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≤33 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