Solu Es Cap Tulo 1 Sinais E Sistemas Discretos
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Introdu¸c˜ao ao Processamento Digital de SinaisSolu¸c˜oes dos Exerc´ıcios Propostos — Cap´ıtulo 1
Jos´e Alexandre Nalon
1. Dados os sinais xc (t) a seguir, encontre as amostras, a representa¸ca˜o em somat´orios de impulsos deslocados, e trace os gr´ aficos de x[n] = xc (nTa ) para Ta = 0, 5, 1 e 2:
a) xc (t) = cos πt
1.0
0.5
0.0
Solu¸ c˜ ao:
-0.5
• Ta = 0, 5s x[n] -1.0
=
. . . + δ[n + 4] − δ[n + 2] + δ[n] −
-10 -9
−δ[n − 2] + δ[n − 4] + . . .
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
0.0
=
. . . − δ[n + 3] + δ[n + 2] − δ[n + 1] + δ[n] −
−δ[n − 1] + δ[n − 2] − δ[n − 3] + . . .
-0.5
-1.0
-5
-4
-3
-2
-1
0
1
2
3
5
4
1.0
• Ta = 2s x[n] -7
0.5
• Ta = 1s x[n] -8
1.0
=
. . . + δ[n + 3] + δ[n + 2] + δ[n + 1] + δ[n] +
0.5
0.0
+δ[n − 1] + δ[n − 2] + δ[n − 3] + . . .
-0.5
-1.0
-2
-1
0
1
2
c) xc (t) = 2−t u(t)
1.0
0.8
0.6
Solu¸ c˜ ao:
0.4
0.2
• Ta = 0, 5s x[n] 0.0
=
δ[n] + 0, 707107δ[n − 1] +
-10 -9
+0, 5δ[n − 2] + 0, 353553δ[n − 3] + . . .
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
0.8
0.4
=
δ[n] + 0, 5δ[n − 1] + 0, 25δ[n − 2] +
+0, 125δ[n − 3] + . . .
0.2
0.0
-5
• Ta = 2s x[n] -7
0.6
• Ta = 1s x[n] -8
1.0
-4
-3
-2
-1
0
1
2
3
4
1.0
=
δ[n] + 0, 25δ[n − 1] + 0, 0625δ[n − 2] +
+0, 015625δ[n − 3] + . . .
0.8
0.6
0.4
0.2
0.0
-2
1
-1
0
1
2
5
2
π π t+
8
4
d) xc (t) = cos
1.0
Solu¸ c˜ ao:
0.5
• Ta = 0, 5s x[n] =
0.0
. . . + 0, 980785δ[n + 3] + 0, 923880δ[n + 2] +
+0, 831470δ[n + 1] + 0, 707107δ[n] +
+0, 555570δ[n − 1] + 0, 382683δ[n − 2] +
+0, 195090δ[n − 3] + . . .
• Ta = 1s x[n] -1.0
-10 -9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
1.0
0.5
0.0
=
. . . + 0, 923880δ[n + 3] + δ[n + 2] +
-0.5
+0, 923880δ[n + 1] + 0, 707107δ[n] +
-1.0
+0.382683δ[n − 1] − 0, 382683δ[n − 3] + . . .
• Ta = 2s x[n] -0.5
-5
-4
-3
-2
-1
0
1
2
3
4
5
1.0
0.5
=
. . . − 0, 707107δ[n + 4] + 0.707107δ[n + 2] +
+δ[n + 1] + 0, 707107δ[n]
−0,