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ANLISIS DE SEALES Y SISTEMAS

DEBER 1

Nombre: Fecha:06-Mayo-2015

Paralelo:GR4

2) Dada x [n] , grafcar:

a) x1 [ n ]=1

2x [n3]

x [n+3]


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x [n3]

1

2x [n3]

b)

x2

[ n ]=k=

n

x [k]

x2=(n+1 )+3 (n )+2(n1 )+4(n2 )(n3)3(n4)


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4) Dada x (t) , encontrar y grafcar:

a) y (t)=x (t)+x (t+1)

x (t)


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X(t+1)

x (t)+x (t+1)

b) y (t)=x ( t) (t1 )x(t+1)(t)

x (t1)


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x (t+1)

x (t+1)(t)

x (t)(t)


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x (t) (t1 )x (t+1)( t)

6) Indique 2 caractertica de lo iguiente itema:

a) y (t)=x (t5)

y (t)=x (t5)=a1x

1(t5 )+a

2x

2(t5)

y (t)=a1 [H{x1(t5 )}]+ a2[H{x2(t5)}]

Entonc! ! "#na" #n$a%#ant n " t#&'o

b) y [n ]=x [ n1 ]x [n+1]


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y (t)=x [ n1 ]x [n+1]=a1x

1[ n1 ]x

1[n+1]+a

2x

2[ n1 ]x

2[n+1]

y (t)=a1 [H{x1[ n1 ]x1[n+1]} ]+a2[H{x2 [ n1 ]x2[n+1]}]

E! "#na" #n$a%#ant n " t#&'o

!) "i e conoce que la re#ueta de un itema lineal e in$ariante con

el tiem#o a la e%al dicreta x [ n ]= [n] e la e%al y [n ]=(n1 )[n] &

'btener la re#ueta del itema a la e%alx

1[n ]

X1=[ n ]+ [ n1 ]+2[n2 ]+2[ n3 ]

[ n ] + [ n1 ]+ [ n2 ] + [n3]

[n ] [ n1 ] + [ n1 ] [ n2 ]+ [ n2 ] [n3 ]+ [ n3 ] [n4 ]

[n ] [ n4 ]

y1

[ n ]=H{X1 [ n ]}

H{ [n ] [n4 ]}

H [ n ]H [ n4 ]


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y [n ]=H{ [ n ]}=(n1 )[n]

R&'"a(ano*

y1

[ n ]=y [ n ]y [n4]

y1

[ n ]=(n1 ) [ n ](n41) [n1 ]

y1

[ n ]=(n1 ) [ n ](n5) [ n1 ]

() *n itema lineal e in$ariante con el tiem#o re#onde de la +orma

y (y )=H{t (t)}=[52 (t+2 ) et

2

](t)ncuentre el modelo a ecuaci-n di+erencial del itema y en bae a

ete determine la re#ueta del itema a la e%al x (t)=(t)

yh=( t+2 ) et2

(+

1

2 )

2

=2

++

1

4

y + y +1

4y =C

1x+C

2x +C

3x

x (t)=t (t)

x (t)= (t)+t ( t)= ( t)

x (t)=(t)

y (t)=g (t)=d

dtr (t)=

d

dtH{t (t)}


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g ( t)+ g (t)+ 14

g ( t)=C1

(t)+C2 (t)+C

3t (t)

g ( t)=d

dt{[52 (t+2) et2 ]}=(t)+t e

t2 (t)

g ( t)= (t)+(112 t)et2 (t)

g ( t)= (t)(114 t)et2 (t)

(t)(114 t)et2 (t)+(t)+(112 t)e

t2 (t)+ 1

4[(t)+t e

t2 (t)]=C

1(t)+C

2 (t)+C

3t (t)

C

(2+C3 t) (t)

(t)+(t)+ 14

(t)=C1

(t)+

C1=

1

4

C2=0

C3=0

y + y +1

4 y =1

4 x +(0)x +(0)x

y + y +1

4y=

1

4x


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si :x (t)=(t)

x (t)=(t)

entonces :y + y + 14

y=14

[ 14 14 + 316 ] +[ 14 14 ] + 14 [ 14 ] , 14 +0

0

y

+0

0

y

K1=1

4, K

2=1

4

y (t)=h (t)=( 14 14 t)et2 (t)+ (t)

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