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is anyone could help understand this 50/60Hz rejection filter?

The basic information:

Fs=1MHz, DOR(data output rate)=25SPS, with both 50Hz & 60Hz rejeciton, settling within 1 cycle, 1/DOR.

Frequency Response, DOR=25SPS

The simplified sturcture: SINC5 + SINC1 + 50/60Hz-Rejection. So how to relize it? What I know is that the SINC5's DOR=31.25KHz, OSR=32. I don't know how to realize both 50/60Hz rejection, it seems to be realized by a "sinc" like filter too.

Simplified Structure

Ring
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3 Answers3

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I don't know how to realize both 50/60Hz rejection, it seems to be realized by a "sinc" like filter too.

The SINC filter, which is typically 3rd order and denoted SINC³, is defined in the frequency domain as:

H(f) = [1/N * Sin(N*π*f/fM) / Sin(π*f/fM)]³

Where: N is the decimation ratio fM is the modulator frequency

You calculate separate filters for 50Hz and 60Hz.

Juha P
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  • Thank you! I know fM/N is the notch frequency... but as response figure show, the notch is not an simple integer multiple of 50Hz nor 60Hz. And if realized in this way, the settling time is not guaranteed...Could you further explain these issues? – Ring Feb 13 '23 at 11:56
  • I'm far from expert (even from noobie) in this subject so my quess is; as the block diagram tells, you "mix" min four SINC filters when 50Hz/60Hz rejection filter stage is enabled so, maybe plot shows response of them all together and not just rejection filter stage. – Juha P Feb 13 '23 at 12:26
  • thank you! yes it do mix sinc5+sinc1+50Hz + 60Hz, my post was not clear. – Ring Feb 13 '23 at 12:45
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As depicted the 50 Hz and 60 Hz rejection filter is not necessarily a CIC filter (but could be constructed with one when the sampling rate is an integer multiple of $50 \times 60$ such that 50 and 60 are factors. However it is much more likely that they would be constructed as the cascade of two 2nd order IIR nulling filters as detailed in this post, and summarized below:

2nd order IIR nulling filter

This is a "2nd Order Biquad" filter, and the frequency response for this structure with the coefficients as given above is:

$$ H(z) = K\frac{z^2-2z\cos\omega_n+1}{(z^2-2az\cos\omega_n+a^2)} $$

Where $K$ is a normalizing gain constant, if desired for adjusting the filter gain in the passband regions to be 1, given as:

$$K = \frac{1+2a\cos(\omega_n)+a^2}{2+2\cos(\omega_n)}$$

$\omega_n$ is the notch frequency in normalized radian frequency in units of radians/sample, determined by dividing the radian frequency by the sampling rate.

$\alpha$ is the radius to the pole for the filter, which for narrow bandwidths is close to 1. The value for $\alpha$ is approximately $1-\omega/2$, where $\omega$ is the 3 dB bandwidth of the notch in radians/sample. The exact solution is detailed in this post. For high-Q designs (tight notch) and for relatively high sampling rates compared to the notch frequencies the 2nd order biquad (as this is called) can be challenging to implement with fixed-point designs, and it is imperative to confirm results with bit accurate simulations. If available precision is an issue, alternate structures exist as suggested in this blog post by Rick Lyons, including the "Coupled Form" by Rader and Gold (1967) and improved structures by Agarwal and Burrus (1975).

For example if the sampling rate was 1200 Hz, the first filter associated with 50 Hz would have:

$$\omega_1 = \frac{(2\pi) 50}{1200}$$

And the second filter (repeating the structure above with the two filters in series) with the notch at 60 Hz would have:

$$\omega_2 = \frac{(2\pi) 60}{1200}$$

If we wanted a 1 Hz bandwidth notch, then using the approximation:

$$\alpha = 1 - \frac{(2\pi) 5}{1200} = 0.99476$$

With the resulting frequency response:

freq response

Dan Boschen
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The most common method to do AC line rejection is to use a comb filter. Set the delay such that the notches occur at 60, 180, etc.

  • Thank you! 2 more quesitions! The Bode Plot in my post shows it has 25SPS output data rate and 1/25=40ms settling time. Fisrt, ignore the SINC5, go through a SINC1 $\frac{1-z^{-osr}}{1-z^{-1}}$, the data rate is $\frac{Fs}{osr}$ and DC settling time is $\frac{osr}{Fs}$. Then I cascade a line rejeciton filter $( 1+z^\frac{-Fs}{osr50} )( 1+z^\frac{-Fs}{osr60} )$ that porvides both 50Hz and 60Hz rejection. Now how could I gurantee the DC response time? What should I do if $\frac{Fs}{osr*60}$ is not an integer? Is any misconception in my comments? thank you! – Ring Feb 15 '23 at 01:43