HY448 Sample Problems 10 November 2014 These sample problems include the material in the lectures and the guided lab exercises. 1 Part 1 1.1 Combining logarithmic quantities A carrier signal with power 7 dbm is added to a noise signal with power 5 dbm. What is the total signal power in mw? in dbm? Solution: Never add logarithmic units of power directly. To find the sum of two logarithmic units of power (such as dbm or dbw), first convert them to linear quantities: 7 dbm = 10 7 dbm 10 dbm W 5.012 mw and 5 dbm 5 dbm = 10 10 dbm W 0.316 mw The sum is then and the equivalent in dbm is 0.316 mw + 5.012 mw = 5.328 mw 10 log 10 (5.328) 7.266 dbm 1.2 Applying gain in db An amplifier has a gain of 3 db. What is the power level at the output of the amplifier is the power of the signal at its input is 6 dbm? 0.25 mw? Solution: If power is given in logarithmic units (e.g., dbm), we can apply gain in db by adding: 6 dbm + 3 db = 3 dbm If power is given in linear units (e.g., mw), we can apply gain in db by converting and multiplying: 0.25 mw 10 3 db 10 = 0.25 mw 2 = 0.50 mw
hy448 sample problems 2 1.3 Signal to noise ratio A signal with power 3 mw is distorted by noise with power 0.001 mw. What is the signal to noise ratio (SNR) in db? Solution: ( ) 3 10 log 10 = 34.77 db 0.001 1.4 Receiver diversity A wireless receiver has a pair of receive antennas A 1 and A 2. They are separated by some physical distance and have different orientations, so that they sense the ambient RF environment differently. A transmitter is sending a signal while moving about the room in a deterministic pattern. The SNR (in db) of the signal received at each antenna is function of time (t), with SNR 1 = 30 cos t and SNR 2 = 20 cos t. The plot in Figure 1 shows the received SNR at each antenna for the time interval 0 t 2π: The receiver in this scenario needs at least 0 db SNR to process the transmitted message. We will calculate the outage probability as the ratio of a time interval in which the the received signal is less than 0 db. SNR db (t) SNR 1 SNR 2 Figure 1: Receive SNR at two antenna elements. Suppose we receive exclusively on A 1. What will be the outage probability for the time interval 0 t 2π? Suppose we receive exclusively on A 2. What will be the outage probability for the time interval 0 t 2π? We will consider different ways of processing the signals received on the two antennas, so as to benefit from their diversity. One post-processing technique is called switching. With this technique, we begin using the signal from the antenna with the highest SNR. We use this antenna as long as its SNR remains above some threshold value; if it falls below this threshold, we switch to the other antenna. Using switching with a threshold of 0 db, what will be the outage probability for the time interval 0 t 2π? Another post-processing technique is selecting. With this technique, we always use the antenna with the highest SNR. Using selecting as a post-processing technique, what will be the outage probability for the time interval 0 t 2π? In this scenario, which post-processing technique would be preferred?
hy448 sample problems 3 Which technique (A 1 exclusively, A 2 exclusively, switching, selecting) has the lowest outage probability? Which technique (A 1 exclusively, A 2 exclusively, switching, selecting) has a highest average SNR? Solution: Using A 1 alone, we have an SNR below 0 db 50% of the time. We therefore have an outage probability of 0.5. Using A 2 alone, we have an SNR below 0 db 50% of the time. We therefore have an outage probability of 0.5. Using switching, we will have an outage probability of 0.25. Using selecting, we will have an outage probability of 0.25. Both switching and selecting have the lowest outage probability, of 0.25. Selecting has the highest average SNR of the post-processing techniques under consideration. 1.5 Noise figure with cascaded components Consider the receiver shown in Figure 2. The antenna is connected to the signal processing chain with a long cable that attenuates the signal by 1.5 db, but adds only thermal noise (no additional noise). The first stage of the DSP chain, an RF amplifier, has a noise figure of 7 db and a gain of 20 db. The second stage, a mixer, has a noise figure and conversion gain (ratio of output power to input power) of 8 db. The third stage is an IF amplifier with a gain of 60 db and a noise figure of 6 db. Figure 2: A multi-stage RF processing chain. Find the noise figure of the entire system shown in Figure 2. Find the noise figure of the entire system if the RF amplifier is placed before the cable with 1.5 db loss. Which system is preferable? Why? Solution: Find the noise factor F and gain G at each stage. Then apply Friis formula. For the cable, G 1 = 10 1.5 10 = 0.7080 F 1 = 1 G 1 = 1.4125
