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Filtering Problem
Goal: design a filter to attenuate the disturbances
SIGNAL
s[n]

x[n]
v[n]
NOISE
y[n]  s[n]
Filter
Define Signal and Noise in the Frequency Domain
n
n
n
v[n] NOISE
SIGNAL
s[n]

y[n]  s[n]
Filter
x[n]
SIGNAL
| S ( ) |

NOISE
| V ( ) |
 P
Mostly NOISE
0
P
Mostly
SIGNAL
  (rad )
Mostly NOISE
IDEAL Filter
Since the filter has real coefficients, we need only the
positive frequencies
H ( )
PASS
Band
 p STOP 
Band

Non-IDEAL Filter
Since the filter has real coefficients, we need only the
positive frequencies
| H () |
PASS
Band
p
S
Trans.
Band
STOP
Band
Design a Low Pass Filter: IDEAL
Hd ( )
1
   P  P


First we can determine an infinite length expansion using
the DTFT:
H d ()  DTFThd [n] 

 j n
h
[
n
]
e
d
n  
1
hd [n]  IDTFTH d () 
2
1
hd [n] 
2
P
j n
e
 d 
 P

j n
H
(

)
e
d
 d

sin  P n   P


sinc  P
n

 

n

Design a Low Pass Filter: IDEAL (continued)
hd [n] 
This means the following. If
then
H d () 

 j n
h
[
n
]
e
d
n  

Hd ( )
P
 
sinc P n 

  
1
   P  P

0.3
hd [n]
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-60
-40
-20
0
20
40
60
n

From IDEAL to FIR
Notice that
lim hd [n]  0
n  
Then we can approximate with a finite sum…
H d () 
L
 j n
h
[
n
]
e
d
n L
… and choose the filter as h[n]  hd [n  L]
2L
H ()   h[n]e
 jn
n 0
actual
2L
  hd [n  L]e
n 0
 jn
 L
 jn   jL
   hd [n]e e  H d ()e jL
 n L

desired
Summary of design of Low Pass FIR
Given the Pass Band Frequency 0  P  
The impulse response of the filter: let N  2 L
P
P

h[n] 
sinc
(n  L) , n  0,...,N

 

2L
H ()   h[n]e jn  H d ( )e jL
n 0
Magnitude and Phase:
Magnitude:
Phase:
H ( )  H d ( )
 H ( )   H d ( )  L
 L in the passband
Example of Freq. Response
P   / 4
h=(wp/pi)*sinc((wp/pi)*(n-L));
freqz(h)
Magnitude (dB)
50
 20dB
0
-50
-100
0
0.1
0
Phase (degrees)
n=0:N;
L  20, N  40
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0.9
1
 20
-500
-1000
-1500
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
Impulse Response
n=0:N;
stem(n,h)
h[n]
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
0
5
10
15
20
25
30
35
40
Example with Hamming window
1
1

 2n  
h[n]  sinc  (n  20)  0.54  0.46 cos
 , n  0,...,40
4
40  
4

 



hamming window
h[n]
n=0:N;
h0=(wp/pi)*sinc((wp/pi)*(n-L));
h=h0.*hamming(N);
stem(n,h)
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
0
5
10
15
20
25
30
35
40
Example with Hamming window
1
1

 2n  
h[n]  sinc  (n  20)  0.54  0.46 cos
 , n  0,...,40
4
40  
4

 



hamming window
Magnitude (dB)
50
0
~ 50dB
-50
-100
-150
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
Phase (degrees)
freqz(h)
H ( )
-500
-1000
-1500
Low Pass Filter Design: Analytical
Transition Region: depends on the window and the filter length N
Attenuation: depends on window only
20
window

Rectangular
4 / N
Hamming
8 / N
Blackman
12 / N
attenuatio n
0
-20
attenuation
-13dB
-40
-43dB
-60
-58dB
-80
-100
-120
0
0.5
1
1.5
2
2.5
 P S
  transition region
3
Example of Low Pass Filter Design
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
02 4 / 20  2 / 5rad
Stop Band
2 5 / 20   / 2 rad
Step 2: from pass band, determine ideal filter impulse
response
hd (n) 
P
  2
 2n 
sinc P n  sinc 

