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Ideal Filters
One of the reasons why we design a filter is to remove disturbances
)(ns
)(nv
)(nx )()( nsny Filter
SIGNAL
NOISE
We discriminate between signal and noise in terms of the frequency spectrum
F
)(FS
)(FV
0F0F 0F F
)(FY
0F0F
Conditions for Non-Distortion
Problem: ideally we do not want the filter to distort the signal we want to recover.
IDEAL
FILTER
)()( tstx )()( TtAsty Same shape as s(t), just scaled and delayed.
0 200 400 600 800 1000-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0 200 400 600 800 1000-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Consequence on the Frequency Response:
otherwise
passbandtheinisFifAeFH
FTj
0)(
2
F
F
|)(| FH
)(FH
constant
linear
For real time implementation we also want the filter to be causal, ie.
0for 0)( nnh
h n( )
n
since
0
)()()(k
knxkhny only spast value
FACT (Bad News!): by the Paley-Wiener Theorem, if h(n) is causal and with finite energy,
dH )(ln
ie cannot be zero on an interval, therefore it cannot be ideal. )(H
dHH )(log)0log()(log1 2 1
2
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
)()(...)1()1()()0()( NnxNhnxhnxhny
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
)(...)1()()(...)1()( 101 NnxbnxbnxbNnyanyany NN
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.
h n( )
n
h n( )x n( ) y n( )
h n( ) 0h n( ) 0
Problem: Given a desired Frequency Response of the filter, determine the impulse response .
Hd ( )h n( )
Recall: we relate the Frequency Response and the Impulse Response by the DTFT:
n
njddd enhnhDTFTH )()()(
deHHIDTFTnh njddd )(
2
1)()(
Example: Ideal Low Pass Filter
c c
Hd ( )
A
csin1
( ) sinc2
c
c
cj n cd
nh n Ae d A A n
n
)(nhd
n
c4
DTFT
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;
• magnitude and phase are not independent, i.e. they cannot be specified arbitrarily
h n( )
h n( ) 0
H( )H( ) 0
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.
We do this by
a) Windowing
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
-100 -50 0 50 1000
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
-100 -50 0 50 1000
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
)(nhd
)(nhw
rectangular window
hamming window )(nhw
infinite impulse response(ideal)
finite impulse response
L L
L L
L L
L L
b) Shifting in time, to make it causal:
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
)(nhw)()( Lnhnh w
Effects of windowing and shifting on the frequency response of the filter:
a) Windowing: since then )()()( nwnhnh dw )(*)(2
1)(
WHH dw
cc
)(dH
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
40
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
40
*
*
|)(| W |)(| wHrectangular window
hamming window
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
attenuation
For different windows we have different values of the transition region and the attenuation in the stopband:
transition region
Rectangular -13dB
Bartlett -27dB
Hanning -32dB
Hamming -43dB
Blackman -58dB
N/4N/8
N/8
N/8
16 / N
nattenuatio
-100 -50 0 50 100-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
L L
12 LN
)(nhw
n
with
Effect of windowing and shifting on the frequency response:
b) shifting: since then )()( Lnhnh w Lj
w eHH )()(
Therefore
phase.in shift )(H)H(
magnitude, on theeffect no )()(
w L
HH w
See what is ).(wH
For a Low Pass Filter we can verify the symmetry Then ).()( nhnh ww
)cos()(2)0()()(1
nnhhenhHn
wwnj
nww
real for all . Then
otherwise ,'
passband; in the 0)(
caretdonHw
The phase of FIR low pass filter:
passband; in the )( LH Which shows that it is a Linear Phase Filter.
0 0.5 1 1.5 2 2.5 3-120
-100
-80
-60
-40
-20
0
20
don’t care
)(H
dB
)(H
degrees
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
Step 1: translate specifications into digital frequency
Pass Band
Stop Band 2 5 20 2 / / rad
0 2 4 20 2 5 / / rad
40dB
F kHz54 10
2
2
5
10Step 2: from pass band, determine ideal filter impulse
response
h n ndc c( )
sinc sinc2n
5
2
5
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
10
N
So choose N=81 which yields the shift L=40.
Finally the impulse response of the filter
h n
nn
( ). . cos , ,
2
50 54 0 46
2
800 80sinc
2(n - 40)
5 if
0 otherwise
The Frequency Response of the Filter:
H( )
H( )
dB
rad
A Parametrized Window: the Kaiser Window
The Kaiser window has two parameters:
N
Window Length
To control attenuation in the Stop Band
0 20 40 60 80 100 1200
0.5
1
1.5
n
][nw 0
1
10
5
There are some empirical formulas:
A
Attenuation in dB
Transition Region in rad
N
Example:
Sampling Freq. 20 kHz
Pass Band 4 kHz
Stop Band 5kHz, with 40dB Attenuation
,5
2 P 2
S
dBA
radPS
4010
3953.3
45
N
Then we determine the Kaiser window
),( Nkaiserw
][nw
n
Then the impulse response of the FIR filter becomes
][
)(
)(sin][ nw
Ln
Lnnh c
ideal impulse response
with 20
921 SPc
221245 LLN
][nh
n
dBH |)(|
(rad)
Impulse Response
Frequency Response
Example: design a digital filter which approximates a differentiator.
Specifications:
• Desired Frequency Response:
kHzF
kHzFkHzFjFH d 5 if 0
44 if 2)(
• Sampling Frequency
• Attenuation in the stopband at least 50dB.
kHzFs 20
Solution.
Step 1. Convert to digital frequency
||2
if 0
5
2
5
2- if 000,20
)()(2/
jFjFHH
s
FFdds
Step 2: determine ideal impulse response
5
2
5
2
000,202
1)()(
dejHIDTFTnh njdd
From integration tables or integrating by parts we obtain
a
xa
edxxe
axax 1
Therefore
0 if 0
0 if 5
2sin
25
2cos
5
4000,20
)( 2
n
nn
n
n
n
nhd
Step 3. From the given attenuation we use the Blackman window. This window has a transition region region of . From the given transition region we solve for the complexity N as follows
N/12
N
121.0
5
2
2
which yields . Choose it odd as, for example, N=121, ie. L=60. 120N
Step 4. Finally the result
120
4cos08.0
120
2cos5.042.0
)60(5
)60(2sin
2605
)60(2cos
5
4000,20)(
2
nn
n
n
n
n
nh
1200for n
0 20 40 60 80 100 120 140-3
-2
-1
0
1
2
3x 10
4
0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
5
0 0.5 1 1.5 2 2.5 3 3.5-250
-200
-150
-100
-50
0
50
100
150
Impulse response h(n)
)(H
dBH )(
Frequency Response