Chapter 2(2e) 2l(a)un]=x[n]+yn]={351-2814o (b)vn]=x[n]wn]={-15 (c)s[n]=yln]-wm]={53-2-999-3} (d)rn]=4.5yn]={031.54.5-13.51840.5-9 2.2 (a) From the figure shown below we obtain vn vIn- B yIn n]=xn]+avn-1] and yIn]=βvn-1]+yvn-1=(β+γvn-1]. Hence, In-1=x[n-1]+av[n-2] and yn-1]=(B+yVn-2. Therefore yn]=(+n-l=(B+)xn-1+a(β+ym-2]=(+y)xn-l+aB+y) (阝+γ)x[n-1]+ay[n-1 (b) From the figure shown below we obtain x[n-3] yIn] yn]=Yx[n-2]+β(xn-1]+xn-3)+(xn]+xn-4]) c) From the figure shown below we obtain vIn xIn -1vn-1
2 Chapter 2 (2e) 2.1 (a) un xn yn [] [] [] { } =+= − 3 5 1 2 8 14 0 (b) vn xn wn [] [] [] { } = ⋅ =− − − 15 8 0 6 20 0 2 (c) sn yn wn [] [] [] { } = − = −− − 53 2 999 3 (d) rn yn [] . [] { . . . . } == − − 4 5 0 31 5 4 5 13 5 18 40 5 9 2.2 (a) From the figure shown below we obtain x[n] y[n] v[n] v[n–1] z –1 α β γ vn xn vn [] [] [ ] =+ − α 1 and y n v n v n v n [] [ ] [ ] ( )[ ] = −+ −= + − β γ βγ 1 1 1 . Hence, vn xn vn [ ][ ] [ ] −= −+ − 11 2 α and y n v n [ ] ( )[ ] −= + − 1 2 β γ . Therefore, yn vn xn vn [] ( )[ ] ( )[ ] ( )[ ] = + −= + −+ + − β γ β γ αβ γ 1 1 2 = + −+ + − + ( )[ ] ( ) [ ] ( ) β γ αβ γ β γ x n y n 1 1 = + −+ − ( )[ ] [ ] βγ α xn yn 1 1. (b) From the figure shown below we obtain x[n] y[n] z –1 α β γ z –1 z –1 z –1 x[n–1] x[n–2] x[n–4] x[n–3] yn xn xn xn xn xn [] [ ] [ ] [ ] [] [ ] = − + −+ − γβ α 2 1 3 4. ( ) + +− ( ) (c) From the figure shown below we obtain v[n] x[n] y[n] z –1 z –1 v[n–1] –1 d1
vIn]=x[n]-dvn-l and yin]=d vn]+vin-I]. Therefore we can rewrite the second equation as yin]=d,(x[nI-d, vIn-1 )+vIn-1=d, x(n]+(l-d2)vIn-1] d;nl+(1-4)×xm-1-d4vn-2)=dnl+(1-42kn-1-d1-cyn-2 From Eq. (1), yIn-1]=d xin-1+(1-d2) Min-2],or equivalently dAyIn-1]=dix(n-1]+d, 1-d)Mn-2. Therefore, +d3ym-1-=dnl+(1-42km-1-4(1-4Mn-21+4xn-1+41(-4va-21 d, x[n]+xn-ll, or yIn]=dx[n]+xn-1]-dyn-l] (d) From the figure shown below we obtain vIn- v[n-2] xn →→y[n [n] vn]=x[n]-wIn], wn]=d, vn-1+d,un, and u[n]=vn-2]+xn]. From these equations we get wIn]=d,x[n]+d, x[n-1]+d,xIn-2]-d, wIn-1]-d, wIn-2]. From the figure we also obtain yn]=vIn-2]+wn]=xIn-2]+wn]-wIn-2], which yields d1yn-l]=dx[n-3]+d win-1]] doyln-2]=doxn-4+d, wIn-2]-d,wIn-4], Therefore, n]+d1yn-1+d2yn-2]=xn-2]+d1xn-3]+d2xn-4 1]+d,wn-2 3]+d,wn-4 yIn]=d,x[n]+dxIn-1+xIn-2]-dyln-1-d2yln-2] 2.3(a)x[n]={3-201452}, x-n]={2 3},-3≤n≤3. n]=-(x[n]+x[-n])={5/23/22123/25/2},-3≤n≤3,and xod[n]=(x[n]-x-n)={1/2-7/2-2027/2-1/2},-3≤n≤3. (b)y[n]={071-349-2} y-n]={-294-3170),-3≤n≤3. n]==(yn]+y-n]) 85/2-35/28-1,-3≤n≤3,and dn]==(yn]-y-n])={1 2013/2-1},-3≤n≤3
