Appendix A PROBABILITY AND RANDOM SIGNALS. A.1 Probability

Appendi A PROBABILITY AND RANDOM SIGNALS Deterministic waveforms are waveforms which can be epressed, at least in principle, as an eplicit function of time. At any time t = t, there is no uncertainty about the value of the waveform. Communication signals, however, are at least partially unpredictable, otherwise the object of communication is defeated. Furthermore, the ever-present agitation of the universe at the atomic level constitutes the noise in communication systems. Such noise waveforms are also unpredictable. Unpredictable waveforms such as a signal s(t) or a noise n(t) are called random processes. While random processes are unpredictable, they can very often be estimated. The estimation of the performance of a random process is generally associated with a certain probability of being correct. A. Probability Relative Frequency Approach: In an eperiment repeated N times, if the event A occurs m times, then P(A) the probability of event A is defined as m P(A) = lim N N (A.) This definition is known as the relative frequency definition. Note that as N, the limit eists. Thus it follows, 0 P(A) (A.2) Such eperiments (trials) are usually done in the mind rather than actually performed. For instance, an unbiased coin is tossed; a head and a tail are equally likely. Thus we reason that if many trials are performed, the probability of having a head is 0.5. i

ii (The relative frequency approach to probability is not rigorous, since there remain several undefined factors, e.g. how large is N in order that Eq. (A.) is true? Thus we use the aiomatic approach for rigour: The probability of an event A is a number assigned to this event such that P(A) 0 P(S) = where S is made up of all mutually eclusive and ehaustive events and if A and B are mutually eclusive. P(A + B) = P(A) + P(B) A.2 Mutually Eclusive Events Two possible outcomes of an eperiment are mutually eclusive if the occurrence of one precludes the occurrence of the other. If the events A and A 2 have probabilities P(A ) and P(A 2 ) respectively, and if A and A 2 are mutually eclusive then P(A A 2 ) = P(A ) + P(A 2 ) (A.3) Etending to n mutually eclusive outcomes, P(A A 2 A n ) = If M is the total number of outcomes, then M P(A i ) = i= n P(A i ) i= (A.4) (A.5) A.3 Joint Probability, Conditional Probability and Independent Events If an eperiment has two sets of outcomes A, A 2,... and B, B 2,..., then the probability of the joint occurrence of A i and B j is written as P(A i B j ) or simply P(A i, B j ). If the outcome B is influenced by the outcome A, then we have to introduce the concept

iii of conditional probability denoted by P(B j A i ), i.e. the probability of B j given that the outcome of A is A i. Suppose the eperiment is performed N times out of which N i is the number of times A i occurs with or without B j, and N j the number of times B j occurs with or without A i, also N ij is the number of times the A i and B j jointly occur, then, P(B j A i ) = N ij = N ij/n N i N i /N = P(A i, B j ) P(A i ) (A.6) Similarly, since N ij = N ji, we have, P(A i B j ) = N ji = P(A i, B j ) N j P(B j ) (A.7) So that Thus, P(A i, B j ) = P(B j A i )P(A i ) = P(A i B j )P(B j ) P(A i B j ) = P(A i) P(B j ) P(B j A i ) (A.8) (A.9) which is known as Bayes theorem. If the outcome of B is not influenced by the outcome of A, then A and B are independent for which } P(A i B j ) = P(A i ) (A.0) P(B j A i ) = P(B j ) and P(A i, B j ) = P(A i )P(B j ) Etending to any arbitrary number of independent outcomes, P(A i, B j, C k,...) = P(A i )P(B j )P(C k )... (A.) (A.2) A.4 Random Variables: Discrete The set of elements consisting of all possible distinct outcomes of an eperiment is called the sampled space. The individual elements of the sample space are called sample points, i.e each sample point corresponds to one distinct outcome of the eperiment. It s mathematically attractive to assign a real number to each sample point in the sample space according to some rule. This number is called a random variable. For eample we can assign a to a head and a to a tail in the tossing of a coin. Then the random

