WIXLIB

9Thus, the signals fn (t) are orthogonal. It is also straightforward to prove that the signals have unitenergy :Z fi (t) 2 dt 1, i 1, 2, 3Hence, they are orthonormal.b. We first determine the weighting coefficientsZ x(t)fn (t)dt,xn n 1, 2, 3 x1 x2 x3ZZ4x(t)f1 (t)dt 04x(t)f2 (t)dt 1212ZZ041dt 12Z12dt 12Z3dt 212Z4dt 03x(t)dt 0ZZZZZ 41 21 31 41 1dt dt dt dt 0x(t)f3 (t)dt 2 02 12 22 3000As it is observed, x(t) is orthogonal to the signal wavaforms fn (t), n 1, 2, 3 and thus it can notrepresented as a linear combination of these functions.Problem 2.11a. As an orthonormal set of basis functions we consider the set 1 0 t 1 f1 (t) f2 (t) 0 o.w 1 2 t 3 f3 (t) f4 (t) 0 o.w 1 1 t 20 o.w1 3 t 40 o.wIn matrix notation, the four waveforms can be represented as s1 (t)2 1 1 1 s (t) 2 110 2 s3 (t) 1 1 1 1 s4 (t)1 2 22f1 (t) f2 (t) f3 (t) f4 (t)Note that the rank of the transformation matrix is 4 and therefore, the dimensionality of thewaveforms is 4PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

10b. The representation vectors arehs1 s2s3s42 1 1 1hi 2 1 1 0hi 1 1 1 1hi 1 2 2 2ic. The distance between the first and the second vector is:rhip2d1,2 s1 s2 4 2 2 1Similarly we find that :d1,3 d1,4 d2,3 d2,4 d3,4 ppppp s1 s3 2 s1 s4 2 s2 s3 2 s2 s4 2 s3 s4 2 rhrhrhrhrhi1 0 2 0i1 1 1 3i 3 2 0 1 3 3 3 20 1 3 3Thus, the minimum distance between any pair of vectors is dmini222i2 25 5 12 142 31 19 5.2Problem 2.12As a set of orthonormal functions we consider the waveforms 1 0 t 1 1 1 t 2f1 (t) f2 (t) 0 o.w 0 o.wThe vector representation of the signals iss1 s2 s3 s4hhh2 2 22 0 0ii0 2 2hi 2 2 0 1 2 t 3f3 (t) 0 o.wiPROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

11Note that s3 (t) s2 (t) s1 (t) and that the dimensionality of the waveforms is 3.Problem 2.131. P (E2 ) P (R2, R3, R4) 3/7.2. P (E3 E2 ) P (E3 E2 )P (E2 ) P (R2)3/7 13 .3. Here E4 {R2, R4, B2, R1, B1} and P (E2 E4 E3 ) P (E2 E3 E4 )P (E3 E4 ) P (R2)P (R2,B2,R1,B1) 14 .4. E5 {R2 , R4 , B2 }. We have P (E3 E5 ) P (R2 , B2 ) 27 and P (E3 ) P (R1, R2, B1, B2) and P (E5 ) 73 . Obviously P (E3 E5 ) 6 P (E3 )P (E5 ) and the events are not independent.47Problem 2.141. P (R) P (A)P (R A) P (B)P (R B) P (C)P (R C) 0.2 0.05 0.3 0.1 0.5 0.15 0.01 0.03 0.075 0.115.2. P (A R) P (A)P (R A)P (R) 0.010.115 0.087.Problem 2.15The relationship holds for n 2 (2-1-34) : p(x1 , x2 ) p(x2 x1 )p(x1 )Suppose it holds for n k, i.e : p(x1 , x2 , ., xk ) p(xk xk 1 , ., x1 )p(xk 1 xk 2 , ., x1 ) .p(x1 )Then for n k 1 :p(x1 , x2 , ., xk , xk 1 ) p(xk 1 xk , xk 1 , ., x1 )p(xk , xk 1 ., x1 ) p(xk 1 xk , xk 1 , ., x1 )p(xk xk 1 , ., x1 )p(xk 1 xk 2 , ., x1 ) .p(x1 )Hence the relationship holds for n k 1, and by induction it holds for any n.PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

