7-FIR

advertisement
Ideal Filters
One of the reasons why we design a filter is to remove disturbances
SIGNAL
s(n)

x(n)
Filter
y(n)  s(n)
v(n)
NOISE
We discriminate between signal and noise in terms of the frequency spectrum
Y (F )
S (F )
V (F )
 F0
F0
F
F0
 F0
F0
F
Conditions for Non-Distortion
Problem: ideally we do not want the filter to distort the signal we want to recover.
x(t )  s(t )
y(t )  As(t  T )
IDEAL
FILTER
2
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
Same shape as s(t),
just scaled and
delayed.
0
200
400
600
800
-2
1000
0
200
400
600
800
1000
Consequence on the Frequency Response:
constant
| H (F ) |
F
linear
 H (F )
F
 Ae j 2FT
H (F )  
0
if F is in the passband
otherwise
For real time implementation we also want the filter to be causal, ie.
h( n)



  






n
h(n)  0 for n  0
since

y (n)   h(k )
x
(n

k)

k 0
past values only
FACT (Bad News!): by the Paley-Wiener Theorem, if h(n) is causal and with finite energy,

 ln H ( ) d   

ie H ( )
cannot be zero on an interval, therefore it
1  2
log H ( )  log( 0)  
cannot be ideal.

2
 log H ( ) d  
1
Characteristics of Non Ideal Digital Filters
| H ( ) |
p
IDEAL

Positive freq. only
NON IDEAL
Two Classes of Digital Filters:
a) Finite Impulse Response (FIR), non recursive, of the form
y(n)  h(0) x(n)  h(1) x(n  1)  ...  h( N ) x(n  N )
With N being the order of the filter.
Advantages: always stable, the phase can be made exactly linear, we can approximate any
filter we want;
Disadvantages: we need a lot of coefficients (N large) for good performance;
b) Infinite Impulse Response (IIR), recursive, of the form
y(n)  a1 y(n 1)  ... aN y(n  N )  b0 x(n)  b1x(n 1)  ... bN x(n  N )
Advantages: very selective with a few coefficients;
Disadvantages: non necessarily stable, non linear phase.
Finite Impulse Response (FIR) Filters
Definition: a filter whose impulse response has finite duration.
x ( n)
y ( n)
h( n)
h( n)

       
h(n)  0






       
n
h(n)  0
Problem: Given a desired Frequency Response Hd ( ) of the filter, determine the impulse
response h(n) .
Recall: we relate the Frequency Response and the Impulse Response by the DTFT:
H d ()  DTFThd (n) 

 h (n)e
n  
1
hd (n)  IDTFTH d ( ) 
2
 j n
d

j n
H
(

)
e
d
 d

Example: Ideal Low Pass Filter
1
hd (n) 
2
 c

Ae j n d   A
c
sin c n 


 A c sinc  c
n

 
hd (n)
Hd ( )
DTFT
A
   c
 c

n


 c

n

4
Notice two facts:
• the filter is not causal, i.e. the impulse response h(n) is non zero for n<0;
• the impulse response has infinite duration.
This is not just a coincidence. In general the following can be shown:
If a filter is causal then
• the frequency response cannot be zero on an interval;
h(n)  0

h( n)





H( )
H( )  0
• magnitude and phase are not independent, i.e. they cannot be specified arbitrarily
As a consequence: an ideal filter cannot be causal.
Problem: we want to determine a causal Finite Impulse Response (FIR) approximation of the
ideal filter.
hw (n)
rectangular window
We do this by
2
0.25
1.8
0.2
a) Windowing
1.6
0.15
1.4

1.2
1
hd (n)
0.8
0.25
0.6
0.1
0.05
0
0.4
0.2
-0.05
0.2
0.15

0.1
0.05
0
-100
-50
L
0
50
-0.1
-100
100
L
-0.05
-50
0
50
100
1.8
100
50
100
L
0.2
1.6
infinite impulse response
(ideal)
50
0.25
2
-0.1
-100
L
0
hw (n)
hamming window
0
-50

1.4
1.2
1
0.8
0.15
0.1
0.05
0.6
0
0.4
-0.05
0.2
0
-100
-50
L
0
50
L
100
-0.1
-100
-50
L
0
L
finite impulse response
b) Shifting in time, to make it causal:
h(n)  hw (n  L)
hw (n)
0.25
0.25
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0
-0.05
-0.05
-0.1
-100
-50
0
50
100
-0.1
-100
-50
0
50
100
Effects of windowing and shifting on the frequency response of the filter:
a) Windowing: since
hw (n)  hd (n)w(n)
| W ( ) |
then H w ( ) 
1
H d ( ) *W ( )
2
| H w ( ) |
rectangular window
40
20
20
0
0

-20
H d ( )
*
-40
-60

-100
-120
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
hamming window
40
20
20
0