iv variable can assume two discrete values and. In general, if there are n outcomes, we assign n discrete values, 2,..., n to these sample points and the random variable can then assume all these n discrete values. (A random variable is actually a function in a conventional sense). Thus, a random variable is a real-valued point function (u) defined over the sample space S where u S. The probability assigned to each value of the random variable is denoted by P ( i ). Here the subscript refers to the random variable whereas the argument i is the particular value of the random variable. Thus P ( i ) is the probability that a random variable assumes a value i. Generally, if only one random variable is involved in the discussion, the omission of the subscript causes no confusion, hence the probability is denoted by P( i ). If there are n mutually eclusive outcomes, then n P ( i ) = i= (A.3) If there are two sets of outcomes described by two random variables and y, then P y ( i, y j ) = (A.4) i j For eample, throwing two dice where represents the outcomes of the first dice, and y those of the second. If and y are two random variables, then the conditional probability of = i and y = y j is denoted by P ( i y = y j ). For a given value y j of y, must assume any of the n values, 2,..., n. Hence, Similarly, Also, and Thus Bayes rule is, n P ( i y = y j ) = i= n P y (y j = i ) = j= P ( i y = y j ) = P y( i, y j ) P y (y j ) P y (y j = i ) = P y( i, y j ) P ( i ) P ( i y = y j ) = P y(y j = i )P ( i ) P y (y j ) (A.5) (A.6) (A.7) (A.8) (A.9)

v A.5 Random Variables: Continuous If the random variable can assume an innumerably infinite possible sample point, then the sample space is continuous and the random variable is thus continuous. For eample, the temperature T of a room is a continuous sample space. In this case, a more meaningful measure is the probability of observing the temperature in some small interval T. Hence we inquire about the probability of observing a random variable below some value. This probability denoted by F () is defined as Thus F () = probability( ) F () = 0 F ( ) = F () is called the cumulative distribution function of. Note that for 2 >, } F ( 2 ) F ( ) (A.20) (A.2) (A.22) Probability Density function In Eq. (A.22), if 2 >, the outcomes that, and that < 2 are mutually eclusive. Hence, probability( 2 ) = probability( ) + probability( < 2 ) Hence If 0, then by Taylor s epansion F ( + ) = F () + probability( < + ) (A.23) Comparing Eqs. (A.23) and (A.24) F ( + ) = F () + df () +... (A.24) d probability( < + ) = lim 0 The derivative of F () w.r.t is denoted by p (), i.e. p () = df () d df () (A.25) d (A.26) p () is called the probability density function of the random variable. From Eq. (A.26), F () = p ()d (A.27)

vi where, of course, F () = 0. Also, Note that and probability( < 2 ) = F ( 2 ) F ( ) = = 2 p ()d 2 p ()d p ()d (A.28) p ()d = p () 0 (A.29) (A.30) For a discrete random variable, the probability density function may be considered as a limiting case of the continuous variable, i.e. the probability density function is concentrated as impulses at some discrete points. Thus if a discrete random variable assumes values, 2,..., n with probabilities a, a 2,..., a n then the probability density function is, n p () = a r δ( r ) (A.3) Observe that since then p ()d = r= n a r = r= n r= a r δ( r )d = (A.32) It is possible to have a probability density function being both continuous and discrete at some points. Eample. A signal may have a probability density function as shown in Fig A.(a). If such a signal is passed through a limiter which clips all the voltage level greater than A, the new probability density function will be as shown in Fig A.(b) where the impulse appearing at = A has a strength K = A p ()d Similarly, if the limiter is used to clip both positive as well as negative amplitudes above A, the new probability density function will appear as in Fig A.(c) where the strength of the impulse at = A is given by K = A p ()d

vii p ( ) p ( ) (a) p ( ) 2 0 A (b) A 0 A (c) Figure A.: Thus, for Fig A.(c), the probability of observing any particular voltage amplitude where A < < A is zero, while the probability of observing the voltage amplitude A is K and that of observing voltage amplitude A is also K. A.6 Joint Distribution We can etend the concept of probability density functions to the outcome of two (or more) random variables which may or may not be independent of each other. Thus correspondingly, for two random variables and y, the probability that + d while y y y + dy is given by probability( + d, y y y + dy) = p y (, y)ddy where p y (, y) is the joint probability density function. Etending Eq. (A.33) to a finite interval, probability( 2, y y y 2 ) = The cumulative distribution function is, y2 2 F y (, y) = probability(, y y) = y y p y (, y)ddy (A.33) p y (, y)ddy (A.34) (A.35)

