Introdução à Análise de Sinais e Sistemas Resposta de Exercícios Selecionados
João Marques de Carvalho Professor Titular Aposentado DEE/CEEI/UFCG Professor Participante COPIN/CEEI/UFCG
Edmar Candeia Gurjão e Luciana Ribeiro Veloso Professores Professores DEE/CEEI/UFCG
Campina Grande, Paraíba, Brasil c Edmar Candeia Gurjão, Luciana Ribeiro Veloso e João Marques de Carvalho, Maio de 2014
Capítulo 2 2.1
x(t ) = t + + 1, y(t ) = t
− 1,
g(t ) = x(t ) + y(t ), h(t ) = x (t ) y(t ), g(t ) = t + + 1 + t h(t ) =
−2 ≤ t ≤ 2
− 1 = 2t (t + + 1)(t − 1) = t − t + + t − 1 = t − 1 2
2
h(t )
g(t )
t
t
(a)
(b)
Figura 0.1: Figuras relativas ao Exercício 2.1
2.2 Gráficos na Figura 0.2
i
Capítulo 2 2.1
x(t ) = t + + 1, y(t ) = t
− 1,
g(t ) = x(t ) + y(t ), h(t ) = x (t ) y(t ), g(t ) = t + + 1 + t h(t ) =
−2 ≤ t ≤ 2
− 1 = 2t (t + + 1)(t − 1) = t − t + + t − 1 = t − 1 2
2
h(t )
g(t )
t
t
(a)
(b)
Figura 0.1: Figuras relativas ao Exercício 2.1
2.2 Gráficos na Figura 0.2
i
g(t
− 1) g(t + + 2)
−3
1
t
3
−6
2
(a)
(b)
g(t /2)
g(2t + + 2)
−8
4
t
−3 −1
(c)
(d)
Figura 0.2: Figuras relativas ao exercício 2.2
2.3
g(t ) = cos(ωit ), h(t ) = cos(ωit + + φ), s(t ) = g(t ).h(t ) = cos(ωit ).cos(ωit + + φ)
=
t
1 2
cos(ωit
− ω t − φ) + cos(2ω + φ) i
i
= cos( φ) + cos(2ωi + φ)
−
Considerando ωi = π a) φ = 0
s(t ) = 12 cos(0) + cos(2ωit ) = 12 + cos(2ωit ) b) φ = π/2
s(t ) = 12 cos(π/2) + cos(2ωit + + pi/2) = cos(2ωit + + π/2)
ii
t
c) φ = 3 π/2
s(t ) = 12 cos(3π/2) + cos(2ωit + + 3 pi/2) =
s(t )
−sen(2ω t ) i
s(t )
s(t )
1.5 1.0
-1
0.5
1
t (s)
1.0
-1 1
t
-1.0 (a) Caso 1
-1 1 -1.0
(b) Caso 2
(c)
Figura 0.3: Figuras relativas ao exercício 2.3
2.4
g(t ) = cos(3t ), h(t ) = cos(3t + + φ), x(t ) = g(t ) + h(t )
• Para φ = 0 h(t ) = cos(3t ), x(t ) = cos(3t ) + cos(3t ) = 2 cos(3t ) Valor máximo de x (t ) é 2 .
• Para φ = π/2 h(t ) = cos(3t + + π/2) = sen(3t ), x(t ) = cos(3t ) + sen(3t )
iii
t
Derivando a função x(t ) com relação a t e igualando a zero tem-se o valor de t para qual a função x (t ) assume o valor máximo.
d (cos3t + sen3t )
= 0
dt sen3t + cos3t = 0
−
3t =
pi/4
t = pi/12 Portanto,
x(π/12) = cos(π/4) + sen(π/4) = Valor máximo de x (t ) é
√
2
√ 2.
• Para φ = π h(t ) = cos(3t + π) = x(t ) = cos(3t )
−cos3t ,
− cos(3t ) = 0
Valor máximo de x (t ) é 0 (constante).
2.5
z(t ) = e j2π100t = cos(200πt ) + jsen(200πt ), x(t ) = Re z(t ) = cos(200πt ) y(t ) =
{ } Im { z(t )} = sen(200πt )
tx(t ) y(t ) = tcos(200πt )sen(200πt ) 1
t [ sen(400πt ) 2
tsen(400πt ) 2
2.6 Figura 0.5 iv
− 12 sen(0)]
Figura 0.4: Figuras relativas ao exercício 2.5
2.7 a)
z(t ) = k (t ) + y(t ) = 2t + 1 + t z(t ) = 3t
−1
−2
b)
f (t ) = k (t ) y(t ) = ( 2t + 1)(t f (t ) = 2t 2
− 3t − 2
− 2)
c)
− 1) = 2(t − 1) + 1 2t − 1
s(t ) = k (t s(t ) =
v
g(t )
g(t
−2
− 2)
g(t + 1)
−2
−1
2
t
−2 1
(a)
t
4
−2
1
(b)
(c)
g(t /2)
g(3t + 2)
−2
−2
−2
4
t
t
−1
(d)
(e)
Figura 0.5: Figuras relativas ao Exercício 2.6
d)
x(t ) = y(t + 2) = t + 2 x(t ) = t
−2
e)
j (t ) = k (2t + 2) = 2 (2t + 2) + 1 j (t ) = 4t + 5 f)
