2.15 ∑x2lm=∑(xsm]+xlmn)2 ∑x2 x2an]+2∑ kevIn]xod[n]=∑xvn]+∑x3dl D=-oo as 2ns- Xevin )odiN =0 since Xev In xoaIn is an odd sequence. 2.16xn]=cos(2πkn/N) 0≤n≤N-1. Hence, E、=∑c2nkn/N=∑ n 1 (1+cos(4rkn/N))=3+3>cos(4Ikn/N) Letc=∑cos(4πkn/N,ands=∑sin(4xkn/N n=0 =0 Therefore C+js ∑ 4]kn/N =O. Thus, C=ReC+jS =0 C=Re C+js=0,it E=N 217(a)xn=n]. Energy=∑u2m=∑ Average power=,m、1ns ∑n=kmn32=m1 K 1 K→∞2K+ K2K+12 (b)x2nl=叫 u[n]. Energy=∑n=c(mpm)=∑a=n k-yoo 2K+12n=K(n(n )2=lim1 yk Average power= lim I K→∞2K+1 (e)xin]=Aoejoan. Energy Jn- oeA t Ens_JAo =oo. Average power lim K lim K→∞2K+1 K→∞2K+1 K Aol K→∞2K+1 (d) xn]=Asi M+o=Aoeo+Aye joon where Oo M ?cie and A=A -jo From Part(c), energy = oo and average power= A6+Af+4A2A2-3 218Now叫sJ1,n≥0 1,n<0, Hence,μ-n-1l l0.n≥o. Thus, x[n=山m]+μ-n-1 2.19(a)Consider a sequence defined by x[n]=>8[k]7 2.15 xn x n x n n ev od n 2 2 [] [] [] =−∞ ∞ =−∞ ∞ ∑ ∑ = + ( ) =++ =+ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑∑ ∑ ∑∑ x n x n x nx n x n x n ev n od n ev od n ev n od n 22 22 [] [] [] [] [] [] 2 as n= ∞ x nx n ev od ∞ ∑ = – [] [] 0 since x n x n ev od [ ] [ ] is an odd sequence. 2.16 x n kn N [ ] cos( / ), = ≤≤ 2π 0 n N –1. Hence, E kn N x n N = = − ∑cos ( / ) 2 0 1 2π = + ( ) = − ∑ 1 2 0 1 1 4 cos( / ) πkn N n N = + = − ∑ N n N kn N 2 1 2 0 1 cos( / ). 4π Let C kn N n N = = − ∑cos( / ), 4 0 1 π and S kn N n N = = − ∑sin( / ). 4 0 1 π Therefore C jS ej kn N n N + = = − ∑ 4 0 1 π / = − − = e e j k j kN 4 4 1 1 0 π π / . Thus, C C jS = + Re 0 . { } = As C C jS = + Re 0 { } = , it follows that Ex N = 2 . 2.17 (a) xn n 1[ ] [ ]. = µ Energy = µ2 2 [] . n 1 n n =−∞ ∞ =−∞ ∞ ∑ ∑= =∞ Average power = lim ( [ ]) lim lim . K n K K K n K K K n K K →∞ + =− →∞ = →∞ K = + = + ∑ ∑ = 1 2 1 1 2 1 1 2 1 1 2 2 2 0 µ (b) xn nn 2[ ] [ ]. = µ Energy = n n n n n ( µ[] . ) = =∞ =−∞ ∞ =−∞ ∞ ∑ ∑ 2 2 Aveerage power = lim ( [ ]) lim . K n K K K n K K n n K n →∞ =− →∞ = + = + ∑ ∑ = ∞ 1 2 1 1 2 1 2 2 0 µ (c) x n Aeo j no 3[] . = ω Energy = Ae A o j n n o n ωo =−∞ ∞ =−∞ ∞ ∫ = =∞ ∑ 2 2 . Average power = lim lim lim . K o j n n K K K o n K K K o o K A e K A K A K o A →∞ =− →∞ + =− →∞ = + = + ∑ ∑ = 1 2 1 1 2 1 2 2 1 2 2 2 ω 2 (d) xn A n M A e Ae o jn jn o o [ ] sin , = π +     = + 2 − φ 1 ω ω where ωo M= 2π , A A e o j = − 2 φ and A A e j 1 2 = − φ. From Part (c), energy = ∞ and average power = A A AA A o o 2 1 2 2 1 2 2 4 3 4 ++ = . 2.18 Now, µ[ ] , , , . n n n = ≥ <    1 0 0 0 Hence, µ[ ] , , , . −− = < ≥    n n n 1 1 0 0 0 Thus, x[n] = µ µ [ ] [ ]. n n + −− 1 2.19 (a) Consider a sequence defined by x n k k n [ ] [ ]. = =−∞ ∑δ