Q4 (b) The TV picture consists of 480 x 500 x 5 pixels per frame. There are 30 frames/sec.

1. 40

2. Q4 (b) The TV picture consists of 480 x 500 x 5 pixels per frame. There are 30 frames/sec. The total bit rate required to transmit the picture is:480x500x5x30 bps. This is 36 Mbps. Consider the maximum bit rate of the channel. A S/N of 35dB implies a ratio of 3162 in absolute units. (35 =10 log10(S/N) =>10 3.5 = S/N =3162). Using Shannon’s Theorem, the max data rate of the channel is found. Max data rate = Bandwidth x log2(1+ S/N) = 4.6 x 106 x log2(3163) = 53.45 Mbps It appears the channel can carry the transmission.

3. Q7 (p + q)n = nC0 p0qn + nC1p1q n-1 + …….nCrprq n-r …….nCnpnq0 The Binomial Theorem. If p denotes the probability of a repeater failure, then the term nCrprq n-r represents the probability of precisely ‘r’ repeater failures in a set of ‘n’ repeaters. The factor nCr refers to the total number of ways that ‘r’ failures could occur. The repeaters are in sequence, if one or more fail, then there is line failure. P(line failure) = 1 – P(no failure) = 1 - nC0 p0qn In this question we have p = 0.01, and n = 50, so q = 0.99 P(line failure) = 1 – (0.99)50 = 0.6 (approx) There is a 60% chance of a line failure in the course of a year.

4. Q8. 8. Suppose it is required to transmit at 64kbps over a 3.4kHz telephone channel. What is the minimum S/N that can accomplish this? 64000 = 3400 x log2(1+S/N) 64000/3400 = log2(1 + S/N) 18.82 = log2(1 + S/N) 2 18.82 = 1 + S/N 2 18.82 –1 = S/N S/N = 451 x 2 10 (approx)

5. 9.A twisted wire pair has an attenuation of 0.7dB per km at 1kHz. (a)What is the maximum repeater-less distance if an attenuation of 20dB can be tolerated? (b)If the wire pair is conditioned with loading coils to achieve an attenuation of 0.2dB/km, how long is the maximum repeater less distance in this case? There is attenuation of 0.7 dB per km. The attenuation is additive when expressed in dBs, so for every 0.7 in 20 that represents a km. (20/0.7 =28.5) So if a 20 dB loss can be tolerated, then the max repeater-less distance is 28.5 km. Part (b) is similar. (20/0.2 =100 km)

6. 10.Suppose a transmission channel operates at 3 Mbps, and has a bit error rate of 0.001. Bit errors occur at random, and occur independently of each other. Suppose the following code is used: to transmit a 1 bit the code word 111 is used, and to transmit a 0 the code word 000 is used. The receiver takes the three received bits and decides which bit was sent by taking a majority vote of the three bits. What is the probability that the receiver makes a decoding error.? If the receiver misreads precisely 0 bits, there will be no decoding error. If the receiver misreads precisely 1 bit within a code word, there will be no decoding error. However, if precisely 2 bits in a code word are misread, then there is a decoding error. And if precisely 3 bits are misread, there is a decoding error. P(decoding error) = P(2 or more bit misreads) = 1- {P(0) + P(1)} Use the Binomial Probability Distribution to work this out. (p + q)3 = p3 + 3 p2q + 3p1q2 + q3 The term q3 represents the probability that no bit error occurs. The term 3p1q2 represents the probability that precisely 1 bit error occurs. p= 0.001, q = 0.999 P(decoding error) = 1 –{0.00299 + 0.99700} = almost zero