l (t ) = y( t + 1) = x(t ) =
− −t − 1 vi
−t + 1 − 2
t
2.8 a)
d (n) = 2a(n) + 3b(n) = 2 (n2 d (n) = 5n2 + 1
− 2) + 3(n
2
b)
c(n) = a(n)b(n) = ( n2 c(n) = n4
−4
− 2)(n
2
+ 2)
c) 2
g(n) = a(n g(n) = n2
− 2) = (n − 2) − 2 − 4n − 6
d)
h(n) = b(n + 2) = ( n2 + 2)2 + 2 h(n) = n2 + 4n + 6 e)
i(n) = a(2n + 1) = ( 2n + 1)2 i(n) = 4n2 + 4n
−1
−2
f)
j(n) = b( n + 1) = ( n + 1)2 + 2 j(n) =
− − n − 2n + 3 2
2.9 Figura 0.6
2.10 x(n) = u(n-1) - u(n-5) vii
+ 2)
2.11
x1 (t ) =
∞
∑ δ(t kT 1 ), x 2(t ) =
k = ∞
−
−
∞
∑ δ(t mT 2 )
−
m= ∞
−
a) y1 (t ) = x1 (t ) + x2 (t ) i) T 1 = T 2 /2 ii) T 1 = T 2
→ T = 2T 2
1
iii) T 1 = 2 T 2 b) y1 (t ) = x1 (t ) x2 (t ) i) T 1 = T 2 /2 ii) T 1 = T 2
→ T = 2T 2
1
iii) T 1 = 2 T 2
2.12
a(n) = (n2 b(n) = (n2
a(n) = (n2 b(n) = (n2
− 2)[u(n + 2) − u(n − 4)] + 2)[u(n + 4) − u(n − 2)]
− 2)[δ(n + 2) + δ( N + 1) + δ(n) + δ(n − 1) + δ(n − 2) + δ(n − 3)] + 2)[δ(n + 4) + δ(n + 3) + δ(n + 2) + δ(n + 1)δ(n) + δ(n − 1)]
d (n) = 2a(n) + 3b(n)
+ 2)[δ(n + 4) + δ(n + 3)]+[2(n2 2) + 3(n2 + 2)] [δ(n + 2) + δ(n + 1) + δ(n) + δ(n 1)] + 2(n2 2)[δ(n 2) + δ(n = 3(n2 + 2)[δ(n + 4) + δ(n + 3)]+(5n2 + 2)[u(n + 2) u(n 2)] + 2(n2 2)[δ(n 2) + δ(n 3)] =
2
3(n
−
−
−
e(n) = (n2
= (n4
−
2
−
−
− −
− 2)(n + 2)[u(n + 2) − u(n − 2] − 4)[u(n + 2) − u(n − 2)] viii
−
− 3)
2.13
x(t ) = 2cos(200t + π/6) y(t ) = x2 (t ) y(t ) = 2cos(200t + π/6)2cos(200t + π/6) y(t ) = 2[1 + cos(400t + π/3)]
→ T = inde f inido cos(400t + π/3) → T = 2 π/400 → T = π/200
z1 (t ) = 1 z2 (t ) =
1
2
2
T = m.m.c(T 1 , T 2 ) = π/200, ω0 = 2π/π/200 = 400 . 2.14 a) s1 (t ) = 10cos(100t + π/3) + 16sin(150t + π/6) 10cos(100t + π/3)
→ T = 2π/100 16sin(150t + π/6) → T = 2 π/150 1 2
2π 2π π , 150 T = mmc( 100 ) = 50
b) s2 (n) = 2cos(πn/2) + 3cos(2πn/5) 2cos(πn/2)
→ mN
1
=
1
3cos(2πn/5)
→ mN
2
=
2
N = m.m.c(4, 5)
2π
N → → N = 4 m π/2 1
1
1
2π
N → m 2π/5
2 2
=
5
→ N = 5
1
2
→ N = 20
2.15 a) x(t ) = 2. Como o sinal x (t ) é de tempo contínuo e constante, ele é periódico para qualquer valore de T R∗+ . Desta forma, não é possível identificar o menor valor de T para o qual a equação x (t ) = x (t + T ) seja verdadeira.
∈
b) x(n) = 2. Como o sinal é uma constante no tempo discreto, o período do sinal (N) tem que ser um número inteiro positivo. Portanto, o menor número inteiro para o qal a equação x(n) = x(n+N) é a unidade ( N = 1 ). ix
2.16 a) Se x 1 (t ) for um sinal periódico com período T 1 e x 2 (t ) for um sinal periódico com período T 2 , a soma x(t ) = x 1 (t ) + x2 (t ) é um sinal periódico se e somente se T 1 /T 2 for um número racional. b) Se x 1 (n) for um sinal periódicos com período N 1 , e se x 2 (n) for um sinal periódico com perido N 2 , o sinal x 1 (n) + x2 (n) será sempre um sinal periódico posi a razão entre N 1 e N 2 sempre será um número racional.
2.17 O período fundamental da soma e o produto de dois sinais periódicos com periódo N 1 e N 2 é N 1 N 2 N = mdc( N 1 , N 1 ) Entretanto, o período da multiplicação pode ser menor.
