Computing analytical signal using FFT in C++


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I’m currently lost trying to figure out how to implement an equivalent version of MATLAB’s hilbert() function in C++. I’m very new to signal processing, but, ultimately, I would like to figure out a way to phase shift any given signal by 90 degrees. I was attempting to follow the method suggested in this question on MATLAB central, which appears to work based on tests using GNU Octave.

I have what I believe to be a working implementation of both FFT and the inverse FFT, and I have tried implementing the method described in the answer to this post in order to compute the analytical signal. I have tried doing this by applying the FFT, setting the upper half of the array to zero, and then applying the inverse FFT, but, based on graphs I made of output from a test, there must be a problem with the way I have implemented finding the analytical signal.

What would be a suitable way to implement the hilbert() function from MATLAB in C++ given a working implementation of FFT and inverse FFT? Is there a better way of achieving the 90 degree phase shift?

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Answer ( 1 )

    January 12, 2017 at 11:00 am

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    Checking the MATLAB implementation the following should return the same result as the hilbert function. You’ll obviously have to modify it to match your specific implementation. I’m assuming a signal class of some sort exists.

    signal hilbert(const signal &x)
        int limit1, limit2;
        signal xfreq = fft(x);
        if (x.numel % 2 == 0) {
            limit1 = x.numel/2;
            limit2 = limit1 + 1;
        } else {
            limit1 = (x.numel + 1)/2;
            limit2 = limit1;
        // multiply the first half by 2 (except the first element)
        for (int i = 1; i < limit1; ++i) {
            xfreq[i].real *= 2;
            xfreq[i].imag *= 2;
        for (int i = limit2; i < x.numel; ++i) {
            xfreq[i].real = 0;
            xfreq[i].imag = 0;
        return ifft(xfreq);

    Edit: Forgot to set the second half to zeros.
    Edit2: Fixed a logical error. I coded the following up in MATLAB which matches hilbert.

    function h = hil(x)
        n = numel(x);
        if (mod(n,2) == 0)
            limit1 = n/2;
            limit2 = limit1 + 2;
            limit1 = (n+1)/2;
            limit2 = limit1+1;
        xfreq = fft(x);
        for i = 2:limit1
            xfreq(i) = xfreq(i)*2;
        for i = limit2:n
            xfreq(i) = 0;
        h = ifft(xfreq);
    Best answer

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