viii Obviously, From Eq. (A.35), we have p y (, y) = p y (, y)ddy = 2 y F y(, y) (A.36) (A.37) The individual probability densities p () and p y (y), usually called the marginal densities, can be obtained from the joint density p (, y). To obtain these densities, we have lim p () = probability( < + d, < y ) 0 = lim 0 = lim 0 + + lim p () = 0 [ p y (, y)dyd ] p y (, y)dy d regardless of where y lies lim 0 (A.38) Since, 0, the integral inside the brackets can be treated as a constant over the range (, + ). Hence, [ + ][ ] d p y (, y)dy Similarly, Thus, = lim 0 p () = p y (y) = [ ] p y (, y)dy p y (, y)dy p y (, y)d F () = probability(, y ) (A.39) (A.40) If and y are independent, then = p ()d = p y(, y)dyd and F y (y) = y p y(y)dy = y p y(, y)ddy probability( 2, y y y 2 ) = [ 2 ] [ y2 ] p ()d p y (y)dy y (A.4) (A.42)

i If there is no confusion, the density functions p (), p y (y) can simply be written as p() and p(y). A.7 Conditional Densities Etending the idea of conditional probability to continuous variables, we define the conditional probability density p ( y = y j ) as the probability density of given that y has a value y j. The probability density p ( y = y j ) is the intersection of a plane y = y j with the joint probability density function p y (, y) as shown in Fig A.2. Similarly, p y (y = i ) p (, y) p (, y) y Figure A.2: is the intersection of a plane = i with p y (, y). From definition, prob( y < y y + y) = = prob(, y < y y + y) prob(y < y y + y) y+ y p y (, y)dyd y y+ y y p y (y)dy As y 0, we have lim y 0 y p y(, y)d prob( y < y y + y) = lim y 0 yp y (y) = p y(, y)d p y (y)

As y 0, L.H.S. becomes prob( y = y) = F ( y = y). Hence, Now, Thus, Similarly, From Eq. (A.46) Substituting this in Eq. (A.45), Similarly and F ( y = y) = p(, y) p(y) d p ( y = y) = df ( y = y) d p ( y = y) = p y (y = ) = p(, y) p(y) p(, y) p() p(, y) = p y (y = )p() p ( y = y) = py(y =)p() p(y) p y (y = ) = p( y=y)p(y) p() p y (y = ) = p(,y) p(,y)dy (A.43) (A.44) (A.45) (A.46) (A.47) Eqs. (A.47) are the Bayes rule for continuous random variables. Note that from Eqs. (A.46) and (A.45) we obtain, p(,y) p ( y = y) = p(,y)d (A.48) Note also that p ( y = y)d = and p y(y = )dy = i.e. observing in the interval (, ) given that y = y is a certainty. The continuous random variables and y are said to be independent if p ( y = y) = p() Hence, p y (y = ) = p(y) p y (, y) = p()p(y) (A.49) (A.50) (A.5)

i A.8 Statistical Average (Mean) If a random variable can assume n values,..., n, and if the eperiment is repeated N times (N ), while m,..., m n be the number of trials favorable to outcomes,..., n respectively, then the mean value of is As N, we have = N (m +... + m n n ) = m N +... + m n N n = n i P ( i ) i= (A.52) where P ( i ) = probability of assuming the value i. The mean value is also called the epected value denoted by E[]. Thus, = E[] = i ip ( i ) and if is continuous = E[] = (A.53) p()d Often, it is desired to find the mean value of a certain function of a random variable. For eample, suppose we have a noise whose amplitude is represented by a random variable. In practice we are more interested in the mean square value of the signal, i.e. we desire E[ 2 ]. In general, if y = g() by definition, E[y] = yp y (y)dy (A.54) But the probability of y falling between y and y + y is p y (y)dy, and this is caused by falling between and +, the probability of which is p ()d. Thus, p y (y)dy = p ()d E[y] = If z is a function of and y such that z = φ(, y), g()p ()d (A.55) then, E[z] = φ(, y)p(, y)ddy (A.56)

