Fundamentals of Wireless Communication

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1 Communication Technology Laboratory Prof Dr H Bölcskei Sternwartstrasse 7 CH-8092 Zürich Fundamentals of Wireless Communication Homework 3 Solutions Handout date: April 27, 2018 Problem 1 Estimation of the delay spread and the mean delay time 1 We start with the I O relation expressed in terms of the delay-doppler spreading function, r(t) = S H (, )s(t )e j2πt dd, and compute the power of the output signal using the WSSUS assumption as follows: E r(t) 2 = E S H (, )s(t )e j2πt SH (, )s (t )e j2π t ddd d = E S H (, )SH (, ) s(t )s (t )e j2πt e j2π t ddd d = C H (, )δ( )δ( )s(t )s (t )e j2πt e j2π t ddd d = C H (, ) s(t ) 2 dd = s(t ) 2 C H (, )dd = s(t ) 2 q()d (11) Now, if we excite the channel by an infinitesimally short input signal with s(t) 2 = δ(t), we obtain E r(t) 2 = s(t ) 2 q()d = δ(t )q()d = q(t) From q(t) the mean delay time can be computed as the first moment of q(t), ie, = 1 P q()d with P = q()d, and the multipath delay spread as the second moment of q(t), ie, ( )2 q()d σ = P

2 2 The power delay profile (PDP) of the wireless channel is taken as { P λe λ, 0, q() = 0, < 0 Note that the PDP is zero for negative delays, because no signal can arrive earlier than it was emitted The first and second moments of q(t) can be computed from the properties of the PDF of an exponentially distributed random variable as follows: = 1 P σ 2 = 1 P 0 0 q(t)d = 1 λ, ( ) 2 q()d = 1 λ 2 The energy received within a window of duration 2ασ around the mean delay time can be computed as E ασ ;+ασ = = +ασ ασ +ασ q(t)dt max0, ασ P λe λt dt = P e λt +ασ max0, ασ = P (e λ(+ασ ) e ( = P e 1 α e max0,1 α) λ max0, ασ ) Finally, the fraction of the total energy that arrives within a window of 2ασ around the mean delay time is given by E ασ ;+ασ P = e max0,1 α e 1 α Setting α = 1, we conclude that more than 86% of the total energy arrives within the interval σ ; + σ When α increases, the percentage of energy arriving within interval ασ ; + ασ approaches 100% very quickly Problem 2 Doppler Spread 1 For the input signal s(t) = e j2πf0t, the corresponding received signal is r(t) = h(t, )e j2πfo(t ) d = e j2πfot h(t, )e j2πfo = e j2πfot L H (t, f o ) 2

3 2 The autocorrelation of r(t) is given by where Er(t + t)r (t) = E e j2πfo(t+ t) L H (t + t, f o )e j2πfot L H (t, f o) = e j2πfo t R H ( t, 0) = e j2πfo t C H (, )e j2π t e j2π0 dd = e j2πfo t C H (, )e j2π t dd = e j2πfo t p()e j2π t d p() = C H (, )d is the power Doppler profile Note that the autocorrelation function of r(t) depends only on t and not on t Therefore, r(t) is WSS 3 We compute the power spectral density of r(t) as S r (f) = F{R r (t)}, where R r (t) = Er(t + t )r (t ) as follows S r (f) = R r (t)e j2πtf dt t = e j2πfot p()e j2πt d e j2πft dt t = p() e j2π( f+fo)t dtd t = p()δ( f + f o )d = p(f f o ) Hence, p(f) describes the spectral broadening of an impulse in frequency Problem 3 WSS and US Assumption Scattered paths with equal delay and different Doppler shifts 1 As the time-varying channel impulse response is given by h(t, ) = δ() ( α 1 e j2π 1t + α 2 e j2π 2t ) 3

4 the time correlation function is given as R h (t + t, t, 1, 2 ) = Eh(t + t, 1 )h (t, 2 ) = E δ( 1 ) (α ) 1 e j2π1(t+ t) + α 2 e j2π 2(t+ t) δ( 2 ) ( α1e j2π1t + α2e j2π 2t ) = δ( 1 )δ( 2 ) E (α ) 1 e j2π1(t+ t) + α 2 e j2π ( 2(t+ t) α 1 e j2π1t + α2e j2π 2t ) = δ( 1 )δ( 2 )E α 1 2 e j2π1 t + α 2 2 e j2π2 t + α 1 α2e j2π( 1 2 )t+j2π 1 t + α 2 α1e j2π( 2 1 )t+j2π 2 t = δ( 1 )δ( 2 ) E α 1 2 e j2π1 t + E α 2 2 e j2π2 t + Eα 1 α2 e j2π( 1 2 )t+j2π 1 t + Eα 2 α1 e j2π( 2 1 )t+j2π 2 t = δ( 1 )δ( 2 ) e j2π1 t + e j2π2 t + Eα 1 α2 e j2π( 1 2 )t+j2π 1 t + Eα 2 α 1 e j2π( 2 1 )t+j2π 2 t 2 When the path gains corresponding to different Doppler shifts are uncorrelated the time correlation function simplifies to R h (t + t, t, 1, 2 ) = δ( 1 )δ( 2 ) e j2π1 t + e j2π 2 t Thus, the time correlation function depends only on the time difference and not on the absolute time Hence, the channel is WSS in time In other words, a channel which is white in Doppler frequency is WSS in time Scattered paths with different delays and no Doppler shift 1 From the time-varying impulse response h(t, ) = α 1 δ( 1 ) + α 2 δ( 2 ) we compute the time-varying channel transfer function L H (t, f) as follows: L H (t, f) = α 1 δ( 1 ) + α 2 δ( 2 ) e j2πf d = α 1 e j2πf 1 + α 2 e j2πf 2 The frequency correlation function of this channel is equal to R H (t 1, t 2, f + f, f) = EL H (t 1, f + f)l H (t 2, f) ( ) ( ) = E α 1 e j2π(f+ f) 1 + α 2 e j2π(f+ f) 2 α1e j2πf 1 + α2e j2πf 2 = E α 1 2 e j2π f 1 + α 2 2 e j2π f 2 + α 1 α2e j2πf( 2 1 ) j2π f 1 + α 2 α1e j2πf( 1 2 ) j2π f 2 = E α 1 2 e j2π f 1 + E α 2 2 e j2π f 2 + Eα 1 α 2 e j2πf( 2 1 ) j2π f 1 + Eα 2 α 1 e j2πf( 1 2 ) j2π f 2 = e j2π f 1 + e j2π f 2 + Eα 1 α 2 e j2πf( 2 1 ) j2π f 1 + Eα 2 α 1 e j2πf( 1 2 ) j2π f 2 4

