Path Loss in dB (L_dB) = 10 × log₁₀(L)

• Definitions of variables:
• L: Linear path loss = 5 × 10¹⁰

d) Based on your results, what does this level of attenuation suggest about the propagation environment (e.g., urban, rural) and the approximate distance between the transmitter and receiver?

This high attenuation (107 dB or 5 × 10¹⁰ linear) suggests an urban environment with many obstacles like buildings and traffic, causing signal blocking and scattering. In rural areas, loss is lower due to open spaces. The distance might be several kilometers (e.g., 5-10 km), as free space loss increases with distance squared, but urban factors make it worse.

2: Free Space Propagation (Friis Equation)

a) State the Friis free space transmission equation and briefly explain the meaning of each term in the formula.

Formula: P_r = (P_t × G_t × G_r × λ²) / ((4πd)²)

• Definitions of variables:
• P_r: Received power (Watts) - power arriving at receiver.
• P_t: Transmitted power (Watts) - power sent by transmitter.
• G_t: Transmitter antenna gain (linear) - how much the antenna focuses the signal.
• G_r: Receiver antenna gain (linear) - how much the receiving antenna captures signal.
• λ: Wavelength (meters) - distance between wave peaks, related to frequency.
• d: Distance between antennas (meters).

b) Describe a practical situation where free space propagation is a reasonable assumption, and one where it might not be appropriate.

Reasonable assumption: Satellite communication in space, where there's no air, buildings, or ground to interfere - just empty space.

Not appropriate: Urban mobile phone calls, where buildings, cars, and people cause reflections, absorptions, and blocking.

c) Suppose two communication antennas have clear line-of-sight at 2.4 GHz, the transmitter radiates 15 W, and both antennas have 6 dB gain. If the antennas are separated by 3 km:

i) Calculate the signal wavelength.

Formula: λ = c / f

• Definitions of variables:
• λ: Wavelength (meters)
• c: Speed of light = 3 × 10⁸ m/s
• f: Frequency = 2.4 × 10⁹ Hz

ii) Using the Friis equation, estimate the received power at the mobile device.

First, convert gains: G_t = G_r = 10^(6/10) = 3.981 (linear)

P_r = (15 × 3.981 × 3.981 × (0.125)²) / ((4π × 3000)²)

iii) Express the received power result in dBm.

P_dBm = 10 × log₁₀(P_r × 1000)

d) Discuss whether the obtained received power would be acceptable for reliable communication in practical cellular systems.

-55.82 dBm is acceptable for many cellular systems, as typical sensitivity is around -100 dBm or lower. It allows good signal for data or voice, but in noisy areas, it might need boosting.

3: Free Space Path Loss in Satellite Communication

a) Explain why a satellite communication link experiences high free space path loss compared to a terrestrial link.

Satellites are far away (e.g., 36,000 km), so distance (d) in FSPL formula is huge, making loss proportional to d² very large. Terrestrial links are shorter (km), so less loss.

b) Give two practical design methods used in satellite systems to compensate for this large path loss.

1. Use high-gain antennas to focus signals like a spotlight.
2. Increase transmit power or use sensitive receivers to detect weak signals.

c) A geostationary satellite is at a distance of 36 000 km and communicates at 4 GHz. The ground station antenna gain is 45 dB and the satellite antenna gain is 40 dB. If the transmitter outputs 250 W:

i) Estimate the free space path loss.

FSPL (dB) = 20 log₁₀(d) + 20 log₁₀(f) + 20 log₁₀(4π/c) but standard: FSPL = 20 log₁₀(d) + 20 log₁₀(f) - 147.55 (for d in km, f in GHz)

ii) Calculate the net path loss once the antenna gains are included.

Net loss = FSPL - G_t - G_r = 195.56 - 45 - 40 = 110.56 dB

iii) Calculate the received power at the satellite.

P_t = 250 W = 23.98 dBW = 53.98 dBm

P_r (dBm) = P_t (dBm) - Net loss = 53.98 - 110.56 = -56.58 dBm

d) Explain why high-gain antennas are essential in such systems and how they affect link reliability

High-gain antennas boost signal directionally, reducing effective loss. They improve reliability by increasing received power, countering noise and fading for clearer links.

4: Reference Distance Path Loss Model

a) Define the reference distance (d_0) used in path loss modeling and explain its purpose.

Reference distance (d_0)

d_0 is a standard close distance where path loss is measured or calculated accurately, used as a starting point to predict loss at farther distances. Purpose: To avoid errors in models at very short distances where free space assumptions fail.

b) Explain why (d_0) is normally set to a small value for indoor systems and a larger value for outdoor cellular systems.

Indoor: Small d_0 (e.g., 1 m) because spaces are small, signals vary quickly with walls. Outdoor: Larger d_0 (e.g., 100 m) as far-field starts later, and environments are open.

c) Measurements show that at a reference distance of 100 m from a base station the received signal is –35 dBm. Assuming free-space path loss:

i) Estimate the received power at 500 m from the base station.