2.18
y(t ) = x (t )[ u(t ) + u(t
−
− kT ]
2.20
x(t ) = impar cos(2πt )u(t )
{
}
x
x(t
− 1)
x(2
−2 −1
− t )
− 1− 2
1
2
3
t
4
−1
1
2
(a)
t
3
(b)
x(2t
− 2)
− 1− 2
−1
1
2
t
3
(c)
x(t /2
− 2)
− 1− 2
0
2
4
8
6
(d)
u(t )[ x(t ) + x(2
− t )]
− 2− 1− 3
0
2
t
(e)
xi
Figura 0.6: Figuras relativas ao Exercício 2.9
10
t
x(n) 1
n 0
1
2
3
4
Figura 0.7: Exercício 2.10
x(n
− 1)
1
n 0
1
3
2
5
4
Figura 0.8: Exercício 2.10, a)
x(2
− n)
1
n -2
0
-1
1
3
2
Figura 0.9: Exercício 2.10, a)
x(2n + 1) 1
n -2
-1
0
1
2
3
Figura 0.10: Exercício 2.10, a)
xii
x(2n + 1)u(2, n) 1
n -2
0
-1
1
2
3
Figura 0.11: Exercício 2.10, a)
x(2n + 1)δ(n
− 3)
1
n -2
-1
0
1
2
3
Figura 0.12: Exercício 2.10, a)
Figura 0.13: Exercício 2.11 a) item i)
xiii
Figura 0.14: Exercício 2.11 a) item ii)
Figura 0.15: Exercício 2.11 a) item iii)
Figura 0.16: Exercício 2.11 b) item i)
xiv
Figura 0.17: Exercício 2.11 b) item ii)
Figura 0.18: Exercício 2.11 a) item iii)
xv
Figura 0.19: Exercício 2.13
Figura 0.20: Exercício 2.14 a)
xvi
Figura 0.21: Exercício 2.14 b)
xvii
1) z(t ) = cos (2πt )u(t ), x (t ) = 12 ( z(t )
− z(−t ))
Figura 0.22: Exercício 2.20 a) 2) x(n) = 2 cos(πn/4)
− 2cos(πn/2 + π/6) sin(πn/8) N N 2π 8 → = = → N = 8 2cos (πn/4) − m m π/4 1 1
1
1
1
1
2cos(πn/2 + π/6)
→ mN
2
=
1
→ mN
3
sin(πn/8)
=
3
2π
N → m π/2
2
= 4
2
2π
N → m π/8
3
= 16
3
N = mmc(8, 4, 16) = 16 . 3) x(n) = cos
8πn 7
+ π6 ,
N m
=
2π 8π/7
→
N m
=
7 4
→ N = 7
4) x(t ) = Par sin(2πt + π/4)u(t ) , z (t ) = sin (2πt + π/4)u(t ), x (t ) = 12 z(t ) + 12 z( t )
{
}
−
5) x(t ) = Impar [sin(2πt + π/4)]u(t ) , x 1 (t ) = sin (2πt + π/4)u(t )
{
6) x(n) = cos( 8π7n ), ω0 = 87π ,
N m
} = ωπ → 2
0
N m
2.21
y(t ) = g (t )π(t
− kT ), T → período de g(t ) xviii
=
2π 8π/7
→
N m
=
7 4
(racional ), N = 7 .
Figura 0.23: Exercício 2.20 d)
Figura 0.24: Exercício 2.20 e)
2.22
x(t ) = cos(2π f 1t ), T = 22ππ f 1 u(t )
→ T = 1/ f , y(t ) = 1
− u(t − 1/ f ), y(t ) = x(t ) p(t ). 1
2.23
p(t ) = u(t
− (k − 1)1/ f ) − u(t − k / f ) 1
1
xix
cos(2π f 1t ), 0,
0
≤ t ≤ 1/ f
1
c.c
, p(t ) =
2.24 K
2 p(t ) = ∑k = K [u(t 1
− K / f ) − u(t − K / f 1)] 1
1
2
2.25 a) y(t ) = t 2 , y p (t ) = 12 [t 2 + ( t )2 ] = t 2 , y i (t ) = 0 .
−
b) z(n) = n, z p (n) = 12 [n
1 2
− n] = 0, y (n) = [n − (−n)] = n. c) w(t ) = t − 1, 0 ≤ t ≤ 3, w (t ) = (t − 1)[u(t ) − u(t − 3)], i
3
3
1
1 w(−t ) − 2 2 1 {(t − 1)[u(t ) − u(t − 1)] − (t + 1)[u(t + 3) − u(t )]} 2 u(t − 3) t u(t − 1) t u(t + 3) u(t + 3) − 2 − 2 − 2 2
w p (t ) =
w(t ) 3
=
3
3
=
wi (t ) =
1
3
{(t − 1)[u(t ) − u(t − 1)]+(t ?3 + 1)[u(t + 3) − u(t )]} 2 t u(t − 3) u(t − 3) t u(t + 3) u (t + 3) 2t u(t ) 2u(t ) − − − 2 + 2 + 2 2 2 2 3
=
3
3
3
1 2
− u(n − 4), x (n) = [u(n + 3) − u(n − 4)] + δ(n), x (n) = 1) − u(n − 4)] − [u(n + 3) − u(n)]
d) x(n) = u (n)
p
i
1 2
1 2
[u(n
−
2.26
− −
−
− −
−
− − →
x(t ) = x( t ), e y(t ) = y( t ). x (t ) y(t ) = x( t ) y( t ), z(t ) = x(t ) y(t ), z( t ) = x( t ) y( t ) z(t ) = z( t ), função par. 2.27
x(t ) = t , y (t ) = sin(2πt ) 2.28
f (t ) =
sin(2πt )/t 0
≥
t 1 Sinal nem par nem ímpar e não periódico. t < 1 xx
2.29
y(t ) = e(σ+ jω)t = eσt e jωt = eσt (cos(ωt ) + j sin(ωt )), Re y(t ) = eσt cos(ωt ), Im y(t ) = eσt sin(ωt ).
{ }
a) σ < 1
Figura 0.25: Exercício 2.29 a)
b) σ > 1
Figura 0.26: Exercício 2.29 b)
c) σ = 0
xxi
{ }
Figura 0.27: Exercício 2.29 c)
2.30 e jt
a) y(t ) = e jt . y p (t ) = 12 y(t ) + 12 y( t ) =
−
2
− jt + e 2 = cos(t )
b) z(n) = u(n), 1
δ(n)
2
2
z p (n) = z (n)/2 + z( n)/2 = u(n)/2 + u( n)/2 = +
−
−
zi (n) = u(n)/2 + u( n)/2
−
c) w(t ) sin(ω0t + π/4)
w p (t ) =
=
1
1
2 1 2
1
1
2
2
w(t ) + w( t ) = sin(ω0t + π/4) + sin( ω0t + π/4) 2
−
sin (ω0t + π/4)
wi (t ) =
1 2
1
− 2 sin(ω t − π/4) = 0
sin(ω0t + π/4) +
1 2
√ 2 2
sin (ω0t + π/4) =
−
cos(ω0t )
√ 2 2
d) x(n) = δ(n), x p (n) = 12 δ(n) + 12 δ( n) = δ(n), x i (n) = 12 δ(n)
sin ω0t
(0.1)
1 2
− δ(−n) = 0. e) y(t ) = t − 2, 0 ≤ t ≤ 4, y (t ) = (t − 2)[u(t ) − u(t − 4)], y (t ) = (t − 1)[u(t ) − u(t − 4)] + (−t − 2)[u(t + 4) − u(t )], y (t ) = (t − 2)[u(t ) − u(t − 4)] + (t + 2)[u(t + 4) − u(t )] f) z(n) = n, −5 ≤ n ≤ 5, z (n) = [ z(n) + z(−n)] = 0 , z (n) = z(n). −
p
1 2
1 2
i
1 2
p
1 2
1 2
i
g) x(t ) = cos(t ) + sin(t ) + sin(t ) cos(t ) x p (t ) = 12 [cos(t ) + sin(t ) + sin(t ) cos(t )] + 1 2
−
−
−
−
[cos( t ) + sin( t ) + sin( t ) cos( t )] = cos(t ). xi (t ) = sin(t ) + sin(t ) cos(t ).
