Module Code: CSU 08112 - Module Name: Wireless Technology - Credits: 12

Tutorial Questions Set 1 - Questions and Answers

This document provides detailed answers to the tutorial questions in simple, understandable English. Each answer includes well-defined terms, descriptions, real-life vivid examples, well-arranged calculations, formulas, and definitions of variables used in the formulas. Relevant pictures are embedded using iframes for each main point.

1. Linear Path Loss

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

Linear path loss is the ratio of the transmitted power to the received power in a wireless communication system, expressed without using logarithms. It measures how much the signal strength decreases as it travels from the transmitter to the receiver due to distance and other factors. In simple terms, it's like how a shout gets weaker the farther you are from the person shouting.

Real-life vivid example: Imagine throwing a ball across a field. The farther you throw it, the less speed (power) it has when it reaches the catcher. Similarly, in wireless signals, like your Wi-Fi router signal getting weaker in a distant room.

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

Engineers prefer decibels because large numbers in linear scale (like millions or billions) become smaller, manageable numbers in dB. Decibels use logarithms, making multiplication (like combining gains and losses) into simple addition and subtraction. This simplifies calculations in complex systems.

Real-life vivid example: Instead of saying a signal is weakened by a factor of 1,000,000 (linear), engineers say 60 dB loss, which is easier to add to other losses, like in planning a cell phone tower where multiple factors add up.

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)
- P_r: Received power (in Watts)
- L: Linear path loss (unitless)

Given:
P_t = 25 W
P_r = 5 × 10^{-7} W

L = 25 / (5 × 10^{-7}) = 25 / 0.0000005 = 50,000,000 = 5 × 10^7

So, the linear path loss is 5 × 10^7.

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

Formula: Path Loss in dB = 10 × log_{10}(L)

Definitions of variables:
- L: Linear path loss (unitless)
- log_{10}: Base-10 logarithm

L = 5 × 10^7

Path Loss (dB) = 10 × log_{10}(5 × 10^7) = 10 × (log_{10}(5) + log_{10}(10^7)) = 10 × (0.6990 + 7) = 10 × 7.6990 = 76.99 dB ≈ 77 dB

So, the path loss in dB is approximately 77 dB.

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 level of attenuation (77 dB) suggests an urban environment with obstacles like buildings causing signal weakening. In rural areas, loss is lower for the same distance. The distance might be around 1-5 km, depending on frequency, as urban areas have higher loss due to reflections and absorptions.

Real-life vivid example: In a city like New York, your cell phone signal drops quickly behind tall buildings, unlike in open countryside where it travels farther.

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 × (λ / (4πd))^2

Definitions of variables:
- P_r: Received power (Watts)
- P_t: Transmitted power (Watts)
- G_t: Transmitter antenna gain (linear)
- G_r: Receiver antenna gain (linear)
- λ: Wavelength (meters)
- d: Distance between antennas (meters)
- π: Pi (3.1416)

This equation calculates received power in free space, assuming no obstacles.

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

Reasonable: Satellite to ground station in open space, like GPS signals from space.
Not appropriate: Inside a building with walls, like Wi-Fi in a house, where reflections occur.

Real-life vivid example: Free space: Communicating with a drone in open air. Not: Cell phone in a tunnel.

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 (m)
- c: Speed of light (3 × 10^8 m/s)
- f: Frequency (Hz)

f = 2.4 GHz = 2.4 × 10^9 Hz
λ = 3 × 10^8 / 2.4 × 10^9 = 0.125 m

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

P_t = 15 W
G_t = 10^{(6/10)} = 3.981 ≈ 4
G_r = 4
d = 3000 m

P_r = 15 × 4 × 4 × (0.125 / (4 × 3.1416 × 3000))^2
4πd ≈ 4 × 3.1416 × 3000 ≈ 37699
λ / (4πd) ≈ 0.125 / 37699 ≈ 3.315 × 10^{-6}
( )^2 ≈ 1.099 × 10^{-11}
P_r ≈ 15 × 16 × 1.099 × 10^{-11} ≈ 240 × 1.099 × 10^{-11} ≈ 2.638 × 10^{-9} W

iii) Express the received power result in dBm.