ii Eample.2 If z = y and if and y are independent, then z = y Eample.3 If z = + y, and, y independent, then z 2 = 2 + y 2 + 2y A.9 Moments The nth moment of a random variable is defined as E[ n ] = n p()d The nth central moment of is its nth moment about its mean m, thus The variance σ 2 E[( m ) n ] = is the second central moment of, i.e. σ 2 = E[( m ) 2 ] = E[( 2 2m + m 2 )] = E[ 2 ] 2m E[] + m 2 If are independent and if and y are independent and if (A.57) ( m ) n p()d (A.58) = E[ 2 ] 2m 2 + m 2 = 2 m 2 (A.59) z = + y then [ σz 2 = ( + y) 2 ( + y) 2] = [ 2 + 2y + y 2 ( + y) 2] ] = [ 2 + 2y + y 2 2 2y y 2 ) )] = [( 2 2 + (y 2 y 2 = σ 2 + σ2 y (A.60) where y = y if and y are independent.

iii A.0 Some Useful Probability Distributions A.0. The Gaussian Probability Density Also called the normal probability density function is very important because many naturally occurring eperiments are characterized by random variables with a Gaussian Density. The Gaussian density is defined as [ ] ( m ) 2 p() = ep (A.6) 2πσ 2 Eample.4 Show that for a Gaussian density, p()d = 2σ 2 = m 2 = σ 2 + m 2 (A.62) A.0.2 The Rayleigh Probability Density Consider a comple phasor re jθ which may be written as where re jθ = n c + jn s n c = a k cosθ k k n s = a k sin θ k k (A.63) (A.64) If all θ k are all independent and are uniformly distributed random variables, also if all a k are random variables of similar magnitudes, then the central limit theorem ensures that n c and n s are Gaussian random variables, i.e. and p nc (n c ) = p ns (n s ) = 2πσ 2 nc e nc2 /2σ 2 nc 2πσ 2 ns e ns2 /2σ 2 ns (A.65) σ 2 n c = σ 2 n s = k σ 2 a k 2 σ2

iv Here, σ 2 a k is the variance of the random variable a k, and n c and n s are independent. The joint density is given by where r 2 = n 2 c + n2 s. But 2 +n 2 s )/2σ 2 p ncn s (n c, n s ) = p nc (n c ) p ns (n s ) = e (nc 2πσ 2 = e r2/2σ2 2πσ 2 (A.66) and transforming differential areas such that p ncn s (n c, n s ) dn c dn s = p rθ (r, θ) dr dθ dn c dn s = rdr dθ, we have Thus, p ncn s (n c, n s ) dn c dn s = e r/2σ2 2πσ 2 r dr dθ = p rθ(r, θ) dr dθ p r,θ (r, θ) = ( ) ( re r2 /2σ 2 2π σ 2 ) = p r (r) p θ (θ) (A.67) The last step in (A.67) is arrived at because we note that the epression of p rθ (r, θ) is free of θ, indicating that r and θ are independent and that the probability density function p θ (θ) must be a constant. Therefore, we conclude that and p r (r) = r σ 2 ep p θ (θ) = 2π [ r 2 2σ 2 ] 0 r = 0 elsewhere (A.68) (A.69) and p r (r) is called the Rayleigh distribution. For a random variable r, having a Rayleigh density, the mean value of r is E[r] = σ 2 0 r 2 e r2 /2σ 2 dr = σ π 2 (A.70)

v The mean square value of r is The variance σ 2 r is E[r 2 ] = r 3 e r2 /2σ σ 2 2 dr 0 = 2σ 2 (A.7) σr 2 = E[ 2 ] (E[]) 2 ( = 2 π ) σ 2 2 (A.72) A.0.3 The Rician Probability Density The Rayleigh distribution is derived from the fact that no one of the n c and n s components in (A.64) predominates. In some cases, it is found that one particular signal does dominate. This dominating sinusoidal wave is called the specular component. In this case, we assume the received signal has a specular component A cosω c t added to the previous sum of random terms. Hence, where σ 2 n c = σ 2 n s = σ 2. s(t) = (n c + A) cosω c t n s sin ω c t (A.73) Considering the term (n c +A) alone, we note that the sum represents a Gaussian variable with A being the average value and σ 2 being the variances; let we have n c = n c + A p n c (n c ) = e (n c A)2 /2σ 2 2πσ 2 (A.74) Now, r 2 = n 2 c + n 2 s = (n c + A) 2 2 + n s θ = tan n ( ) s = tan ns (A.75) n c n c + A with n c = r cosθ and n s = r sin θ, thus we have p rθ (r, θ) dr dθ = p n c n s (n c, n s ) dn c dn s { } = e (n c A) 2 +n 2 s /2σ 2 dn 2πσ 2 c dn s = e A2 /2σ 2 re (r2 2rAcos θ)/2σ 2 2πσ 2 dr dθ (A.76)