xxii
Capítulo 3 3.1 a) com memória, causal, variante no tempo, instável, linear. b) Sem memória, causal, não linear, instável, variante ao deslocament0. c) Sem memória, causal, estável, variante no tempo, linear. d) y(n) = 12 x(n no tempo.
1 2
− 1) − x(−n + 1), com memória, não causal, estável, linear, variante
e) causal, estável, com memória, não linear, invariante no tempo. f) y(n) = 22 x(n + 1) = 12 x( n ante no tempo.
− − 1). Com memória, não causal, estavel, linear, vari-
g) Linear, não causal, não estável, com memória, variante ao deslocamento. h) Causal, sem memória, estável, linear, invariante no tempo. i) Causal, com memória, estável, linear, invariante no tempo. j) Causal, com memória, não estável, linear, variante no tempo. k) Causal, sem memória, não estável, não linear, variante no tempo. l) Não causal, com memória, estável, invariante no tempo, linear considerando n 0 < ∞.
3.2 Não é inversível.
xxiii
3.3 a) Não linear e invariante ao deslocamento. b) Não linear e variante ao desclocamento. c) Lienar e invariante ao tempo.
3.4 a) instável b) Estável c) Instável d) Instável e) Instável f) Estável g) Instável
xxiv
Capítulo 4 4.1
∗ − 3) y(n) = 0, n < 0 y(n) = ∑ ==
n 3 k 3 1 = n + 1, n
a) y(n) = u (n + 3) u(n
≥ 0, y(n) =
Figura 0.28: Exercício 4.1 a)
(n + 1)u(n). b) y(n) = 3n u( n + 3) u(n
−
∗ − 2)
Figura 0.29: Exercício 4.1 b)
= ∑∞ k =0 3−k 3n w−2 = 3 n−2 ∑∞ k =0
• p/ n ≤ 5, y(n) = ∑ =−−∞ 3 n 1 k
k
−1
n 3 2
3
• p n > 5, y(n) = ∑ =− k
k
infty 3
= ∑∞ k =0 xxv
1 3
k
3
3
1 3
k
= 3 3 1−11/3 = 3 4/2.
=
−2
n 3
−
1
1 3
=
y(n) =
−
n 1
/2. n 5 4 n > 5 3 /2,
3
≤
c) y(n) = cos π2n u(n) u(n
∗ − 1)
Figura 0.30: Exercício 4.1 c)
• y(n) = 0, n < 2 • −
n 1
y(n) =
∑ cos
k =0
= =
1 2 1 2
∑− πn 2
− −−
n 1
=
1
e jπn/2
1
e jπ/2
1
e jπn/2
1
− j
jπn/2
e
k =0
+
2
− e− π / + 1 − e− π/ −π/ 1 − e + 1
j n 2 j
2
j n 2
1 + j
4.2
x1 (n) = u(n)
− u(n − 4), x (n) = u(n) − u(n − 6) a) y (n) = x (n) ∗ x (n) • p/ n < 0, y(n) = 0 • p/ n ≥ 0 e n − 3 ≤ 0, 0 ≤ n ≤ 3, y(n) = ∑ = = n + 1 1
2
1
2
n n 0
xxvi
e − jπn/2 2
Figura 0.31: Exercício 4.2
Figura 0.32: Exercício 4.2 a) n k n 3
• p/ 3 ≤ n ≤ 5, y(n) = ∑ = − 1 = n − n + 3 + 1 = 4 • p/ n > 5 e n − 3 ≤ 5, 5 ≤ n ≤ 8, y(n) = ∑ − − 1 = 5 − n + 3 + 1 = 9 − n • y(n) = 0, n > 9 b) y (n) = x (n) ∗ x (n − 3) 5
k n 3
2
1
2
Figura 0.33: Exercício 4.2 b)
• y(n) = 0, n < 3 • p/ 3 ≤ n < 6, y(n) = sum = = n − 3 + 1 = n − 2 • p/ 6 ≤ n < 8, y(n) = ∑ − = n − n + 3 + 1 = 4 n k 3
n n 3
xxvii
8
• p/ 8 ≤ n ≤ 11, y(n) = ∑ − • p/ n > 11, y(n) = 0
n 3 1 = 8
− n + 3 + 1 = 12 − n
4.3
x(n) = u(n)
− u(n − 50), v(n) = u(n − 31)u(n − 42), w(n) = u(n + 51) − u(n − 17) 1. ]a)] y (n) = x (n) ∗ v(n), n = 31 b) y (n) = v(n) ∗ w(n), n − 31 = −51, n = −20 c) y (n) = x(n) ∗ w(n), n = 51 1
2
3
xxviii
Capítulo 5 5.1 a) x(t ) = e−5t u(t ), h (t ) = e−2t [u(t ) = u(t
− 3)]
Figura 0.34: Exercício 5.1 a)
• p/ t < 0, y(t ) = 0 • p/ 0 ≤ t ≤ 3,
t
y(t ) =
=
• t ≥ 3 y(t ) = −
0
e−5t e−2(t −τ) d τ = e−2t
t
