Respostas - Livro

Respostas - Livro: Digital Signal Processing
1
Capı́tulo 2
• 2.1 a) Não é periódico b)N=20 amostras, c)N=5 amostras d)14 amostras.
• 2.2 a)0,6 rad/s, −0, 2π rad/s, −0, 15π rad/s e 3π/14 rad/s.
b)2,4 rad/s, −0, 8π rad/s, −0, 6π rad/s e 6π/7 rad/s.
• 2.3 −π/3 rad/s, 0 rad/s, −π rad/s e π rad/s.
• 2.4 ωo = 4 rad/s, θ = 1, 49 rad e A = 4, 99.
• 2.12 cod = 4/3, c1d = 0, 33, c2d = 1/3
◦
◦
• 2.13 c0d = 3/4, c1d = 0, 56e−j63,4 , c2d = 3/4 e c3d = 0, 56ej63,4
• 2.24 a) Para N > 2, ωo = 2π/N
b) ωo = 2M π/N para −N/2 ≤ M < N/2
c) N-1 freqüências.
2
Capı́tulo 3
• 3.12 Xd (ω) = 1, Limitado e contı́nuo.
• 3.13
a) Xd (ω) = 2jsen(ωT )
b) e−j0,5ω (2 cos 0, 5ω − 2)
c) e−j0,75ω (−2 cos 0, 75ω + cos 0, 25ω)
d) A seqüência x[n] diverge porque cresce sem limites quando n→ − ∞
• 3.14 Xd (ω) =
1
1−0,9e−jω
• 3.15 Xd (ω) =
1
1+0,9e−jω
• 3.16
1
X(ω) = jω+0,2
1
Xd (ω) = 1−0,9048e
−j0,5ω
A sobreposição na freqüência (aliasing) é significativa.
• 3.17
1
X(ω) = jω+0,2
Xd (ω) = 1−0,99e1−j0,05ω
A sobreposição na freqüência (aliasing) é insignificante.
• 3.18
• 3.20 w =
• 3.21 Sim
2π
NT
3
Capı́tulo 4
• 4.1 Xd [0] = 4, Xd [1] = 1 e Xd [2] = 1
• 4.2 Xd [0] = 3, Xd [1] = 1 − 2j, Xd [2] = 3 e Xd [3] = 1 + 2j
• 4.3 a) Xd [0] = 0, Xd [1] = 1, 732j4 e Xd [2] = −1, 732j
b) Xd [0] = 0, Xd [1] = 1.5 + j2, 598 e Xd [2] = 1, 5 − j2, 598
c) Xd [0] = 2, Xd [1] = −3(1 + j), Xd [2] = 0 e Xd [3] = −3(1 − j)
• 4.20 x1 = [1 − 2 1]
x2 = [1 − 2 1 0]
x = [1 − 2 1]
• 4.21 x1 = [0 − 1 1]
x2 = [0 − 1 0 1]
x[−1] = 1, x[0] = 0, x[1] = −1
4
Capı́tulo 5
• 5.1 y[n] =
Pn
k
k=0 (1, 0001) x[n
− k], y[n] = 1, 0001y[n − 1] + x[n]
• 5.2 h[n] = (1, 0001)n
• 5.3 y[n] = 0, 2(x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4] + x[n − 5]) e y[n] = y[n − 1] + 0, 2 ∗
(x[n] − x[n − 5])
• 5.4
• 5.5 h[n] = (−1)n u[n], IIR, h[0] = 1, h[1] = −1 e h[n] = 0 para n ≥ 2, FIR
z
z
+ 0.5 z−1
, y[n] = 0, 5 − 0, 5(−1)n
• 5.8 Y (z) = −0.5 z+1
• 5.9 H(z) =
b1 z 2 +b2 z+b3
a1 z 2 +a2 z+a3 ,
• 5.10
• 5.12 y[n] = 1 + n, A saı́da não é limitada.
P
P
• 5.13 y[n] = −2x[n] + 4/3 nk=0 x[k] + 5/3 nk=0 (−0.5)n−k x[k], y[n + 2] − 0, 5y[n + 1] − 0, 5y[n] =
x[n + 2] + x[n], y[n] − 0, 5y[n − 1] − 0, 5y[n − 2] = x[n] + x[n − 2]
• 5.15 a) estável, b)Instável, c)Estável, d)Estável, e)Estável, f)Estável
5
Capı́tulo 6
• 6.1 hhp [n] =
−sin[(n−no )ωc ]
(n−no )
e hhp [no ] = 1 − ωc /π
• 6.2
• 6.3 hbp [no ] = ωcuπ−ωcl e
o )ωcu ]
−
hbp [n] = sin[(n−n
(n−no )π
• 6.4 hbs [no ] = 1 −
• 6.5
ωcu −ωcl
π
sin[(n−no )ωcl ]
(n−no )π
e hbs [n] =
sin[(n−no )ωcl ]
(n−no )π
−
sin[(n−no )ωcu ]
(n−no )π
• 6.6 H(z) =
0.045(z+1)
z−0.91
• 6.7 H(z) =
z−1
10z+8
• 6.8 H(z) =
0.0105(z 2 −1)
z 2 −1.8586z+0.9801
• 6.9 H(z) =
z 2 −1.9999z+1
z 2 −1.9799z+0.9801
• 6.10 H(z) =
6
(z−1)(z−ej2.1 )(z−e−j2.1 )
(z−0.8)(z−0.8ej2.1 )(z−0.8e−j2.1 )
Capı́tulo 7
• 7.1 Retangular: H(z) = 0, 1447 + 0, 2678z −1 + 0, 3183z −2 + 0, 2678z −3 + 0, 1447z −4
Hamming: H(z) = 0, 0116 + 0, 1446z −1 + 0, 3183z −2 + 0, 1446z −3 + 0, 0116z −4
• 7.2 Retangular: H(z) = 0, 0762(1 + z −5 ) + 0, 2117(z −1 + z −4 ) + 0, 3052(z −2 + z −3 )
Hamming: H(z) = 0, 006(1 + z −5 ) + 0, 082(z −1 + z −4 ) + 0, 28(z −2 + z −3 )
• 7.4