FDMA Quantitative Test - Satellite Communication

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Instructions: This test contains 10 quantitative questions on Frequency Division Multiple Access (FDMA) for satellite communication. Attempt all questions. Each question is worth 1 point. Use the button below to reveal answers and detailed explanations.

Question 1

A satellite transponder has a total bandwidth of 36 MHz. If FDMA is used with 6 MHz guard bands between channels and each channel requires 4 MHz bandwidth, what is the maximum number of channels that can be accommodated?

Question 2

A satellite communication system uses FDMA with carriers spaced 5 MHz apart. If the total available bandwidth is 60 MHz and guard bands of 0.5 MHz are required at each end of the spectrum, how many carriers can be supported?

Question 3

In an FDMA satellite system, each earth station transmits a carrier with bandwidth of 2 MHz. If the satellite transponder has a bandwidth of 40 MHz and requires 1 MHz guard bands between carriers, calculate the maximum number of earth stations that can access the transponder simultaneously.

Question 4

An FDMA satellite system operates at 6 GHz uplink with carriers spaced 7.5 MHz apart. If the system has a total bandwidth allocation of 67.5 MHz, how many carriers can be accommodated considering no guard bands at the edges?

Question 5

A satellite transponder with 54 MHz bandwidth is to be shared by 12 stations using FDMA. If guard bands of 1.2 MHz are required between channels, what is the maximum bandwidth available for each station's carrier?

Question 6

In an FDMA system, the carrier-to-interference ratio (C/I) is measured as 18 dB. If the interference increases by a factor of 4, what is the new C/I ratio in dB? Assume all other factors remain constant.

Question 7

An FDMA satellite network has 8 channels, each with bandwidth of 5 MHz. If the channels are spaced with 0.5 MHz guard bands between them, what is the total bandwidth occupied by the system?

Question 8

A satellite transponder operating at 14 GHz downlink has an effective isotropic radiated power (EIRP) of 46 dBW. If the system noise temperature is 500 K and the bandwidth per FDMA channel is 4 MHz, calculate the carrier-to-noise density ratio (C/N₀) in dB·Hz for a single channel.

Question 9

In an FDMA system with 15 channels, the total bandwidth is 80 MHz. If guard bands between channels are equal and each channel has equal bandwidth, and the guard bands at the extremes are half of those between channels, calculate the bandwidth of each channel if the guard band between channels is 1 MHz.

Question 10

An FDMA satellite system shares a 72 MHz transponder among multiple stations. If each station requires 6 MHz for its carrier and 0.8 MHz guard bands are placed between carriers, what is the spectral efficiency in bits/Hz if each carrier supports a data rate of 10 Mbps?

Answers and Explanations

Question 1 Answer: 6 channels

Calculation:

Let N be the number of channels.

Total bandwidth = N × (channel bandwidth) + (N-1) × (guard band)

36 MHz = N × 4 MHz + (N-1) × 6 MHz

36 = 4N + 6N - 6

36 = 10N - 6

10N = 42

N = 4.2

Since we can only have whole channels, maximum N = 4

Wait, let's check: For 4 channels: 4×4 + 3×6 = 16 + 18 = 34 MHz (fits within 36 MHz)

For 5 channels: 5×4 + 4×6 = 20 + 24 = 44 MHz (exceeds 36 MHz)

So maximum is 4 channels.

Explanation: In FDMA, guard bands are placed between channels to prevent interference. The total bandwidth required is the sum of all channel bandwidths plus all guard bands. We solve for N using the equation: Total BW = N×B_ch + (N-1)×B_guard. The calculation shows that 4 channels can be accommodated within the 36 MHz transponder.

Question 2 Answer: 11 carriers

Calculation:

Total available bandwidth after edge guard bands = 60 MHz - 2×0.5 MHz = 59 MHz

Number of carriers = Floor(59 MHz / 5 MHz) = Floor(11.8) = 11

Check: 11 carriers occupy 11×5 MHz = 55 MHz

Plus guard bands at edges: 55 + 0.5 + 0.5 = 56 MHz (within 60 MHz)

Explanation: First subtract the guard bands at both ends of the spectrum. Then divide the remaining bandwidth by the carrier spacing to find how many carriers can fit. We take the floor since we can't have a fraction of a carrier. Each carrier is assumed to be centered in its allocated slot with the spacing defining the center-to-center distance.

Question 3 Answer: 13 earth stations

Calculation:

Let N be the number of earth stations.

Total bandwidth = N × (carrier bandwidth) + (N-1) × (guard band)

40 MHz = N × 2 MHz + (N-1) × 1 MHz

40 = 2N + N - 1

40 = 3N - 1

3N = 41

N = 13.67 ≈ 13 (maximum integer)

Check: For 13 stations: 13×2 + 12×1 = 26 + 12 = 38 MHz (fits within 40 MHz)

Explanation: With N carriers, there are (N-1) guard bands between them. The equation accounts for both the carrier bandwidths and the guard bands. Solving gives N = 13.67, but we can only have whole stations, so the maximum is 13 stations, using 38 MHz of the 40 MHz available.

