Digital data (a sequence of binary bits) can be transmitted by various pule waveforms.

Chapter 2 Line Coding Digital data (a sequence of binary bits) can be transmitted by various pule waveforms. Sometimes these pulse waveforms have been called line codes. 2.1 Signalling Format Figure 2.1 shows a number of line codes for transmission of binary data 10110001. In Unipolar non-return-to-zero or On-off keying One is represented by one physical level (such as a DC bias on the transmission line). Zero is represented by zero voltage as shown in Figure 2.2. This format allows for long series without change, which makes synchronization difficult. Disadvantages of this codding are the waste of power due to the transmitted DC level and the power spectrum of the transmitted signal does not approach zero at zero frequency. In Bipolar non-return-to-zero One is represented by a positive voltage. Zero is represented by a negative voltage. In this format the voltage swings from positive to negative on the trailing edge of the previous bit clock cycle as shown in Figure 2.3. In Unipolar return to zero RZ, when the signal is 1, it drops (returns) to zero 26

Figure 2.1: Different signalling formats. Figure 2.2: Unipolar non-return-to-zero. between each pulse. While the level 0 remains at zero value in the whole pulse duration. See Figure 2.4. A separate clock does not need to be sent alongside the signal, but suffers from using twice the bandwidth to achieve the same data-rate as compared to non-return-to-zero format. Although RZ contains a provision for synchronization, it still has a DC component. In Bipolar return to zero RZ, the level 1 is represented by a positive value and returns to zero between each pulse. While the level 0 is represented by a negative value and also returns to zero before the pulse deration finishes. see Figure 2.5. #Spring 2014# 27

Figure 2.3: Bipolar non-return-to-zero. Figure 2.4: Unipolar return to zero RZ. Alternate Mark Inversion AMI is a kind of bipolar encoding where the level 0 is encoded as zero volts, as in unipolar encoding, whereas the level 1 is encoded alternately as a positive voltage or a negative voltage. see Figure 2.6. Finally Manchester coding in which the name comes from its development at the University of Manchester. Many versions are there, one of them is shown below. A 0 is expressed by a low-to-high transition, a 1 by high-to-low transition The transitions which signify 0 or 1 occur at the midpoint of a period. Manchester code always has a transition at the middle of each bit period. The existence of transitions allows the signal to be self-clocking, and also allows the receiver to align correctly. The price of these benefit is a doubling of the bandwidth requirement compared to simpler N RZ coding schemes. Manchester code is used in LAN IEEE 802.3 (Ethernet). There are many other signalling formats which are discussed in the literatures. There are many formats because the channel characteristics vary from application to another. For example; if the channel is (AC) coupled, a format with a large DC component should not be #Spring 2014# 28

Figure 2.5: Bipolar return to zero RZ. Figure 2.6: Alternate Mark Inversion AM I. chosen. Some of the important parameters to be considered in selecting a signalling format are: the spectral characteristics, immunity of the format to noise, bit synchronization, cost and complexity of the implementation. 2.1.1 Some Notes on Signalling Formats - The unipolar NRZ is simple to implement. There are no pulse transition for long sequence of 0s or 1s, which are necessary for synchronization of the receiver. There is no way to detect when an error has occurred at the receiver side. - The bipolar RZ formats guarantee the synchronization, but there is no capability for error detection. - The AMI RZ format has an error detection property, if two sequential pulses are received with the same polarity, it is evident that an error has occurred. Example 1 The AM I RZ signalling waveform representing the binary se- #Spring 2014# 29

