1、CHAPTER 2Problem 2.1 :P(Ai) =3?j=1P(Ai,Bj),i = 1,2,3,4Hence :P(A1) =3?j=1P(A1,Bj) = 0.1 + 0.08 + 0.13 = 0.31P(A2) =3?j=1P(A2,Bj) = 0.05 + 0.03 + 0.09 = 0.17P(A3) =3?j=1P(A3,Bj) = 0.05 + 0.12 + 0.14 = 0.31P(A4) =3?j=1P(A4,Bj) = 0.11 + 0.04 + 0.06 = 0.21Similarly :P(B1) =4?i=1P(Ai,B1) = 0.10 + 0.05 +
2、0.05 + 0.11 = 0.31P(B2) =4?i=1P(Ai,B2) = 0.08 + 0.03 + 0.12 + 0.04 = 0.27P(B3) =4?i=1P(Ai,B3) = 0.13 + 0.09 + 0.14 + 0.06 = 0.42Problem 2.2 :The 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|xk1,.,x1)p(xk1|xk2,.,x1) .p(x1)Then f
3、or n = k + 1 :p(x1,x2,.,xk,xk+1)=p(xk+1|xk,xk1,.,x1)p(xk,xk1.,x1)=p(xk+1|xk,xk1,.,x1)p(xk|xk1,.,x1)p(xk1|xk2,.,x1) .p(x1)Hence the relationship holds for n = k + 1, and by induction it holds for any n.1Problem 2.3 :Following the same procedure as in example 2-1-1, we prove :pY(y) =1|a|pX?y ba?Proble
4、m 2.4 :Relationship (2-1-44) gives :pY(y) =13a(y b)/a2/3pX?y ba?1/3X is a gaussian r.v. with zero mean and unit variance : pX(x) =12ex2/2Hence :pY(y) =13a2(y b)/a2/3e12(yba)2/3108642024681000.050.10.150.20.250.30.350.40.450.5ypdf of Ya=2b=3Problem 2.5 :(a) Since (Xr,Xi) are statistically independent
5、 :pX(xr,xi) = pX(xr)pX(xi) =122e(x2r+x2i)/222Also :Yr+ jYi= (Xr+ Xi)ejXr+ Xi= (Yr+ jYi)ej= Yrcos + Yisin + j(Yrsin + Yicos) ?Xr=Yrcos + YisinXi=Yrsin + Yicos?The Jacobian of the above transformation is :J =?XrYrXiYrXrYiXiYi?=?cossinsincos?= 1Hence, by (2-1-55) :pY(yr,yi)=pX(Yrcos + Yisin),(Yrsin + Y
6、icos)=122e(y2r+y2i)/22(b) Y = AX and X = A1YNow, pX(x) =1(22)n/2ex?x/22(the covariance matrix M of the random variables x1,.,xnisM = 2I, since they are i.i.d) and J = 1/|det(A)|. Hence :pY(y) =1(22)n/21|det(A)|ey?(A1)?A1y/22For the pdfs of X and Y to be identical we require that :|det(A)| = 1 and (A
7、1)?A1= I = A1= A?Hence, A must be a unitary (orthogonal) matrix .Problem 2.6 :(a)Y(jv) = E?ejvY?= E?ejv?ni=1xi?= E?n?i=1ejvxi?=n?i=1E?ejvX?=?X(ejv)?nBut,pX(x) = p(x 1) + (1 p)(x) X(ejv) = 1 + p + pejv Y(jv) =?1 + p + pejv?n3(b)E(Y ) = jdY(jv)dv|v=0= jn(1 p + pejv)n1jpejv|v=0= npandE(Y2) = d2Y(jv)d2v
8、|v=0= ddv?jn(1 p + pejv)n1pejv?v=0= np + np(n 1)p E(Y2) = n2p2+ np(1 p)Problem 2.7 :(jv1,jv2,jv3,jv4) = E?ej(v1x1+v2x2+v3x3+v4x4)?E (X1X2X3X4) = (j)44(jv1,jv2,jv3,jv4)v1v2v3v4|v1=v2=v3=v4=0From (2-1-151) of the text, and the zero-mean property of the given rvs :(jv) = e12v?Mvwhere v = v1,v2,v3,v4?,M
9、 = ij.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 :(jv)vi= ?ive12v?Mvwhere ?i=i1,i2,i3,i4 . Also note that :?jvvi= ij= jiHence :4(jv1,jv2,jv3,jv4)v1v2v3v4|V=0= 1234+ 2314+ 2413Proble
