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DEBER N°01
TEMA: EJERCICIOS CAPITULO 5 LIBRO DE SEÑALES Y SISTEMAS DE OPPENHEIM
5.1 Use la ecuación de análisis (5.9) de la transformada de Fourier
para calcular las transformadas de:
a) ( 12 )n−1
u [n−1]
e
e
x [n ] (¿¿− jwn)
(¿¿ jw )= ∑n=−∞
∞
¿
X ¿
e¿
jw¿¿¿
X ¿
e
e
1−1
2(¿¿− jw)
(¿¿− jw)¿
¿
b) ( 12 )|n−1|
e¿ jw¿¿¿e¿
X ¿
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e¿
jw¿¿¿e¿
− jw¿¿¿¿
X ¿
5.2 Use la ecuación de análisis (5.9) de la transformada de Fourier
para calcular las transformadas de:
a) δ [ n−1 ]+δ [ n+1 ]
x ( e jw )=¿ (e− jw )+(e jw )
x ( e jw)=¿ 2(e− jw )+(e jw )
2
x ( e jw )=2cosw
b) δ [ n+2 ]−δ [n−2]
x ( e jw
)=¿
(e j2w
)− (e− j2w
)cos2w− jsen2w x ( e jw )=cos2w+ jsen2w−¿
)
x ( e jw)=2 jsen2w5.3 Determine la transformada de Fourier para −π ≤ ω
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a−1=−( 12 j )e− j π
4
X (e jω )=2 π a1δ (ω− π 3 )+2π a−1δ (ω+ π 3 )
X (e jω )=( π j ){e j
π
4 δ (ω− π 3 )−e− j π
4 δ (ω+ π 3 )}
b) 2+cos( π 6 n+π
8 ) N =12
x [ n ]=2+(1
2 )e
j( π 6 n+ π
8 )+(
1
2 )e
− j(π 6 n+ π
8)
a0=2
a1=( 12 )e
j π
8
a−1=( 12 )e− j π
8
X (e jω )=2 π a0δ (ω )+2π a
1δ (ω−2π 12 )+2π a−1δ (ω+
2π
12 ) X (e jω )=4 πδ (ω)+π {
e j π
8 δ
(ω−π 6 )+
e− j π
8 δ
(ω+ π 6 )}
5." Use la ecuación de s#ntesis (5.$) de la transformada de Fourier
para calcular las transformadas in%ersas de Fourier de:
a) X 1 (e jω )=∑
k =∞
∞
[2πδ (ω−2 πk )+πδ (ω−π 2−2kπ )+πδ (ω+ π 2−2πk )]
x [ n ]= 1
2π
∫−π
π
X 1 (e jω ) e jωn dω
x [ n ]= 12π ∫−π
π
(2πδ (ω−2πk )+πδ (ω−π 2−2πk )+πδ (ω+ π 2−2πk ))e jωn dω x [ n ]= 1
2π ∫−π
π
(2πδ (ω )+πδ (ω−π 2 )+πδ (ω+ π 2 ))e jωn dω
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x [ n ]=e j0+ 12
e j(−π 2 )n+ 1
2e
j ( π 2 )n
x [ n ]=1+cos π 2
n
b) X 2 (e jω )={ 2 j ,0
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b) x2 [ n ]= x ' [ n ]+ x [n]
2
x ' (−n )= X (e jw ) X (e jw )+ X ' (e jw )
x (e jw )=12¿
c) x3 [ n ]=(n−1)2 x [n ]
x3 [ n ]¿n2 x [ n ]−2nx [ n ]+1
5.- ,ara cada una de las siuientes transformadas de Fourier use
las propiedades de la transformada de Fourier (tabla 5.1) para
determinar si la se!al correspondiente en el dominio del tiempo es:
(i) real imainaria o ni lo uno ni lo otro/ (ii) par impar o ninuna de
las dos. 0aa esto sin e%aluar la in%ersa de las trasformadas dadas.
a) X 1 (e jω )=e− jω∑
k =1
10
(senkω )
Ia!"#a$"%& #"#!'#%(
b) X 2 (e jω )= jsen(ω)cos (5ω)
Ra*& "+a$(
c) X 1 (e jω )= A (ω )+e jB(ω) ,%#,
A (ω)=
{
1,0≤|ω|≤ π 8
0, π
8
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impulso h2 [ n ] . a interconeión en paralelo 'ue resulta tiene larespuesta en frecuencia
H ( e jω)= −12+5e− jω
12−7e− jω+e− j2ω
Determine h2 [ n ] .