Figure 2.7: Manchester coding. quence: 0100101011 is transmitted over a noisy channel. The received waveform is shown in the following figure, which contains a single error. Locate the position of this error. Justify your answer. Figure 2.8: Solution: Clearly, the error is located at the bit position 7, where we have a negative pulse. This bit is in error because with AMI, positive and negative pulses are used alternatively for bit 1, and no pulse is used for bit 0. The pulse in position 7 representing the third digit 1 in the data stream should have had positive polarity. 2.2 Digital Carrier Modulation Systems Because base band digital signals have low power at law frequencies, these signals are suitable for transmission over a pair of copper wires or coaxial cables. Base band signals can not be transmitted over a radio link because this would require impractically large antennas. Hence #Spring 2014# 30

we use analog modulation techniques. In binary modulation schemes, the modulation process corresponds to switching (or keying) the amplitude, frequency or phase of the continuous wave signal (the carrier) between either of two values corresponding to binary bits 0 and 1. The three main types of digital modulation are: 2.2.1 Amplitude Shift Keying (ASK) In this scheme, the modulated signal can be expressed as A cos ω c t, for bit 1; x c (t) = 0, for bit 0. That is why this scheme is known as On-Off Keying (OOK). The transmission bandwidth of ASK signal is B T = 2B, where B is the baseband bandwidth B = f s = 1/T s. In Additive White Gaussian Noise (AW GN) channel, the bit error probability for coherent detection in P SK is given by: ( A P e = Q 2 T ) ( s E ) b = Q, 4N o N o where E b is the average signal energy per bit, N o is the noise power and Q is Q function which is defined as: Q(z) = P (X z) = z 1 2π e x2 /2 dx. This function is widely used in bit error calculations. Use tables of Q f unction To find Q(x). While for non-coherent detection, the bit error probability can be approximated as: P e = 1 2 e 1 2 (E b/n o ). #Spring 2014# 31

Example 2 Find the minimum E b N o needed for ASK receiver to have P e < 1.5 10 2 for a) coherent and b) non-coherent detectors. Solution: a) For coherent detector, P e = Q ( E b N o ) from which E b N o table of Q function; E b N o = [Q 1 (1.5 10 2 )] 2 = [2.17] 2 = 4.7089 = 6.7dB. b) For non-coherent detector, P e = 1 2 e 1 2 (E b/n o ), from which E b N o = 2 ln(2p e ) = 2 ln(3 10 2 ) = 7.0092 = 8.47dB. = [ Q 1 (P e ) ] 2. Using the 2.2.2 Binary Frequency Shift Keying (BF SK) In this scheme, the modulated signal can be expressed as A cos ω 1 t, for bit 1; x c (t) = A cos ω 2 t, for bit 0. Currently, F SK is most popular format in practice. The minimum transmission bandwidth of F SK is B t = f 2 f 1 + 2B = f + 2B, where f 1 = 2πω 1. It is clear that the transmission bandwidth of F SK is higher that that of ASK. It can be shown that by properly choosing of ω 1 and ω 2, this deviation can be eliminated. The bit error probability for F SK has similar formula as ASK in both coherent and non-coherent detectors given above. The average signal energy per bit is given as E b = A2 T s N o. 2.2.3 Binary Phase Shift Keying (BP SK) In this scheme, the modulated signal can be expressed as A cos ω c t, for bit 1; x c (t) = A cos(ω c t + π) = A cos ω c t, for bit 0. #Spring 2014# 32

The BP SK can be considered as a superposition of two ASK waves. The transmission bandwidth of P SK signal is B T = 2B. In BP SK, there is only coherent detection and for AW GN channel, the bit error probability is given as: ( A P e = Q 2 T ) ( s = Q 2 E ) b, N o N o The following figure illustrates these digital modulation schemes for the case in which the data bits are represented by the bipolar N RT waveform. 2.3 Multi-Level Digital Modulation Schemes The concept of digital modulations discussed above can be extended to multi-level to increase spectral efficiency by simply converting the binary signal to multi-level signal before modulation. These schemes are known as: #Spring 2014# 33