Formula: P_dBm = 10 × log_{10}(P_r × 1000)

P_r = 2.638 × 10^{-9} W = 2.638 × 10^{-6} mW
10 log_{10}(2.638 × 10^{-6}) ≈ 10 × (-5.578) ≈ -55.78 dBm

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

-55.78 dBm is acceptable for many cellular systems, where sensitivity is around -100 dBm or better. It's good for data transfer, but marginal if noise is high.

Real-life vivid example: Like having enough signal bars on your phone for clear calls.

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 very far (36,000 km), so distance causes high loss, as loss increases with distance squared. Terrestrial links are shorter (km).

Real-life vivid example: Yelling to someone across a street vs. to space.

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

1. High-gain antennas to focus signal.
2. Higher transmit power or sensitive receivers.

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.

Formula: FSPL (dB) = 32.44 + 20 log_{10}(f_{MHz}) + 20 log_{10}(d_{km})

f = 4000 MHz, d = 36000 km
FSPL = 32.44 + 20 log_{10}(4000) + 20 log_{10}(36000) = 32.44 + 72.04 + 90.10 = 194.58 dB

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

Net path loss = FSPL - G_t - G_r = 194.58 - 40 - 45 = 109.58 dB

iii) Calculate the received power at the satellite.

P_t = 250 W = 53.98 dBm
P_r (dBm) = 53.98 - 109.58 = -55.6 dBm

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

They boost signal directionally, reducing effective loss, improving reliability by increasing SNR, like using a megaphone.

4: Reference Distance Path Loss Model

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

d_0 is a close distance where path loss is measured or calculated accurately, used as starting point for predicting loss at larger distances. Purpose: To account for near-field effects and provide a baseline.

Real-life vivid example: Measuring signal at 1 m from router, then predicting at 10 m.

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 spaces, d_0=1-10 m to avoid walls. Outdoor: Larger, d_0=100-1000 m for free space assumption.

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.

For free space, P_r(d) = P_r(d_0) × (d_0 / d)^2

d = 500 m, d_0 = 100 m
P_r(500) = -35 + 20 log_{10}(100/500) = -35 + 20 log_{10}(0.2) = -35 + 20*(-0.699) = -35 -13.98 = -48.98 dBm

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

d = 1000 m
P_r(1000) = -35 + 20 log_{10}(100/1000) = -35 + 20 log_{10}(0.1) = -35 + 20*(-1) = -35 -20 = -55 dBm

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

Higher loss means smaller cells in high-loss areas, closer base stations for coverage.

Real-life vivid example: More towers in cities than rural areas.

(Hint noted: n=2 for free space.)

5. Path Loss Exponent and Environment

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

Path loss exponent (n) is the rate at which signal power decreases with distance, in model P_r ∝ 1/d^n. Role: Adjusts model for different environments.

Real-life vivid example: n=2 in open air, like light spreading.

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

Free space: n=2
Urban: n=3-5
Indoor: n=1.5-4

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 log_{10}(d_0/d)

P_r(400) = -60 + 10*4*log_{10}(200/400) = -60 +40*log_{10}(0.5) = -60 +40*(-0.301) = -60 -12.04 = -72.04 dBm

ii) Estimate the received power at 800 m.

P_r(800) = -60 +40*log_{10}(200/800) = -60 +40*log_{10}(0.25) = -60 +40*(-0.602) = -60 -24.08 = -84.08 dBm

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

Rapid drop means smaller coverage, need more base stations for good signal.

Real-life vivid example: Dense towers in downtown areas.

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 signal from heat, N = k T B.
Noise figure: Measure of receiver's added noise, NF = 10 log_{10}(F), F = total noise / kTB.

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

Higher NF adds more noise, reducing sensitivity (ability to detect weak signals).

Real-life vivid example: Noisy earphones make quiet sounds hard to hear.

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.