10、m 2.8 :For the central chi-square with n degress of freedom :(jv) =1(1 j2v2)n/24Now :d(jv)dv=jn2(1 j2v2)n/2+1 E (Y ) = jd(jv)dv|v=0= n2d2(jv)dv2=2n4(n/2 + 1)(1 j2v2)n/2+2 E?Y2?= d2(jv)dv2|v=0= n(n + 2)2The variance is 2Y= E (Y2) E (Y )2= 2n4For the non-central chi-square with n degrees of freedom :(
11、jv) =1(1 j2v2)n/2ejvs2/(1j2v2)where by definition : s2=?ni=1m2i.d(jv)dv=?jn2(1 j2v2)n/2+1+js2(1 j2v2)n/2+2?ejvs2/(1j2v2)Hence, E (Y ) = jd(jv)dv|v=0= n2+ s2d2(jv)dv2=?n4(n + 2)(1 j2v2)n/2+2+s2(n + 4)2 ns22(1 j2v2)n/2+3+s4(1 j2v2)n/2+4?ejvs2/(1j2v2)Hence,E?Y2?= d2(jv)dv2|v=0= 2n4+ 4s22+?n2+ s2?and2Y=
12、 E?Y2? E (Y )2= 2n4+ 42s2Problem 2.9 :The Cauchy r.v. has : p(x) =a/x2+a2, x (a)E (X) =?xp(x)dx = 0since p(x) is an even function.E?X2?=?x2p(x)dx =a?x2x2+ a2dxNote that for large x,x2x2+a2 1 (i.e non-zero value). Hence,E?X2?= ,2= 5(b)(jv) = E?jvX?=?a/x2+ a2ejvxdx =?a/(x + ja)(x ja)ejvxdxThis integra
13、l can be evaluated by using the residue theorem in complex variable theory. Then,for v 0 :(jv) = 2j?a/x + jaejvx?x=ja= eavFor v W, then X(f)ej2fais also bandlimited. The corresponding autocor-relation function can be represented as (remember that X(f) is deterministic) :X( a) =?n=X(n2W a)sin2W? n2W?
14、2W? n2W?(1)Let us define :X(t) =?n=X(n2W)sin2W?t n2W?2W?t n2W?We must show that :E?|X(t) X(t)|2?= 0orE?X(t) X(t)?X(t) ?m=X(m2W)sin2W?t m2W?2W?t m2W?= 0(2)First we have :E?X(t) X(t)?X(m2W)?= X(t m2W) ?n=X(n m2W)sin2W?t n2W?2W?t n2W?9But the right-hand-side of this equation is equal to zero by applica
15、tion of (1) with a = m/2W.Since this is true for any m, it follows that E?X(t) X(t)?X(t)?= 0. AlsoE?X(t) X(t)?X(t)?= X(0) ?n=X(n2W t)sin2W?t n2W?2W?t n2W?Again, by applying (1) with a = t anf = t, we observe that the right-hand-side of the equationis also zero. Hence (2) holds.Problem 2.18 :Q(x) =12
16、?xet2/2dt = P N x, where N is a Gaussian r.v with zero mean and unitvariance. From the Chernoffbound :P N x e vxE?e vN?(1)where v is the solution to :E?NevN? xE?evN?= 0(2)Now :E?evN?=12?evtet2/2dt=ev2/212?e(tv)2/2dt=ev2/2andE?NevN?=ddvE?evN?= vev2/2Hence (2) gives : v = xand then :(1) Q(x) ex2ex2/2
17、Q(x) ex2/2Problem 2.19 :Since H(0) =?h(n) = 0 my= mxH(0) = 010The autocorrelation of the output sequence isyy(k) =?i?jh(i)h(j)xx(k j + i) = 2x?i=h(i)h(k + i)where the last equality stems from the autocorrelation function of X(n) :xx(k j + i) = 2x(k j + i) =?2x,j = k + i0,o.w.?Hence, yy(0) = 62x, yy(
18、1) = yy(1) = 42x, yy(2) = yy(2) = 2x, yy(k) = 0 otherwise.Finally, the frequency response of the discrete-time system is :H(f)=?h(n)ej2fn=1 2ej2f+ ej4f=?1 ej2f?2=ej2f?ejf ejf?2=4ejfsin2fwhich gives the power density spectrum of the output :yy(f) = xx(f)|H(f)|2= 2x?16sin4f?= 162xsin4fProblem 2.20 :(k