h [ n ]=h1 [ n ]+h
2 [ n ]
H ( e jw)= H 1 ( e jw )+ H 2 (e
jw )
H 1(e jw)= 1
1−0.5 (e− jw )
H 2 (e jw
)= −12+5− jw
12−7 (e− jw )+e−2 jw− 1
1−0.5e− jw = −2
1−0.25e− jw
h2 [ n ]=−2(14 )
n
u [n]
5.214alcule la transformada de Fourier de las siuientes se!ales:
a) x [ n ]=u [n−2 ]−u[n−6]
x [n ]=δ [n−2 ]+δ [n−3 ]+δ [n−4 ]+δ [n−5 ]
X (e jw )=e−2 jw+e−3 jw+e−4 jw+e−5 jw
b) x [ n ]=( 12 )−n
u [−n−1 ]
X (e jw )= ∑n=−∞
−1
( 12 )−n
e− jw
X (e jw )=∑n=1
∞
( 12 e jw)n
X (e jw
)=e
jw
2
1
(1−1
2e
jw )
c) x [ n ]=( 13 )|n|
u [−n−2 ]
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e
X (¿¿ jw)=∑n=−∞
−2
(1
3
)
−n
e
− jwn
¿e
X (¿¿ jw)=∑n=2
∞
( 13 e jw)n
¿e
X (¿¿ jw)=e
j2w
9
1
(1−1
3e
jw)
¿
d) x [ n ]=2n sen( π 4 n)u[−n ]
X (e jw )= ∑n=−∞
0
2n
sen(πn /4)e− jwn
X (e jw )=−∑n=0
∞
2−n
sen (πn/4)e jwn
X (e jw )=−12 j∑n=0
∞
[(12 )n
e jπn/ 4
e jw−( 12 )
n
e− jπn/4
e jwn]
X (e jw )=−12 j
[ 1
1−( 12 )e jπ /4 e jw−
1
1−(12 )e− jπ / 4 e jw]
e) x [ n ]=( 12 )|n|
cos( π 8 (n−1)) X (e jw )= ∑
n=−∞
∞
(12 )
|n|cos(
π 8(n−1))e
− jw
X (e jw )=12[
e− jπ /8
1−( 12 )e jπ /8 e jw+
e jπ /8
1−( 12 )e− jπ /8e− jw]
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+14[
e jπ /4
e jw
1−(1
2 )e
jπ /8
e
jw
+ e
jπ /8e
jw
1−(1
2 )e
− jπ /8
e
jw
]
f) x [ n ]={ n ,−3≤ n ≤30,conotro vaor x [n ]=−3δ [n+3 ]−2δ [n+2 ]−δ [n+1 ]+δ [n−1 ]+2δ [n−3 ]+δ 3[n−3]
X (e jw )=−3e3 jw−2e2 jw−e jw+e− jw+2e−2 jw+3e−3 jw
) x [ n ]=sen
(π
2
n
)+cos (n )
x [ n ]= 12 j
[e jπn/2−e− jπn /2 ]+12[e jn+e− jn]
X (e jw )= π j [δ (w− π 2 )−δ (w+ π 2 )]+π [δ (w−1 )+δ (w+1)]
0≤∨w∨¿ π
) x [ n ]=sen (5 π 3 n)+cos( 7π 3 n)
x [ n ]=sen (5 πn3 )+cos( 7πn3 ) x [ n ]=−sen( πn3 )+cos( πn3 ) x [ n ]=−1
2 j [e jπn/3−e− jπn/3 ]+ 1
2[ e jπn/3+e− jπn/3 ]
X (e jw )=−π j [δ (w−π 3 )−δ (w+ π 3 )]+π [δ (w− π 3 )−δ (w+π 3 )]
0≤∨w∨¿ π
i) x [ n ]= x [ n−6 ] , y x [ n ]=u [ n ]−u [n−5 ] !ara0≤ n ≤5
N =6
ak =1
6∑n=0
5
x [n]e− j (2π " 6 )kn
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ak =1
6∑n=0
5
e− j (2π " 6 ) kn
ak =1
6 [ 1−e− j5kπ " 3
1−e− j (2π " 6 ) k ] X (e jw )=∑
=−∞
∞
2π ( 16 )[ 1−e− j5 kπ /3
1−e− j (2π /6) k ]δ (ω−2π 6 −2 π)
6) x [ n ]=(n−1)( 13 )|n|
( 13 )|n|
#( $%& ) 4
5−3cosw
n( 13 )|n|
#( $%& )− j 12 senw
(5−3cosw )2
x [ n ]=n( 13 )|n|
−( 13 )|n|
# ( $%& ) 4
5−3cosw− j
12 senw
(5−3cosw )2
7) x [ n ]=( sen (πn/5)πn )cos ( 7 π 2 )n
x1 [n ]=sen (πn/5)
πn # ( $%& ) X 1 (e jw
)=
{ 1,|w|
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5.22 8 continuación se muestran las transformadas de Fourier de las
se!ales discretas. Determine la se!al correspondiente a cada
transformada.a)
X (e jω )={1, π
4≤|ω|≤ 3π
4
0, 3 π
4≤|ω|≤ π ,0≤|ω|≤ π
4
x [ n ]= 12π ∫−π
π