   5
 5 
4 5
2 
5 2
 

10
10

F kHz

Example of Low Pass Filter Design (continued)
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 even number N such that:
8 

N 10
So choose N=80 which yields the shift L=40.
Example of Low Pass Filter Design (continued)
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

Example of Low Pass Filter Design (continued)
The Frequency Response of the Filter:
H( )
dB

 H( )
rad

Design Parameters
Pass Band Frequency
Stop Band Frequency
Pass Band Ripple
Stop Band Attenuation
 p rad.
S rad.
20log10  A/ B  dB
20log10  A/ C  dB
H ()
A
B
C
p
S


Computer Aided Design of FIR Filters
Best Design tool for FIR Filters: the Equiripple algorithm. It
minimizes the maximum error between the frequency
responses of the ideal and actual filter.
Step 1: define the desired filter with pairs of frequencies and
values. For a Low Pass Filter:
f=[0, f1, f2, 1];
A=[1, 1,
0,
0];
ωPASS
π
ω
f 2  STOP
π
f 1
where
Linear Interpolation
1
0
PASS
STOP
 (rad)
Computer Aided Design of FIR Filters
Step 2: Choose filter length N from desired attenuation as
Attenuation (dB)
N~
11 STOP  PASS / 
Step 3: call “firpm” (pm = Parks, McClellan)
h = firpm(N, f, A);
which yields the desired impulse response
h  h[0], h[1]  h[ N ]
Example: Low Pass Filter
Passband: 3kHz
Stopband: 3.5kHz
Attenuation: 60dB
Sampling Freq: 15 kHz
Then compute:
3,000 2

15,000 5
3,500 7
STOP  2

15,000 15
60
N
 82
7
2
11 15  5 
 PASS  2
h = firpm(82,[0,2/5,7/15,1], [1,1,0,0]);
Frequency Response
freqz(h)
Magnitude (dB)
50
0
50dB, not quite
yet!
-50
Increase N.
-100
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
Phase (degrees)
0
-1000
-2000
-3000
-4000
Increase Filter Length N
h=firpm(95, [0, 2/5, 7/15, 1], [1,1,0,0]);
freqz(h)
Magnitude (dB)
50
0
-50
-100
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
Phase (degrees)
0
-1000
-2000
-3000
-4000
Impulse Response of Example
stem(h)
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
0
10
20
30
40
50
60
70
80
90
100
Typical Applications
In applications such as Radar, Sonar or Digital Communications we transmit a
Pulse or a sequence of pulses.
For example, we transmit a short sinusoidal burst:
x(t )  A sin(2F0t ), 0  t  T1
2
F0  500Hz
1.5
1
T1  8 m sec
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
sec
5
6
7
8
-3
x 10
Typical Applications
For example in a communications signal we send “0” and “1”.
1  x(t )
0   x(t )
For example let:
4
Transmitted:
3
2
1
0
-1
-2
-3
-4
0
1
0
0.005
0.01
1
0.015
millisec
0.02
0
0.025
0.03
Received Signal with Noise
Suppose you receive the signal with 0dB SNR:
8
6
4
2
0
-2
-4
-6
-8
0
0.005
0.01
0.015
millisec
0.02
0.025
0.03
DTFT of Signal
Magnitude of the DFT of the Pulse (in dB):
50
dB
40
30
20
10
0
-10
-20
-30
-40
-50
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
Hz
DTFT of Signal
Bandwidth of the Signal (take about -20dB from max) is about 1.0kHz
50
40
~ 20dB
30
20
10
0
-10
-20
-30
-40
-50
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
Design the Low Pass Filter
Pass 0 to 1kHz;
Stop 1.5kHz to 10.0kHz
N=81
Magnitude (dB)
Attenuation 40dB
50
0
-50
-100
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Normalized Frequency ( rad/sample)
0.9
1
0
Phase (degrees)
Sampling Freq 20.0kHz
-500
-1000
-1500
Filtered Data
10
5
Received signal:
0
-5
-10
0
5
10
15
20
25
30
35
millisec
4
2
… and Filtered:
0
-2
-4
0
5
10
15
20
millisec
25
30
35
4
3
Compare to
the original:
2
1
0
-1
-2
-3
-4
0
0.005
0.01
0.015
millisec
0.02
0.025
0.03
40