(d) gIn])=4sin(o, n). Since -4sgIn]s4 for all values of n, igIn]) is a bounded (e)(v()=3 cos2(0, n2). Since -3< v(n]s3 for all values of n, (v(n]) is a bounded 2. 8(a)Recall, xe[n]=5(x[n]+x[-n]). Since x[n]is a causal sequence thus x[-n]=0 Vn>0. Hence x[n]=xe[n]+xe,[-n]= 2xe [n, Vn>0. For n=0, x[0]=Xev[O] Thus x[n] can be completely recovered from its even part Likewise, xod[n]=3(x[n] -xF-n)2xn), n>0, 0, 0. Thus x[n] can be recovered from its odd part Vn except n =0. (b)2ya[n]=y[n]-y*[-n]. Since y[n] is a causal sequence yIn]= 2y[n] Vn>0 For n=0, Imly[0Jl=y[O]. Hence real part of y[0] cannot be fully recovered from yaIn Therefore y[n] cannot be fully recovered from yIn] 2y [n]=yIn]+y*[-n]. Hence, y[n]= 2ys[n] Vn>0 For n=0, Rely[O])=y [0]. Hence imaginary part of y[O] cannot be recovered from yes[n] Therefore y[n] cannot be fully recovered from y In] 2.9 xev[n]=5(x[n]+x[-n)). This implies, xev[-n]=5(x[-n]+x(n)=xen Hence even part of a real sequence is even xd[n]=5(x[n]-x[-n)). This implies, xd[-n]=5(x[-n]-x[n])=-x[n] Hence the odd part of a real sequence is odd. 2.10 RHS of Eq. Since x[n]=0 n<0, Hence xes[n]+xcs[n-N]=(x[n]+x*IN-n)=xpcs[n], 0snsN-I RHS of Eq. (2. 176b)is Xca[n]+xca[n-N]=3(x[n]-x*[-n]+,(xn-NI-x*[n-NI (x[n]-x*[N-n])=x。n],0≤n≤N-15 (d) { [ ]} sin( ). gn na = 4 ω Since −≤ ≤ 4 4 g n[ ] for all values of n, {g[n]} is a bounded sequence. (e) { [ ]} cos ( ). vn nb = 3 2 2 ω Since −≤ ≤ 3 3 v n[ ] for all values of n, {v[n]} is a bounded sequence. 2.8 (a) Recall, x n xn x n ev[ ] [ ] [ ]. = +− ( ) 1 2 Since x[n] is a causal sequence, thus x[–n] = 0 ∀ > n 0. Hence, xn x n x n ev ev [] [] [ ] = +− = 2x n ev[ ], ∀ > n 0. For n = 0, x[0] = xev[0]. Thus x[n] can be completely recovered from its even part. Likewise, x n xn x n xn n n od[ ] [ ]– [ ] [ ], , , . = − ( ) = > =     1 2 1 2 0 0 0 Thus x[n] can be recovered from its odd part ∀n except n = 0. (b) 2y n y n y n ca[ ] [ ] *[ ] = −− . Since y[n] is a causal sequence yn y n ca [] [] = ∀ 2 n > 0. For n = 0, Im{ [ ]} [ ] y yca 0 0 = . Hence real part of y[0] cannot be fully recovered from y n ca [ ]. Therefore y[n] cannot be fully recovered from y n ca [ ]. 2y n y n y n cs[ ] [ ] *[ ] = +− . Hence, yn y n cs [] [] = ∀ 2 n > 0 . For n = 0, Re{ [ ]} [ ] y ycs 0 0 = . Hence imaginary part of y[0] cannot be recovered from y n cs[ ]. Therefore y[n] cannot be fully recovered from y n cs[ ]. 2.9 x n xn x n ev[ ] [ ] [ ]. = +− ( ) 1 2 This implies, x n x n xn x n ev ev [– ] [– ] [ ] [ ]. = + ( ) = 1 2 Hence even part of a real sequence is even. x n xn x n od[ ] [ ] – [– ] . = ( ) 1 2 This implies, x n x n xn x n od od [– ] [– ] – [ ] – [ ]. = ( ) = 1 2 Hence the odd part of a real sequence is odd. 2.10 RHS of Eq. (2.176a) is x n x n N xn x n xn N x N n cs cs [ ] [ ] [ ] *[ ] [ ] *[ ] . + −= + − ( ) + −+ − ( ) 1 2 1 2 Since x[n] = 0 ∀ < n 0, Hence x n x n N xn x N n x n cs cs pcs [ ] [ ] [ ] *[ ] [ ], + −= + − ( ) ≤ ≤ 1 2 = 0 n N –1. RHS of Eq. (2.176b) is x n x n N xn x n xn N x n N ca ca [ ] [ ] [ ] *[ ] [ ] *[ ] + −= − − ( ) + −− − ( ) 1 2 1 2 = −− ( ) = ≤≤ 1 2 xn x N n x n pca [ ] *[ ] [ ], 0 n N –1