d τ
0
− τ t e−2t −3t 0 e2t − 2t e 3 | = (e − e ) = (1 − e−3t ) e
−3
0
T 5τ e 2(t τ) d τ t 3 e
− − −
b) x(t ) = e−|t |[u(t +1)−u(t −1)] , h (t ) = u(t
• p/ t < 0, y(t ) = 0 • p/ 0 ≤ t < 1, y(t ) = −−
−3
=
e−2t
−3t −3(t −3) ) −3 ( e − e
− 1), y(t ) = x(t ) ∗ h(t )
t 1 τ 1
3
e d τ = eτ xxix
|−−
t 1 1
= et −1
− e−
1
= e−1(et
− 1)
Figura 0.35: Exercício 5.1 b)
• p/ 1 ≤ t < 2, y(t ) =
0
−1
= eτ
0
−
t 1
τ
e d τ +
0
e−τ d τ
e −τ t −1 | +
|− −1 = 2 − e− − e− e [ ( + )− ( − )] , h (t ) = u (t ) − u(t − 3) 1
0
1
c) x(t ) = e−3t u
t 2
t 1
u t 4
Figura 0.36: Exercício 5.1 c)
• p/ t < −2, y(t ) = 0 • p/ −2 ≤ t < 1, y(t ) = − e− τd τ = − e− τ | = − (e− − e ) • p/ 1 leqt <, y(t ) = − e− τd τ = − e− τ| − = − f rac13(e− − e− ( − )) • p/ 4 ≤ t < 7, y(t ) = − e− τd τ = − e− τ| − = − f rac13(e− − e− ( − )) d) x(t ) = e− ( − ) , h (t ) = u (t − 1) − u(t − 5) • p/ t < 3, y(t ) = 0 t
1 3
3
2
t t 3 4
1 3
3
3tu
t 2
xxx
t
2
3
1 3
3
t 3
3
t t 3
3
4
t 3
1 3
3t
6
3t
12
3
t 3
3
t 3
Figura 0.37: Exercício 5.1 d)
• p/ 3 ≤ t < 7, y(t ) = − e− τd τ = − f rac13(e− ( − ) − e− ) • p/ t ≥ 7, y(t ) = −− e− τd τ = − f rac13(e− ( − ) − e− ( − )) e) x(t ) = e− [u(t ) − u(t − 5)], h (t ) = u(t − 1) t 1
3
3
2
t 1 t 5
3
3
t 1
t 1
6
3
t 5
3t
Figura 0.38: Exercício 5.1 e)
• p/ t < 0, y(t ) = 0 • p/ 0 ≤ t < 5 y(t ) = e τd τ = − (e− − e ) = − (e − 1) • p/ t ≥ 5, y(t ) = e− τd τ = − (e− − e ) = − f rac13(e− − 1) t 3
1 3
0
5 0
3
1 3
xxxi
3t
15
0
0
1 3
3t
15
Capítulo 6 6.1 a) x(t ) = sin (4t ) + cos(6t ), T 1 = 24π = π2 , T 2 = 26π = π3 , T = π. ∞
∑
x(t ) =
ak e
j 2πkt T
∞
k = ∞
= a+2 =
1 2 j
, a −2 =
1 2 j
1
k = ∞
−
2 j
e j4t
− 21 j e−
j 4t
−
1 2 j 1 2 j 1 2
,
ak =
= =
−1 ≤ t ≤ 1, T = 2. T 0
1
e−
2
1
2
2
,
x(t )e− j
2πkt
T
dt =
1 2
1
−1
te− j 1
1
k =
−2
k = 3 e 3 c.c.
2πkt 2
dt =
e− jk π
1 2
1
−1
te− jk πt dt 1
e jk π
− jk π) (− jk πt − 1) |− = 2 (− jk π) (− jk π − 1) + 2 ( jk π) ( jk π − 1) − jk 1π 2 j cos(k π) − ( jk 1π) sin(k π) − j(−k π1)
2( 1
0
T 0 jk πt
1
+ e j6t + e− j6t
k = 2
0,
1
−
, a 3 = 12 , a −3 = 12 .
ak =
b) x(t ) = t ,
ak e j2kt
∑
=
2
1
2
2
xxxii
2
k
c) x(t ) = e−3t ,
−1 ≤ t ≤ 1, T = 2
ak = = 1
=
2
1
1
e3t e− jk ω0t dt
−−
1
=
0
2
1
1
e
−1
jk πt
dt =
1
1 2
−3 − k jπ 1 [e−( − −6 + 2 jk π 3
e−(3− jk π)t dt
−1
e−(3− jk π)t
2
=
1
jk π)t
1
|−
1
+ e(3− jk π) =
1
−6 + 2 jk π
[e3 e− jk π
jk π
− e− e 3
]
6.2
f (t ) = π82 (sin(t )
π/2)
−
1 sin 32
− −
−
− π ), f (t ) = 1 + π (sin(2t −
( t ) +...), f (t ) = 1 + f (2t ( t π/2) + ....)
1 sin 7 72 1 sin7 2 72
(5t ) 1 sin (5(2t π/2)) 32
−
−
8
2
2
6.3 a) T 0 = 2 , ω0 = 22π = π4.
ak =
=
1 2
0
T 0 jk ω0t
e−
2
e− jk ω0t
4 ∑∞ k =−∞, k impar k π sin(k πt ) x (t ) =
a0 = x(t ) =
1 2
0
2
2 1
jk
0
k
b) T 0 = 2
− jk ω t dt + 1 2e
1
2
0
1
1dt +
8
2
2 jk π
dt
+ e− jk π )
ak = Bk + jC k ., Bk = 0, C k =
π sin (πt ) +
1
jk ω0 t
−2e−
| − − jk ω | = − jk 1π (e− π − 1 − e 1 0
2
k
1
0
− jk ω a = − π , k ímpar. 0
ak = 0, k par.
1 e− jk ω t dt =
2
4
2 ,6667
π
k
1,6 sin(3πt ) + π sin(5πt ) + ....