vi Now, p rθ (r, θ) cannot be written as a product p r (r)p θ (θ) since a term in the equation with both variables multiplied together as r cos θ. This indicates that r and θ are dependent variables. p r (r) can be found by integrating overall values of θ, i.e., p r (r) = e A2 /2σ 2 re r2 /2σ 2 2πσ 2 2π 0 e racos θ/σ2 dθ The integral cannot be evaluated in terms of elementary functions. The integral I o (z) 2π 2π 0 e z cos θ dθ is called the modified Bessel function of the first kind and zero order. Thus, { r +A p r (r) = 2 )/2σ 2} ( ) r A I σ 2 e (r2 o σ 2 (A.77) (A.78) (A.79) and is called the Rician distribution. The Rician distribution is plotted in Fig. A.3. Figure A.3: A. The Error Function For a Gaussian probability density with m = 0, the cumulative density is given by prob( ) = F() = 2πσ 2 e 2 /2σ 2 d (A.80) This integral cannot be easily evaluated but is directly related to the error function thoroughly tabulated. The error function of u is defined as erf(u) 2 π u 0 e 2 d (A.8)

vii Note that erf(0) = 0 and erf( ) =. The complementary error function erfc(u) is defined as erfc(u) = erf(u) 2 e 2 d (A.82) π To epress F() in Eq. (A.80) in terms of erf(u) and erfc(u), we have [ u F() = d = ] e 2 /2σ 2 d 2πσ 2 2πσ 2 = e 2 /2σ 2 2πσ 2 d u e 2 /2σ 2 d ( = ) 2 e u2 du where u = 2 π σ 2 σ 2 = 2 erfc( σ 2 ) = { } + erf( 2 σ 2 ) Since tabulated values are only obtained for positive u in erfc(u), thus for 0 F() = F( ) = = σ 2 e u2 du π { = } 2 e ζ2 dζ 2 π σ 2 e 2 /2σ 2 2πσ 2 d = erfc( 2 σ 2 ) (A.83) (A.84) Eample.5 Consider the transmission of a signal which can assume two values 0 and E. Due to the noise present during transmission, the signal received, v, is no more 0 or E but rather v = n when 0 is transmitted v = n + E when E is transmitted where n is a random variable representing the noise. We assume the noise has a Gaussian distribution and zero mean. Thus when 0 is transmitted, the received signal v has a Gaussian distribution centered at 0 (Fig A.4(a)). When E is transmitted, the received signal has a Gaussian distribution centered at E (Fig A.4(b)). A threshold level a has to be set such that if the received signal v is below a, we decide that the transmitted signal is 0; if the received signal v is above a, then we decide the transmitted signal is E. Thus if 0 is transmitted and the received signal v > a, then it constitute an error. The probability of this is given by, (Fig A.4(a)) e v2 /2σ 2 prob(v > a)) = dv = 2πσ 2 2 erfc( a σ 2 ) a

viii p( v) 0 a v (a) p( v) 0 a E v (b) Figure A.4: If E is transmitted and v < a, then an error is received, the probability of which is, (Fig A.4(b)) a e (v E)2 /2σ 2 prob(v < a) = dv 2πσ 2 Letting u = (E v)/σ 2 E a σ 2 prob(v < a) = = E a σ 2 e u2 π du e u2 π du = 2 erfc(e a σ 2 ) The combined probability of error is shown in Fig A.5. The shaded area constitutes the total probability of error if the decision threshold is chosen to be a. Obviously, the combined prob- 0 a E v Figure A.5:

i ability of error is a minimum when the decision threshold is chosen to be at the intersection of the two curves, i.e. a = E 2. A.2 The Transform of Probability Density Functions From Eq. (A.54), the mean of a function g() of the random variable is E[g()] = Let g() = e jζ, thus E[e jζ ] = g()p ()d p ()e jζ d = P (ζ) (A.85) Thus, the epected value of e jζ is the Fourier transform of the probability density function. Other transforms of the probability density function are also used: (a) The characteristic function E[e jζ )] = (b) The moment generating function E[e s ] = p()e jζ d = 2πF [p()] p()e s d = 2πjL [p()] Note that p () = P (ζ)e jζ dζ 2π A.3 Probability Density Function of a Sum of Two Independent Random Variable Let and y be two statistically independent random variables. Let z = + y

