MATLAB: Frequency and Sampling Frequency

audiofrequencyMATLABsignal processing

Which is the frequency of a tone heard when played with twice the original sampling frequency?

For example, we have this tone:

>> fs=24000; % Sampling frequency
>> t=0:1/fs:1;
>> x=cos(2*pi*1000*t); % frequency = 1000 Hz
>> sound(x, fs);

There we see that the frequency is 1000 Hz, but if we play the tone with twice its sampling frequency:

>> sound(x, 2*fs);

Which is the frequency of the tone then?

Best Answer

This is more a philosophical question than a mathematical one.

The frequency of the original signal remains ‘fs’ regardless, The frequency of the generated tone is whatever is passed to the sound function, here either ‘fs’ or ‘2*fs’.

At least that is the way I interpret it.

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MATLAB: How to Butterworth Filter with Bandpass [10 500] with sampling rate 1000

You have to divide your passband and stopband frequencies by the Nyquist frequency to normalise them:

Fs = 1000;                                          % Sampling Frequency
Fn = Fs/2;                                          % Nyquist Frequency
Wp = [10  495]/Fn;                                  % Norrmalised Passband Frequencies
Ws = [5   499]/Fn;                                  % Normalised Stopband Frequencies
Rp =  1;                                            % Passband Ripple (dB)
Rs = 25;                                            % Stopband Ripple (dB)
[n,Wn] = buttord(Wp, Ws, Rp, Rs);                   % Optimal Filter Order
[b,a] = butter(n, Wn);                              % Calculate Filter Coefficients
[sos,g] = tf2sos(b,a);                              % Convert To Second-Order Sections For Stability
figure(1)
freqz(sos, 4096, Fs)                                % Filter Bode Plot

You cannot have any frequency exactly at zero or exactly at the Nyquist frequency, so I designed your passband to be close enough to them to be stable. In your filter, you can also use a high-pass design with a lower cutoff of 10 Hz, since your upper passband is at the Nyquist frequency. The passband and stopband ripple values are necessary to specify in the design but irrelevant to a Butterworth filter. These are acceptable values for most filter designs (for example Chebyshev) that require them.

You would use the ‘sos’ and ‘g’ variables in your actual filter:

filtered_signal = filtfilt(sos, g, signal);

(This design works well in R2016a and should also work in earlier releases.)

Best Answer

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