Filtering Problem

Filtering Problem

][ns

][nv

][nx ][][ nsny Filter

SIGNAL

NOISE

Goal: design a filter to attenuate the disturbances

Define Signal and Noise in the Frequency Domain

][ns

][nv

][nx

][][ nsny Filter

SIGNALNOISE

Mostly SIGNAL

n nn

|)(| S|)(| V

)(rad 0

SIGNALNOISE

P

Mostly NOISEMostly NOISE

P

IDEAL Filter

Since the filter has real coefficients, we need only the positive frequencies

)(H

p PASS Band

STOP Band

Non-IDEAL Filter

Since the filter has real coefficients, we need only the positive frequencies

PASS Band

STOP Bandp S

Trans. Band

|)(| H

Design a Low Pass Filter: IDEAL

First we can determine an infinite length expansion using the DTFT:

n

njddd enhnhDTFTH ][][)(

deHHIDTFTnh njddd )(

21)(][

nn

ndenh PPPnjd

P

P

sincsin21][

PP

Hd ( )

1

Design a Low Pass Filter: IDEAL (continued)

nnh PP

d

sinc][This means the following. If

then

n

njdd enhH ][)(

PP

Hd ( ) 1

-60 -40 -20 0 20 40 60-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

][nhd

n

From IDEAL to FIR

0][lim

nhdnNotice that

Then we can approximate with a finite sum…

L

Ln

njdd enhH ][)(

desiredactual

… and choose the filter as ][][ Lnhnh d

Ljd

LjL

Ln

njd

L

n

njd

L

n

nj eHeenheLnhenhH

)(][][][)(

2

0

2

0

Summary of design of Low Pass FIR

Given the Pass Band Frequency

The impulse response of the filter: let

P0

NnLnnh PP ,...,0,)(sinc][

LN 2

Magnitude and Phase:

Ljd

L

n

nj eHenhH

)(][)(2

0

)()(

)()(

LLHH

HH

d

d

Magnitude:Phase:

in the passband

Example of Freq. Response

4/ P 40,20 NL

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

Normalized Frequency ( rad/sample)

Pha

se (d

egre

es)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50

Normalized Frequency ( rad/sample)

Mag

nitu

de (d

B)

dB20

n=0:N; h=(wp/pi)*sinc((wp/pi)*(n-L)); freqz(h)

20

Impulse Response

0 5 10 15 20 25 30 35 40-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3][nh

n=0:N;

stem(n,h)

Example with Hamming window

][nh

0 5 10 15 20 25 30 35 40-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

40,...,0,40

2cos46.054.0)20(41sinc

41][

nnnnh

n=0:N;

h0=(wp/pi)*sinc((wp/pi)*(n-L));

h=h0.*hamming(N);

stem(n,h)

hamming window

Example with Hamming window

40,...,0,40

2cos46.054.0)20(41sinc

41][

nnnnh

hamming window

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

Normalized Frequency ( rad/sample)

Pha

se (d

egre

es)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-150

-100

-50

0

50

Normalized Frequency ( rad/sample)

Mag

nitu

de (d

B)

)(Hfreqz(h)~ 50dB

Low Pass Filter Design: Analytical

Transition Region: depends on the window and the filter length NAttenuation: depends on window only

0 0.5 1 1.5 2 2.5 3-120

-100

-80

-60

-40

-20

0

20

attenuation

transition region

Rectangular -13dB

Hamming -43dB

Blackman -58dB

N/4N/8

N/12

nattenuatio

P S

window

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 rad 2/20/52

rad5/220/420

10

40dB

F kHz54 10

225

Step 2: from pass band, determine ideal filter impulse response

52sinc

52sinc)( nnnh PP

d

Example of Low Pass Filter Design

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:

10

8

N

So choose N=80 which yields the shift L=40.