Let p (), p y (y), p z (z) be the probability density functions of, y and z respectively and P (ζ), P y (ζ), P z (ζ) be the respective Fourier transforms. By definition, Thus, by Eq. (A.56) P z (ζ) = P z (ζ) = E[e jζz ] p y (, y)e jζ(+y) ddy But and y are independent, thus the joint-probability density function is p y (, y) = p()p(y) Hence, substituting this joint density function into P z (ζ), we have P z (ζ) = Using the convolution theorem, p()e jζ p(y)e jζy ddy = P (ζ)p y (ζ) (A.86) p z (z) = p () p y (y) = p (λ)p y (z λ)dλ (A.87) Etending to n independent random variables having probability density functions p( ), p( 2 ),..., p( n ), and if z = + 2 +... + n then, p z (z) = p( ) p( 2 )...p( n ) (A.88) Eample.6 Two independent random variables and y both have Gaussian probability density functions given by ( m)2 /2σ2 p () = e σ 2π If z = + y, then But p y (y) = P (ζ) = σ 2π = (y my)2 /2σ2 y e σ y 2π P z (ζ) = P (ζ)p y (ζ) { e σ2 ζ 2 /2 e ( m)2 /2σ 2 e jζ d } e jζm (A.89)

i and similarly, { P y (ζ) = e ζ2} σ2 y e jζmy P z (ζ) = [e )ζ2] [ ] (σ2 +σ2 y e jζ(m +m y) Hence p z (z) = F [P z (ζ)] = e [z (m+my)]2/2(σ2 2π(σ 2 + σ2 y ) = (z mz)2 /2σ2 e z σ z 2π +σ2 y ) (A.90) where σ 2 z = σ2 + σ2 y m z = m + m y From Eq. (A.90), we see that the probability density function of a sum of two independent Gaussian random variables is also Gaussian. This result can be etended to any number of normally distributed random variables. A.4 The Central Limit Theorem The theorem states that if p( ), p( 2 ),..., p( n ) are the probability density functions of n random variables, then the probability density function of the sum of these n random variables tends to be Gaussian as n. The mean and variance of this gaussian density are respectively the sum of the means and the sum of the variances of the independent variables. The proof of the theorem will be omitted but rather demonstrated with a square pulse function p() in Fig A.6. p( ) p( ) p( ) p( ) p( ) p( ) 0 2 0 2 3 0 3 Figure A.6

ii A.5 Correlation Between Random Variables The covariance r of two random variables and y is defined as r E[( m )(y m y )] (A.9) If and y are independent random variables, then r = E[( m )(y m y )] = = ( m )(y m y )p(, y)ddy ( m )p()d (y m y )p(y)dy = (m m )(m y m y ) = 0 (A.92) On the other hand, if and y are dependent, say if increases, y increases; if decreases, y decreases, then we epect that E[( m )(y m y )] > 0. Similarly, if and y are related such that when increases, y decreases, and when decreases y increases, then we anticipate E[( m )(y m y )] < 0. If maimum possible relation between and y is assumed, i.e. = y, then, with m = m y = 0, we have E{y} = E[ 2 ] = E[y 2 ] = σ 2 = σ2 y = σ σ y (A.93) Or, if = y E{y} = E[ 2 ] = E[ y 2 ] = σ 2 = σ2 y = σ σ y We define the correlation coefficient ρ between and y as Since ma r = σ σ y, we have ρ r σ σ y = E{y} σ σ y ρ (A.94) (A.95) (A.96) If and y are independent then ρ = 0. Conversely, if ρ = 0, and y are said to be uncorrelated. But ρ = 0 does not necessarily mean that and y are independent. Thus Independent uncorrelated ρ = 0. Eample.7 is a random variable with p() =,. Let y = 2 2. Thus and y are not independent; however, and y are uncorrelated. E[] = 2 d = 0