12Problem 2.161. Let T and R denote channel input and outputs respectively. Using Bayes rule we havep(T 0)p(R A T 0)p(T 0)p(R A T 0) p(T 1)p(R A T 1)0.4 16 0.4 61 0.6 131 4p(T 0 R A) and therefore p(T 1 R A) 43 , obviously if R A is observed, the best decision wouldbe to declare that a 1 was sent, i.e., T 1, because T 1 is more probable that T 0.Similarly it can be verified that p(T 0 R B) 47 and p(T 0 R C) 14 . Therefore,when the output is B, the best decision is 0 and when the output is C, the best decision isT 1. Therefore the decision function d can be defined as(1, R A or Cd(R) 0, R BThis is the optimal decision scheme.2. Here we know that a 0 is transmitted, therefore we are looking for p(error T 0), this isthe probability that the receiver declares a 1 was sent when actually a 0 was transmitted.Since by the decision method described in part 1 the receiver declares that a 1 was sent whenR A or R C, therefore, p(error T 0) p(R A T 0) p(R C T 0) 31 .3. We have p(error T 0) 31 , and p(error T 1) p(R B T 1) 31 . Therefore, by thetotal probability theoremp(error) p(T 0)p(error T 0) p(T 1)p(error T 1)11 0.4 0.6 331 3Problem 2.17Following the same procedure as in example 2-1-1, we prove : y b1pXpY (y) a aPROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

13Problem 2.18Relationship (2-1-44) gives :pY (y) 13a [(y b) /a]2/3pX" y ba 1/3 #X is a gaussian r.v. with zero mean and unit variance : pX (x) 12π e xHence :2/31 21 ( y ba )epY (y) 3a 2π [(y b) /a]2/32 /2pdf of Y0.50.450.4a 20.35b 30.30.250.20.150.10.050 10 8 6 4 20y246810Problem 2.191) The random variable X is Gaussian with zero mean and variance σ 2 10 8 . Thus p(X x) Q( σx ) and 4 10 4p(X 10 ) Q Q(1) .15910 4 4 10 4 Q(4) 3.17 10 5p(X 4 10 4 ) Q10 4p( 2 10 4 X 10 4 ) 1 Q(1) Q(2) .81822)p(X 10 4 X 0) p(X 10 4 , X 0)p(X 10 4 ).159 .318p(X 0)p(X 0).5PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

14Problem 2.201) y g(x) ax2 . Assume without loss of generality that a 0. Then, if y 0 the equation2 has no real solutions and f (y) 0. If y 0 there are two solutions to the system, namelyy axpYx1,2 y/a. Hence,fY (y) fX (x2 )fX (x1 ) ′′ g (x1 ) g (x2 ) ppfX ( y/a) fX ( y/a)pp 2a y/a2a y/ay1e 2aσ2 2ay 2πσ2) The equation y g(x) has no solutions if y b. Thus FY (y) and fY (y) are zero for y b. If b y b, then for a fixed y, g(x) y if x y; hence FY (y) FX (y). If y b then g(x) b yfor every x; hence FY (y) 1. At the points y b, FY (y) is discontinuous and the discontinuitiesequal toFY ( b ) FY ( b ) FX ( b)andFY (b ) FY (b ) 1 FX (b)The PDF of y g(x) isfY (y) FX ( b)δ(y b) (1 FX (b))δ(y b) fX (y)[u 1 (y b) u 1 (y b)] y2b1 Qe 2σ2 [u 1 (y b) u 1 (y b)](δ(y b) δ(y b)) σ2πσ 23) In the case of the hard limiterp(Y b) p(X 0) FX (0) 12p(Y a) p(X 0) 1 FX (0) 12Thus FY (y) is a staircase function andfY (y) FX (0)δ(y b) (1 FX (0))δ(y a)PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