3 vn xn dvn [] [] [ ] =− − 1 1 and y n d v n v n [] [] [ ] = +− 1 1 . Therefore we can rewrite the second equation as y n d x n d v n v n d x n d v n [] [] [ ] [ ] [] [ ] = −− ( ) + −= +− ( ) − 11 1 1 2 1 1 1 1 (1) = +− dxn d xn dvn ( )( −− − ) 1 1 2 1 [ ] [ ] [ ] 1 12 = +− dxn d xn d d vn ( ) −− − ( ) − 1 1 2 1 1 2 [] [ ] [ ] 1 11 2 From Eq. (1), y n d x n d v n [] [] [ ] −= −+− 1 11 2 ( ) − 1 1 2 , or equivalently, dyn d xn d d vn 1 1 2 1 1 2 [] [] [ ] −= −+ − 1 11 2 ( ) − . Therefore, yn dyn dxn d xn d d vn d xn d d vn [] [ ] [] [ ] [ ] [ ] [ ] + −= +− ( ) −− − ( ) − + −+ − ( ) − 11 1 2 1 1 2 1 2 1 1 2 1 1 11 2 11 2 = +− dxn xn 1 [ ] [ ], or y n d x n x n d y n 1 [] [] [ ] [ ] = + −− − 1 1 1 1. (d) From the figure shown below we obtain v[n] x[n] y[n] v[n–1] –1 d1 z –1 z –1 v[n–2] d 2 w[n] u[n] vn xn wn [ ] [ ] [ ], = − wn dvn d un [ ] [ ] [ ], = −+ 1 2 1 and u n v n x n [ ] [ ] [ ]. = −+ 2 From these equations we get w n d x n d x n d x n d w n d w n [] [] [ ] [ ] [ ] [ ] = + −+ − − −− − 21 2 1 2 1 2 1 2 . From the figure we also obtain y n v n w n x n w n w n [ ] [ ] [ ] [ ] [ ] [ ], = −+ = −+ − − 2 2 2 which yields dyn dxn dwn dwn 111 1 [ ] [ ] [ ] [ ], −= −+ −− − 1313 and d yn d xn d wn d wn 222 2 [ ] [ ] [ ] [ ], −= −+ −− − 2424 Therefore, yn dyn d yn xn dxn d xn [] [ ] [ ] [ ] [ ] [ ] + −+ − = − + −+ − 12 1 2 1 2 2 3 4 + + −+ − (wn dwn d wn wn dwn d wn [] [ ] [ ] [ ] [ ] [ ] ) − −+ −+ − ( ) 12 1 2 1 22 3 4 = −+ + − xn d xn dxn [ ] [] [ ] 2 1 2 1 or equivalently, yn d xn dxn xn dyn d yn [ ] [ ] [ ] [ ] [ ] [ ]. = + −+ − − −− − 21 1 2 12 1 2 2.3 (a) x n[ ] { }, = − 3 2 0 1 4 5 2 Hence, x n n [ ] { }, . − = − −≤≤ 25410 23 3 3 Thus, x n xn x n n ev[ ] ( [ ] [ ]) { / / / / }, , = + − = −≤≤ 1 2 5 2 3 2 2 1 2 3 2 5 2 3 3 and x n xn x n n od[ ] ( [ ] [ ]) { / / / / }, . = − − = − − − −≤≤ 1 2 12 72 2 0 2 72 12 3 3 (b) y n[] { } = −− 0 7 1 3 4 9 2 . Hence, y n n [ ] { }, . − =− − − ≤ ≤ 2 9 4 3 1 7 0 3 3 Thus, y n yn y n n ev[ ] ( [ ] [ ]) { / / }, , = + − =− − − − ≤ ≤ 1 2 1 8 5 2 3 5 2 8 1 3 3 and y n yn y n n od[ ] ( [ ] [ ]) { / / }, = − − = − − − −≤≤ 1 2 1 1 3 2 0 1 3 2 1 3 3
(c)w[n]={-5436-501}, Hence,wf-n]={0-5634-5},-3≤n≤3 Thus, WevIn]=-(wIn]+ wl-nD=(-2 2-1 6-1 2-2),-3sns3, and Wod[n]==(wn]-w-n)={-3240-4-23},-3≤n≤3, 2.4(a) xIn]=gIn]gln]. Hence x[-n]=g-nIgl-n]. Since g[n] is even, hence g[-n]=gIn Therefore x[-n]=g-n]g-n] =gnIgIn]=x[n]. Hence x[n] is even (b)u[n]=gIn]h[n]. Hence, u[-n]=g-n]h[-n]= gIn](-h(n])=-gInhn]=-un]. Hence un is odd (c)vn= hohn. Hence v-n= hohn=(-hnDchnd=hohn= vn. Hence vIn] is even 2.5 Yes, a linear combination of any set of a periodic sequences is also a periodic sequence and the period of the new sequence is given by the least common multiple(Icm)of all periods. For our example, the period lcm(N1, N2, N3). For example, if N1=3, N2=6, and N3=12, then N=lcm(3,5,12)=60. 