I have 2 ADC channels of constant voltage measurement with small amount of high frequency noise only and no low frequency oscillations. The final signal should be the simple sum of those and denoised with the moving average in the end. Does the moving average denoising of each separate channel before final summation be more noise effective than applying summation first and do the final moving average later?

channel_1_noisy = channel_1_clean + noise_1
channel_2_noisy = channel_2_clean + noise_2

sum_noisy = channel_1_noisy + channel_2_noisy
sum_denoised = moving_average(sum_noisy)

or sacrificing computer time for 2 moving average filtering operations?

channel_1_denoised = moving_average(channel_1_noisy)
channel_2_denoised = moving_average(channel_2_noisy)
sum_denoised = channel_1_denoised + channel_2_denoised

Having more channels will the effect be exaggerated?


1 Answer 1


According to your processing chains (filter --> sum vs sum --> filter) the two approaches, by being LTI systems, should produce the same results under infinite precision arithmetic. You can show this mathematically by using the properties of convolution operation.

Note that a moving average filter is just an ordinary LTI system.

Note also that nonlinearities such as quantization operations may affect the above fact's validity.

  • $\begingroup$ yes, I also compared the actual results on the test data $\endgroup$ Commented Jun 25, 2015 at 10:11

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