Matched filter

How can I write a mtached filter in Matlab? I have the filter design and signal processing tool boxes.

Answer 1

[Honglei Chen]

If you have a signal, x, then the matched filter's coefficients is given by time reverse of x, i.e., x(end:-1:1). If your signal is complex, you also need to to use complex conjugate. You can then use it just as an FIR filter. For example,

>> x = ones(10,1);
>> b = x(end:-1:1);
>> y = filter(b,1,x);

[vsee]

Thanks. I entered the same code that you provided. x is a column vector of 10 one's. b seems to be the same. Is b really the mirror image of x? In other words can you explain what x(end:-1:1) is doing? Thanks

[Honglei Chen]

I was just giving an example. x and b are the same because x happens to be the same as its time inverse. You can replace x with a vector representing your particular signal. The key here is "end:-1:1" which flips the signal. You can also use flipud or fliplr depending on your vector's orientation.

[Nuzat Alam]

Hii i was wondering what is the impulse response of matched filter here?I think it is related to b. Do you think if i want to select a specific sample number for b, i also need to change the length of input signal x?

[Mustahsan Ali Khan Khanzada Rajput]

How to apply matched filter if the signal is complex?

[Honglei Chen]

It's the same, use the filter function. The only difference is when you come up with the filtering coefficient, you need to add a conjugate, like

b = conj(x(end:-1:1))

HTH

Answer 2

[Andrew Newell]

Check out MATLAB Simulations for Radar Systems Design from the File Exchange.

Answer 3

[Mathuranathan Viswanathan]

Hi You can try the model given here

Square Root raised Cosine - Matched filter implementation http://www.gaussianwaves.com/2011/04/square-root-raised-cosine-filter-matchedsplit-filter-implementation-2/

Regards Mathuranathan

Answer 4

[Bhaskar]

To elaborate on Honglei's answer. If the signal you are trying to design a matched filter for, is x, then --

>> b = x(end-1:1);
>> y = filter(b,1,x);

You should already have x.

[mmm ssss]

how can implement the same filter on image ?

how i should represent x.

Answer 5

[Nicole Bienert]

The built in xcorr function can be used - you just need to normalize by fs and square the output. See below for an ideallized example:

%Purpose: Demonstrate how to match filter correctly

BW=3.84e6;
fs = 50*BW; %sample rate
T= 1/fs; %sample period
fc = 330e6; %center freq
chirpLen=0.075; %chirp length
A=3; %amplitude of voltage signal (normally unknown)
Ar=2; %amplitude of reference voltage signal (normally unknown)

%create the signal withought noise and zero padded on either side (zero
%padding not necessary because xcorr does that, I'm just demonstrating that
%signals don't need to be the same length.)
sig=[zeros(1,ceil(chirpLen*fs)),A*chirp(t,0,t(end),BW),zeros(1,ceil(chirpLen*fs))];

%create the reference chirp
ref_chirp=Ar*chirp(t,0,t(end),BW);
t=[0:T:(length(ref_chirp)-1)*T];
%normalize reference chirp: The reference chirp needs to have energy of 1 
%so that it doesn't bias the output of the match filter. A filter shouldn't
%be applying gain to the signal or changing the units. The signal is in
%volts, so we divide by the square root of the energy to normalize it.

%If you know the signal's amplitude (for CW or FMCW):
energy=Ar^2/2*chirpLen;
%If you don't know the signal's amplitude, integrate to find energy (if it is noiseless):
    %energy=trapz(t,ref_chirp.^2)
ref_chirp=ref_chirp/sqrt(energy);


% perform match filtering
[R,lags] = xcorr(sig,ref_chirp); %signals don't need to be the same length

%R is the sum of each data sample as the signals are shifted past
%eachother, so to make the numerical integration correct, you need to
%multiply by dx which is T in this case. Then to get the filtered voltage
%signal in units of energy, you need to square it.
R=(abs(R*T)).^2; %absolute value only necessary if signals are complex
% take only positive side
R = R(lags>=0);
lags=lags(lags>=0);

[matchFiltPeak,index]=max(R);

figure()
plot(lags*T,R)
xlim([index-250 index+250]*T)

display(['Energy in signal was: ',num2str(A.^2/2*chirpLen)])
display(['which is the same as the peak of the match filter: ',num2str(matchFiltPeak)])

[Andreas Blom]

I tried running this code but cant. Unrecognized function or variable 't'.

[NoYeah]

you may missed below line

t=[0:T:(length(ref_chirp)-1)*T];

[Anastasia Kov]

I faced the same code error as Andreas talking about.

I guess, the issue is here:

ref_chirp=Ar*chirp(t,0,t(end),BW);

t=[0:T:(length(ref_chirp)-1)*T];

"ref_chirp" referes to "t" and vise versa.

There must be some additional definition for "ref_chirp" or "t", isn't it?

[Anastasia Kov]

As well as I do not understand this line:

R=(abs(R*T)).^2; %absolute value only necessary if signals are complex

It will not work unfortunately, R is not defined.

Could somebody help please?

Would be very usefull to have this code working.

Thank you!

[MUKESH KUMAR]

how to use this code in my problem

[Walter Roberson]

@Bojan comments to @Nicole Bienert

code not working, t is undefined and later circular reference to refchirp

[Charles Sutherland]

This worked for me:

Change

t=[0:T:(length(ref_chirp)-1)*T];

to

t=[0:T:((chirpLen*fs)-1)*T];

and then move it to -

BW=3.84e6;
fs = 50*BW; %sample rate
T= 1/fs; %sample period
fc = 330e6; %center freq
chirpLen=0.075; %chirp length
A=3; %amplitude of voltage signal (normally unknown)
Ar=2; %amplitude of reference voltage signal (normally unknown)
t=[0:T:((chirpLen*fs)-1)*T];

Worked for me anyway...

Cheers