2.6(a)xpcs[n]={xln]+x*-n]}={Aa+A*(*)},and pcan=lxn -x [nll=laa-a()l, -NsnsN (b)h[n]= -2+j5 4-j3 5+j6 3+j-7+j2) -2sns2, and hence, h*F-n]=[-7-j2 3-j 5-j6 4+j3 -2-j51,-2sns2. Therefore, hpcs=:{hm+h“[-n}=(4.5+j.535-j253.5+p2-4.5-n.5}and pan]=;{hn-h*[-n}=2.5+j350.5-jj-0.5-j-25+35}-2sns2 2.7(a)x[n])=Aa where A and a are complex numbers, with la<1 Since for n <0, a" can become arbitrarily large hence [x[n] is not a bounded sec (b)y[n]=Aa"un] where A and a are complex numbers, with a <1 In this case by[nsA Vn hence (y[n]) is a bounded sequence (c)h(n]=CB"u[n] where C and B are complex numbers, with B>1 Since B becomes arbitrarily large as n increases hence (h[n]) is not a bounded sequence
4 (c) wn[ ] { }, =− − 5 4 3 6 5 0 1 Hence, w n n [ ] { }, . − = − − −≤≤ 1 0 5 6 3 4 5 3 3 Thus, w n wn w n n ev[ ] ( [ ] [ ]) { }, , = + − =− − − − − ≤ ≤ 1 2 2 2 1 6 1 2 2 3 3 and w n wn w n n od[ ] ( [ ] [ ]) { }, , = − − =− − − − ≤ ≤ 1 2 3240 4 23 3 3 2.4 (a) xn gngn [ ] [ ][ ] = . Hence x n g ng n [ ] [ ][ ] −=− − . Since g[n] is even, hence g[–n] = g[n]. Therefore x[–n] = g[–n]g[–n] = g[n]g[n] = x[n]. Hence x[n] is even. (b) u[n] = g[n]h[n]. Hence, u[–n] = g[–n]h[–n] = g[n](–h[n]) = –g[n]h[n] = –u[n]. Hence u[n] is odd. (c) v[n] = h[n]h[n]. Hence, v[–n] = h[n]h[n] = (–h[n])(–h[n]) = h[n]h[n] = v[n]. Hence v[n] is even. 2.5 Yes, a linear combination of any set of a periodic sequences is also a periodic sequence and the period of the new sequence is given by the least common multiple (lcm) of all periods. For our example, the period = lcm ( , , ) NN N 123 . For example, if N1 = 3, N2 = 6, and N3 = 12, then N = lcm(3, 5, 12) = 60. 2.6 (a) x n xn x n A A pcs n n [ ] { [ ] *[ ]} { *( *) }, = + −= + 1 − 2 1 2 α α and x n xn x n A A pca n n [ ] { [ ] *[ ]} { *( *) }, = − −= − 1 − 2 1 2 α α −≤≤ N n N. (b) hn j j j j j [] { } =− + − + + − + 2 5 4 3 5 6 3 7 2 −≤ ≤ 2 2 n , and hence, hn j j j j j *[ ] { } − =− − − − + − − 7 2 3 5 6 4 3 2 5 , −≤ ≤ 2 2 n . Therefore, h n hn h n j j j j pcs[ ] { [ ] *[ ]} { . . . . . . } = + − =− + − + − − 1 2 4 5 1 5 3 5 2 5 3 5 2 4 5 1 5 and h n hn h n j j j j j pca[ ] { [ ] *[ ]} { . . . . . . } = − −= + − −−−+ 1 2 25 35 05 6 05 25 35 −≤ ≤ 2 2 n . 2.7 (a) xn A n { } [] . = { } αα α where A and are complex numbers,with 1 < Since for n < 0, α n can become arbitrarily large hence {x[n]} is not a bounded sequence. (b) yn A n { } [] . = < αµ α α [n] where A and are complex numbers,with 1 In this case yn A [ ] ≤ ∀ n hence {y[n]} is a bounded sequence. (c) hn C n { } [] . = > βµ β β [n] where C and are complex numbers,with 1 Since β n becomes arbitrarily large as n increases hence {h[n]} is not a bounded sequence
(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-1
5 (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