iii The covariance is and y are uncorrelated. E[y] = E[ 2 ] = 2 2 d = 3 { ( r = E{( m )(y m y )} = E y )} 3 { = E 3 } ( = 3 ) d 3 2 3 For two random variables and y both having Gaussian probability density functions, if and y are uncorrelated, i.e. if ρ = 0, then they are independent. A.6 Random Processes To obtain the probabilities of various possible outcomes of an eperiment, we can either repeat the eperiment many times, or we can perform a large collection of the eperiment simultaneously. Such a collection of eperiments is called an ensemble, and the individual outcome called the sample function. For eample, consider a noise waveform n(t). We may measure repeatedly the output of a single noise source, or we may make simultaneous measurements of the output of an ensemble of statistically identical noise sources. Suppose we want to measure the mean square of the noise source: The average obtained from measurements on a single sample function at successive times is called a time average represented by n 2 (t). The average obtained from measurements on an ensemble of noise sources is called an ensemble average represented by E[n 2 (t)] t=t where t is the instant at which the measurements are performed. In general, the time average and the ensemble average will not be the same. For eample, if the statistical characteristics of the sample functions change with time, then the ensemble average will be different at different instants. When the statistical characteristics of the sample functions do not change with time, the random process is called stationary. Even if the random process is stationary, the ensemble average may not be the same as the time average for while the individual sample functions are stationary, they may differ in statistical characteristics to each other. If a random process is such that the ensemble average is identical to the time average, the process is called ergodic. An ergodic process is stationary, but a stationary process is not necessarily ergodic. Throughout this course, we deal only with ergodic random processes, i.e. n 2 (t) = E{n 2 (t)} t

iv A.7 Auto-Correlation of a Random Process The autocorrelation of ergodic random process n(t) is defined as R n (τ) = lim T T T 2 T 2 n(t)n(t + τ)dt (A.97) Since the auto-correlation of a deterministic signal is the inverse transform of the power spectral density, the power spectral density of a random process is defined similarly as in the case of a deterministic process such that S n (ω) = F[R n (τ)] = R n (τ)e jωτ dτ (A.98) In order to define the power spectral density of a deterministic function f(t) which etend from to, we select a section of this waveform which etends from T/2 to T/2. This waveform f T (t) = f(t) within this range, outside this range f T (t) = 0. Now, f T (t) F T (ω) and over the interval T, the normalized power density is F T (ω) 2 /T. As t, f T (t) f(t), thus the power spectral density of a deterministic waveform is given by S f (ω) = lim T T F T(ω) 2 (A.99) Correspondingly the power spectral density of a sample function of a random process is defined as S n (ω) = lim T T N T(ω) 2 But since n(t) is a random process, it will be meaningful to define only the ensemble average of the power density spectrum, i.e. [ ] S n (ω) = E lim T T N T(ω) 2 [ ] = lim E T T N T(ω) 2 (A.00) where E[ ] represents the ensemble average and N T (ω) is the Fourier transform of a truncated section of a sample function of the random process n(t). The auto correlation function R n (τ) in Eq. (A.97) is a time average. For an ergodic process, we can change this into an ensemble average, thus R n (τ) = E[n(t)n(t + τ)] (A.0)

v Now n(t) is a random variable at t, n(t + τ) is another random variable at t + τ, R n (τ) is the covariance of two different random variables. Suppose for some τ, we find R n (τ) = 0, then the random variables n(t) and n(t + τ) are uncorrelated, and if these random processes have Gaussian distributions, then they are also independent. In fact, as far as actual noise is concerned the covariance R n (τ) is zero ecept when τ = 0, i.e. n(t) and n(t + τ) are uncorrelated unless τ = 0. This is because noise in communication systems are generally regarded as white, i.e., their spectral density function is a constant throughout the whole spectrum. i.e. Now, since R n (τ) S n (ω) and if S n (ω) = k, then R n (τ) = kδ(τ) R n (τ) = 0 for τ 0. Thus n(t) and n(t+τ) are uncorrelated, and if they have Gaussian probability density functions, they are independent. - -

