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.

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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:

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University True statements: b, c, and e False statements: a and d For the true statements: We know that for the same distance between constellation points in a 16QAM and 16PSK system, the 16QAM system uses roughly 1.6 dB less peak power, and the converse is true as well where if we set the peak power of each of these the same, the distance between constellation points for the 16QAM system is larger than it is for 16PSK meaning the 16QAM system is more immune to noise (specifically AWGN). This is also essentially a geometry problem that you can solve as well. For c, if you look at the BER graph from the lecture notes, you’ll see that for the same signal and noise power, the BER for 8QAM is lower than that for 8PSK. The reasoning is the same as that for statement b being true. For e, the peak output power has nothing to do with the bit rate, but if know the symbol rates for a 16QAM and QPSK system are the same, the bit rate for 16QAM will be twice as large since it uses 4 bits per symbol while QPSK uses just 2. For the false statements: For a, we know that if the peak output power for 16QAM and 64QAM is the same, this means the distance from the origin to the furthest constellation point is the same, so we’re packing 64 points in the same area as 16. Thus, the distance between constellation points in the 64QAM system must be closer than those in the 16QAM system and not the same. For d, the peak output power is independent of the bandwidth as the bandwidth only depends on the type of pulse shaping (if any) and the symbol period or symbol rate of the baseband data. Thus, we don’t have enough information to determine if the statement in d is true or not. 9. The following signals are multiplied together in the time domain as follows: ࠵?(࠵?) = cosZ2࠵?350࠵? + ࠵? 4 \ ] ࠵?(࠵?) = cos(2࠵?125࠵?) ࠵?(࠵?) = ࠵?(࠵?)࠵?(࠵?) = cosZ2࠵?350࠵? + ࠵? 4 \ ] cos(2࠵?125࠵?) Draw a rough sketch of the magnitude frequency spectrum of the resulting signal, ࠵?(࠵?) . Include both positive and negative frequency components. Sketch does not need to be to scale in terms of the exact magnitude of each frequency component, but you must label all frequencies on the x-axis. Solution: This can be done in 2 ways. We know the spectrum of a single cosine, so we can draw out the spectrum for each signal ( x(t) and y(t) ) and then perform convolution of those spectra to determine the final spectrum of m(t) . Or, we can use trigonometric identities to rewrite m(t) as follows: ࠵?(࠵?) = 1 2 ‘cosZ2࠵?350࠵? + ࠵? 4 \ − 2࠵?125࠵?] + cosZ2࠵?350࠵? + ࠵? 4 \ + 2࠵?125࠵?]b

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University ࠵?(࠵?) = 1 2 ‘cosZ2࠵?225࠵? + ࠵? 4 \ ] + cosZ2࠵?475࠵? + ࠵? 4 \ ]b Note, the additional phase shift on x(t) does propagate through but since we’re drawing the magnitude spectrum, the phase won’t appear in our sketch. Keep in mind that if you were to draw this in complex frequency space, the phase would appear. So, our sketch of m(t) in the frequency domain looks like this: 10. Why do we often use a raised cosine pulse instead of a Nyquist pulse? Select all applicable reasons. a. A raised cosine pulse has a narrower bandwidth than a Nyquist pulse. b. A raised cosine pulse is easier to implement and more practical than a Nyquist pulse. c. A raised cosine pulse allows us to increase the number of symbols in our constellation more than a Nyquist pulse. d. A raised cosine pulse allows us to transmit more power than a Nyquist pulse. Solution: Correct reasons: b Incorrect reasons: a, c, and d For the correct reason: Nyquist pulses are very difficult to implement and in most situations are impractical. This is why we often use a raised cosine pulse, as it is much simpler to implement than a Nyquist pulse and approximates a Nyquist pulse nicely. For the incorrect reasons: A Nyquist pulse has the narrowest bandwidth of all pulses, and it is the absolute theoretical minimum bandwidth required to transmit a given amount of pulses (digital data) per second. A raised cosine filter has a larger bandwidth than a Nyquist pulse for all practical purposes, since we can technically make a Nyquist pulse from a raised cosine pulse by setting a =0, but this is not f (Hz) | m(f) | 225 475 -225 -475

