ECE424_Practice_Exam1_Solutions-1

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ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University This practice exam reflects the topics and types of questions you’re likely to see on the first exam. This exam covers the first 4 weeks of class so it does not include radio non-linearities. This practice exam includes some conceptual questions, some multiple choice, and some that require calculation using concepts and formulas discussed in lecture and lecture videos. The exam will be open book, so feel free to use your notes and other resources. Topics include: 1. Power, dB, and dBm related to RF signals 2. Analog modulation (AM, FM, PM) 3. Digital modulation (QAM, PSK, QPSK, FSK, ASK, etc.) 4. QAM constellations, bandwidth, and bitrate 5. Pulse shaping 6. Transmitters & transmitter architecture 1. Refer to the diagram below. The two signals, A and B , are summed together. What is the power of the output signal in dBm? Solution: Signal powers in dBm cannot be summed as in the above diagram due to properties of logarithms. So to determine the final power in dBm, we must take each input signal and find the power in linear units (Watts or mW), add these together for the resulting output signal in linear units and then convert this value back to dBm. This is done as follows: dBm = 10log !" * # ! ! mW + ® ࠵? mW = 10 ’ dBm 10 , -

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University So for 7.9 dBm, we get: ࠵? mW = 10 ..0 !" , ≈ 6.17 mW And for 1.2 dBm, we get: ࠵? mW = 10 !.1 !" , ≈ 1.32 mW Adding the powers together in the linear domain gives us: ࠵? Out,mW = 6.17 mW + 1.32 mW = 7.49 mW Converting the output power to dBm gives us: ࠵? Out,dBm = 10 log !" * ..60 mW ! mW + ≈ ࠵?. ࠵?࠵? dBm 2. The I and Q baseband signals (not pulse shaped) of a QAM system appear as shown below. How many symbols is this system capable of transmitting? You may assume all possible signal levels are shown in the baseband pulses. Solution : In an IQ modulation scheme, each symbol comes from the combination of an I amplitude and a Q amplitude. Therefore, we can find all the combinations of the number of I levels and Q levels to determine how many symbols this system can transmit. Here, since we can assume all the possible signal levels of I and Q are shown, we can see that there are 4 distinct amplitudes for both I and Q . All possible combinations of these levels of I and Q give us: 16 symbols; so this is a 16QAM system 3. Using the baseband IQ data from question 2, what is the symbol rate? Solution: 0 4 8 12 16 20 24 28 32 Time ( s) -2 -1 0 1 2 Baseband I Data 0 4 8 12 16 20 24 28 32 Time ( s) -2 -1 0 1 2 Baseband Q Data

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University Observing the data from the baseband plots of I and Q , we can see that the smallest pulse for each is 4 μ s. So, each symbol period is 4 μ s long. The symbol rate is the inverse of the symbol period so we have: Symbol rate = 1 Symbol period = 1 4 × 10 78 s = 250,000 Hz = ࠵?࠵?࠵? kHz 4. A 64QAM transmitter occupies a bandwidth of 35 MHz. If raised cosine pulse shaping was used with a =0.4, what is the bit rate? Solution: Since we know the occupied bandwidth of the transmitter and the roll-off factor of the raised cosine pulse shaping of the IQ data, we can determine the symbol rate and then the bit rate as follows. Note, since we’re talking about a transmitter, we can assume the bandwidth given is the RF bandwidth. Also note, there is no information given about the carrier frequency and we don’t need to know what it is to solve the problem since the bit rate is independent of the carrier frequency and only depends on the symbol rate and the bandwidth/spectral efficiency of the modulation scheme. ࠵?࠵? RF = !;< = " = ࠵? > (1 + ࠵?) ® ࠵? > = ?@ RF !;< = AB×!" % Hz !;".6 = 25 MHz (same as 25 MSym/sec, since we have 1 symbol/s/Hz, that’s 1 symbol per second per Hz of the pulse rate) Since we’re using 64QAM, we have 6 bits/s/Hz (that’s 6 bits per second per Hz of the pulse rate, same as 6 bits/symbol) and we get: Bit rate = (25 × 10 8 Hz) × 6 bits s∙Hz = ࠵?࠵?࠵? Mbps 5. A low-pass filter (LPF) used to shape digital baseband data before upconversion onto a carrier will both reduce our transmitted bandwidth and decrease intersymbol interference (ISI). True or False? Solution: False. While a LPF does decrease our transmitted bandwidth by removing the sharp edges from the digital data resulting in removing the sidelobes and attenuating higher frequencies in the frequency domain, it does so at the cost of spreading out each symbol pulse causing previous symbols to bleed into the current symbol which is called intersymbol interference or ISI. This can cause detection errors at our receiver. If we want reduced transmission bandwidth and decreased ISI, we must use a raised cosine filter (there are other types that work as well but this is the most common). 6. Which of the following statements about a SSB transmitter are true? Select all that apply. a. We must phase shift our original signal by 90 degrees in one branch. b. Each branch uses the same carrier signal. c. We can only transmit the lower sideband of our signal.

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University d. We are able to save power by transmitting only one sideband. e. Both sidebands – upper and lower – are needed to reproduce the baseband signal. Solution: True statements: a, d False statements: b, c, and e For the true statements: In order to create a SSB signal, we need to shift our original signal by 90 degrees in one branch so that it will later cancel one of the sidebands when combined with the other branch. Since we’re transmitting only one sideband, we don’t waste power transmitting both when only one copy is needed. You can think of it this way: for the same transmitter output power as a DSB transmitter, since we only have 1 sideband we can boost the power in that sideband higher than the equivalent in a DSB signal while keeping the overall output power the same. Similarly, to keep the sideband power in a SSB signal the same as the corresponding sideband in a DSB signal, we can lower our output power in our SSB transmitter because it’s all dedicated to one sideband and not spread out over two. Power savings in a SSB transmitter versus DSB is approximately 3 dB. For the false statements: Each branch uses the same carrier frequency but one is sine and the other is a cosine, so they are different signals. Essentially, both signals must be orthogonal and sine and cosine are readily available. We can transmit either sideband by adjusting the phase shift of the original signal to be +/- 90 degrees. Only a single sideband is needed to properly reproduce the original baseband signal. The baseband signal will have a mirror copy in the negative frequency domain and when the baseband signal is upconverted, this copy appears as the other sideband. They both contain the same baseband information, so we only need one of them to reproduce our baseband signal. 7. The phase of the carrier signal changes in FSK (frequency shift keying). True or False? Solution: True. Frequency is the derivative of phase, so the phase of the carrier changes in FSK. 8. For the same peak output power, which of the following statements hold true: a. The distance between constellation points in a 16QAM and 64QAM system are the same. b. The distance between constellation points in a 16QAM system are further apart than those in a 16PSK system. c. The bit error rate (BER) using 8QAM is lower than 8PSK (assume the same noise power in each). d. The bandwidth used by a QPSK system is less than one using 64QAM. e. The bit rate of a 16QAM system is greater than that of a QPSK system for the same symbol rate. Solution:

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