154) The random variable y g(x) takes the values yn xn with probabilityp(Y yn ) p(an X an 1 ) FX (an 1 ) FX (an )Thus, FY (y) is a staircase function with FY (y) 0 if y x1 and FY (y) 1 if y xN . The PDFis a sequence of impulse functions, that isfY (y) NXi 1[FX (ai 1 ) FX (ai )] δ(y xi )N h a iXaii 1Q Qδ(y xi ) σσi 1Problem 2.21For n odd, xn is odd and since the zero-mean Gaussian PDF is even their product is odd. Sincethe integralodd function over the interval [ , ] is zero, we obtain E[X n ] 0 for n odd.R of annLet In x exp( x2 /2σ 2 )dx. Obviously In is a constant and its derivative with respect to xis zero, i.e., Z x2x2d1In nxn 1 e 2σ2 2 xn 1 e 2σ2 dx 0dxσ which results in the recursionIn 1 nσ 2 In 1This is true for all n. Now let n 2k 1, we will have I2k (2k 1)σ 2 I2k 2 , with the initialcondition I0 2πσ 2 . Substituting we have I2 σ 2 2πσ 2 I4 3σ 2 I2 3σ 4 2πσ 2 I6 5 3σ 2 I4 5 3σ 6 2πσ 2 I8 7 σ 2 I6 7 5 3σ 8 2πσ 2. . . and in general if I2k (2k 1)(2k 3)(2k 5) · · · 3 1σ 2k 2πσ 2 , then I2k 2 (2k 1)σ 2 I2k (2k 1)(2k 1)(2k 3)(2k 5) · · · 3 1σ 2k 2 2πσ 2 . Using the fact that E[X 2k ] I2k / 2πσ 2 ,we obtainIn 1 3 5 · · · (n 1)σ nfor n even.PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

16Problem 2.22a. Since (Xr , Xi ) are statistically independent :pX (xr , xi ) pX (xr )pX (xi ) 1 (x2r x2i )/2σ2e2πσ 2Also :Yr jYi (Xr Xi )ejφ Xr Xi (Yr jYi ) e jφ Yr cos φ Yi sin φ j( Yr sin φ Yi cos φ) X Y cos φ Y sin φ rri Xi Yr sin φ Yi cos φ The Jacobian of the above transformation is :J Xr Yr Xr Yi Xi Yr Xi Yi cos φ sin φsin φcos φ 1Hence, by (2-1-55) :pY (yr , yi ) pX ((Yr cos φ Yi sin φ) , ( Yr sin φ Yi cos φ))2221e (yr yi )/2σ 2πσ2b. Y AX and X A 1 YNow, pX (x) ′21e x x/2σ(2πσ2 )n/2(the covariance matrix M of the random variables x1 , ., xn isM σ 2 I, since they are i.i.d) and J 1/ det(A) . Hence :pY (y) 1′ 1 ′ 121e y (A ) A y/2σ2n/2 det(A) (2πσ )For the pdf’s of X and Y to be identical we require that : det(A) 1 and (A 1 )′ A 1 I A 1 A′Hence, A must be a unitary (orthogonal) matrix .Problem 2.23Since we are dealing with linear combinations of jointly Gaussian random variables, it is clearthat Y is jointly Gaussian. We clearly have mY E[AX] AmX . This means that Y mY A (X mX ). Also note that CY E (Y mY )(Y mY )′ E A (X mX ) (X mX ) A′PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

17resulting in CY ACX A′ .Problem 2.24a.#"nnh PniYY n jvY E ejvX ψX (ejv )ejvxi ψY (jv) E e E ejv i 1 xi Ei 1i 1But,pX (x) pδ(x 1) (1 p)δ(x) ψX (ejv ) 1 p pejv ψY (jv) 1 p pejv nb.E(Y ) jdψY (jv) v 0 jn(1 p pejv )n 1 jpejv v 0 npdvandE(Y 2 ) d2 ψY (jv)d jv n 1 jv jn(1 p pe)pe np np(n 1)pv 0v 0d2 vdv E(Y 2 ) n2 p2 np(1 p)Problem 2.251. In the figure shown belowPROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

18vx R x 2uxlet us consider the region u x, v x shown as the colored region extending to infinity, callu2 v 2this region R, and let us integrate e 2 over this region. We haveZZZZu2 v 2r2e 2 r dr dθe 2 du dv R ZR 2 r2 rex 2drZ 2 π r2 e 2x 2π2 e x2π2dθ0where we have used the fact that region R is included in the region outside the quarter circleas shown in the figure. On the other hand we haveZ Z ZZ222v2 u2 u v2ee 2 dvdu dv duexR Zx 2 u2edux 22πQ(x) 2 2π (Q(x))2From the above relations we conclude that2π (Q(x))2 and therefore, Q(x) 12 e x22π x2e2.PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

