Midterm Examination Solutions In class, closed book, 80 minutes. All problems carry equal credit. Do any four of the five problems. Begin each problem on a separate page. 1. Consider a communication network for a city of one million inhaants. a. Suppose that each inhaant, during the busiest hour of the day, is involved in an average of 4 data transactions per hour that use this network. Each transaction, on average, causes 4 packets of 1000 s each to be sent. Each packet travels over an average of 3 links. What is the aggregate average number of s per second carried by this network? How many 64 k/sec voice telephone links are required to carry this tra c? b. Suppose that the inhaants use their telephones an average of 10% of the time during the busy hour. How many voice telephone links are required for this? Assume that all calls are within the city and travel over an average of 3 links. Solution. a. Data are generated at an average rate of 10 6 4 transactions hour 4 packets transaction 1000 s packet 1 3600 hour seconds 4.44 10 6 s second. As each has to travel over three links, the tra c supported by the network on average is approximately 1.332 10 7 s per second. The number of 64 kbps voice telephone links needed to carry this tra c is hence 1.332 10 7 64 10 3 = 209. b. On average, at any given instant there are 10 6 0.1 = 10 5 individuals on the phones. This requires 10 5 /2 = 5 10 4 end-to-end voice links as all calls are between two people (assuming no conference calls). In circuit switching all voice links are dedicated for the duration of the call. As each such call requires three links, it follows that 1.5 10 5 voice links are needed. Compare with the answer for data transfer. 2. In the Fibre Distributed Data Interface (FDDI) every 4- block of data is encoded using a 5- code, sometimes called a 4B/5B code, which is constructed so that there is never more than one leading zero and no more than two trailing zeros in each 5- codeword. The resulting 5- codewords are then transmitted using NRZ-I encoding. 1
The 4B/5B code table is shown below. 4- data 5- codeword 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 a. The data sequence 0010011110000100 is to be transmitted over an FDDI network. Sketch the transmitted waveform. b. What are the advantages of the FDDI encoding technique? Comment on its bandwidth e ciency vis à vis Manchester encoding and on whether long strings of 0s in the data stream will cause synchronisation problems. What if there is a long string of 1s in the data stream? Solution. a. Breaking up the data sequence into four blocks, the corresponding coded sequence can be written down directly by inspection from the 4B/5B code table. Spaces have been introduced for reading clarity. Data sequence: 0010 0111 1000 0100 Coded sequence: 10100 01111 10010 01010 The corresponding NRZ-I waveform is shown below. b. The FDDI scheme almost doubles the bandwidth e ciency compared to Manchester encoding. (Recall that NRZ requires only one-half the bandwidth of Manchester.) The improvement in bandwith e ciency over Manchester encoding is not 2
quite 100% because five s are transmitted for every four data s. Thus, the improvement in bandwidth e ciency over Manchester encoding is 80%. Long strings of 1s do not pose a synchronisation or DC problem in FDDI which inherits its immunity from the di erential encoding NRZ-I. In addition, unlike plain vanilla NRZ-I, FDDI also enjoys immunity from synchronisation or DC problems caused by long strings of 0s by virtue of the 4B/5B encoding observe that the encoding guarantees that there will be no more than 3 contiguous 0s in the coded sequence. Thus, neither long strings of 1s nor long strings of 0s cause synchronisation or DC di culties in FDDI. This improvement over NRZ-I is at a cost of just 20% in bandwidth. 3. A linear, time-invariant channel has the following response to a pulse input. a. Sketch the output waveform y(t) of the channel when the input waveform x(t) is as shown below. Hint: What would be the output waveform if p(t τ 1 ) + p(t τ 2 ) is input for any fixed τ 1 and τ 2? b. Suppose NRZ-L signalling is used to encode 3- sequences of data using the pulse p(t). Are the resulting waveforms received distortion-free by the receiver? Can the receiver decode the original stream without errors? Assume the receiver has perfect synchronisation and that the channel is noise-free. Solution. a. The input waveform x(t) is just the superposition of shifted versions of the pulse p(t): x(t) = p(t) p(t T) + p(t 2T). As the system is LTI, it follows that the output waveform is given by and sketched below. y(t) = q(t) q(t T) + q(t 2T) 3
