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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def calc_autocorrelation(Signal, FFT=False, PyCUDA=False):
 """ Calculates the autocorrelation from a given Signal via using Parameters Signal : array-like Array containing the signal to have the autocorrelation calculated for FFT : optional, bool Uses FFT to accelerate autocorrelation calculation, but assumes certain certain periodicity on the signal to autocorrelate. Zero-padding is added to account for this periodicity assumption. PyCUDA : bool, optional If True, uses PyCUDA to accelerate the FFT and IFFT via using your NVIDIA-GPU If False, performs FFT and IFFT with conventional scipy.fftpack Returns ------- Autocorrelation : ndarray Array containing the value of the autocorrelation evaluated at the corresponding amount of shifted array-index. """

    if FFT==True:
        Signal_padded = scipy.fftpack.ifftshift((Signal-_np.average(Signal))/_np.std(Signal))
        n, = Signal_padded.shape
        Signal_padded = _np.r_[Signal_padded[:n//2], _np.zeros_like(Signal_padded), Signal_padded[n//2:]]
        if PyCUDA==True:
            f = calc_fft_with_PyCUDA(Signal_padded)
        else:
            f = scipy.fftpack.fft(Signal_padded)
        p = _np.absolute(f)**2
        if PyCUDA==True:
            autocorr = calc_ifft_with_PyCUDA(p)
        else:
            autocorr = scipy.fftpack.ifft(p)
        return _np.real(autocorr)[:n//2]/(_np.arange(n//2)[::-1]+n//2)
    else:
        Signal = Signal - _np.mean(Signal)
        autocorr = scipy.signal.correlate(Signal, Signal, mode='full')
        return autocorr[autocorr.size//2:]/autocorr[autocorr.size//2]