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ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University trivial and is not done. For c, pulse shaping has nothing to do with the constellation size. And for d, again pulse shaping has nothing to do with our transmit power. 11. In a two-stage transmitter (heterodyne), carrier (LO) leakage is more of a problem than it is for a direct conversion (single-stage) transmitter. True or False? Solution: False. In a two-stage transmitter, neither LO – the IF or the final RF LO – appears in between our desired data sidebands. A two-stage transmitter eliminates the effect of LO or carrier leakage that we see in direct conversion transmitters. 12. What is the RF bandwidth of a signal that was pulse shaped using a raised cosine filter with a =0.35 if the symbol rate is 2.25 MHz? Solution: The formula for the RF bandwidth of a signal that was pulse shaped with a raised cosine filter is: ࠵?࠵? ?@ = 1 + ࠵? ࠵? F = ࠵? > (1 + ࠵?) = 2.25 × 10 8 Hz(1 + 0.35) ≈ ࠵?. ࠵?࠵? MHz 13. A new system was just designed that has 975 MHz of bandwidth. It utilizes FDMA to allow multiple users at the same time, and supports 65 users. Each channel operates with 64QAM at a symbol rate of 12 MHz. If raised cosine pulse shaping is used for each user in this system, what is the roll-off factor a ? You may assume that each users’ channel has a sharp cutoff and doesn’t interfere with adjacent channels (in the real world, guard bands are typically used around each channel). Solution: We know that our FDMA system has 975 MHz of total bandwidth for 65 users. This means that each user is allotted 975 MHz/65 = 15 MHz for their channel. Within each channel, we know our symbol rate is given as 12 MHz. Since we’re talking about channels in an FDMA system, we can assume we’re dealing with RF bandwidth so we can solve for a as follows: ࠵?࠵? ?@ = ࠵? > (1 + ࠵?) → ࠵? = ࠵?࠵? ?@ ࠵? > − 1 = 15 × 10 8 Hz 12 × 10 8 Hz − 1 = ࠵?. ࠵?࠵? 14. In an IQ modulator, the phase shift between the LO for the I branch and the LO for the Q branch is not exactly 90 degrees. You can assume this is the only non-ideality we’re dealing with. Which of the following statements about the constellation are true? Select all that apply. a. The entire constellation will shift from the origin. b. The constellation will rotate around the origin but remain square. c. The constellation will expand. d. The constellation will shear (go from a square shape to a parallelogram).

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University e. The constellation will change from a square to a rectangle. Solution: True statements: d False statements: a, b, c, and e For the true statement: A phase error in the I or Q branch will cause the constellation to shear and change from a square to a parallelogram shape since there will be cross-talk between the in-phase and quadrature data due to the LO signals not being orthogonal. That is, some of the I channel will leak into Q and vice-versa. For the false statements: The constellation will only shift from the origin if there is a DC offset in the receiver. The constellation will rotate around the origin and remain square due to an equivalent phase shift in both branches which can be caused by RF components in the transmitter or receiver. The constellation will expand if there is a gain imbalance in the I and Q branches. If the gain imbalance is uniform, it will expand in size and remain square, and if the gain imbalance is not uniform, the constellation will transform from a square to a rectangle. 15. In a two-stage upconverting transmitter, how many copies of the original baseband signal spectrum appear in the final output before filtering? Original baseband signal spectrum refers to only the portion appearing in the positive frequency domain. Count the copies in the negative frequency domain in your answer. Solution: 8 copies. In a two-stage upconverting transmitter, after the first stage the DC point of the baseband signal is shifted to the positive and negative IF frequency. The baseband signal has a mirror image in the negative frequency domain, so after the first stage there are now 4 copies. After the second mixing stage, the DC point is now shifted to the positive and negative final RF frequency, which results in 8 copies of the original baseband signal spectrum (a copy being just the portion originally in the positive frequency domain). 16. An envelope/power detector can be used to receive DSB-SC and SSB-SC signals and reproduce the original message accurately. True or False? Solution: False. An envelope/power detector requires a sufficient amount of DC bias in the baseband signal to accurately reproduce the original message. Without this DC bias, which doesn’t exist in DSB-SC or SSB-SC since the carrier is not present, an envelope/power detector does not operate properly and produces a

ECE 424 (524) - Radio System Design Practice Exam 1 Department of Electrical and Computer Engineering NC State University distorted/incorrect version of the original message. A coherent detector must be used in DSB-SC and SSB-SC systems. 17. A SSB transmitter utilizes 2 mixers and orthogonal carrier signals along with a 90 degree phase shift of the original signal to only 1 of the upconverting mixers to produce the SSB output signal to transmit. A SSB receiver also needs 2 mixers to bring the SSB signal back to baseband for processing. Is the previous statement about SSB receivers true or false? Solution: False. While a SSB transmitter does use 2 mixers with orthogonal carriers and a phase shift of the original baseband signal by 90 degrees to one of the mixers, the receiver only needs a single mixer to bring the SSB signal back to baseband. SSB transmits just a single message and needs the 2 mixers in the transmitter to cancel one of the upconverted sidebands, leaving a SSB signal that can be downconverted with just a single mixer.

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