Cap 5 Oppenheim

8/16/2019 Cap 5 Oppenheim

1/14

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 ¿

DANIEL CASTILLOQUINTO ELECTRÓNICA


2/14

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


3/14

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



4/14

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


5/14

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 ]



7/14

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]



8/14

+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



9/14

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|


10/14

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



11/14

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



12/14

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



13/14

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)



14/14

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]


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