使用CUDA的希尔伯特变换

Question

为了在1D阵列上进行希尔伯特变换,必须:

• FFT数组
• 数组的两半,另一半为零
• 反FFT结果

我正在使用PyCuLib进行FFT.我的代码到目前为止

def htransforms(data):
    N = data.shape[0]
    transforms       = nb.cuda.device_array_like(data)          # Allocates memory on GPU with size/dimensions of signal
    transforms.dtype = np.complex64                             # Change GPU array type to complex for FFT 

    pyculib.fft.fft(signal.astype(np.complex64), transforms)    # Do FFT on GPU

    transforms[1:N/2]      *= 2.0      # THIS STEP DOESN'T WORK
    transforms[N/2 + 1: N]  = 0+0j     # NEITHER DOES THIS ONE

    pyculib.fft.ifft_inplace(transforms)                        # Do IFFT on GPU: in place (same memory)    
    envelope_function      = transforms.copy_to_host()          # Copy results to host (computer) memory
    return abs(envelope_function)

我觉得它可能与Numba的CUDA接口本身有关......它是否允许像这样修改数组(或数组切片)的各个元素？我认为它可能,因为变量transforms是a numba.cuda.cudadrv.devicearray.DeviceNDArray,所以我想也许它有一些与numpy相同的操作ndarray.

简而言之,使用Numba device_arrays,在片上进行简单操作的最简单方法是什么？我得到的错误是

不支持的操作数类型*=''DeviceNDArray'和'float'

Answer 1

我会使用pytorch

您使用 pytorch 的函数（并且我删除了abs以返回复数值）

def htransforms(data):
    N = data.shape[-1]
    # Allocates memory on GPU with size/dimensions of signal
    transforms       = torch.tensor(data).cuda()
    
    torch.fft.fft(transforms, axis=-1)
    transforms[:, 1:N//2]      *= 2.0      # THIS STEP DOESN'T WORK
    transforms[:, N//2 + 1: N]  = 0+0j     # NEITHER DOES THIS ONE

    # Do IFFT on GPU: in place (same memory)
    return torch.abs(torch.fft.ifft(transforms)).cpu() 

但你的转换实际上与我在维基百科上找到的不同

维基百科版本

def htransforms_wikipedia(data):
    N = data.shape[-1]
    # Allocates memory on GPU with size/dimensions of signal
    transforms       = torch.tensor(data).cuda()
    
    transforms = torch.fft.fft(transforms, axis=-1)
    transforms[:, 1:N//2]      *= -1j      # positive frequency
    transforms[:, (N+2)//2 + 1: N] *= +1j # negative frequency
    transforms[:,0] = 0; # DC signal
    if N % 2 == 0:
        transforms[:, N//2] = 0; # the (-1)**n term
    
    # Do IFFT on GPU: in place (same memory)
    return torch.fft.ifft(transforms).cpu() 

def htransforms_wikipedia(data):
    N = data.shape[-1]
    # Allocates memory on GPU with size/dimensions of signal
    transforms       = torch.tensor(data).cuda()
    
    transforms = torch.fft.fft(transforms, axis=-1)
    transforms[:, 1:N//2]      *= -1j      # positive frequency
    transforms[:, (N+2)//2 + 1: N] *= +1j # negative frequency
    transforms[:,0] = 0; # DC signal
    if N % 2 == 0:
        transforms[:, N//2] = 0; # the (-1)**n term
    
    # Do IFFT on GPU: in place (same memory)
    return torch.fft.ifft(transforms).cpu() 

data = torch.zeros((1, 2**10))
data[:, 2**9] = 1;
tdata = htransforms(data).data;
plt.plot(tdata.real.T, '-')
plt.plot(tdata.imag.T, '-')
plt.xlim([500, 525])
plt.legend(['real', 'imaginary'])
plt.title('inpulse response of your version')

您的版本的脉冲响应是1 + 1j * h[k]维基h[k]百科版本的脉冲响应。如果您正在使用真实数据，维基百科版本很好，因为您可以使用rfft和irfft生成一个线性版本

data = torch.zeros((1, 2**10))
data[:, 2**9] = 1;
tdata = htransforms_wikipedia(data).data;
plt.plot(tdata.real.T, '-');
plt.plot(tdata.imag.T, '-');
plt.xlim([500, 525])
plt.legend(['real', 'imaginary'])
plt.title('inpulse response of Wikipedia version')