Formula: N = k T B F, F = 10^{(NF/10)}

k = 1.38 × 10^{-23} J/K
T = 290 K
B = 200000 Hz
NF = 6 dB, F = 10^{0.6} ≈ 3.98

k T B = 1.38e-23 * 290 * 200000 ≈ 8.004e-16 W
N = 8.004e-16 * 3.98 ≈ 3.186e-15 W

ii) Convert the noise power into dBm.

N_dBm = 10 log_{10}(3.186e-15 * 1000) = 10 log_{10}(3.186e-12) ≈ -115 dBm

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

Increases noise, lowering SNR, reducing performance unless signal increases too.

7: Signal-to-Noise Ratio (SNR)

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

SNR = P_signal / P_noise, in dB: 10 log_{10}(P_s / P_n)

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

Higher SNR means clearer signal, lower errors.

Real-life vivid example: Clear conversation in quiet room vs. noisy party.

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.

SNR = -85 - (-105) = 20 dB

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

Yes, 20 > 12.

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

Good voice quality, low bit errors.

8: System Gain and Fade Margin

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

System gain: Total power budget, P_t - P_r_min.
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.

Compensates for signal drops from obstacles or weather.

Real-life vivid example: Extra battery for phone in bad coverage areas.

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.

System gain = 33 - (-97) = 130 dB

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

Fade margin = -80 - (-97) = 17 dB

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

Allows signal to dip without loss of connection.

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.
Cluster size (K): Number of cells per group.
Co-channel reuse ratio (D/R): Distance to reuse / 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.

K = i^2 + i j + j^2 = 4 + 2 + 1 = 7

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

D = R √(3 K) = 1.5 √21 ≈ 1.5 * 4.583 = 6.874 km

iii) Determine the frequency reuse ratio.

1/K = 1/7 ≈ 0.143

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

Smaller K increases capacity but interference.

10: Signal-to-Interference Ratio (SIR)

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

SIR = P_signal / P_interference. SNR includes noise, SIR only interference from other signals.

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

Increases both signal and interference equally.

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.

SIR ≈ (D/R)^{-n} / 6, with n=4, D/R = √(3*7) = √21 ≈4.583

(D/R)^4 ≈ (4.583)^4 ≈ 441
SIR = 441 / 6 ≈ 73.5 linear ≈ 18.66 dB

ii) Express the SIR result in dB.

18.66 dB

d) If the minimum acceptable SIR requirement is 18 dB, evaluate whether this system would meet that requirement and suggest two practical approaches to improve network performance.

Yes, slightly above. Improve: Sectoring antennas, increase cluster size.

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 and losses in dB. Important to check if received power > sensitivity.

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.

f=2400 MHz, d=4 km
FSPL = 32.44 + 20 log10(2400) + 20 log10(4) = 32.44 + 67.6 + 12.04 = 112.08 dB

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

RSL = P_t - L_tx + G_tx - FSPL + G_rx - L_rx = 23 -2 +12 -112.08 +15 -2 = 50 -116.08 = -66.08 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.

Feasible, -66 > -85, reliable with margin.

12. Link Margin and Reliability

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

Sensitivity: Minimum P_r for operation. Link margin: Actual P_r - sensitivity.

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.

f=5000 MHz, d=6 km
FSPL = 32.44 + 20 log10(5000) + 20 log10(6) = 32.44 + 73.98 + 15.56 = 121.98 dB

ii) Calculate the received signal level at the receiver.

RSL = 20 -3 +10 -121.98 +14 -3 = 44 -127.98 = -83.98 dBm

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

Margin = -83.98 - (-90) = 6.02 dB

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 is marginal, may not handle heavy fading; better with 10-20 dB.

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 has low power, downlink: Base has high power; different antennas, sensitivities.

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: AP to client = 20 -2 +10 -114 +14 -2 = 44 -118 = -74 dBm

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

Uplink: Client to AP = 15 -2 +14 -114 +10 -2 = 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, margin 8 dB. Uplink: -79 > -89, margin 10 dB. Downlink more critical (less margin? Wait, 8 vs 10, downlink less). Implication: Balance power or antennas.