Homework Ch.3
12. Suppose that a link between two telephone offices has 50 repeaters. Suppose that the probability that a repeater fails during a year is 0.01, and that repeaters fail independently of each other.
a. What is the probability that the link does not fail at all during one year?
Let p be the probability that a repeater fails during a year, then 1 – p is the probability that it does not fail, and the probability that all 50 repeaters do not fail is (1 – .01) 50 ¡Ö e–50(.01) = 0.605 where we have used the approximation (1 – p)n ¡Ö e–np which is valid for large n and small p.
b. Repeat (a) with 10 repeaters; with 1 repeater.
The probability that all 10 repeaters do not fail is (1 – .01) 10 ¡Ö e–10(.01) = 0.905, and the probability that a single repeater does not fail is 0.99.
The moral of the calculations is that a system that requires the functioning of a large number of relatively reliable components may be fairly unreliable. In terms of repeaters, this implies that minimizing the number of repeaters needed in a link is important from the point of view of reliability. Of course this also reduces the cost expended to install and maintain the repeaters.
19. An analog television signal is a lowpass signal with a bandwidth of 4 MHz. What is the bit rate required if we quantize the signal and require an SNR of 60 dB?
Solution:
W = 4 x 106 Hz
2W = 8 x 106 samples/sec
SNR = 60 dB = 6m – 7.2dB
6m = 67
m ¡Ö 11 bits
R = 2W x 11 = 88 Mbps
22. A telephone office line card is designed to handle modem signals of the form x(t) = Acos(2ðfct + ö(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.
Solutions follow questions:
a. How many levels must the quantizer have to produce the desired SNR? mxAAASNRmm64076.1)1000025.1(log10)12/)2/2(2/)100/((log10)12/2/)100/((log104021022102210+-===.== 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.
29. What is the maximum reliable bit rate possible over a telephone channel with the following parameters?
Solutions follow questions:
a. W = 2.4 kHz SNR = 20 dB
An SNR of 20 dB corresponds to a value of 100. The channel capacity formula then gives
C = 2400 log2 (1 + 100) = 15979 bps.
b. W = 2.4 kHz SNR = 40 dB
C = 2400 log2 (1 + 10000) = 31890 bps.
c. W = 3.0 kHz SNR = 20 dB
C = 3000 log2 (1 + 100) = 19974 bps.
d. W = 3.0 kHz SNR = 40 dB
C = 3000 log2 (1 + 10000) = 39863 bps.
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 log2 (1 + SNR), C ¡Ý = 64 kbps
= W log2 (1 + SNRmin) Ë log2 (1 + SNRmin) = 64/3 Ë 1 + SNRmin = 3
Ë SNRmin = 2.64 x 106
in dB: SNRmin = 10 log10 (2.64 x 106) = 64.2 dB ˆ a very clean channel
31. Suppose that a low-pass communications system has a 1 MHz bandwidth. What bit rate is attainable using 8-level pulses? What is the Shannon capacity of this channel if the SNR is 20 dB? 40 dB?
Solution:
Nyquist pulses can be sent over this system at a rate of 2 million pulses per second. Eight-level signaling carries 3 bits per pulse, so the bit rate is 6 Mbps.
The Shannon capacities are:
C = 1000000 log2 (1 + 100) = 6.6 Mbps.
C = 1000000 log2 (1 + 10000) = 13.3 Mbps.
33. Consider a baseband transmission channel with a bandwidth of 10 MHz. What bit rates can be supported by the bipolar line code and by the Manchester line code?
Solution:
From Figure 3.26 we see that a bipolar code with pulses T-seconds wide occupies a bandwidth of W = 1/T Hz. Therefore a 10 MHz bandwidth allows a signaling rate of 10 megabits/second.
From the figure it can also be seen that a Manchester code occupies twice the bandwidth. Hence a 10 MHz bandwidth allows a signaling rate of 5 megabits/second.
44. Use Figure 3.47 and Figure 3.50 to explain why the bandwidth of twisted-wire pairs and coaxial cable decreases with distance.
Solution:
The bandwidth is the range of frequencies where the channel passes a significant proportion of the power in the input signal. Both figures show that the attenuation when measured in dB/km increases with higher frequency. For example, the attenuation for 19-gauge wire at 100 kHz is about 5 dB/km and at 1MHz it is about 15 dB/km. This implies that the relative attenuation between a lower frequency and a higher frequency increases with distance. Using the same example, we have that at 1 km and at 10 km, the attenuation at 100 kHz is 5 dB and 50 dB respectively, but at 1 MHz it is 15 dB and 150 dB respectively. Thus at longer distances higher frequencies are attenuated much more severely and consequently the bandwidth decreases with increasing distance.
45. Calculate the bandwidth of the range of light covering the range from 1200 nm to 1400 nm. How many Hz per person are available if the population of the world is six billion people? Repeat for 1400 nm to 1600 nm. Keep in mind that the speed of light in fiber is approximately 2 x 108 m/sec.
Solution:
Frequency and wavelength are related by the expression: f = v/ë, where v = 2x108. The frequencies corresponding to 1200 nm and 1400 nm are f1 = v/ë1 = 2(108)/1.2(10–6) and f2 = v/ë2 = 2(108)/1.4(10–6) so the bandwidth is BW = f1 – f2 = 23.8 THz.
The Hz/person is 23.8 x1012 / 6 x109 = 5.95 x103 Hz or approximately 6 KHz per person.
For the band from 1400 nm to 1600 nm we get BW = f1 – f2 = 2(108)(1/1.4 – 1/1.6)106 = 17.8 THz. The Hz per person is 17.8 x1012 / 6 x109 = 2.96 x103 Hz or approximately 3 KHz per person.
57. An early code used in radio transmission involved using codewords that consist of binary bits and contain the same number of 1s. Thus, the 2-out-of-5 code only transmits blocks of 5 bits in which 2 bits are 1 and the others 0.
Solutions follow questions:
a. List the valid codewords.
11000
10100
10010
10001
01100
01010
01001
00110
00101
00011
b. Suppose that the code is used to transmit blocks of binary bits. How many bits can be transmitted per codeword?
There are 10 possible codewords. Three bits per codeword can be transmitted if eight codewords are used.
c. What pattern does the receiver check to detect errors?
Each received codeword should have exactly two bits that are ones and three bits that are zeros to be a valid codeword.
d. What is the minimum number of bit errors that cause a detection failure?
A valid codeword can be changed into another valid codeword by changing a 1 to a 0 and a 0 to a 1. Therefore, two bit errors can cause a detection failure.
59. Suppose that two check bits are added to a group of 2n information bits. The first check bit is the parity check of the first n bits, and the second check bit is the parity check of the second n bits.
Solutions follow questions:
a. Characterize the error patterns that can be detected by this code.
If we rearrange the 2n information bits and the 2 parity bits into two codewords, each consisting of n information bits and a parity bit, we see that in effect we have divided the overall codeword into two subcodewords of length n + 1. An error pattern is detectable if the error pattern in the first codeword and in the second codeword are both detectable. An error pattern in each subcodeword is detectable if the number of errors in the subcodeword is odd. Therefore an error pattern is detectable if the number of errors in both subcodewords is odd.
b. Find the error detection failure probability in terms of the error-detection probability of the single parity check code.
An error detection failure occurs is either the first subcodeword or the second subcodeword or both fail. Let Pdetect (n + 1) be the probability of successful error detection in the single parity code of length n + 1, then
P[detection failure in code of length 2n + 2] =
= 1 – P[successful detection in both codes of length n + 1]
= 1 – Pdetect (n + 1) Pdetect (n + 1)