If n <0 then k=0 is not included in the sum and hence xn=0, whereas for n>0, k=0 is included in the sum hence x[n]=1 V n20. Thus x[n]= >8Ik]= 1,n≥0 l0.n<0 u[n] n≥0. (b) Since u[n] it follows that u[n-l n< n≤0 Hene,-p-={0.m≠0=an 2.20 Now x[n]=Asin(@on+o) (a)Given x[n]=0 -12-2-V2 0 v2 2 v2. The fundamental period is N=4 hence o=2兀/8=元/4. Next from x0=Asin(φ)=0 we get o=0, and solving x[l]=Asin(+o)=Asin(/4)=-v2 we get A=-2 (b) Given xn The fundamental period is n= 4, hence ①0=2T4=π/2. Next from x[O]=Asin()=√2 and x[1]=Asin(π/2+9)=Acos()=√2it can be seen that A=2 and o= T/4 is one solution satisfying the above equations (c)x[n]=3 -3. Here the fundamental period is N= 2, hence Oo=T. Next from x[O] Asin(o)=3 and x[l]= Asin(o+I)=-Asin(o)=-3 observe that A=3 and o=t/2 that A=3 and o= T/2 is one solution satisfying these two equations (d)Given x[n]=0 1.5 0-1.5, it follows that the fundamental period of x[n] isN ence @0=2T/4=T/2. Solving x[0]=Asin(o)=0 we get 9=0, and solving x[l=Asin (T/2)=1.5, we get A= 1.5 2.21(a)xI[n]=e Ju 4n. Here, 0=0.4T. From Eq (2.44a), we thus get N- 2rr 2Ir =5r=5 forr=l 0.4兀 (b)x2[n]=sin(0.6n+0.6T). Here, Oo=0.6T. From Eq(2.44a), we thus get 2兀r2πr10 =-r=10 forr= 3 0.6兀3 (c)x3[n]=2cos(1.Intn-05)+2sin(O.7Tn). Here, @1=l.lT and 2=0.7T. From Ec (2. 44a), we thus get Ni ar In 20 2πr2兀 r?. To be periodic l兀11 rI and n. 020.7兀8 If n < 0 then k = 0 is not included in the sum and hence x[n] = 0, whereas for n ≥ 0 , k = 0 is included in the sum hence x[n] = 1 ∀ ≥ n0. Thus x n k n n n k n [] [] , , , , = = [ ]. ≥ <    = =−∞ ∑δ µ 1 0 0 0 (b) Since µ[ ] , , , , n n n = ≥ <    1 0 0 0 it follows that µ[ –] , , , . n n n 1 1 1 0 0 = ≥ ≤    Hence, µµ δ [n] – [ ] , , , , n [ ]. n n − = n = { ≠ 1 = 1 0 0 0 2.20 Now xn A n [ ] sin = + (ω φ 0 ). (a) Given x n[ ] = − −− { } 0 2 2 2 0 2 2 2 . The fundamental period is N = 4, hence ωo = π =π 28 4 / / . Next from x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving xA A [ ] sin( ) sin( / ) 1 4 = 4 2 π + = π =− φ we get A = –2. (b) Given x n[ ] = −− { } 22 2 2 . The fundamental period is N = 4, hence ω0 = 2π/4 = π/2. Next from x A [ ] sin( ) 0 2 = = φ and xA A [ ] sin( / ) cos( ) 12 2 = += = πφ φ it can be seen that A = 2 and = /4 is one solution satisfying the above equations. φ π (c) x n[ ] = − { } 3 3 . Here the fundamental period is N = 2, hence ω π 0 = . Next from x[0] = Asin( ) φ = 3 and x A A [ ] sin( ) sin( ) 1 3 = + =− =− φπ φ observe that A = 3 and = /2 φ π that A = 3 and φ = π/2 is one solution satisfying these two equations. (d) Given x n[] . . = − { } 0 15 0 15 , it follows that the fundamental period of x[n] is N = 4. Hence ωπ π 0 = = 24 2 / / . Solving x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving x A [ ] sin( / ) . 1 2 15 = = π , we get A = 1.5. 2.21 (a) ˜ [] . . xn e j n 1 0 4 = − π Here, ωo = π 0 4. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 4 5 5 ω . for r =1. (b) xn n ˜ [ ] sin( . . ). 2 = π+ π 06 06 Here, ωo = π 0 6. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 6 10 3 10 ω . for r = 3. (c) xn n n ˜ [ ] cos( . . ) sin( . ). 3 = π − π+ π 2 11 05 2 07 Here, ω1 = π 1 1. and ω2 = π 0 7. . From Eq. (2.44a), we thus get N r r r 1 1 1 1 1 2 2 1 1 20 11 = π = π π = ω . and N r r r 2 2 2 2 2 2 2 0 7 20 7 = π = π π = ω . . To be periodic