0
*
-60
-80
-120
c
-40
-80
-100
 c
-20
-20
-40
-60
-40
-60
-80
-80
-100
-100
-120
-20
0
0.5
1
1.5
2
2.5
3
-120
For different windows we have different values of the transition region and the attenuation in the
stopband:

attenuatio n
20
Rectangular
4 / N
-13dB
Bartlett
8 / N
0
-27dB
-20
Hanning
8 / N
-32dB
-40
Hamming
8 / N
-43dB
-60
Blackman
16 / N
-58dB
-80
attenuation
-100
hw (n)
-120
N  2L  1
with
0
0.5
1
0.25

transition
region
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-100
-50
L
0
50
L
100
n
1.5
2
2.5
3
Effect of windowing and shifting on the frequency response:
b) shifting: since
Therefore
h(n)  hw (n  L)
then
H ( )  H w ( )e jL
H ( )  H w ( ) no effect on the magnitude,
 H( )   H w ( )  L shift in phase.
See what is  H w ( ).
For a Low Pass Filter we can verify the symmetry
H w ( ) 
real for all
.

h
n  
hw (n)  hw (n).
Then

 j n
(
n
)
e
 hw (0)  2 hw (n) cos( n)
w
n 1
Then
in thepassband;
0
 H w ( )  
don' t care, otherwise
The phase of FIR low pass filter:
 H ( )  L in thepassband;
Which shows that it is a Linear Phase Filter.
20
0
-20
H ( )
-40
-60
dB
-80
-100
-120
0
0.5
1
1.5
2
2.5
3

 H ( )
degrees
don’t care
Example of Design of an FIR filter using Windows:
Specs:
Pass Band 0 - 4 kHz
Stop Band
> 5kHz with attenuation of at least 40dB
Sampling Frequency 20kHz
 40dB
Step 1: translate specifications into digital frequency
Pass Band
Stop Band
0  2 4 / 20  2 / 5 rad
2 5 / 20   / 2   rad
Step 2: from pass band, determine ideal filter impulse
response
c
 c  2
 2n 
hd (n) 
sinc
n  sinc 

 5

  5
4 5
2 
5 2
 

10
10

F kHz

Step 3: from desired attenuation choose the window. In this case we can choose the hamming
window;
Step 4: from the transition region choose the length N of the impulse response. Choose an odd
number N such that:
8 

N 10
So choose N=81 which yields the shift L=40.
Finally the impulse response of the filter
2
 2(n - 40)  
 2 n  
  0.54  0.46 cos
  , if 0  n  80,
 sinc

 80  
5 
h( n)   5

0
otherwise

The Frequency Response of the Filter:
H( )
dB

 H( )
rad

A Parametrized Window: the Kaiser Window
The Kaiser window has two parameters:
N
Window Length

To control attenuation in the Stop Band
1.5
w[n]
 0
1
 1
 5
0.5
  10
0
0
20
40
60
80
100
120
n
There are some empirical formulas:
A
Attenuation in dB
Transition Region in rad


 

N


 
Example:

Sampling Freq. 20 kHz
Pass Band 4 kHz
2
P 
,
5
S 
Stop Band 5kHz, with 40dB Attenuation
   S   P 
A  40dB

10
rad

N  45
  3.3953

2
Then we determine the Kaiser window
w  kaiser( N ,  )
w[n]
n
Then the impulse response of the FIR filter becomes
sin  c (n  L) 
h[n] 
w[n]

(n  L)


ideal impulse
response
with
9
 c   P   S  
20
1
2
N  45  2 L  1  L  22
h[n]
Impulse Response
n
| H ( ) |dB
Frequency Response
 (rad)
Example: design a digital filter which approximates a differentiator.
Specifications:
• Desired Frequency Response:
 j 2 F if  4kHz  F  4kHz
H d (F )  
if
F  5kHz
0
• Sampling Frequency Fs  20kHz
• Attenuation in the stopband at least 50dB.
Solution.
Step 1. Convert to digital frequency
H d ( )  H d ( F ) F F / 2
s
2
2

 jFs  j 20,000 if - 5    5


0
if
|  | 

2
Step 2: determine ideal impulse response
hd (n)  IDTFTH d ( ) 
1
2
From integration tables or integrating by parts we obtain
2
5


2
5
j 20,000 e j n d
ax
e
1

ax
xe
dx

x




a
a
Therefore


 2n 
 2n  
cos
sin






4

5
5

 2 
  if n  0


20,000

hd (n)  
5
n
n2







if n  0
0
Step 3. From the given attenuation we use the Blackman window. This window has a transition
region region of 12 / N . From the given transition region we solve for the complexity N as
follows
 
which yields

2

2
12
 0.1 
5
N
N  120. Choose it odd as, for example, N=121, ie. L=60.
Step 4. Finally the result

 2 (n  60) 
 2 (n  60)  
cos
sin 



4

5
5

 2 
   0.42  0.5 cos 2n   0.08cos 4n  
h(n)  20,000




 5
 
n  60
(n  60) 2
120
120








for 0  n  120
Frequency Response
H ( )
2
x 10
5
1.8
1.6
Impulse response h(n)
3
x 10
1.4
1.2
4
1
0.8
2
0.6
0.4
1
0.2
0
0
0
0.5
1
1.5
2
2.5
3
3
3.5
-1

-2
-3
0
20
40
60
80
100
120
140
H ( ) dB
150
100
50
0
-50
-100
-150
-200
-250
0
0.5
1
1.5
2
2.5

Download
Related flashcards
Create Flashcards