b. In addition to a T-unit delay, the channel causes a spreading of each pulse waveform so that successive s encoded via NRZ-L are subject to intersymbol interference. Thus, all output waveforms are distorted. Nonetheless, in the absence of noise and assuming perfect synchrony, the receiver can ambiguously decode the original stream without errors. To see this observe that each input string results in a unique output waveform so that the receiver could actually maintain a look-up table which maps each possible output waveform to the unique input string that causes it. Much less e ort is needed, however, and the receiver can actually decode the input string sequentially from the output waveform. How would he go about doing so? 4. Consider the periodic rectangular pulse train x(t) shown below. Suppose T = 1 ms. The pulse train x(t) is input to a linear, time-invariant transmission medium which introduces a delay of 1/6 ms. The medium has a flat passband between 2000 Hz and 4000 Hz and blocks all other frequencies. Determine the channel output signal y(t) and sketch it. Hint: Use Euler s formulæ for the trigonometric functions to put the complex exponential Fourier series in a familiar trigonometric form first. Solution. As usual, set f 0 = 1/T. The DC (or zero frequency) coe series is then given by cient of the Fourier a 0 = 1 T T/2 x(t)dt = 1 T 0 dt = 1 2, while, for n 0, the Fourier coe cients are given by a n = 1 T T/2 x(t)e j2πnf 0t dt = 1 T 0 e j2πnf 0 t dt = e j2πnf 0 t j2πnf 0 T T/2 1 = j2πnf 0 T (1 e jπnf 0T ) = 1 cos(nπ). j2πn 0 4
Thus, when n 0, a n = { 1/jπn if n is odd, 0 if n is even. Thus, the pulse train has the Fourier series representation x(t) = 1 2 + n= n odd e j2πnf 0 t jπn = 1 2 + n=1 n odd 2 πn sin(2πnf 0t) by combining the corresponding negative and positive terms in the complex exponential Fourier series and using Euler s formula. This looks very familiar. And indeed it is. Look up the odd square wave we d analysed in class. How is it related to the pulse train shown here? If we pass x(t) through an LTI system with transfer function H(f) we obtain the output waveform y(t) = 1 2 H(0) + n= n odd H(nf 0 )e j2πnf 0 t. jπn Write H(f) = A(f)e jθ(f). We re given { 1 if 2000 < f < 4000, A(f) = 0 otherwise, and, as the medium introduces a constant delay τ = T/6 = 1/6 ms, θ(f) = 2πfτ. Observe, in particular, as f 0 = 1/T = 1000 Hz, { 1 if n = 3 or n = 3, A(nf 0 ) = 0 otherwise. so that it follows that all harmonics nf 0 for n 3 are blocked and the only terms in the Fourier series that survive are the terms n = 3 and the term n = 3. Thus, y(t) = 1 ( H(3f0 )e j6πf 0 t H( 3f 0 )e j6πf 0 t ) = 1 ( e j6πf 0 t j6πf 0 τ e j6πf 0t+j6πf 0 τ ) j3π j3π = 2 3π sin(6πf 0t 6πf 0 τ) = 2 3π sin(6πf 0t π) as f 0 τ = 1/6. It follows that y(t) = 2 3π sin(6πf 0t). 5. A particular noisy communication link has a length of 10 km. In a digital signalling scheme, each d km length of the link may be modelled as a binary symmetric channel with raw error probability p = p(d) for any 0 d 10. Let P = P (n) denote the end-to-end error probability when n uniformly spaced repeaters are deployed in the link. For our purposes the receiver himself counts as a repeater. 5
a. Determine P (1), P(2), and P(3) when p(d) = d 2 /1000 for 0 d 10. b. Repeat when p(d) = d/100 for 0 d 10. Comment on the two cases. Solution. For any n, the direct approach shows us that P (n) = n k=0 k odd ( ) n p k (1 p n k ) = 1 k 2( 1 (1 2p) n ), where p = p(d) = p(d/n) where D = 10 km is the end-to-end distance along the link, and d = D/n is the link length between successive repeaters. a. When p(d) = d 2 /1000, direct substitution shows that P (1) = 1 2 ( 1 (1 2 10 2 /1000) 1) = 0.1, P (2) = 1 2 ( 1 (1 2 5 2 /1000) 2) = 0.04875, P (3) = 1 2 ( 1 (1 2 10 2 /3 2 1000) 3) = 0.0325981. b. When p(d) = d/100, direct substitution again yields P (1) = 1 2( 1 (1 2 10/100) 1 ) = 0.1, P (2) = 1 2( 1 (1 2 5/100) 2 ) = 0.095, P (3) = 1 2( 1 (1 2 10/3 100) 3 ) = 0.0934815. It is clear that increasing the number of repeaters brings more benefit when the raw error probability increases faster than linearly with link length. 6