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)
采纳的回答
Wayne King
2011-10-9
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);
1 个评论
Walter Roberson
2011-10-9
h = (2/L)*cos(2*pi*fb*(0:L-1)/fs);
for the 0 <= n < L case.
更多回答（10 个）
Walter Roberson
2011-10-9
0 个投票
The description says 0 < n < L but you have n going from 1 to L which is one step too far (n == L)
3 个评论
Walter Roberson
2011-10-9
Notice the strict inequality: 0 < n < L . That means that n cannot be 0 and cannot be L, so there is no need to use "for n=0:L" (not unless the instructions are wrong.)
If the instructions had said 0 <= n <= L, then you would code as
for n=0:L
h(n+1)=(2/L)*cos(2*pi*fb*n/fs);
end
moonman
2011-10-9
Walter Roberson
2011-10-9
for n=0:L-1
h(n+1)=(2/L)*cos(2*pi*fb*n/fs);
end
moonman
2011-10-9
0 个投票
1 个评论
Wayne King
2011-10-9
Yes, you can confirm by looking at the frequency response
fvtool(h,1,'Fs',fs);
moonman
2011-10-9
1 个评论
Wayne King
2011-10-9
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.
Wayne King
2011-10-9
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.
Wayne King
2011-10-9
[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
2011-10-17
5 个评论
moonman
2011-10-17
Wayne King
2011-10-17
It is often desirable to obtain power estimates, which is the magnitude squared.
moonman
2011-10-17
Jan
2013-9-17
@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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