192. InR xuv Re y22dyy2define u e y22and dv dyy2and use the integration by parts relation2 y2v du. We have v y1 and du yeZNow note that xu dv y2x2Z 2 dy e 2 e 2 y2 2πQ(x)e dy 2yyxx2 y2exR dy. ThereforeRxe y22dyy2 0 which results inx2 x2e 21e 2 2πQ(x) 0 Q(x) x2πxOn the other hand, note thatZ xwhich results inor, 22π 1 xQ(x) x22 y2e1dy 22yxZ 2 y2edy x2πQ(x)x2 x2 e 22π 2πQ(x) 2 Q(x)xxx2e 2xwhich results inx2xe 2Q(x) 2π(1 x2 )3. From we havex2x2x1e 2 Q(x) e 22π(1 x2 )2πx2211 x2 x2 e Q(x) e2π( x1 x)2πxAs x becomes large x1 in the denominator of the left hand side becomes small and the twobounds become equal, therefore for large x we haveQ(x) x21e 22πxProblem 2.26PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

201. FYn (y) P [Yn y] 1 P [Yn y] 1 P [x1 y, X2 y, . . . , Xn y] 1R (P [X y])nAwhere we have used the independence of Xi ’s in the last step. But P [X y] y A1 dy A yA .Therefore, FYn (y) 1 (A y)nAn ,and fYn (y) ddy FYn (y)n 1 n (A y)An, 0 y A.2.y n 1n 1 AAny nλ 1 1 AynA λy nλ1 λe λy y 01 Aynf (y) Problem 2.27ihψ(jv1 , jv2 , jv3 , jv4 ) E ej(v1 x1 v2 x2 v3 x3 v4 x4 )E (X1 X2 X3 X4 ) ( j)4 4 ψ(jv1 , jv2 , jv3 , jv4 ) v1 v2 v3 v4 0 v1 v2 v3 v4From (2-1-151) of the text, and the zero-mean property of the given rv’s :1′ψ(jv) e 2 v Mvwhere v [v1 , v2 , v3 , v4 ]′ , M [µij ] .We obtain the desired result by bringing the exponent to a scalar form and then performingquadruple differentiation. We can simplify the procedure by noting that :1 ′ ψ(jv) µ′i ve 2 v Mv viwhere µ′i [µi1 , µi2 , µi3 , µi4 ] . Also note that : µ′j v viHence : µij µji 4 ψ(jv1 , jv2 , jv3 , jv4 ) V 0 µ12 µ34 µ23 µ14 µ24 µ13 v1 v2 v3 v4PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

21Problem 2.281) By Chernov bound, for t 0,P [X α] e tα E[etX ] e tα ΘX (t)This is true for all t 0, henceln P [X α] min [ tα ln ΘX (t)] max [tα ln ΘX (t)]t 0t 02) Hereln P [Sn α] ln P [Y nα] max [tnα ln ΘY (t)]t 0where Y X1 X2 · · · Xn , and ΘY (t) E[eX1 X2 ··· Xn ] [ΘX (t)]n . Hence,ln P [Sn α] max n [tα ln ΘX (t)] nI(α) t 01P [Sn α] e nI(α)nR 1dΘX (t) 0 etx e x dx 1 tas long as t 1. I(α) maxt 0 (tα ln(1 t)), hence dt(tα ln(1 α 1α 1 t)) 0 and t α . Since α 0, t 0 and also obviously t 1. I(α) α 1 ln 1 α α 1 ln α, using the large deviation theoremln P [Sn α] e n(α 1 ln α) o(n) αn e n(α 1) o(n)Problem 2.29For the central chi-square with n degress of freedom :ψ(jv) Now :1(1 j2vσ 2 )n/2jnσ 2dψ(jv)dψ(jv) v 0 nσ 2 E (Y ) jn/2 1dvdv(1 j2vσ 2 ) d2 ψ(jv) 2nσ 4 (n/2 1)d2 ψ(jv)2 EY v 0 n(n 2)σ 2dv 2dv 2(1 j2vσ 2 )n/2 2 The variance is σY2 E Y 2 [E (Y )]2 2nσ 4For the non-central chi-square with n degrees of freedom :ψ(jv) 1(1 j2vσ 2 )n/222ejvs /(1 j2vσ )PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond thelimited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

22where by defi

Solutions Manual For Digital Communic