Free space: P_r(d) = P_r(d_0) × (d_0 / d)^2

ii) Estimate the power at 1000 m (1 km).

d) Based on these values, discuss how path loss with distance influences cell planning and base station spacing.

Power drops quickly with distance (as d^2), so cells must be closer in high-loss areas to maintain signal. Planning: Space base stations to overlap coverage, avoiding dead zones.

5. Path Loss Exponent and Environment

a) Define the path loss exponent and explain its role in wireless propagation models.

Path loss exponent (n)

n is a number showing how fast signal weakens with distance; higher n means faster loss. Role: Adjusts models to fit environments, like n=2 for free space.

b) Compare typical values of the path loss exponent in free space, urban, and indoor environments.

Free space: n=2 (open air). Urban: n=3-5 (buildings block). Indoor: n=1.5-4 (walls vary).

c) In an urban area, a user receives –60 dBm at a location 200 m from a base station. If the path loss exponent is taken as 4:

i) Estimate the received power at 400 m.

P_r(d) = P_r(d_0) - 10 n log₁₀(d/d_0)

ii) Estimate the received power at 800 m.

d) Explain what these results imply for coverage and the need for additional base stations in dense urban deployments.

Power drops fast (12 dB per double distance for n=4), so coverage shrinks, needing more bases to fill gaps, increasing costs but improving service.

6: Noise Power and Noise Figure

a) Define thermal noise power and noise figure in the context of a mobile receiver.

Thermal noise power:

Random power from heat in electronics, limiting weak signal detection.

Noise figure:

Measure of how much extra noise a receiver adds, in dB; lower is better.

b) Explain how the noise figure affects the sensitivity of a receiver.

Higher noise figure adds more noise, reducing sensitivity (ability to detect weak signals), like making ears less sharp.

c) A mobile receiver has a bandwidth of 200 kHz, a noise figure of 6 dB, and a standard system temperature of 290 K.

i) Calculate the total noise power at the receiver input.

N = k T B F (linear)

• k: Boltzmann's constant = 1.38 × 10⁻²³ J/K
• T: Temperature = 290 K
• B: Bandwidth = 200,000 Hz
• F: Noise factor = 10^(NF/10) = 10^(0.6) ≈ 4

ii) Convert the noise power into dBm.

d) What is the practical impact of increasing receiver bandwidth on noise power and system performance?

More bandwidth lets in more noise, raising N, which worsens SNR and reduces range or data rate.

7: Signal-to-Noise Ratio (SNR)

a) Define Signal-to-Noise Ratio (SNR) in wireless communications.

SNR

SNR is the ratio of signal power to noise power, showing how much stronger the useful signal is over background noise.

b) Explain why SNR is a key indicator of link quality and data reliability.

High SNR means clear signal, low errors; low SNR causes distortions, drops.

c) A mobile phone receives a signal at –85 dBm while the noise power is –105 dBm:

i) Calculate the SNR in dB at the receiver.

ii) Determine whether the SNR meets a minimum requirement of 12 dB for acceptable quality.

Yes, 20 dB > 12 dB, so acceptable.

d) Interpret the result with regard to voice quality and bit error performance.

20 dB SNR gives clear voice, low bit errors (e.g., <10^-5), reliable for calls/data.

8: System Gain and Fade Margin

a) Define system gain and fade margin in mobile communication systems.

System gain:

Total power advantage from transmit to minimum receive level.

Fade margin:

Extra signal above minimum to handle fading.

b) Explain why fade margin is crucial when designing links that must tolerate fading and shadowing.

It buffers against signal drops from obstacles/weather, keeping link up.

c) A base station transmits with 33 dBm and the minimum received power required for acceptable performance is –97 dBm:

i) Calculate the system gain.

ii) If the actual measured received power is –80 dBm, calculate the fade margin.

d) Using practical reasoning, discuss how fade margin improves link reliability in real cellular systems.

17 dB margin absorbs 10-20 dB fades from rain/trees, reducing outages, like extra battery in phone for long days.

9: Frequency Reuse and Co-Channel Distance

a) Define frequency reuse, cluster size, and co-channel reuse ratio in a cellular network.

Frequency reuse:

Using same frequencies in different cells to increase capacity.

Cluster size (K):

Number of cells before frequencies repeat.

Co-channel reuse ratio (D/R):

Distance to same-frequency cell over cell radius.

b) In a standard hexagonal cellular layout with cell radius 1.5 km and a reuse pattern defined by (i = 2) and (j = 1):

i) Calculate the cluster size.

ii) Estimate the distance between the centers of nearest co-channel cells.

iii) Determine the frequency reuse ratio.

c) Discuss how these parameters influence system capacity and interference levels.

Smaller K increases capacity (more channels/cell) but raises interference; larger K reduces interference but lowers capacity.