Example of Low Pass Filter Design (continued)

Finally the impulse response of the filter

h nn

n( )

. . cos , ,

25

0 54 0 46280

0 80sinc2(n - 40)

5 if

0 otherwise

Example of Low Pass Filter Design (continued)

The Frequency Response of the Filter:

H( )

H( )

dB

rad

Example of Low Pass Filter Design (continued)

Design Parameters

Pass Band Frequency rad.

Stop Band Frequency rad.

Pass Band Ripple dB

Stop Band Attenuation dB

p

S

BA /log20 10

CA /log20 10

AB

C

p S

)(H

Best Design tool for FIR Filters: the Equiripple algorithm. It minimizes the maximum error between the frequency responses of the ideal and actual filter.

Computer Aided Design of FIR Filters

PASS STOP

Linear Interpolation

1

0 )(rad

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]; πω

πω

STOP

PASS

2 f

1 fwhere

Computer Aided Design of FIR Filters

Step 3: call “firpm” (pm = Parks, McClellan)

h = firpm(N, f, A);

which yields the desired impulse response

][]1[],0[ Nhhhh

Step 2: Choose filter length N from desired attenuation as

/11(dB)n Attenuatio~

PASSSTOP

N

Passband: 3kHz

Stopband: 3.5kHz

Attenuation: 60dB

Sampling Freq: 15 kHz

Example: Low Pass Filter

Then compute:

8211

60157

000,15500,32

52

000,15000,32

52

157

N

STOP

PASS

h = firpm(82,[0,2/5,7/15,1], [1,1,0,0]);

Frequency Response

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0

Normalized Frequency ( rad/sample)

Pha

se (d

egre

es)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50

Normalized Frequency ( rad/sample)

Mag

nitu

de (d

B)

freqz(h)

50dB, not quite yet!

Increase N.

Increase Filter Length N

h=firpm(95, [0, 2/5, 7/15, 1], [1,1,0,0]);

freqz(h)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4000

-3000

-2000

-1000

0

Normalized Frequency ( rad/sample)

Pha

se (d

egre

es)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50

Normalized Frequency ( rad/sample)

Mag

nitu

de (d

B)

Impulse Response of Example

stem(h)

0 10 20 30 40 50 60 70 80 90 100-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

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:

Typical Applications

10 0),2sin()( TttFAtx

0 1 2 3 4 5 6 7 8

x 10-3

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

sec

sec8500

1

0

mTHzF

For example in a communications signal we send “0” and “1”.

For example let:

Typical Applications

)(0)(1tx

tx

0 0.005 0.01 0.015 0.02 0.025 0.03-4

-3

-2

-1

0

1

2

3

4

millisec

1 0 1 0

Transmitted:

Suppose you receive the signal with 0dB SNR:

Received Signal with Noise

0 0.005 0.01 0.015 0.02 0.025 0.03-8

-6

-4

-2

0

2

4

6

8

millisec

Magnitude of the DFT of the Pulse (in dB):

DTFT of Signal

0 1000 2000 3000 4000 5000 6000 7000 8000 9000-50

-40

-30

-20

-10

0

10

20

30

40

50

dB

Hz

Bandwidth of the Signal (take about -20dB from max) is about 1.0kHz

DTFT of Signal

0 1000 2000 3000 4000 5000 6000 7000 8000 9000-50

-40

-30

-20

-10

0

10

20

30

40

50

dB20~

Pass 0 to 1kHz;

Stop 1.5kHz to 10.0kHz

Sampling Freq 20.0kHz

Attenuation 40dB

N=81

Design the Low Pass Filter

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1500

-1000

-500

0

Normalized Frequency ( rad/sample)

Pha

se (d

egre

es)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100

-50

0

50

Normalized Frequency ( rad/sample)M

agni

tude

(dB

)

Filtered Data

Received signal:

… and Filtered:

0 0.005 0.01 0.015 0.02 0.025 0.03-4

-3

-2

-1

0

1

2

3

4

millisec

Compare to the original:

0 5 10 15 20 25 30 35 40-4

-2

0

2

4

millisec

0 5 10 15 20 25 30 35-10

-5

0

5

10

millisec