5 2 When the path gains corresponding to different delays are uncorrelated, the frequency correlation function simplifies to R H (t 1, t 2, f + f, f) = e j2π f 1 + e j2π f 2 Thus, the frequency correlation function depends only on the frequency difference and not on the absolute frequency and hence the channel is WSS in frequency In other words, a channel that exhibits uncorrelated scattering (US) for different delays, is WSS in frequency Problem 4 Left Eigenvectors and Principle of Biorthogonality 1 As y is a right eigenvector of A corresponding to the eigenvalue λ, it satisfies y H A = λy H (41) We now prove that A H y = λ y (42) and that A T y = λy (43) We start with (42) Take the Hermitian conjugate of both sides in (41) to obtain (y H A) H = (λy H ) H (44) Now note that the left-hand side of (44) is equal to (y H A) H = A H y and that the right-hand side of (44) is equal to (λy H ) H = λ y Hence, A H y = λ y That is, y is a right eigenvector of A H corresponding to λ To prove (43), take the transpose of both terms in (41) (y H A) T = (λy H ) T (45) Now, note that the left-hand side of (45) is equal to (y H A) T = A T y and that the right-hand side of (45) is equal to (λy H ) T = λy 5

6 Hence, A T y = λy That is, y is a right eigenvector of A T corresponding to the eigenvalue λ For the last point, consider the following 2 2 real matrix A = The eigenvalues of this matrix are readily found as λ = 1 with multiplicity 2 The right eigenvectors corresponding to λ = 1 are found by solving the linear system which has solution x = a 0, where a R Ax = x The left eigenvectors associated to λ = 1 are found by solving the linear system y T A = y T 0 which has solution y =, where b R Hence, for the matrix A right and left b eigenvectors are not the same 2 The vectors x and y satisfy Ax = λx (46) y H A = µy H (47) where λ µ We prove that x and y are orthogonal, ie, that x, y = 0, by manipulating y H Ax in the following two ways As x satisfies (46), then On the other hand, as y satisfies (47), Therefore, y H Ax = λy H x = λ x, y y H Ax = µy H x = µ x, y λ x, y = µ x, y As λ µ, the inner product x, y must be zero, and the vectors x and y must be orthogonal 6

7 Problem 5 Detection in Gaussian Noise 1 We can obtain a scalar sufficient statistic y (for x on the basis of the observation of r), by projecting r on the direction of h Hence, the sufficient statistic y is given by where z = h T w, and hence z N (0, N 0 /2) y = h T r = E s x h 2 + h T w = E s x + z As the probability density function of y given x is equal to f Y X (y x) = 1 exp ( (y E s x) 2 ), πn0 the log-likelihood ratio corresponding to y is given by N 0 Es LLR(y) = log f Y X(y x = 1) f Y X (y x = 1) = 4y N 0 Furthermore, the threshold η = P(x = 1)/P(x = 1) equals 1 and log η = 0 Hence, the MAP rule can be expressed as follows: Choose x = 1 if LLR(y) > 0 Otherwise, choose x = 1 Using the isotropic property of Gaussian noise, we readily find the probability of error: P(e) = P(e x = 1) = P(LLR(y) 0 x = 1) = P(y 0 x = 1) = P(z + E s 0) = P(z ( ) 2Es E s ) = Q 2 Consider the random vector Aw, where w CN (0, I) The random vector Aw has zero mean and covariance given by N 0 E Aw(Aw) H = E Aww H A H = AA H = K Hence, z = Aw Since A is invertible, y = A 1 r is a sufficient statistic, and we can consider the equivalent detection problem y = E s A 1 hx + A 1 z = E s gx + w (51) where g = A 1 h Note that the detection problem in (51) is a binary detection problem in additive white Gaussian noise The multiplication by A 1 is called noise whitening filtering Repeating the same steps as in Appendix D of the lecture notes, we obtain a sufficient statistic for x by projecting y on v = g/ g The sufficient statistic y is given by y = v H y = E s g x + w where w = v H w, and, as v has unit norm, w CN (0, 1) Since E s g x is real and w is circularly symmetric (ie, R(w) and I(w) are independent), we can further extract a sufficient statistic by taking the real part of y R(y) = E s g x + R(w) 7

8 where R(w) N (0, 1/2) Now, compute the log-likelihood ratio LLR(R(y)) as LLR(R(y)) = log exp (R(y) E s g ) 2 exp (R(y) + E s g ) 2 = 4R(y) E s g The probability of error is given by P(e) = P(e x = 1) = P(LLR(R(y)) 0 x = 1) = P(R(y) 0 x = 1) = P( E s g + R(w) 0) = P(R(w) E s g ) ( ) ( ) = Q 2Es g 2 = Q 2Es A 1 h 2 8