M aryask. If M = 2 it is BASK, M could be 16 or 32. This scheme is rarley used in practice. M aryf SK. It is a good example of power efficient modulation scheme so it is very much interest for increasing the noise immunity of modulation format compared with BF SK. In general the relation between bit error probability and symbol error probability for M aryf SK is given as P s = log 2 (M P e ). M aryp SK. As in all M ary signalling techniques, M aryp SK is used to increase the spectral efficiency of BP SK. Several types of M aryp SK are there as: Quadrature Phase Shift Keying (QP SK), it is the more common type that is used extensively in applications including cellular services and satellite systems. Bandwidth of QP SK is twice that for BP SK. π/4 QP SK. It is so called because the four symbols set is rotated by π/4 or by 45 at every new symbol transition. The bandwidth of π/4 QP SK is the same as QP SK. and same bit error probability assuming ideal coherent detection. Offset QP SK (OQP SK). It is used in cellular CDMA system for the reverse link (mobile to base station). Both bandwidth and performance of OQP SK is same as ordinary QP SK. Minimum Shift Keying (MSK). It is defined in different forms and has different versions. It is easily generated and has constant amplitude signal, so it can be amplified with class C apmlifiers without distortion. Gaussian Minimum Shift Keying (GMSK). It is derivative of MSK. It is very power efficient and it is adopted by the Global System for Mobile (GSM) and second generation digital cordless telephone applications. GM SK has very good #Spring 2014# 34

bit error probability performance. #Spring 2014# 35

Chapter 3 Concepts of Probability and Statistics In this chapter we will give a brief discussion in probability theory and some statistical terms that are needed to follow the material in digital communication topics. Chance behavior is unpredictable in the short run, but has a regular and predictable pattern in the long run. The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions. 3.1 Some Terminologies Sample Space: the set of all possible outcomes of a random phenomenon. Event: any set of outcomes of interest. Probability of an event: the relative frequency of this set of outcomes over an infinite number of trials. P r(a) is the probability of event A. Sometimes it is written as P (A). Example 1 Suppose we roll two die and take their sum. 36

Sample space S = {2, 3, 4, 5,, 11, 12} P r(sum = 5) = 4 36 Because we get the sum of two die to be 5 if we roll a (1, 4), (2, 3), (3, 2) or (4, 1). 3.2 Operations in Probability Let A and B denote two events. A B is the event that either A or B or both occur. A B is the event that both A and B occur simultaneously. The complement of A is denoted by A. A is the event that event A does not occur. Note that P r(a) = 1 P r(a). A and B are mutually exclusive if both events cannot occur at the same time. A and B are independent events if and only if P r(a B) = P r(a) P r(b). Multiplication Law: If A 1,, A k are independent events, then P r(a 1 A2 Ak ) = P r(a 1 )P r(a 2 ) P r(a k ). Addition Law: If A and B are any events, then P r(a B) = P r(a) + P r(b) P r(a B) Note: This law can be extended to more than 2 events. The conditional probability of B given A P r(b A) = P r(a B) P r(a) #Spring 2014# 37

A and B are independent events if and only if P r(b A) = P r(b) = P r(b A) 3.3 Random Variable A random variable is a variable whose value is a numerical outcome of a random phenomenon Usually denoted by X, Y or Z. It can be: Discrete: a random variable that has finite or countable infinite possible values. Example: the number of days that it rains yearly. Continuous: a random variable that has an (continuous) interval for its set of possible values. Example: amount of preparation time to do something. 3.4 Probability Distributions The probability distribution for a random variable X gives the possible values for X, and the probabilities associated with each possible value (i.e., the likelihood that the values will occur). The methods used to specify discrete probability distributions are similar to (but slightly different from) those used to specify continuous probability distributions. f(x) is the probability mass function for a discrete random variable X having possible values x 1, x 2,. f(x i ) = P r(x = x i ) is the probability that X has the value x i. Properties of f(x): #Spring 2014# 38

