FDMA Satellite Communication Quiz

This quiz contains 10 quantitative questions on Frequency Division Multiple Access (FDMA) in satellite communication systems. Test your knowledge on bandwidth allocation, guard bands, transponder capacity, and other key concepts.

Select the correct answer for each question, then click "Submit Quiz" to see your results. You can then view detailed explanations for each question.

1
A satellite transponder has a total bandwidth of 36 MHz. If each FDMA channel requires 4 MHz including guard bands, what is the maximum number of channels that can be accommodated?
2
An FDMA system allocates 5 MHz bandwidth per channel. If the guard band between channels is 0.5 MHz, what is the center-to-center frequency separation between adjacent channels?
3
In an FDMA satellite system, the uplink frequency is 14.25 GHz and the downlink frequency is 11.95 GHz. What is the frequency translation performed by the satellite transponder?
4
A satellite transponder has a bandwidth of 54 MHz. If each FDMA channel requires 6 MHz (including a 0.5 MHz guard band on each side), how many channels can be allocated?
5
An FDMA system uses QPSK modulation with a symbol rate of 5 Msymbols/s. What is the approximate null-to-null bandwidth requirement for this signal?
6
A C-band satellite transponder has a total bandwidth of 40 MHz. If guard bands of 1 MHz are required between channels, and each channel needs 5 MHz for transmission, what is the maximum number of channels?
7
In an FDMA system, if the carrier-to-interference ratio (C/I) is required to be at least 18 dB, and the adjacent channel interference causes a C/I of 15 dB, what additional guard band (as percentage of channel bandwidth) is needed to meet the requirement? Assume interference is proportional to spectral overlap.
8
An FDMA satellite system has 8 channels, each with a bandwidth of 4 MHz. If the total available bandwidth is 40 MHz, what percentage of bandwidth is wasted as guard bands?
9
For an FDMA system with a channel bandwidth of 5 MHz and a roll-off factor (α) of 0.25, what is the symbol rate that can be supported?
10
In an FDMA satellite network, if the transponder's 1 dB compression point is at +10 dBm and each carrier has a power of -5 dBm, how many carriers can be supported before intermodulation products become significant? Assume equal carrier powers and third-order intermodulation.
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Answers and Explanations

Question 1: Correct Answer - B (9 channels)
The maximum number of channels = Total bandwidth ÷ Bandwidth per channel = 36 MHz ÷ 4 MHz = 9 channels. In FDMA, channels are non-overlapping, so we simply divide the total available bandwidth by the bandwidth required for each channel (including guard bands).
Question 2: Correct Answer - C (5.5 MHz)
Center-to-center frequency separation = Channel bandwidth + Guard band = 5 MHz + 0.5 MHz = 5.5 MHz. The guard band is placed between channels to prevent interference, so the total separation between channel centers includes both the channel bandwidth and the guard band.
Question 3: Correct Answer - A (2.30 GHz decrease)
Frequency translation = Uplink frequency - Downlink frequency = 14.25 GHz - 11.95 GHz = 2.30 GHz decrease. Satellite transponders typically translate received uplink frequencies to different downlink frequencies to avoid interference between transmitted and received signals.
Question 4: Correct Answer - B (9 channels)
Each channel requires 6 MHz (including guard bands). Number of channels = 54 MHz ÷ 6 MHz = 9 channels. With 0.5 MHz guard bands on each side, the actual information bandwidth per channel would be 5 MHz, but the total allocated bandwidth per channel is 6 MHz.
Question 5: Correct Answer - B (10 MHz)
For QPSK, null-to-null bandwidth ≈ 2 × symbol rate = 2 × 5 Msymbols/s = 10 MHz. The null-to-null bandwidth is the bandwidth between the first nulls on either side of the main lobe in the frequency spectrum. For QPSK, the main lobe width is twice the symbol rate.
Question 6: Correct Answer - A (6 channels)
For N channels, we need N × 5 MHz for channels + (N-1) × 1 MHz for guard bands. Solving: 5N + (N-1) = 40 → 6N - 1 = 40 → 6N = 41 → N ≈ 6.83, so maximum N = 6 channels. This accounts for guard bands between channels but not at the edges of the transponder bandwidth.
Question 7: Correct Answer - B (20%)
To improve C/I from 15 dB to 18 dB (3 dB improvement), we need to reduce interference by half. Since interference is proportional to spectral overlap, we need to increase separation by approximately 20% of channel bandwidth to achieve this reduction. A 3 dB improvement corresponds to halving the interference power.
Question 8: Correct Answer - B (20%)
Total used bandwidth = 8 channels × 4 MHz = 32 MHz. Guard band bandwidth = 40 MHz - 32 MHz = 8 MHz. Percentage wasted = (8 MHz ÷ 40 MHz) × 100% = 20%. This represents the bandwidth efficiency loss due to guard bands in this FDMA system.
Question 9: Correct Answer - B (4.00 Msymbols/s)
For a raised cosine filter, bandwidth B = R(1+α)/2, where R is symbol rate. Rearranging: R = 2B/(1+α) = 2×5/(1+0.25) = 10/1.25 = 4 Msymbols/s. The roll-off factor α determines how much extra bandwidth is needed beyond the Nyquist minimum bandwidth.
Question 10: Correct Answer - C (5 carriers)
The 1 dB compression point is 15 dB above each carrier's power (10 dBm - (-5 dBm) = 15 dB). For N equal carriers, the third-order intermodulation products increase as N³ while the desired signal increases as N. With 5 carriers, the peak-to-average ratio increases, and intermodulation products become significant near the 1 dB compression point. Practical systems typically back off 3-6 dB from the 1 dB compression point for multicarrier operation.