logosite.iugaza.edu.ps/hghurab/files/2017/09/ch7part2-.pdf · using linear convolution yn( ) ......
TRANSCRIPT
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Use the DFT in linear filter
Suppose thar we have a finite-duration sequence x(n) of length L ,
which excites an FIR filter ,with impulse response h(n) of length M
The output sequence y(n) of the FIR filter can be expressed in time
domain as the linear convolution of x(n) and h(n).
If size of x(n)=
If size of h(n)=
( ) ( ) ( )
To avoid aliasing , we need to take DFT of N points
where: N L+M 1
L
M
Y k X k H k
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2
2
( ) ( ) |
( ) ( ) |
k
N
k
N
Y k Y k = 0,1,.......,N 1
X H k = 0,1,.......,N 1
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Zero padding
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• Zero padding of analyzed sequence results in “approximating”
its DTFT better
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Determine the response of
FIR filter with impulse response: h(n)={ ,2,3}
to the input x(n)={ ,2,
1
1 2,1}
Ex:
(a) Using linear convolution(chapter 2)
] ;
] ;
1221
0321
0000
03663
002442
0001221
038119411
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h(n)={ ,2,3} , 1 1x(n)={ ,2,2,1}
(b) Using DFT&IDFT (without zero padding )
2 2
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2 3
4 4 4
4 2 4 6
4 4 4
3 6 9
4 4 4
1 1 1 1 1 1 1 1
1 1
1
1
j jN
NW e W e j
W W WW
W W W
W W W
2 3
2 4 6
3 6 9
1 1 1 1
( ) ( ) 1 1
1 11 ( ) ( ) ( )
1 ( ) ( ) ( )
j j j j j
j j j
j j j
1 1
1 1
1 1
1 1 1 1
1 1
1 1 1 1
1 1
N N N
j j
j jX W x
j
1
1 6
2 2 2( )
3 2
0 2 2
jX k
j j
12
2 2 2
1 1 1 1 1 6
1 1 2 1 ( )
1 1 1 1 2 0
1 1 1 1
N N N
j j jX W x X k
j j j
1 2
1 2
( ) ( ) ( )
( 0) (0) (0) (6)(6) 36
( 1) ( 1 )( 2 2 ) 4
( 2) (0)(2) 0
( 1) ( 1 )( 2 2 ) 4
( ) {36,4 ,0, 4 }
Y k X k X k
Y k X X
Y k j j j
Y k
Y k j j j
Y k j j
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1
1
4 4 ,4
1 1 1 1 36
1 1 4
1 1 1 1 0
1 1 4
m
j j jx W X
j j j
1/ 4 1/ 4 1/ 4 1/ 4
1/ 4 / 4 1/ 4 / 4 =
1/ 4 1/ 4 1/ 4 1/ 4
1/ 4 / 4 1/ 4 / 4
j j
j j
3 1 2
36 9
4 7
0 9
4 11
( ) ( ) ( ) {9,7,9,11}
j
j
x m x n x n
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h(n)={ ,2,3} , 1 1x(n)={ ,2,2,1}
(c) Using circular convolution formula
1 * 1 2 2 1 1 2 2 1
2 * 1 1 2 2 2 2 4 4
3 * 2 1 1 2 6 3 3 6
0 * 2 2 1 1 0 0 0 0
9 7 9 11
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1* 1 2 2 1 0 0 1 2 2 1 0 0
2* 0 1 2 2 1 0 0 2 4 4 2 0
3* 0 0 1 2 2 1 0 0 3 6 6 3
0* 1 0 0 1 2 2 0 0 0 0 0 0
0* 2 1 0 0 1 2 0 0 0 0 0 0
0* 2 2 1 0 0 1 0 0 0 0 0 0
1 4 9 11 8 3
(d) Using circular convolution (with zero padding)
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L=3,M=4 N=3+4 1 6 N at least must be 6 take N=8
h(n)={ ,2,3,0,0,0} , x(n)={ ,2,2,11 1 ,0,0}
(e) Using DFT&IDFT (with zero padding)
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0
3
4 2 4( ) 1 2 2 , k=0,1,2,...
: ( ) ( ) , k=0,1,2,.......,7
6 , 1 1.707 4.21
2 1 , 3 0.29 0.1
...
2132
4
.,7k k k
j j
nj
n
j
k
DFT X k x n e
X(k = 0) = X(k = ) = j
X(k = ) = j X
X k e e
(k = ) =
X )
e
j
(k =
0 , 5 0.29 0.12132
6 1 , 7 1.707 4.21
= X(k = ) = j
X(k = ) = j X(k = ) = j
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4
0
2
: ( ) ( ) , k=0,1,2,.......,7
6 , 1 2.414 4.414
2 2 2 , 3 0.4
( ) 1 2 3 , k=0,1
14 1.585
,2,......
8
4
.
2
,7
nkj
n
k kj j
DFT H k h n e
H(k = 0) = H(k = ) = j
H(k = ) = j H(k = ) = j
X k e e
H(k = ) =
, 5 0.414 1.5858
6 2 2 , 7 2.414 4.414
H(k = ) = j
H(k = ) = j H(k = ) = j
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( ) ( ) ( )
( 0) (0) (0) 36 , (1) 14.07 17.48
(2) 4 , (3) 0.07 0.515
(4) 0 , (5) 0.07 0.515
(6) 4
Y k X k H k
Y k X H Y j
Y j Y j
Y Y j
Y j
, (7) 14.07 17.48 Y j
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8
0
1: y(n)= ( ) , n=0,1,2,.......,7
8
( ) {1, 4,9,11,8,3,0,0}
nkj
k
IDFT X k e
y n
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Using linear convolution
( ) { ,4,9,11, }1 8,3y n
Using DFT&IDFT or circular Con.(without zero padding)
( ) { ,7,9 11}9 ,y n
Using DFT&IDFT or circular Con.(with zero padding)
( ) { ,4,9,11, }1 8,3y n
(0) : (0) (4) 1 8 9
(1) : (1) (5) 4 3 7
(2) : (2) (6) 9 0 9
(3) : (3) (7) 11 0 11
index y y
index y y
index y y
index y y
( ) { ,7,9 11}9 ,y n
( ) { ,4,9,11,8,3,1 0,0}y n
Aliasing occurs when the size of DFTs is smaller than L+M-1
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DFT repeats itself every N points (Period = N) but we usually display it for n = 0 ,…, N-1
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