0 f(x) 1 i f(x i) = f(x 1 ) + f(x 2 ) + = 1 Example 2 Suppose the random variable X is the number of rooms in a randomly chosen owner-occupied housing unit in a hotel. The distribution of X is: Rooms X 1 2 3 4 5 6 7 Probability 0.083 0.071 0.076 0.139 0.210 0.224 0.197 3.4.1 Expected Value Expected Value of X or mean µ = E(X) = R x i P r(x = x i ) = i=1 R x i f(x i ), i=1 where the sum is over R possible values. R may be finite or infinite. 3.4.2 Variance It represents the spread, relative to the expected value, of all values with positive probability and it is defined as: σ 2 = V ar(x) = R (x i µ) 2 P r(x = x i ) i=1 R σ 2 = x 2 i P r(x = x i ) µ 2 i=1 The standard deviation of X, denoted by σ, is the square root of its variance. #Spring 2014# 39

Example 3 For the Room example given above, find the following: E(X), V ar(x), and P r [a unit has at least 5 rooms]. Solution: Try to solve this example. E(X) = 4.782,. 3.4.3 Binomial Distribution - Two possible outcomes: Success (S) and Failure (F ). - Repeat the situation n times (i.e., there are n trials). - The probability of success, p, is constant on each trial. - The trials are independent. Let X = the number of S s in n independent trials. (X has values x = 0, 1, 2,, n) Then X has a binomial distribution with parameters n and p. The binomial probability mass function is P r(x = x) = n x p x (1 p)n x, x = 0, 1, 2,, n where n x is the binomial coefficient given by n x = n! (n x)!x! Expected Value: µ = E(X) = np Variance: σ 2 = V ar(x) = np(1 p) #Spring 2014# 40

Example 4 Each child born to a particular set of parents has probability 0.25 of having blood type O. a) If these parents have 5 children, what is the probability that exactly 2 of them have type O blood? b) What is the expected number of children with type O blood? c) What is the probability of at least 2 children with type O blood? Solution: a) Let X = the number of boys P r(x = 2) = f(2) = 5 2 0.25 2 (1 0.25) 5 2 = 0.2637 b) µ = 5(.25) = 1.25 c) 5 P r(x 2) = 5 k=2 k 1 = 1 5 k=0 k (0.25) k (0.75) 5 k (0.25) k (0.75) 5 k = 0.3671875 3.5 Continuous Random Variable f(x) is the Probability density function for a continuous random variable X. Properties - f(x) 0 - f(x)dx = 1 - P r[a X b] = b f(x)dx = P r[a < X < b] a #Spring 2014# 41

- Mean or Expected Value of X µ = E(X) = - Variance σ 2 = V ar(x) = xf(x)dx (x µ) 2 f(x)dx = x 2 f(x)dx µ 2 3.5.1 Normal Distribution Most widely used continuous distribution. Also known as Gaussian distribution. It has the following density function: f(x) = 1 ( (x µ) 2 ), 2πσ 2 2σ 2 where µ is its mean, and σ 2 ia its variance. Standard Normal Distribution A normal distribution with mean 0 and variance 1 is called a standard normal distribution. Standard normal probability density function: f(x) = 1 ( x 2 ) 2π 2 Standard normal cumulative probability function (Φ(z)): Let Z N(0, 1) Φ(z) = P r(z z) Summery property: Φ( z) = 1 Φ(z) how to make standardization: Suppose X N(µ, σ 2 ) and let Z = X µ. Then Z N(0, 1). σ #Spring 2014# 42

If X N(µ, σ 2 ), what is P r(a < X, b)? Form equivalent probability in terms of Z : ( a µ P r(a < X < b) = P r σ < Z < b µ ) σ Use standard normal tables to compute latter probability. Example 5 Suppose the distribution of heights of young women are normally distributed with µ = 64 and σ 2 = 2.72. What is the probability that a randomly selected young woman will have a height between 60 and 70 inches? ( 60 64 P r(60 < X < 70) = P r 2.7 < Z < = P r( 1.48 < Z < 2.22) 70 64 ) 2.7 = Φ(2.22) Φ( 1.48) = 09868 0.0694 = 0.9174 3.6 Central Limit Theorem Let X 1,..., X n be be a random sample from any independent and identically distributed (i.i.d.) random variables with mean µ and variance σ 2 each. Then the sample mean X is approximately normally distributed with mean µ and variance σ 2 /n. #Spring 2014# 43