Kindly verify my code of DTMF decoding step

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I want to decode DTMF tones by using FIR Filter. The filters that are required to be used in filter bank are cnstructed by Sinusoidal impulse response of the form h[n]= (2/L)cos(2*pi*fb*n/fs) where 0<n<L
L is filter length fb defines the freq location of passband e.g we can pick 697Hz
the book says to generate bandpass filter for 770 Hz component with L=50 and fs=12000. This has to be done by creating a vector of filter co-efficients ,h770 which are determined by evaluatiing above stated equation. plot the filter coefficients using stem().
I have done it in this way. Is it ok
h=[];
L=50;
fs=12000;
fb=770;
for n=1:L
h(n)=(2/L)*cos(2*pi*fb*n/fs);
end
stem(h)

Accepted Answer

Wayne King
Wayne King on 9 Oct 2011
Hi, Notice how Walter has indexed the for loop from 0, but was careful to index the vector h from 1.
h=[];
L=50;
fs=12000;
fb=770;
for n=0:L
h(n+1)=(2/L)*cos(2*pi*fb*n/fs);
end
stem(h)
So that it is legal MATLAB indexing, however, I would say to avoid the for loop entirely. This is cleaner MATLAB.
L=50;
fs=12000;
fb=770;
% if you really want 0<n<L
h = (2/L)*cos(2*pi*fb*(1:L-1)/fs);
stem(h);

More Answers (10)

moonman
moonman on 9 Oct 2011
I dont have test data. i just want somebody to read my statement and see my code i hope it is ok but just want to cross check from experts

Walter Roberson
Walter Roberson on 9 Oct 2011
The description says 0 < n < L but you have n going from 1 to L which is one step too far (n == L)
  3 Comments
moonman
moonman on 9 Oct 2011
The instructions say
0<=n < L (n is greater than or equal to 0)
How can i code this
???
Walter Roberson
Walter Roberson on 9 Oct 2011
for n=0:L-1
h(n+1)=(2/L)*cos(2*pi*fb*n/fs);
end

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moonman
moonman on 9 Oct 2011
Yes, this issue i know but when i tell matlab to do
for n=0:L
It gives error
??? Attempted to access h(0); index must be a positive integer or logical.
How to remove this error

moonman
moonman on 9 Oct 2011
ok thanks king
kindly confirm me is my code ok as per book question
  1 Comment
Wayne King
Wayne King on 9 Oct 2011
Yes, you can confirm by looking at the frequency response
fvtool(h,1,'Fs',fs);

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moonman
moonman on 9 Oct 2011
Thanks King, can u explain 'Fs' and fs in this command
The book also tells to plot magnitude response by this
h770=[];
L=50;
fs=12000;
fb=770;
for n=1:L
h770(n)=(2/L)*cos(2*pi*fb*n/fs);
end
fs=8000;
ww=0:(pi/256):pi;
ff=ww/(2*pi)*fs;
H=freqz(h770,1,ww);
plot(ff,abs(H));
grid on;
Although the shape is same for both plots but magnitude varies
  1 Comment
Wayne King
Wayne King on 9 Oct 2011
The big difference is that you have not plotted yours in dB
plot(ff,abs(H));
% if you plot
plot(ff,20*log10(abs(H)));
You'll see. I think it is much more common to plot these in dB.

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Wayne King
Wayne King on 9 Oct 2011
fvtool() is doing that under the hood and much more, I question why your book constructs a frequency axis in angular frequencies, when in this application Hertz makes much more sense:
[H,F] = freqz(h,1,[],fs);
plot(F./1000,20*log10(abs(H)));
grid on;
xlabel('kHz'); ylabel('Magnitude-squared (dB)');
Again, I would recommend you avoid a for loop to calculate your FIR filter coefficients and just use the vector approach I showed above.

moonman
moonman on 9 Oct 2011
Yes u are absolutely right I question why your book constructs a frequency axis in angular frequencies, when in this application Hertz makes much more sense:
My last query , the question says to
indicate the location of each of the DTMF frequencies(697,770,852,941,1209,1336 and 1477 Hz) on the plot generated by
h770=[];
L=50;
fs=12000;
fb=770;
for n=1:L
h770(n)=(2/L)*cos(2*pi*fb*n/fs);
end
fs=8000;
ww=0:(pi/256):pi;
ff=ww/(2*pi)*fs;
H=freqz(h770,1,ww);
plot(ff,abs(H));
grid on;
How can i do that, kindly make me clear

Wayne King
Wayne King on 9 Oct 2011
[H,F] = freqz(h,1,[],fs);
plot(F,20*log10(abs(H)));
grid on;
xlabel('Hz'); ylabel('Magnitude-squared (dB)');
set(gca,'xtick',[697,770,852,941,1209,1336,1477]);
set(gca,'xlim',[500 2000]);
Note that you really need to limit your x-axis in order to make the labels reasonable. If you use the whole frequency axis from 0 to the Nyquist, they bunch up.

moonman
moonman on 9 Oct 2011
Wonderful Help Wayne King Thanks a lot for making me understand this issue

moonman
moonman on 17 Oct 2011
Wayne King u wrote this code for me
[H,F] = freqz(h,1,[],fs);
plot(F,20*log10(abs(H)));
grid on;
xlabel('Hz'); ylabel('Magnitude-squared (dB)');
set(gca,'xtick',[697,770,852,941,1209,1336,1477]);
set(gca,'xlim',[500 2000]);
Can u kindly explain me why u wrote Magnitude-Squared (db) in y axis, why squared
  5 Comments
praveen
praveen on 17 Sep 2013
Edited: praveen on 17 Sep 2013
kindly send total program in a single page of dtmf decoder using fir filters try to send quickleyyyyyy please
Jan
Jan on 17 Sep 2013
@praveen: This request is such unfriendly, that it is funny, that you hope to be successful. It is definitely your turn to pick from these many pieces of code a program, that is working for you. Trying to push us to do your work quickly is really the wrong approach.

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