HOMEWORK I SOLUTIONS 3. 14. A way of visualizing the Nyquist theorem is in terms of periodic sampling of the second hand of a clock which makes one revolution around the clock every 60 seconds. The Nyquist sampling rate here should correspond to 2 samples per cycle, that is, sampling should be done at least every 30 seconds. a. Suppose we begin sampling when the second hand is at 12 o’clock and that we sample the clock every 15 seconds. Draw the sequence of observations that result. Does the second hand appear to move forward? 0 15 30 45 0 15 30 45 0 ...Yes, the second hand appears to move forward. b. Now suppose we sample every 30 seconds. Does the second hand appear to move forward or backward? What if we sample every 29 seconds? 0 30 0 30 0 30 0 30 0 30 ... No. It's hard to tell if it moves forward or backward because either direction will give the same sequence. 0 29 58 27 56 25 54 23 ... Yes, the second hand appears to move forward. c. Explain why a sinusoid should be sampled at a little more than twice its frequency. When a sinusoid is being sampled at exactly twice its frequency, it is possible that all samples have value of zero. Therefore, the sampling rate should be a little more than twice its frequency. d. Now suppose that we sample every 45 seconds. What is the sequence of observations of the second clock hand? One sampling every 45 seconds -> 0 45 30 15 0 45 30 15 0 … . The second hand appears to be moving backward. e. Motion pictures are made by taking a photograph 24 times a second. Use part (c) to explain why car wheels in movies often appear to spin backward while the cars are moving forward! The spinning wheel is being sampled at 24 times per seconds. Depending on the rotation speed of the wheel, the wheel may appear either as spinning backward while actually spinning forward. 18. A high-quality speech signal has a bandwidth of 8 kHz. a. Suppose that the speech signal is to be quantized and then transmitted over a 28.8 kbps modem. What is the SNR of the received speech signal? W = 8 kHz and R = 28.8 x 103 bits/sec 2W = 16 K samples/sec m = R/2W = 28.8/16 ˜ 2 bits/sample SNR ˜ 6m – 7.2dB ˜ 5 dB (very approximate) b. Suppose that instead a 64 kbps modem is used? What is the SNR of the received speech signal? R/2W = 64/16 = 4 bits/sample SNR ˜ 24 – 7.2db ˜ 17 dB (very approximate) c. What modem speed is needed if we require an SNR of 40 dB? SNR = 40 dB = 6m – 7.2 dB 6m = 47.2 m = 8 R = 8 bits/sample x (16 x 103 samples/sec) = 128 kbps 22. A telephone office line card is designed to handle modem signals of the form x(t) = Acos(2pfct + f(t)). These signals are to be digitized to yield an SNR of 40 dB using a uniform quantizer. Due to variations in the length of lines and other factors, the value of A varies by up to a factor of 100. ^ refers to exponentiation a. How many levels must the quantizer have to produce the desired SNR? SNR = 40 = 10 log 10(((A/100)^2/2) / (delta^2/2)) substitute in the above formula delta = 2A/(2^m) SNR = 10 log 10( 1.5 * (2^(2*m)) / 10000) = 1.76 - 40 + 6m m=13 b. Explain how an adaptive quantizer might be used to address this problem. An adaptive quantizer may use larger intervals when the signal level is high and smaller intervals when the signal level is low. This way the number of bits will be less. 30. Suppose we wish to transmit at a rate of 64 kbps over a 3 kHz telephone channel. What is the minimum SNR required to accomplish this? Solution: We know that R = 64 kbps and W = 3 kHz. What we need to find is SNRmin. The channel capacity is: C = W log (1 + SNR), C >= Cmin = 64 kbps Cmin = W log2 (1 + SNRmin) implies log2 (1 + SNRmin) = 64/3 implies 1 + SNRmin = 2^(64/3) hence, SNRmin = 2.64 x 10^6 in dB: SNRmin = 10 log10 (2.64 x 10^6) = 64.2 dB 32. Most digital transmission systems are “self-clocking” in that they derive the bit synchronization from the signal itself. To do this the systems use the transitions between positive and negative voltage levels. These transitions help define the boundaries of the bit intervals. Solutions follow questions: a. The nonreturn-to-zero (NRZ) signaling method transmits a 0 with a +1 voltage of duration T, and a 1 with a -1 voltage of duration T. Plot the signal for the sequence n consecutive 1s followed by n consecutive 0s. Explain why this code has a synchronization problem. A long sequence of 1s or a long sequence of 0s produces a long period during which there is no change in the signal level. Consequently, there are no transitions (“zero crossings”) that help a synchronization circuit determine where the boundary of each signaling interval is located. b. In differential coding the sequence of 0s and 1s induces changes in the polarity of the signal; a binary 0 results in no change in polarity, and a binary 1 results in a change in polarity. Repeat part (a). Does this scheme have a synchronization problem? The occurrence of a “1” induces a transition and helps synchronization. However sequences of “0s” still result in periods with no transitions. c. The Manchester signaling method transmits a 0 as a +1 voltage for T/2 seconds followed by a -1 for T/2 seconds; a 1 is transmitted as a -1 voltage for T/2 seconds followed by a +1 for T/2 seconds. Repeat part (a) and explain how the synchronization problem has been addressed. What is the cost in bandwidth in going from NRZ to Manchester coding? Every T-second interval now has a transition in the middle, so synchronization is much simpler. However, the bandwidth of the signal is doubled, as pulses now are essentially half as wide, that is, T/2 seconds.