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Filtering Problem

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Filtering Problem

Goal: design a filter to attenuate the disturbances

Filter

SIGNAL

NOISE

SIGNAL

NOISE

Mostly NOISE

Mostly NOISE

Define Signal and Noise in the Frequency Domain

NOISE

Filter

SIGNAL

Mostly SIGNAL

PASS Band

STOP Band

IDEAL Filter

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

PASS Band

STOP Band

Trans. Band

Non-IDEAL Filter

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

Design a Low Pass Filter: IDEAL

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

Design a Low Pass Filter: IDEAL (continued)

This means the following. If

then

From IDEAL to FIR

Notice that

Then we can approximate with a finite sum…

… and choose the filter as

actual

desired

Summary of design of Low Pass FIR

Given the Pass Band Frequency

The impulse response of the filter: let

Magnitude and Phase:

Magnitude:

Phase:

in the passband

Example of Freq. Response

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

Impulse Response

n=0:N;

stem(n,h)

Example with Hamming window

hamming window

n=0:N;

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

h=h0.*hamming(N);

stem(n,h)

Example with Hamming window

hamming window

freqz(h)

~ 50dB

attenuation

Rectangular-13dB

Hamming-43dB

Blackman-58dB

transition region

Low Pass Filter Design: Analytical

Transition Region: depends on the window and the filter length N

Attenuation: depends on window only

Example of Low Pass Filter Design

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

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

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:

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

Example of Low Pass Filter Design (continued)

Finally the impulse response of the filter

Example of Low Pass Filter Design (continued)

The Frequency Response of the Filter:

Design Parameters

Pass Band Frequencyrad.

Stop Band Frequencyrad.

Pass Band Ripple dB

Stop Band AttenuationdB

Linear Interpolation

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];

where

Computer Aided Design of FIR Filters

Step 2: Choose filter length N from desired attenuation as

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

h = firpm(N, f, A);

which yields the desired impulse response

Example: Low Pass Filter

Passband: 3kHz

Stopband: 3.5kHz

Attenuation: 60dB

Sampling Freq: 15 kHz

Then compute:

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

Frequency Response

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)

Impulse Response of Example

stem(h)

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:

0

1

1

0

Typical Applications

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

For example let:

Transmitted:

Received Signal with Noise

Suppose you receive the signal with 0dB SNR:

dB

Hz

DTFT of Signal

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

DTFT of Signal

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

Design the Low Pass Filter

Pass 0 to 1kHz;

Stop 1.5kHz to 10.0kHz

Sampling Freq 20.0kHz

Attenuation 40dB

N=81

Filtered Data

Received signal:

… and Filtered:

Compare to the original: