ELCT562_PA6

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Clemson University *

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562

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Electrical Engineering

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Apr 3, 2024

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ELCT 562 PA#6 1. Assume that we have the following set of signals: 𝑠𝑖 ( 𝑡 ) = cos(2 𝜋𝑓𝑐𝑡 + 𝜃𝑛 ), 𝑛𝑇𝑠 ≤ 𝑡 ≤ ( 𝑛 + 1) 𝑇𝑠 where 𝑖𝑖 ∈ {0, 1, … 𝑀𝑀 − 1}, and 𝜃𝑛 ∈ 2 𝜋 / 𝑀 . A platform on Mars uses 𝑀 = 2 and the data rate at 𝑅𝑏 = 2 kbps over a satellite link with the following parameters: • Distance between Tx & Rx = 56×106 km • Carrier frequency = 1 GHz • Tx power = 100 milliwatts • Tx antenna gain = 47 dBi • Rx antenna gain = 30 dBi • Receiver noise figure = 2 dB (a) What kind of modulation (or signaling) does this communication system use? BinaryPhase Shift Keying (BPSK) due to the equation having a changing phase term depending on symbol i which can only be two symbols. (b) What is the pulse shape utilized for this communication system? The pulse shape in baseband should be a rectangular pulse. (c) What is the value of 𝑇𝑇𝑠𝑠 ? T s = 1 R b = 1 2000 = 500 μs (d) What is the received signal power in dBm based on the free-space path loss model? L p = 20 log 10 ( 56 ∗ 10 6 km ) + 20log 10 1 GHz + 20log 10 4 π c P r = P t + G t + G r − L p =− 150 dBm (e) What is the energy per data symbol (i.e., 𝐸𝐸𝑠𝑠 ) in dBm (at the receiver)? E s = P r + 10 log 10 T s =− 183.4 dBm (f) What is the energy per bit (i.e., 𝐸𝐸𝑏𝑏 ) in dBm (at the receiver)? The energy would be the same as data symbol because each is carrying one bit, -183.4dBm. (g) What is the complex noise variance (i.e., 𝑁𝑁 0) in dBm (at the receiver)? (kT = -174 dBm/Hz) N 0 = k ∗ T + N ∗ F =− 172 dBm (h) Approximating the link as an AWGN channel,

P e = Q √ 2 ∗ E b N 0 = 0.352 This means that the BER must be higher than 10E-2 which means that it is not possible with parameters. (i) Can we achieve a bit-error rate of 10−2? Why? (Note that Pe = Q �� 2 𝐸𝐸 b 𝑁𝑁 0 � for binary antipodal signaling) R b ' = R b ∗ E b ' N 0 ' E b N 0 = 56 bps This meets the BER requirements of 10E-2 (j) If not, what is the bit rate we can use to obtain a 0.01 error probability (Q �� 2 𝐸𝐸 b 𝑁𝑁 0 � = 0.01 when 𝐸𝐸 b/ 𝑁𝑁 0 = 4.1 dB)? 2. Assume that the frequency range of the passband signal is defined as of | 𝑓𝑓 | ≤ 10 MHz around 2.45GHz. (a) How much bandwidth needs to be allocated in the spectrum for this signal? Total BW = 2.46 GHz − 2.44 GHz = 20 MHz (b) Sketch the spectrum of signals on I-& Q-branches before up-conversion and the signal over the air, i.e., passband signal.

(c) Determine the minimum sample rates on I and Q branches. MinimumSample Rate = 2 ∗ 10 MHz = 20 MHz (d) Assume the modulation is set to BPSK. What is the maximum data rate you can achieve? Maximumdata Rate = 10 Mbps This is because each symbol would represent one bit which is 10MHz on the bandwidth of baseband. 3. Consider a communication system with a very large 𝐾 factor for the Rician channel model. a. Is it likely that there is a line-of-sight path or not? It is likely that there is a strong line of sight between the transmitter and receiver due the K factor representing a ratio of power with line of sight paths. b. Assume that the phase of the received signal increases by 𝜋𝜋 rad/s and the transmitter’s location is fixed. If the carrier frequency is 3 GHz, what is the speed of the receiver? f d = ∆ϕ 2 π ∗ f s = v / λ λ = speed of lighjt / f c v = 0.5 ∗ 0.1 = 0.05 m s

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4. Assume that the communication system A achieves reliable communication (i.e., no error) at 𝐸𝑠 / 𝑁 0= 10 dB by using a modulation where its constellation size is 16 while the communication system B claims reliability at 𝐸𝑠 / 𝑁 0 = 5 dB with a constellation size 𝑀 of 4 for the 10 Mbps over 10 MHz. a. What are the spectral efficiencies of these systems? System 1: log 2 16 = 4 bits per symbol System 2: log 2 4 ¿ 2 bits per symbol b. What are the gaps in dB based on Shannon’s bound? System 1: C 1 = 10log 2 ( 1 + 10 1 ) = 34.6 Mbps Ga p 1 = 34.6 − 10 = 24.6 Mbps System 2: C 2 = 10log 2 ( 1 + 10 0.5 ) = 20.6 Mbps Ga p 2 = 20.6 − 10 = 10.6 Mbps c. Which system is designed better? Why? System 2 (B) is better due to a higher efficiency in more complex environments due to its lower E b /N 0 and is closer to the shannon limit since it has a smaller gap in dB.