X (e jω ) e jωn dω
x [ n ]= 12π ∫−3π 4
−π 4
e jωn
dω+ 1
2π ∫
π
4
3π 4
e jωn
dω
x [ n ]=( 12π )( 1 jn)e jωn| −π 4
−3π 4
+( 12π )( 1 jn )e jωn|3 π
4
π
4
x [ n ]= 1πn [sen(3 πn4 )−sen( πn4 )]
b) X (e jω )=1+3e− jω+2e− j 2ω−4 e− j3ω+e− j10ω
x [ n ]=δ [ n ]+3δ [ n−1 ]+2δ [ n−2 ]−4 δ [ n−3 ]+δ [ n−10 ]
c) X (
e jω
)=e− jω2 !ara
−π ≤ ω ≤ π
x [ n ]= 12π ∫−π
π
X (e jω ) e jωn dω
x [ n ]= 12π ∫−π
π
e
− jω2 e
jωndω
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x [ n ]=( 12π )( 22 jn− j )e(2n−1 )
2 jω| π −π
x [ n ]= (−1 )n+1
π (n−12 )d) X (e jω )=cos2 ω+sen23ω
X (e jω )=1+cos (2ω )2
+1−cos (6ω )
2
X (e jω )=12+1
4e
j2ω+1
4e− j2ω+
1
2−
1
4e6 jω−
1
4e−6 jω
X (e jω )=1+14
e j2ω+
1
4e− j2ω−
1
4e6 jω−
1
4e−6 jω
x [ n ]=δ [ n ]+14
δ [ n−2 ]+14
δ [n+2 ]−14
δ [ n−6 ]−14
δ [n+6 ]
e) X (e jω )= ∑k =−∞
∞
(−1 )k δ (ω−π 2 k )
x [ n ]=∑k =0
3
(−1 )k e jk (π /2 )n
x [ n ]=1−e j
πn
2 +e jπn−e j 3 πn
2
f) X (e jω )=e− jω−
1
5
1−1
5e− jω
X (e jω )= e− jω
1−1
5e− jω
−15
1−1
5e− jω
X (e jω )=e− jω∑n=0
∞
( 15 )n
e− jωn−( 15 )∑n=0
∞
( 15 )n
e− jωn
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X (e jω )=5∑n=1
∞
( 15 )n
e− jωn−( 15 )∑n=0
∞
( 15 )n
e− jωn
x [ n ]=( 15 )n−1
u [n−1 ]−( 15 )n+1
u [ n ]
) X (e jω )=1−
1
3e− jω
1−1
4e− jω−
1
8e−2 jω
X (e jω )=
2
9
1−1
1 e− jω
+
7
9
1+1
4 e− jω
x [ n ]=29 ( 12 )
n
u [ n ]+ 79 (−14 )
n
u [ n ]
) X (e jω )=1−( 13 )
6
e− j6ω
1−1
3e− jω
X (e jω )=1+ 13 e− jω+ 1
32 e− j2ω+ 1
33 e− j3ω+ 1
34 e− j4 ω+ 1
35 e− j5ω
x [ n ]=δ [ n ]+13
δ [ n−1 ]+ 19
δ [ n−2 ]+ 127
δ [ n−3 ]+ 181
δ [ n−4 ]+ 1243
δ [ n−5 ]
5.3". 4onsidere un sistema 'ue consiste en la cascada de dos
sistemas con respuesta en frecuencia
H 1(e jw)= 2−e
− jw
1+1
2
e− jw
H 2 (e
jw)= 1
1−1
2e− jw+
1
4e− j2w
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a) E#c'#-$ *a c'ac".# , ,"/$#c"a ' ,c$"b a* "-a c%+*-%(
H ( e jw )=
2−e− jw
1+1
2e− jw
∗1
1−1
2e− jw+
1
4e− j 2w
H (e jw)= (2−e− jw )
(1+ 12 e− jw)(1−1
2e− jw+
1
4e− j2w)
H ( e jw )= (2−e− jw )
1+1
8e−3 jw
H ( ' )=(2−e− jw )
1+1
8 '−3
( ( )
X ( )=
(2−e− jw)
1+ 18
−3
( ( )+1
8( ( ) −3=2 X ( )− X ( ) −1
y [n ]+ 18
y [ n−3 ]=2 x [ n ]− x [ n−1 ]
b) D-$"# *a $+'-a a* "+'*% ,* "-a c%+*-%(
H (e jw )= (2−e− jw )
(1+ 12 e− jw)(1−12 e− jw+ 14 e− j2w)
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H ( ' )= (2− '−1 )
(1+
1
2 '−1
)(1−
1
2 '−1
+ 1
4 '−2
) H ( )=
A
(1+ 12 −1)+
B
1+1
4(1+ j√ 3)
−1+
)
1+1
4(1− j√ 3)
−1
A=4
3
B=1+ j √ 3
3
) =1− j√ 3
3
H ( )=
4
3
(1+ 12 −1)+
1+ j√ 33
1+1
4(1+ j√ 3)
−1+
1− j √ 33
1+1
4(1− j√ 3)
−1
h [ n ]=43 ( 12 )
n
u [ n ]+1+ j√ 33 [ 14 (1+ j√ 3 )]
n
u [ n ]+1− j √ 33 [14 (1− j √ 3 )]
n
u [n]
DANIEL CASTILLOQUINTO ELECTRÓNICA