hy448 sample problems 4 For the RF amplifier, G 2 = 10 20 10 = 100 F 2 = 10 7 10 = 5.0119 For the frequency mixer, G 3 = 10 8 10 = 6.3096 F 3 = 10 8 10 = 6.3096 For the IF amplifier, G 4 = 10 60 10 = 10 6 F 4 = 10 6 10 = 3.9811 Then, use Friis formula to find the combined noise factor for the original system: or F = 1.4125 + 5.0119 1 0.7080 then F = F 1 + F 2 1 G 1 + F 3 1 G 1 G 2 + F 4 1 G 1 G 2 G 3 + 6.3096 1 0.7080 100 + 3.9811 1 0.7080 100 6.3096 = 7.1607 NF = 10 log 10 (7.1607) = 8.5495 db For the system with the RF amplifier as the first stage, we have F = 5.0119 + 1.4125 1 100 then + 6.3096 1 0.7080 100 + 3.9811 1 0.7080 100 6.3096 = 5.0977 NF = 10 log 10 (5.0977) = 7.0737 db The second system is preferred because it has a lower noise figure. 1.6 Forward error correction with XOR The logical XOR operation, denoted, works as follows: 0 0 = 0 0 1 = 1 1 0 = 1 1 1 = 0 Using XOR, we can create a forward error correction code that will allow us to potentially recover from some bit errors. We will
hy448 sample problems 5 divide our message into packets of length l; for each two consecutive packets, we will transmit the packets, followed by their XOR. For example, if l = 4, and we have the message {10001010}, we can send the packets p 1 = {1000}, p 2 = {1010}, and p 3 = {0010}. At the receiver, if any two packets are received correctly, we can XOR them to recover the third, enabling us to reconstruct the message. With this FEC scheme, what is the ratio of message bits to total bits sent over the link? Suppose we receive the following three-packet FEC group using the scheme described above: 1101, 1010, and 0011. Furthermore, suppose we know that there will be at most one bit error per FEC group. Can we identify whether an error has occurred in transmission? If so, how? Can we recover the correct 8-bit message? Suppose the transmitter adds a parity check bit at the end of each packet, set to 0 if there are an even number of 1s in the packet and 1 if there are an odd number of 1s in the packet (making the total count of 1s in the packet, including the parity bit, even). As before, suppose we know that there will be at most one bit error per FEC group. Using the combined FEC and parity check scheme, we receive the following packets: 11011, 10101, and 00110. With this combined parity check and FEC scheme, what is the ratio of message bits to total bits sent over the link? Can we recover the correct 8-bit message? Solution: With the FEC scheme described in this problem, we must send one extra bit for every two bits of message data. The ratio of message bits to total bits is 2 3. We know that an error has occurred because 1101 1010 = 0011 We cannot recover the correct 8-bit message. It s impossible to distinguish between these three cases valid of FEC packet groups with one bit different from the received packets:
hy448 sample problems 6 Message Packets sent Error 10011010 1001, 1010, 0011 Second bit of first packet 11011110 1101, 1110, 0011 Second bit of second packet 11011010 1101, 1010, 0111 Second bit of third packet With the parity bit and FEC, the ratio of message bits to total bits is 8 15. From the parity bit, we can see that the second packet (10101) has an odd number of ones, and therefore has a bit error. Furthermore, we know that the second bit of this packet is in error (from the XOR). Therefore, we can recover the correct message: 11011110. 2 Part 2 2.1 Frequency mixers A transmitter is operating with an intermediate frequency f IF = 39 MHz. It is mixed with a local oscillator at f LO = 400 MHz, before transmitting (as in Figure 3). We are interested in measuring the output of the transmitter, separately from any issues that may be introduced by an imperfect receiver. To do this, we use a spectrum analyzer tuned to the range from 300 500 MHz. Spectrum analyzers are special-purpose devices designed to show the true RF spectrum as accurately as possible. The display of the spectrum analyzer is shown in Figure 4. In particular, we measure power levels above the ambient noise power at the following frequencies: f RF f LO Figure 3: Frequency mixer. f IF Frequency (MHz) Amplitude (dbm) 322-72.62 361-17.54 400-32.59 439-17.80 478-68.61 Figure 4: Display of spectrum analyzer with f IF = 39 MHz. Categorize each of the signals in the table above as one of the following: Desired mixer product LO-to-RF leakage IF-to-RF leakage Harmonic of LO (specify which harmonic) Harmonic of IF (specify which harmonic)