10: Signal-to-Interference Ratio (SIR)

a) Define Signal-to-Interference Ratio (SIR) and distinguish it from SNR.

SIR:

Ratio of desired signal to interfering signals from other sources.

Distinction: SNR is vs. noise (natural), SIR vs. man-made interference.

b) Explain why increasing transmit power alone does not resolve co-channel interference in cellular systems.

Increasing power boosts both signal and interference equally, so SIR stays same.

c) Consider a hexagonal cellular layout where the cell radius is 1 km, cluster size 7, and the mobile experiences interference from six equidistant co-channel neighbors:

i) Estimate the Signal-to-Interference Ratio (SIR) at the edge of the serving cell.

ii) Express the SIR result in dB.

d) If the minimum acceptable SIR requirement is 18 dB, evaluate whether this system would meet that requirement and suggest two practical approaches (such as sectoring or adjusting reuse patterns) to improve network performance.

Yes, 18.66 > 18 dB, meets. Approaches: 1) Sectoring (divide cells into sectors) reduces interferers. 2) Increase cluster size for larger D, better SIR.

11. Basic Link Budget Concept

a) Define a link budget and explain its importance in determining whether a wireless communication link is feasible.

Link budget:

Accounting of all gains/losses from transmitter to receiver.

Importance: Predicts if received power > sensitivity for feasible link.

b) A wireless base station transmits a signal with an output power of 23 dBm. The signal passes through a transmitter cable with 2 dB loss before reaching an antenna with a gain of 12 dBi. The receiver uses an antenna with 15 dBi gain, followed by a receiver cable with 2 dB loss. The communication distance between the transmitter and receiver is 4 km, and the system operates at 2.4 GHz.

i) Calculate the free space path loss (FSPL) for this link.

ii) Determine the received signal level (RSL) in dBm using the link budget equation.

RSL = P_t - Tx_cable + Tx_ant - FSPL + Rx_ant - Rx_cable = 23 - 2 + 12 - 112.04 + 15 - 2 ≈ 23 - 2 + 12 + 15 - 2 - 112.04 = 46 - 116.04 ≈ -70.04 dBm

c) If the receiver sensitivity is –85 dBm, interpret the result by determining whether the link is feasible and comment on the reliability of the communication link.

-70.04 > -85 dBm, feasible with 15 dB margin, reliable against minor fades.

12. Link Margin and Reliability

a) Define receiver sensitivity and link margin, and explain how they are related in wireless system design.

Receiver sensitivity:

Minimum power for acceptable performance.

Link margin:

Extra power above sensitivity.

Related: Margin = RSL - sensitivity; designs aim for positive margin.

b) A point-to-point wireless link is designed to operate over a distance of 6 km at a frequency of 5 GHz. The transmitter outputs 20 dBm, the transmitting antenna gain is 10 dBi, and the receiving antenna gain is 14 dBi. Both the transmitter and receiver cables introduce 3 dB loss each.

i) Calculate the free space path loss for the link.

ii) Calculate the received signal level at the receiver.

RSL = 20 - 3 + 10 - 121.94 + 14 - 3 = 20 + 10 + 14 - 3 - 3 - 121.94 = 44 - 127.94 ≈ -83.94 dBm

iii) Determine the link margin if the receiver sensitivity is –90 dBm.

c) Based on the calculated link margin, explain whether the link would be considered robust enough for reliable outdoor communication, considering fading and environmental variations.

6 dB margin is marginal; okay for clear days but not robust against 10-20 dB fades from rain/trees - may need improvements.

13. Uplink and Downlink Link Budget Comparison

a) Explain why uplink and downlink link budgets are often different in cellular and wireless access networks.

Uplink (mobile to base): Lower mobile power, smaller antennas. Downlink (base to mobile): Higher base power, better antennas. So uplink often weaker.

b) Consider a wireless access point communicating with a remote client over a distance of 5 km at 2.4 GHz. The access point transmits at 20 dBm using a 10 dBi antenna, while the client transmits at 15 dBm using a 14 dBi antenna. Both sides have 2 dB cable loss, and the free space path loss at this distance is assumed to be 114 dB.

i) Calculate the received signal level at the client (downlink).

Downlink RSL = 20 - 2 + 10 - 114 + 14 - 2 = 20 + 10 + 14 - 2 - 2 - 114 = 44 - 118 = -74 dBm

ii) Calculate the received signal level at the access point (uplink).

Uplink RSL = 15 - 2 + 14 - 114 + 10 - 2 = 15 + 14 + 10 - 2 - 2 - 114 = 39 - 118 = -79 dBm

c) If the receiver sensitivity of the client is –82 dBm and that of the access point is –89 dBm, interpret the results by identifying which link direction is more critical and explain the practical design implication for system planning.

Downlink: -74 > -82 (8 dB margin). Uplink: -79 > -89 (10 dB margin). Downlink more critical (less margin). Implication: Boost mobile power or use better antennas for balanced links.