Question 4 Answer: 9 carriers

Calculation:

Carrier spacing = 7.5 MHz

Total bandwidth = 67.5 MHz

Number of carriers = Total bandwidth / Carrier spacing = 67.5 / 7.5 = 9

Explanation: When carrier spacing is given and no guard bands at edges are mentioned, the number of carriers is simply the total bandwidth divided by the carrier spacing. Each carrier occupies a slot equal to the spacing, which typically includes both the carrier bandwidth and any necessary guard band.

Question 5 Answer: 3.6 MHz

Calculation:

Total bandwidth = 54 MHz

Number of stations (N) = 12

Guard bands between channels = 1.2 MHz each

Number of guard bands = N - 1 = 11

Total guard band bandwidth = 11 × 1.2 = 13.2 MHz

Bandwidth available for all carriers = 54 - 13.2 = 40.8 MHz

Bandwidth per carrier = 40.8 / 12 = 3.4 MHz

Explanation: First calculate the total guard band bandwidth needed between the 12 stations (11 guard bands). Subtract this from the total transponder bandwidth to get the bandwidth available for the carriers themselves. Divide by the number of stations to get the bandwidth per carrier.

Question 6 Answer: 12 dB

Calculation:

Original C/I = 18 dB

Convert to ratio: C/I (ratio) = 10^(18/10) = 10^1.8 ≈ 63.1

Interference increases by factor of 4 → I_new = 4I

New C/I ratio = C/(4I) = (1/4) × (C/I) = (1/4) × 63.1 ≈ 15.78

New C/I in dB = 10 log₁₀(15.78) ≈ 12.0 dB

Explanation: Carrier-to-interference ratio (C/I) is a key parameter in FDMA systems. When interference increases by a factor of 4 (linear scale), the C/I ratio decreases by the same factor. Converting to dB: 10 log₁₀(1/4) = -6.02 dB, so the new C/I = 18 dB - 6.02 dB ≈ 12 dB.

Question 7 Answer: 43.5 MHz

Calculation:

Number of channels (N) = 8

Bandwidth per channel = 5 MHz

Guard band between channels = 0.5 MHz

Number of guard bands = N - 1 = 7

Total channel bandwidth = 8 × 5 = 40 MHz

Total guard band bandwidth = 7 × 0.5 = 3.5 MHz

Total system bandwidth = 40 + 3.5 = 43.5 MHz

Explanation: In an FDMA system with N channels, there are (N-1) guard bands between them. The total bandwidth is the sum of all channel bandwidths plus all guard band bandwidths. No edge guard bands are mentioned in this problem.

Question 8 Answer: 92.8 dB·Hz

Calculation:

We need additional information to solve this: path loss, receiver gain, and Boltzmann's constant.

However, assuming we're only calculating from the given:

EIRP = 46 dBW

Assume typical geostationary satellite path loss at 14 GHz ≈ 206 dB

Assume receiver G/T = 20 dB/K (typical for earth station)

C/N₀ = EIRP - Path loss + G/T + 228.6 dB (where 228.6 dB is 10 log₁₀(k), k = Boltzmann's constant)

C/N₀ = 46 - 206 + 20 + 228.6 = 88.6 dB·Hz

Explanation: The carrier-to-noise density ratio (C/N₀) is calculated using the link budget equation. EIRP is the effective isotropic radiated power, path loss depends on frequency and distance, G/T is the receiver figure of merit, and 228.6 dB is the Boltzmann constant in dB. The actual value would depend on specific system parameters not provided in the question.

Question 9 Answer: 4.5 MHz

Calculation:

Number of channels (N) = 15

Total bandwidth = 80 MHz

Guard band between channels (B_g) = 1 MHz

Guard bands at extremes = 0.5 MHz each (half of B_g)

Let B_c be the bandwidth of each channel.

Total bandwidth = 2×(0.5) + 15×B_c + 14×1 = 80

1 + 15B_c + 14 = 80

15B_c + 15 = 80

15B_c = 65

B_c = 65/15 ≈ 4.33 MHz

Explanation: The system has guard bands at both extremes (each 0.5 MHz), 15 channels, and 14 guard bands between them (each 1 MHz). The equation sums all these components to equal the total 80 MHz bandwidth. Solving for channel bandwidth gives approximately 4.33 MHz per channel.

Question 10 Answer: 1.67 bits/Hz

Calculation:

First, find how many stations can be accommodated:

Let N be the number of stations.

Total bandwidth = N × 6 MHz + (N-1) × 0.8 MHz ≤ 72 MHz

6N + 0.8N - 0.8 ≤ 72

6.8N ≤ 72.8

N ≤ 10.7 → Maximum N = 10

Total data rate = 10 stations × 10 Mbps = 100 Mbps

Total bandwidth used = 10×6 + 9×0.8 = 60 + 7.2 = 67.2 MHz

Spectral efficiency = Total data rate / Total bandwidth used

= 100 Mbps / 67.2 MHz = 1.488 ≈ 1.49 bits/Hz

Explanation: Spectral efficiency measures how efficiently bandwidth is used to transmit data. First determine how many stations can fit in the transponder, then calculate the total data rate those stations can support, and finally divide by the total bandwidth they occupy (including guard bands).