fft normalization vs the norm-option

i have to often normalize the result of circular-convolution, so I 'borrowed'-and-modfied the following routine :

def fft_normalize(x):# Normalize a vector x in complex domain.
    c = np.fft.rfft(x)
    # Look at real and image as if they were real
    ri = np.vstack([c.real, c.imag])
    # Normalize magnitude of each complex/real pair
    norm = np.linalg.norm(ri, axis=0)
    if np.any(norm==0): norm[norm == 0] = np.float64(1e-308) #!fixme
    ri= np.divide(ri,norm)
    c_proj = ri[0,:] + 1j * ri[1,:]
    rv = np.fft.irfft(c_proj, n=x.shape[-1])
    return rv

def fft_convolution(a, b):
    return np.fft.irfft(np.fft.rfft(a) * np.fft.rfft(b))

so that I do this :

fft_normalize(fft_convolution(a,b))

i see in the numpy docs there is a 'norm' option. Is this equivalent to what i'm doing ? And if yes, which option should I use ?

def fft_convolution2(a, b):
    return np.fft.irfft(np.fft.rfft(a) * np.fft.rfft(b), norm='ortho')

When I test it it behaves better when I do fft_normalize()

Second, i had to add this line, but it does not seem, right. any ideas ?

    if np.any(norm==0): norm[norm == 0] = np.float64(1e-308) #!fixme

As a side note, if you know !! numpy docs mentions that they promote float32 to float64 and that scipy.fftpack does not !! Would fftpack be faster ! because scipy says fftpack is obsolete and there is no info on the new scipy ?

@cris are you sugesting i do it this way :

def fft_normalize(x):# Normalize a vector x in complex domain.
    c = np.fft.rfft(x)
    ri = np.vstack([c.real, c.imag])
    norm = np.abs(c)
    if np.any(norm==0): norm[norm == 0] = MIN #!fixme
    ri= np.divide(ri,norm)
    c_proj = ri[0,:] + 1j * ri[1,:]
    rv = np.fft.irfft(c_proj, n=x.shape[-1])
    return rv