Problem 4.1 :(a)ˆx(t)=1π∞−∞x(a)t − adaHence :−ˆx(−t)=−1π∞−∞x(a)−t−ada= −1π−∞∞x(−b)−t+b(−db)= −1π∞−∞x(b)−t+bdb=1π∞−∞x(b)t−bdb =ˆx(t)where we have made the change of variables : b = −a and used the relationship : x(b)=x(−b).(b) In exactly the same way as in part (a) we prove :ˆx(t)=ˆx(−t)(c) x(t)=cosω0t, so its Fourier transform is : X(f)=12[δ(f − f0)+δ(f + f0)] ,f0=2πω0.Exploiting the phase-shifting property (4-1-7) of the Hilbert transform :ˆX(f)=12[−jδ(f − f0)+jδ(f + f0)] =12j[δ(f − f0) − δ(f + f0)] = F−1{sin 2πf0t}Hence, ˆx(t)=sinω0t.(d) In a similar way to part (c) :x(t)=sinω0t ⇒ X(f)=12j[δ(f − f0) − δ(f + f0)] ⇒ˆX(f)=12[−δ(f − f0) − δ(f + f0)]⇒ˆX(f)=−12[δ(f − f0)+δ(f + f0)] = −F−1{cos 2πω0t}⇒ˆx(t)=−cos ω0t(e) The positive frequency content of the new signal will be : (−j)(−j)X(f)=−X(f),f>0,while the negative frequency content will be : j · jX(f)=−X(f ),f<0. Hence, sinceˆˆX(f)=−X(f ), we have :ˆˆx(t)=−x(t).(f) Since the magnitude response of the Hilbert transformer is characterized by : |H(f )| =1,we have that : ˆX(f) = |H(f)||X(f)| = |X(f)|. Hence :∞−∞ ˆX(f) 2df =∞−∞|X(f)|2dfand using Parseval’s relationship :∞−∞ˆx2(t)dt =∞−∞x2(t)dt(g) From parts (a) and (b) above, we note that if x(t)iseven,ˆx(t) is odd and vice-versa.Therefore, x(t)ˆx(t) is always odd and hence :∞−∞x(t)ˆx(t)dt =0.Problem 4.2 :We have :ˆx(t)=h(t) ∗ x(t)where h(t)=1πtand H(f )=−j, f > 0j, f < 0. Hence :Φˆxˆx(f)=Φxx(f) |H(f)|2=Φxx(f)and its inverse Fourier transform :φˆxˆx(τ)=φxx(τ)Also :φxˆx(τ)=E [x(t + τ)ˆx(t)]=1π∞−∞E[x(t+τ)x(a)]t−ada=1π∞−∞φxx(t+τ−a)t−ada= −1π−∞∞φxx(b)b−τdb=1π∞−∞φxx(b)τ−bdb = −φxx(τ)Problem 4.3 :(a)E [z(t)z(t + τ)] = E [{x(t + τ)+jy(t + t)}{x(t)+jy(t)}]= E [x(t)x(t + τ)] − E [y(t)y(t + τ)] + jE [x(t)y(t + τ)]+E [y(t)x(t + τ)]= φxx(τ) −φyy(τ)+j [φyx(τ)+φxy(τ)]But φxx(τ)=φyy(τ)and φyx(τ)=−φxy(τ). Therefore :E [z(t)z(t + τ)] = 0(b)V =T0z(t)dtEV2=T0T0E [z(a)z(b)] dadb =0from the result in (a) above. Also :E (VV∗)=T0T0E [z(a)z∗(b)] dadb=T0T02N0δ(a − b)dadb=T02N0da =2N0TProblem 4.9 :The energy of the signal waveform sm(t)is:E=∞−∞|sm(t)|2dt =∞−∞ sm(t) −1MMk=1sk(t) 2dt=∞−∞s2m(t)dt +1M2Mk=1Ml=1∞−∞sk(t)sl(t)dt−1MMk=1∞−∞sm(t)sk(t)dt −1MMl=1∞−∞sm(t)sl(t)dt= E +1M2Mk=1Ml=1Eδkl−2ME= E +1ME−2ME =M − 1METhe correlation coefficient is given by :ρmn=1E∞−∞sm(t)sn(t)dt =1E∞−∞sm(t) −1MMk=1sk(t)sn(t) −1MMl=1sl(t)dt=1E∞−∞sm(t)sn(t)dt +1M2Mk=1Ml=1∞−∞sk(t)sl(t)dt−1E1MMk=1∞−∞sn(t)sk(t)dt +1MMl=1∞−∞sm(t)sl(t)dt=1M2ME−1ME−1MEM−1ME= −1M −1Problem 4.17 :The first basis function is :g4(t)=s4(t)√E4=s4(t)√3=−1/√3, 0 ≤ t ≤ 30, o.w.Then, for the second basis function :c43=∞−∞s3(t)g4(t)dt = −1/√3 ⇒ g3(t)=s3(t) − c43g4(t)=2/3, 0 ≤ t ≤ 2−4/3, 2 ≤ t ≤ 30, o.wHence :g3(t)=g3(t)√E3=1/√6, 0 ≤ t ≤ 2−2/√6, 2 ≤ t ≤ 30, o.wwhere E3denotes the energy of g3(t):E3=30(g3(t))2dt =8/3.For the third basis function :c42=∞−∞s2(t)g4(t)dt =0 and c32=∞−∞s2(t)g3(t)dt =0Hence :g2(t)=s2(t) − c42g4(t) − c32g3(t)=s2(t)andg2(t)=g2(t)√E2=1/√2, 0 ≤ t ≤ 1−1/√2, 1 ≤ t ≤ 20, o.wwhere : E2=20(s2(t))2dt =2.Finally for the fourth basis function :c41=∞−∞s1(t)g4(t)dt = −2/√3, c31=∞−∞s1(t)g3(t)dt =2/√6,c21=0Hence :g1(t)=s1(t) − c41g4(t) − c31g3(t) − c21g2(t)=0⇒ g1(t)=0The last result is expected, since the dimensionality of the vector space generated by thesesignals is 3. Based on the basis functions (g2(t),g3(t),g4(t)) the basis representation of thesignals is :s4=0, 0,√3⇒E4=3s3=0,8/3, −1/√3⇒E3=3s2=√2, 0, 0⇒E2=2s1=2/√6, −2/√3, 0⇒E1=2Problem 4.18 :s1=√E, 0s2=−√E, 0s3=0,√Es4=0, −√E✻✲f2f1s1s2s4s3ooooæAs we see, this signal set is indeed equivalent to a 4-phase PSK signal.Problem 4.19 :(a)(b) The signal space diagram, together with the Gray encoding of each signal point is givenin the following figure :00011110The signal points that may be transmitted at times t =2nT n =0, 1, ... are given with blankcircles, while the ones that may be transmitted at times t =2nT +1,n=0, 1, ... are givenwith filled