hy448 sample problems 7 Harmonic of IF mixed with LO (specify which harmonic) Harmonic of LO mixed with IF (specify which harmonic) Mixture of harmonic of LO and harmonic of IF (specify which) If we repeat this with an IF frequency of 34 MHz, which of these signals will still appear at the same frequency? Which will appear, but at a different frequency (specify the new frequency)? Which will no longer appear at all? Why? Solution: For upconversion from IF to RF, the output at the RF port of the mixer is f RF = f LO ± f IF. The desired mixer products are at 361 MHz and 439 MHz. The power observed at 400 MHz is LO-to-RF leakage. The peaks observed at 322 MHz and at 478 MHz are the products of the 2nd harmonic of the IF frequency, mixed with the LO frequency (400 2 39 = 322 and 400 + 2 39 = 478). Figure 5 shows the display of the spectrum analyzer when the IF frequency is changed to 34 MHz. When we change the IF frequency to 34 MHz, the peak at 400 MHz will remain at the same frequency, since it is not a product of the IF frequency or any of its harmonics. The peaks at 361 MHz and 439 MHz will move; the new mixer products will appear at f RF = f LO ± f IF, now 366 MHz and 434 MHz. The products of the second IF harmonic and the LO frequency will also move as a result of the change in f IF. These products will now appear at 332 MHz and 468 MHz (400 2 34 = 332 and 400 + 2 34 = 468). Figure 5: Display of spectrum analyzer with f IF = 34 MHz. 2.2 Bandwidth requirements for digital signals An analog speech signal is sampled at a rate of 8 khz and digitized using 8 bits per sample. What is the bitrate of the digital signal? In a software radio competition, all teams are required to use no more than 20 khz of bandwidth in their design. Suppose a team was to transmit the digital signal described above using M-PSK digital modulation. Given the bandwidth constraint, can the full data rate be supported Using M = 4? M = 16? Using M = 16 with a code that adds redundancy to protect against error? What is the highest code rate that can be used?
hy448 sample problems 8 Solution: The bitrate is R b = 8 10 3 samples second 8 bits bits = 64 103 sample second The (modulation) spectral efficiency for 4-PSK is log 2 4 = 2 and the required transmission bandwidth is 64 10 3 2 = 32 khz This is more than the maximum allowed bandwidth, so the data rate of the signal cannot be supported. The (modulation) spectral efficiency for 16-PSK is log 2 16 = 4 and the required transmission bandwidth is 64 10 3 4 = 16 khz This is less than the maximum allowed bandwidth, so the data rate of the signal can be supported. With 16-PSK and coding, the required bandwidth will be 1 64 10 3 R 4 where R is the code rate. For a maximum of 20 khz of bandwidth, the highest code rate possible is 4 5, which means we can send only one bit of redundant information for four message bits. 2.3 ARQ In an ARQ system, data is divided into blocks of size S B bytes. The receiver and transmitter each keep a buffer that can store up to W blocks (W is called the window size). When a block is sent, the transmitter keeps it in its buffer until it receives an acknowledgment from the receiver. If no acknowledgment is received within a specified timeout period, the block is re-transmitted. If the buffer becomes full, the transmitter must wait for acknowledgment from the receiver, at which point it can remove the acknowledged block(s) from the buffer. Consider a wireless system, where ARQ acknowledgments for blocks transmitted in frame i are received at the beginning of frame i + 1. The frame length L F is the length of the frame, and its inverse is the frame rate (number of frames per second).
hy448 sample problems 9 Suppose we have continuous, errorless data transmission in an ARQ system with block size S B, window size W, and frame length L F, What is the maximum throughput (in terms of S B, W, and L F ) that the connection can achieve? For S B = 32 B, W = 512, and L F = 5 msec, what is the maximum throughput in B/s that the connection can achieve? For S B = 16 B, W = 128, and L F = 5 msec, what is the maximum throughput in B/s that the connection can achieve? In the previous example, the acknowledgments for blocks in frame i arrive at the beginning of frame i + 1. What if there is some delay, so that acknowledgments for frame i arrive at the beginning of frame i + D? What is the maximum throughput (in terms of S B, W, L F, and D), that the connection can achieve? Solution: The maximum throughput a connection can achieve is limited by the expression: S B W L F For S B = 32 B, W = 512, and L F = 5 msec, the maximum throughput is 3.277 MB/s. For S B = 16 B, W = 128, and L F = 5 msec, the maximum throughput is 409.6 kb/s. With delayed acknowledgments, the maximum throughput the connection can achieve is limited by the expression: S B W L F D