−3dt = 12 + 12 (−3) = −1
1 , 6 −1 π8 sin(πt ) + 2, 667 sin(3πt ) + sin(5πt ) + .... π π
xxxiii
4
− π x(t ) =
Capítulo 7 7.1 a) x(t ) = 2 [u(t ) 2(2
− 0) = 4
2 5
− u(t − 1)], X (s) = −
2 5
e−2s
−
−e− 2s) , ∀s.
2(1
2
Figura 0.39: Exercício 7.1 a)
xxxiv
X (0) = 2
e 2 0
0t
dt =
b) x(t ) = e−t u(t
− 2) + e−( − )u(t − 3). t 3
− 2) + e−( − )u(t − 3) e−( − ) e− u(t − 2) + e−( − ) u(t − 3) e− e−( − ) u(t − 2) + e−( − ) u(t − 3) 1 e− u(t ) ↔ , ℜ(s) > −1 s+1
x(t ) = e−t e−2 e2 u(t t 2
=
2
=
2
t 3
t 2
t 3
t
z(t ) = z(t
t 3
− α) = e−α s +1 1 , Re(s) > −1 s
−2s
−3s
X (s) = es+1 + es+1 =
e2s +e−3s , s +1
Re(s) >
−1
Figura 0.40: Exercício 7.1 b) 2
c) x(t ) = e−(t −2) u(t ) = e −t u(t )e2 , X (s) = se+1 , Re(s) >
−1.
7.4 s+1/2
H (s) = (s+a)(s+b) , causal e estável. a, b > 0 pólos em s = 1/2 = 2 ab = 14 , a = 3 /4, b = 1 /3. a.b
←
H (s) =
s + 1/2
(s + 1/3)(s + 3/4) xxxv
−a e s = −b.
H (0) = 2,
Figura 0.41: Exercício 7.1 c) causal e estável, pólos no semiplano esquerdo. s = ROC Re (s) > 1/3
−
−1/3 e s = −3/4, zero s = −1/2
Figura 0.42: Exercício 7.4)
H (s) = s+1/2
A s + 1/3
+
b s + 3/4
=
2/5
s + 1/3
s+1/2
A = s+3/4 s=−1/3 = 25 . B = s+1/3 s=−3/4 . 3 h(n) = 25 e−1/3t u(t ) + 35 e 4 t u(t )
|
|
e− jω +1/2
e− jω +1/2
H (ω) = (e− jω 1/3)(e− jω +3/4) = −2 jω 13 − jω + 12 e +1/4 e xxxvi
=
3/5
s + 3/4
H (s) = d 2 y(t ) dt 2
s+1/2 Y (s) s + 1/4) = X (s)(s + 1/2). = , Y (s)(s2 + 13 13 12 X (s) s + 12 s+1/4 13 dy (t ) + 14 y(t ) = dxdt (t ) + 12 x(t ) 12 dt 2
+
Todo sistema causal e estável está em repouso inicial.
7.5 G(s)
H (s) = 1+C (s)G(s) Q(s) = s−b a e C (s) (k ) b/s−a b/s−a H (s) = 1+kb/s−a = s−a+kb = s−ab+kb
−
−
s a
s = a kb é um pólo. Se b>0 a kb < 0 , a < kb , k > ba Se b<0 a kb < 0 , a > kb , k < ba .
−
− −
7.6
x(t ) = [ e−t + e−3t ]u(t ), y (t ) = 2 [e−t
− e−
4t
]u(t )
a)
X (s) = Y (s) =
1
s+1
+
1
s+3
=
2(s + 2)
(s + 1)(s + 3)
, Re(s) >
2
−1
2 6 − , R(s) > −1 = s+1 s+4 (s + 1)(s + 4)
H (s) =
Y (s) X (s)
=
3(s + 3)
(s + 4)(s + 2)
, Re(s) >
Figura 0.43: Exercício 7.6 a)
xxxvii
−2
b) A resposta em frequência existe, pois a ROC de H (s) contém o eixo imaginário.
H (ω) =
3( jω + 3)
( jω + 4)( jω + 2)
c)
A
H (s) =
s+4
+
B s+2
3(2 + 3)(s + 4)
, A=
(s + 4)(s + 2) 3
h(t ) =
2
| =− = 3/2, B = 3((ss++43)()(ss++22)) | =− = 3/2 4
s
s
2
3
e 4tu(t ) + e−2t u(t )
−
2
O sistema é causal e estável: todos os pólos no semi plano esquerdo do plano s . d)
Y (s) X (s)
=
3s + 9 2
s
2
→ s Y (s) + 6sY (s) + 8Y (s) = 3sX (s) + 9 X (s) + 6s + 8 d 2 y(t ) dt 2
+
6 dy (t )
dt
+ 8 y(t ) =
dx (t ) dt
+ 9 x(t )
7.7 d 2 y(t ) dt 2
−
1 3
dy (t ) dt
−
2 3
y(t ) = x (t ), s 2Y (s)
H (s) =
−
1 3
sY (s)
−
2 3
Y (s) = X (s)
A B 1 → H (s) = = (s − 1)(s + 2/3) s − 1 s − 1 − s− 1
s2
A =
1 3
2 3
1
1 | | =− / = −3/5 = = 3 /5, B = s + 2/3 s−1 s 1
2 3
s
a) Estável h (t ) é absolutamente integrável, ROC cont[em s = j ω
h(t ) =
3
[ et u(t )
− 5
H (ω) =
− e−
2 3
t
u(t )]
1 2
1 3
−ω − jω − 2/3 b) Causal, h (t ) é lateral à direita, ROC Re {s} > 1 h (t ) = existe H (ω), ROC não contém s = jω.