0≤n≤N-1, Since,x[< that ≤n≤N-1 [0]=(x[0]+x*[O=Re{xO]} Similarly pca 0]=(xO]-x*[])=jIm{x[O 21(a)Gven∑knl<∞, Therefore. by Schwartz inequality x[n (b)Consider xn/= 1/n, n21, The convergence of an infinite series can be shown gral test. Let an=f(x), where f(x)is a co function for all x21. Then the series 2ns an and the integral f(x)dx both converge or both diverge. For an=1/n, f(x)=l/n. But -dx=(In ∞-0=∞. Hence, 2n=_o x[n]=nai n does not converge, and as a result, x[n] is not absolutely summable. To show ( x[n]) is square-summable, we observe here ap D2,and thus Now -(3=+1m22mma words, x[n]= 1/n is square-summable 2.13 See Problem 2. 12, Part(b) solution. 2.14x,[n= coSCo ,1≤n≤∞.Now, ecoSoc Since n=1 2n2 ∑。1 6 ∑m casoni remore 6 x,In] is square-summable Using integral test we now show that x,In] is not absolutely summable x cos int(o x) where cosint is the cosine integral function TT COSO x dx diverges also diverges n=1n
6 2.11 x n xn x n pcs N [ ] [ ] *[ ] = + <− > ( ) 1 2 for 0 n N –1 ≤ ≤ , Since, x n xN n N [ ][ ] <− > = − , it follows that x , 1 n N –1. pcs[ ] [ ] *[ ] n xn x N n = +− ( ) ≤ ≤ 1 2 For n = 0, x = Re{x[0]}. pcs[ ] [ ] *[ ] 0 00 1 2 = + (x x ) Similarly x n xn x n pca N [ ] [ ] *[ ] = − <− > ( ) 1 2 = −− ( ) ≤ ≤ 1 2 xn x N n [ ] *[ ] , 1 n N –1. Hence, for n = 0, x = jIm{x[0]}. pca[ ] [ ] *[ ] 0 00 1 2 = − (x x ) 2.12 (a) Given x n n [] . =−∞ ∞ ∑ < ∞ Therefore, by Schwartz inequality, xn xn xn n nn [] [] [] . 2 =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑ ∑∑ ≤ < ∞ (b) Consider x n n n otherwise [ ] /, , , . = ≥ { 1 1 0 The convergence of an infinite series can be shown via the integral test. Let a f x n = ( ), where f(x) is a continuous, positive and decreasing function for all x ≥ 1. Then the series an n= ∞ ∑ 1 and the integral f x dx ( ) 1 ∞ ∫ both converge or both diverge. For a n n = 1 / , f(x) = 1/n. But 1 0 1 1 x dx x ∞ ∞ ∫ = (ln . Hence, ) =∞− =∞ x n n n n [ ] =−∞ ∞ = ∞ ∑ ∑= 1 1 does not converge, and as a result, x[n] is not absolutely summable. To show {x[n]} is square-summable, we observe here a n n = 1 2 , and thus, f x x () . = 1 2 Now, 1 1 11 1 1 1 2 x 1 dx x ∞ ∞ ∫ = − = − ∞ + = . Hence, 1 1 2 n n= ∞ ∑ converges, or in other words, x[n] = 1/n is square-summable. 2.13 See Problem 2.12, Part (b) solution. 2.14 x n n n n c 2[ ] 1 cos = , . π ≤ ≤∞ ω Now, cos . ωc n n n πn n ≤ = π ∞ = ∞ ∑ ∑ 2 1 2 2 1 1 Since, 1 6 2 1 2 n= n ∞ ∑ = π , cos . ωc n n πn ≤ = ∞ ∑ 2 1 1 6 Therefore x n 2[ ] is square-summable. Using integral test we now show that x n 2[ ] is not absolutely summable. cos cos cos cosint( ) ω ω ω ω c c c c x x dx x x x x x π = π ⋅ ⋅ ∞ ∞ ∫1 1 1 where cosint is the cosine integral function. Since cosωcx x dx π ∞ ∫1 diverges, cosωc n n = πn ∞ ∑ 1 also diverges