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7/23/2019 Filt Ros fir matlab
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2.1) Pasa-Bajas
( ) ( ) n H z h n Z ∞
−
−∞
= ∑
2
2
2 2
2 2
( ) ( )
1( ) ( )
1,( )
0,
1 1( ) (1)
2
1 1(2 )
2
(22
jwT j fnT
n
j fnT
jwT
fp
j fnT j fnT
fp
j fpn j fpn
F F
H e h n e
h n h n e F
fp fp H e
otra
fph n e f e
fp F Fj nT
fpe e sen n
j n n F
sennfpT
n
parte
π
π
π π
π π
π
π
π π
π
π
∞−
=−∞∞
−
−∞
−
−
=
=
− ≤ =
= ∂ = −
= − =
=
∑
∫
∫
)2
(2 )
2
fpn
fp fp F Sa n fp F F
n F
π
π
π
=
f
F-
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b) Pasa-bandas
F=20khzfp=3khzFc1=Fc2=
1 2
2 2 2
2 1
1( ) (1) (1) )
fp fp
j fnT j fnT j fnT
fp fp
h n e e e f
F
π π π
−∞
−∞ −
= ∂ = + ∂
∫ ∫ ∫ 2 1 2 2 2 2 2 1
2 2 2
21
1 1( )
2 2
j nfp j nfp j fp n j nfp fp j fnT j fnT fp F F F F
fp fp
h n e e e e e e Fj nT j n
π π π π
π π
π π
− −−
−
= + = − + −
2 1 2 2 2 1 2 1
1 1 2 2 (2 1)
2 2
j fp j fp j fp j fp
F F F F e e e e nFc sen nFc sen
n j j n F F
π π π π
π π
π π
− − − − = − = − ÷
[ ]2
(0) 2 1h Fc Fc F
= −
F/2Fp fc2FpFc1Fs1---Fp2 - 0
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a Pasa-altas
F=20khzfp=3khzFc1=Fc2=
( ) ( ) n H z h n Z ∞
−
−∞
= ∑
22 2 2
2
1 1 1( ) ( ) (1) ( ) (1)
F fp
jwT jwT j fnT jwT j fnT j fnT
F fp
H e h e e f h e e f e f F F F
π π π
−∞ ∞
−∞ −∞−
= ∂ = ∂ + ∂∫ ∫ ∫ ∫
2 2 2 22
2
1 1..
2 2
fp F
j fnT j fnT j fnT j fnT
F fp
e e e e Fj nT j n
π π π π
π π
−− −
−−
= + = + +
2
F
2
F −
FpFs-Fp -Fs 0
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2 2
2 2
2 2 2 11 1
..2 2
j fnT j fnT j fp n j nfp
j n j n F F e e e e e e j n j n
π π π π
π π
π π
−−
+ − = + − +
212 cos 2 (2
22
fpn fp F n jsen n fp j n F
n F
π
π π
π π
= + ÷
221
2
nFc sen Fc F nFc F
F
π
π
÷= − ÷
÷
:
2
2
2 2
Donde
fs Fc fp
Fc fp fs fp
Fc fs fp
Fc fp
Fc fp
= −= − −− =
== −