xxxviii
3 5
[et
− e−
f rac23t
]u(t ), Não
Figura 0.44: Exercício 7.7 a)
Figura 0.45: Exercício 7.7 b) c) não causal, não estável, h(t ) lateral à esquerda, ROC não contém s = j ω, não existe H (ω)
h(t ) = X (s) =
s2
3 5
2
(e−t + e− 3 t )u( t )
−
− 3s + 2 , Re{s} > −1 s+1
X (s) =
(s
− 2)(s − 2)
xxxix
s+1
Figura 0.46: Exercício 7.7 c)
Y (s) =
(s
− 2)(s − 1) ×
1
s
=
−2
=
A
− 1)(s + 2/3) (s + 1)(s + 2/3) s + 1 s−2 s−2 | | =− / = −8 A = =− = 9 , B = s + 2/3 s+1 Estável ROC de H (s) −2/3 < Re{s} < 1 , Y (s) → Re{s} > −2/3 s+1
(s
s
1
s
+
B s + 2/3
2 3
− 8e− 23 t )u(t ) causal H (s) → Re{s} > 1, Y (s) → {s} > −2/3 y(t ) = ( 9e− − 8e− )u(t ) nem causal, nem estável H (s) → R{s} < −2/3, Y (s) → −1 < Re{s} < −2/3 y(t ) = 9 e− u(t ) − 8e− u(−t ) y(t ) = ( 9e−t
t
t
xl
2 3
t
2 3
t
Capítulo 8 8.1
f (n) = cos
2nπ
N
,
N 2π = Ω m 0
• N = 2 f (n) = cos(nπ)
=
2π
π
= 2
ak e− jk ω0 n =
∑
f (n) =
N m
k =< N >
2πn 3 1 2
temos a 1 = a−1 =
ak e− jk πn = e− jπn =
k =< N >
ak =
• N = 3, f (n) = cos
∑
1,
k =
0,
2π
1 jπn 2
e
1
+ e− jπn 2
± 1 ± 2l c.c.
2π
2π
= 12 e j 3 n + 12 e− j 3 n comparando com f (n) = sumk =< N > ak e− j 3 kn
ak =
1/2,
k =
0,
±1 ± 3l, l inteiro c.c.
8.2 a) x(n) = 2 + cos
meiro cosseno),
2nπ) 3
N 3 m
=
+ 3cos
nπ
2
, N 1 = 1 (termo constante),
N 2 m
=
2π 3
= 3 (pri-
2π (segundo cosseno), logo, N = mmc(4, 3) = 12 . π/2 = 4
1
1
3
3
2
2
2
2
x(n) = 2 + e j2nπ/3 + e− j2nπ/3 + e jnπ/2 + e− jnπ/2 x(n)
∑
ak e jkn2π/12 =
k =< N >
∑
ak e jknπ/6
k =< N >
e jknπ/6 = e jknπ/3
xli
→ 6k = 23 = 4
a4 = 12 , a−4 = 1 /2, a−3 = 3 /2, a0 = 2, a3 = 3 /2
ak =
b) x(n) = sin
2nπ 3
x(n) =
1 2 j
2 j
2π/3
±3 ± 12l
k =
− 21 j e
j 2nπ/3
+
1 4 j
e jn7π/6 +
2π
= 3 , N 2 =
7π/6
12
=
7
1 2 j
− sin
nπ
4
, 0
2
1 4 j
e jnπ/6 + 2π
π/6
1 4 j
e− jnπ/6
− 41 j e−
j7nπ/6
= 12, logo N = 12
ak e− jknπ/6
∑
k =< N > 7
7
1 4
1
1 2 j 1 4 j
1 4 j
0,
1
c.c.
3
3
1 ∑ x(n)e− j2knπ/4 = ∑ x(n)e− jknπ/3
4 n=0
4 n=0
x(0) = 1 , x(1) = 1 ak =
1
≤ n ≤ 3, N = 4
ak =
−
√ 2 2
, x(2) = 0 , x(3) = 1
√ 2 2
[ x(0)e0 + x(1)e− jk π/2 + x(2)e− jk π + x(3)e− j3k π/2 ]
√ − 2 2 − a = frac14[1 + e k
2
− 21 j , a = 41 j , a− = − 41 j , a = 41 j , a− = − 41 j k = ±4 ± 12l k = ±7 ± 12l a = k = ±1 ± 12l
, a−4 =
k
c) x(n) = 1
1
, N 3 =
ak e− jkn2π/12 =
∑
1
[1 + e jnπ/2 + e− jnπ/2 ]
k =< N >
a4 =
c.c.
2
e j2nπ/3
x(n)
±4 ± 12l
nπ
j2nπ/3
2π
k =
0,
− 21 j e
e j2nπ/3
N 1 +
1
k = 0
1/2,
3/2,
[1 + cos
x(n) =
2,
√ − 2 2 − π/ e + 2 √ 2 − 2
2
ak =
1 4
j 3k π/2
2
jk
[1 +
xlii
2
cos(k π/2)]
]e− j3k π/2 = e jk π/2
8.3 a) ak = 5cos( jπ/4) + 3sin(3k π/4), N 1 = 2π/π/4 = 8, N 2 = 2π/4/3π = 8/3, N = 8.
ak =
ak = 1
x( 1) =
8
−
1
∑
x(n)e− j2knπ/8 =
8 n=< N > 5 jk π/4 2
e
5
3
2
2 j
+ e− jk π/4 +
1
x(n)e− jknπ/4
∑
8 n=< N >
e j3k π/4
− 23 j e−
5
j 3k π/4
1 5 12 12 x(−1) = 20, x(1) = → x(1) = 20, x(−3) = → = −12 j, x(3) = − = 12 j j j 2 8 2 n = ±1 ± 8l 20 n = 3 ± 8l 12 j x(n) = l inteiro −12 j k = ±3 ± 8l
0,
c.c.
