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-12 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