19、) =?12?|k|The power density spectrum is(f)=?k=(k)ej2fk=?1k=?12?kej2fk+?k=0?12?kej2fk=?k=0(12ej2fk)k+?k=0(12ej2f)k 1=11ej2f/2+11ej2f/2 1=2cos2f5/4cos2f 1=354cos2f11Problem 2.21 :We will denote the discrete-time process by the subscript d and the continuous-time (analog)process by the subscript a. Als
20、o, f will denote the analog frequency and fdthe discrete-timefrequency.(a)d(k)=E X(n)X(n + k)=E X(nT)X(nT + kT)=a(kT)Hence, the autocorrelation function of the sampled signal is equal to the sampled autocorrelationfunction of X(t).(b)d(k)=a(kT) =?a(F)ej2fkTdf=?l=?(2l+1)/2T(2l1)/2Ta(F)ej2fkTdf=?l=?1/
21、2T1/2Ta(f +lT)ej2FkTdf=?1/2T1/2T?l=a(f +lT)?ej2FkTdfLet fd= fT. Then :d(k) =?1/21/21T?l=a(fd+ l)/T)ej2fdkdfd(1)We know that the autocorrelation function of a discrete-time process is the inverse Fouriertransform of its power spectral densityd(k) =?1/21/2d(fd)ej2fdkdfd(2)Comparing (1),(2) :d(fd) =1T?
22、l=a(fd+ lT)(3)(c) From (3) we conclude that :d(fd) =1Ta(fdT)iff:a(f) = 0, f : |f| 1/2T12Otherwise, the sum of the shifted copies of a(in (3) will overlap and aliasing will occur.Problem 2.22 :(a)a()=?a(f)ej2fdf=?WWej2fdf=sin2WBy applying the result in problem 2.21, we haved(k) = fa(kT) =sin2WkTkT(b)
23、 If T =12W, then :d(k) =?2W = 1/T,k = 00,otherwise?Thus, the sequence X(n) is a white-noise sequence. The fact that this is the minimum value ofT can be shown from the following figure of the power spectral density of the sampled process:WWfs Wfsfs+ Wfs Wfsfs+ WWe see that the maximum sampling rate
24、fsthat gives a spectrally flat sequence is obtainedwhen :W = fs W fs= 2W T =12W(c) The triangular-shaped spectrum (f) = 1 |f|W, |f| W may be obtained by convolv-ing the rectangular-shaped spectrum 1(f) = 1/W, |f| W/2. Hence, () = 21() =131W?sinW?2.Therefore, sampling X(t) at a rate1T= W samples/sec
25、produces a white sequencewith autocorrelation function :d(k) =1W?sinWkTkT?2= W?sinkk?2=?W,k = 00,otherwise?Problem 2.23 :Lets denote : y(t) = fk(t)fj(t).Then :?fk(t)fj(t)dt =?y(t)dt = Y (f)|f=0where Y (f) is the Fourier transform of y(t). Since : y(t) = fk(t)fj(t) Y (f) = Fk(f)Fj(f).But :Fk(f) =?fk(
26、t)ej2ftdt =12Wej2fk/2WThen :Y (f) = Fk(f) Fj(f) =?Fk(a) Fj(f a)daand at f = 0 :Y (f)|f=0=?Fk(a) Fj(a)da=?12W?2?ej2a(kj)/2Wda=?1/2W,k = j0,k ?= j?Problem 2.24 :Beq=1G?0|H(f)|2dfFor the filter shown in Fig. P2-12 we have G = 1 andBeq=?0|H(f)|2df = BFor the lowpass filter shown in Fig. P2-16 we haveH(f) =11 + j2fRC |H(f)|2=11 + (2fRC)214So G = 1 andBeq=?0|H(f)|2df=12?|H(f)|2df=14RCwhere the last integral is evaluated in the same way as in problem P-2.9 .15