xliii
Capítulo 9 9.1 a) x(n) = δ(n) + 3δ(n
− 1) X ( z) = 1 + 3 z− , | z| > 0 1
b) x(n) = n (0.5)n u(n)
−
X ( z) = (1−00.5. z5 z−11 )2 , z > 0.5
||
c) x(n) = ( 1 + n)u(n) = u (n)
X ( z) =
− nu(n)
z−1 , z > 1 + 1− z−1 1− z−1 1
||
9.3
− z−− , | z| > a, a > 0 1−az z− √ A B √ √ X ( z) = (1−√ az−1−)( = + − − 1+ az ) (1− az ) (1+ az− ) √ √ a−1 1−1/ a 1− z− √ √ A = 1+ az− z− 1= √ = 1+ a/√ a = 2√ a
b) X ( z) =
1
1
2
1
1
1
1
1
1
1
1
a
√ √ a+1 1+1/ a −√ z− − − √ √ √ = 2√ a = z 1= −a 1+ a/ a 1− az √ a−1 √ √ a+1 √ − an √ x(n) = 2 a e u(n) + 2√ a e an u(n) B =
1
1
1
1
9.6
y(n)
− 13 y(n − 1) − 29 y(n − 2) = x(n) − 12 x(n − 1)
Y ( z)(1
− 13 z− − 29 z− ) = X ( z)(1 − 12 z− ) 1
2
xliv
1
H ( z) =
(1 (1
−
1 3
−
z−1 )
1 2
z−1
−
(1
=
2 9
z−2 )
(1
−
z−1 )
1 2
− 2/3 z− )(1 + 1/3 z− ) 1
1
Figura 0.47: Exercício 9.6 Causal
⇒ h(n) lateral à direita, ROC: | z| > H ( z) =
A 2 3
1 2 1 3
1 2 2 3
h(n) =
B
−1 +
− z 1 − z− A = 1 − z− − 1 − z B = 1 − z− 1
2 3
1 1
1 1
1 2 n
1+
z−1 1
z−1 = 32 =
6
z−1 =−3 =
5
( ) u(n) + (
6 3
1 3
6
5 6
−1 ) u(n) n
3
Estável porque h (n) é absolutamente somável. Estável porque a ROC de H ( z) contém z = 1 , portanto
||
H (Ω) = H ( z) z=e jΩ =
1 1
xlv
−
1 3
−
1 2
e− jΩ
e− jΩ
−
2 9
e−2 jΩ
H (0) =
1
1 2 1 3
− −
2 9
= 1 , 16
9.9
P1 = 1/6 e P2 = 1/7 z1 = 3 j e z 2 = 3 j
− √
−√
√ − √ || − √ √ √ √ ( z − 3 j)( z + 3 j) ( z + 3 j) H ( z) = = ( z − 3 j) ( z + 1/6)( z − 1/7) ( z + 1/6)( z − 1/7) √ A B H ( z) = ( z − 3 j )( + ) ( z + 1/6) ( z − 1/7) √ 6 − 42 j ( z + 3 j)( z + 1/6) A = | =− / = ( z + 1/6)( z − 1/7) 13 √ ( z + 3 j)( z − 1/7) 7 − 42 j | | B = = / = ( z + 1/6)( z − 1/7) 13 √ 6 − 42 j 1 7 − 42 j 1 H ( z) = ( z − 3 j)( + ) ( z + 1/6) ( z − 1/7) 13 13 √ 3 j 7 − 42 j z √ 3 j z 6 − 42 j 6 − 42 j 7 − 42 j H ( z) = − 13 ( z + 1/6) + 13 ( z − 1/7) − 13 ( z − 1/7) ( z + 1/6) 13 √ 3 jz− 7 − 42 j 1 6 − 42 j 1 6 − 42 j − 13 (1 + z− /6) + 13 (1 − z− /7) H ( z) = 13 (1 + z− /6) √ 3 jz− 7 − 42 j − 13 (1 − z− /7) , | z| > 1/6 ( z 3 j )( z + 3 j ) H ( z) = , z > 1 /6 ( z + 1/6)( z 1/7)
1 6
z
z 1 7
1
1
1
1
1
1
− 42 j (−1/6) u(n) − 6 − 42 j 13 13 − 7 −1342 j (3) j(1/7) − u(n − 1)
h(n) =
6
n
n 1
−1/6) − u(n − 1) + 7 −1342 j (1/7) u(n)
(3) j(
xlvi
n 1
n
H (Ω) = H ( z) z=e jΩ
− √
√
j Ω 3 j)(e (e jΩ + 3 j) H (Ω) = (e jΩ + 1/6)(e jΩ 1/7)
||
Estável porque a ROC de H ( z) contém z = 1
Y ( z) X ( z) y(n) + Se x (n) = δ(n) + δ(n
1
=
z2 + 3 1 z2 + 42 z
y(n
42
−
1 42
=
−
−2
1 + 3 z 1+
1 42
z−1
−
1 42
z−2
− 1) − 421 y(n − 2) = x(n) + 3 x(n − 2)
− 1)/6
y(n) = x(n) h(n) = h(n) + h(n
∗
− 1)/6.
9.10
√
√ − √ − √ ( z − a − b j)( z − a + b j) z + 2az + a + b H ( z) = = ( z − c − d j )( z − c + d j ) z + 2cz + c + d √ z − z/3 + 1/9 H ( z) = , | z| > (1/4 + j 3/4) = 1 /2 z − z/2 + 1/4 √ √ − (1 − (1/4 − j 3/6) z−1) √ √ H ( z) = ( 1 − (1/6 + j 3/6) z ) (1 − (1/4 + j 3/4) z− )(1 − (1/4 − j 3/4) z− ) √ H ( z) = ( 1 − (1/6 + j 3/6) z− 1)( −( / + √ / ) − + −( / − √ / ) − √ 1 − (1/4 − j 3/6) z −−1 √ / ) = 0, 8333 − 0, 0962 j √ | A = =− /( / + 1 − (1/4 − j 3/4) z −−1 √ 1 − (1/4 − j 3/6) z −−1 √ / ) = 0, 1667 + 0, 0962 j √ | B = =− /( / − 1 − (1/4 + j 3/4) z −−1
P1 = 1 /4 + j 3/4 e p2 = 1 /4 j 3/4 z1 = 1 /6 + j 3/6 e z 2 = 1 /4 j 3/6
2
2
2
2
2
2
2
2
1
1
1
1 4
A j 3 4 z 1
z
1
1 4
j 3 4
z
1
1 4
j 3 4
xlvii
1
1
1 4
B j 3 4 z 1
