Sinais discretos

5) Dado o sinal x[n] = 3(u[-n] –u[-n-8) + 6(u[n+1] – u[n-6), determinar e plotar o sinal y[n] dado por: y[n] = x[-3n-1] x[n] = 3(u[-n]-u[-n-8]) + 6(u[n+1]-u[n-6]) x[n] = -3(u[n]-u[n+8]) + 6(u[n+1]-u[n-6])

Y[n] = x[-3n-1]

6) N | X[n] | | 1 | 2 | 2/3 | 2 | 0 | 2 | 3 | -2 | -10/3 |
a)
x1 [n] = 1,5 x3 [-n-2] + 3 x1 [n] = -1,5 x3 [n+2] + 3 x1 [n] - 3= 1,5 x3 [n+2]
-(1/1,5)x1 [n] + 2 = x3 [n+2]
X3 [n] - 3= (-2/3) x1 [n-2] + 2

7)
X[n] = x[-n]
Par

X[n] = -x[-n]
Ímpar

8)
Xa [n] = xa[-n]

Xa [n] = -Xa[-n]

9)
i)
x[n] = 6u[n-3] x[n] = (1/2)(x[n] + x[-n]) + (1/2)(x[n] – x[-n]) x[n] = (1/2)(6u[n-3] + 6u[-n-3]) + (1/2)(6u[n-3] – 6[-n-3]) x[n] = (1/2)(6u[n-3] - 6u[n+3]) + (1/2)(6u[n-3] + 6[n+3]) x[n] = (3u[n-3] – 3u[n+3) + (3u[n-3] + 3u[n+3)
Par

Ímpar

ii) x[n] = -n x[n] = (1/2(x[n] + x[-n]) + (1/2)(x[n]-x[-n]) x[n] = (1/2)(-n+n) + (1/2)(-n-n) x[n] = 0 –n
Par

Ímpar

Sinais e