vi Problems. Si dice are thrown simultaneously. What is the probability that at least die shows a 3? 2. A card is picked from each of four cards in a 52-card deck. (a) What is the probability of selecting at least one 6 of spades? (b) What is the probability of selecting at least card larger than an 8? 3. A card is drawn from a 52-card deck, and without replacing the first card a second card is drawn. The first and second cards are not replaced an a third card is drawn. (a) If the first card is a heart, what is the probability of the second card being a heart? (b) If the first and second cards are hearts, what is the probability that the third card is the king of clubs? 4. A binary data system uses two symbols 0 and, transmitted with probabilities P 0 and P. Owing to transmission errors a 0 maybe changed to a at the receiver with probability p 0, and similarly for p. Obtain epressions for the following: (i) the approimate umber of errors in a sequence of n digits; (ii) the probability that a received digit is a 0; (iii) the probability that an error has occurred given that a 0 is received. 5. (a) An important probability density function is the Rayleigh density { e 2 /2 0 p() = 0 < 0 i) Prove that p() satisfies p() 0 for all, and p()d =. ii) Find the distribution function F(). (b) Refer to the Rayleigh density function given in (a). Find the probability P( < 2 ), where 2 =, so that P( < 2 ) is a maimum. Hint: Find P( < 2 ); replace 2 by +, and maimize P with respect to. (c) Refer to the Rayleigh density function given in (a). Find i) P(0.5 < 2). ii) P(0.5 < 2). 6. The joint probability density of the random variables and y is p(, y) = ke (+y) in the range 0, 0 y, and p(, y) = 0 otherwise. (a) Find the value of the constant k.

vii (b) Find the probability density p(), the probability density of independently of y. (c) Find the probability P(0 2; 2 y 3). (d) Are the random variables dependent or independent? 7. is a random variable having a gaussian density. E() = 0, σ 2 variable having the values or -, each with probability. 2 =. v is a random (a) Find the joint density p(, v). (b) Show that p(v) = p(, v)d. 8. The joint probability density of the random variables and y is p(, y) = ke (y+) in the range 0, 0 y, and p(, y) = 0 otherwise. (a) Find p() and p(y), the probability density of independently of y and y independently of. (b) Are the random variables dependent or independent? 9. Compare the most probable [p() is a maimum] and the average value of when (a) p () = 2π e ( m)2/2 for all { e 2 /2 for 0 (b) p 2 () = 0 elsewhere 0. Calculate the variance of the random variables having densities: (a) The gaussian density p () = 2π e ( m)2 /2 for all. (b) The Rayleigh density p 2 () = e 2 /2, 0. (c) The uniform density p 3 () = /a, a/2 a/2.. (a) A voltage v is a function of time t and is given by v(t) = cos ωt + y sin ωt in which ω is a constant angular frequency and and y are independent gaussian variables each with zero mean and variance σ 2. Show that v(t) may be written v(t) = r cos(ωt + θ) in which r is a random variable with a Rayleigh probability density and θ is a random variable with uniform density. (b) If σ 2 =, what is the probability that r?

viii 2. The random signal v in Problem (a) is transmitted through a channel in which the channel noise power is a constant η. The average signal to noise ratio is thus given by Γ = E[r2 ] η where from Problem, r is the magnitude of the random signal and has a Raleigh distribution. (a) Show that the signal to noise ratio has also a Rayleigh distribution given by p(γ) = Γ e γ Γ (b) A certain detector for the above random signal has a characteristic such that the probability of error is dependent on the signal to noise ratio. For a given γ, the probability of error is given by P(e γ) = 2 e γ Show that the total probability of error in the detector is given by P(ε) = 2(Γ + ) 3. The independent random variables and y are added to form z. If p () = e 2 /2 0 and (A.02) p y (y) = 2 e y y <, (A.03) find p z (z). 4. The function of time z(t) = cosω 0 t 2 sin ω 0 t is a random process. If and 2 are independent gaussian random variables each with zero mean and variance σ 2, find (a) E(z), E(z 2 ), σ 2 z, and (b) p z (z), (c) R z (τ). 5. A random process n(t) has a power spectral density S n (ω) = η/2 for ω. The random process is passed through a low-pass filter which has a transfer function H(ω) = 2 for ω m ω ω m and H(ω) = 0 otherwise. find the power spectral density of the waveform at the output of the filter.

i 6. White noise n(t) with S n (ω) = η/2 is passed through a low-pass RC network with a 3-dB frequency ω c. (a) Find the autocorrelation R(τ) of the output noise of the network. (b) Sketch ρ(τ) = R(τ)/R(0). (c) Find ω c τ such that ρ(τ) 0..