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Q&A

Low-pass filter after the output DAC in CD players

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[Disclaimer. This is not for an academic class. I'm self-studying.]

I’m reading an introductory book on DSP for audio and computer music [Steiglitz 1996, ISBN 0-8053-1684-1 p. 287].

One of the sections discusses oversampled D/A conversion at the output of the CD players. The audio stream is at 44.1ksps sampling frequency. It's digitally interpolated to 176.4ksps (4x the original sampling frequency) before the D/A. The higher sample rate allows for a simpler and cheaper analog filter downstream of the D/A.

That image is centered at 176.4kHz, way beyond the range of hearing, and it doesn’t take much of an analog filter to do a good job removing it (but see Problem 4).

Problem 4: The first substantial image of the baseband signal in a CD player after oversampled D/A conversion is well beyond the range of human hearing. Why is it still important to filter it out? (Hint: It helps to know something about electronics here.)

My initial ideas for the answer:

  • The audio amplifiers may be nonlinear at high frequencies, and the nonlinearities from the high frequency components may show up in the audible band.
  • EMC. But I doubt that EMC is the answer.
  • Power savings. Why waste power driving the speaker with high frequency, if it isn’t audible.

What do you think?

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[In addition to what Olin wrote earlier.]

I looked-up one of the references from the textbook, and I may have found the answer which the author had in mind.

This entire spectrum must not be passed on to the player amplifier and loudspeaker. Even though the frequencies above 20 kHz are inaudible, they would overload the player amplifier and set up intermodulation products with the baseband frequencies or possibly with the high-frequency bias current of a tape recorder. Therefore all signals at frequencies above the baseband should be attenuated by at least 50 dB.

The article also described the DC reconstruction filter. It's a 3rd order RC filter.

[Here baseband is the audible frequency range between 20 Hz and 20 kHz.]

D. Goedhart, R. J. van de Plassche, and E. F. Stickvoort, “Digital-to-Analog Conversion in Playing a Compact Disc,” Philips Technical Review, vol. 40, no. 6, 1982, pp. 174-179

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I don't know what the book has in mind, but your first point is the main reason I would want to filter out the high frequencies with passive analog components.

Low noise and low distortion are important in audio. It makes a real difference when the amplifier has to still have low noise and distortion to 200 kHz instead of 20 kHz. The former needs 10x the gain bandwidth product to do the same thing, for example.

We often forget that opamps and the like don't really have infinite gain, although the simplified opamp equations usually assume that. The assumption is valid enough as long as the actual closed loop gain is significantly less than the amplifiers open loop gain. The closed loop gain of an opamp circuit is:

    Gain = F / (1 + F/G)

where G is the open loop gain of the opamp and F is the feedback factor, which is the inverse of the gain from the output to the negative input. For example, if ¼ of the output is fed back into the negative input, then F = 4. As long as G >> F, the gain is close enough to F.

Usually we want G to be at least 10x F, which means the final gain can be from F to 0.91 F, or up to 0.28 dB below F. An absolute error of 3 dB in an audio circuit is usually of little consequence since the overall volume is probably user-adjustable anyway.

However, if the gain varies over frequency then it matters. "Good" audio circuits are expected to have a flat gain within 3 dB over the 20 Hz to 20 kHz range. The open loop gain of amplifiers can vary significantly over frequency. For opamps, there is a single dominant pole at a low frequency. The result is characterized as a minimum guaranteed gain⋅bandwidth product. Another way to look at it is that the gain⋅bandwidth product is the frequency at which the gain drops to 1.

Let's say we have an opamp with 1 MHz gain⋅bandwidth. To use our 10x rule of thumb, we want the gain at 20 kHz to be 10x the closed loop gain. That means this opamp can't be used to amplify audio signals by more than 5x. To get the same 5x and be able to handle frequencies up to 200 kHz without causing trouble, we'd need a 10 MHz gain⋅bandwidth opamp. That's going to have more noise than the 1 MHz gain⋅bandwidth opamp when both are designed for low noise.

And, it isn't all about flat gain. To handle higher frequencies, you need higher slew rate to keep the system linear. If the system becomes non-linear, then those high frequencies your ears will ignore can cause noise components at frequencies you won't ignore. They can also cause noise at even higher frequencies, which then cause even more artifacts at audible frequencies.

Consider the case where the amplifier is presented with a sine wave past its design point. That could turn into a triangle wave, for example. That means lots of high frequencies were created by the distortion. Some of those frequencies will be high enough so that individual components don't act linearly anymore, causing artifacts that ultimately become audible noise. It gets messy.

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Closed-loop gain formula (4 comments)
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The oversampling is said to simplify the LPF yet the better reason is that it improves the SNR.

Lower noise comes from better image rejection, lower filter ringing, and significantly lower group delay distortion in the audio band as the band edge is moved up.

The majority of phase shift traverses over +/-1 frequency decade from the breakpoint and group delay is the rate of change of phase shift, you can imagine that shifting the rate to 176.4 kHz eliminates the group delay distortion 1 decade down from there while providing adequate image rejection with a linear phase FIR filter.

Filter skirts only attenuate -6dB per 2f octave per order of filter thus the process of oversampling provides the greatest benefit to SNR.

https://www.digikey.in/en/articles/why-and-how-to-use-digital-filters-for-analog-to-digital-conversions

https://www.technics.com/uk/support/discontinued-products/premium-class/sl-c700.html Image alt text

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One of the main reasons was hinted by Olin's answer, at the end, when he spoke about non-linearities. The same is hinted in the excerpt in Nick Alexeev's answer ("they would overload the player amplifier").

I want to be explicit and mention a very easy example: consider a simple amplifier with a voltage gain of 10, bandwidth of 200kHz (even if it is intended only for audio signals), and an output dynamic range of 10Vpp.

If you feed it with a signal composed with two sine waves, one at 1kHz with amplitude 100mVpp, the other at 100kHz with amplitude 100mVpp, you get both signal amplified at the output, with 1Vpp amplitude for both. The speaker will filter out the 100kHz component and all is fine.

Now increase the amplitude of the 100kHz input component to 2Vpp. If the amplifier were ideal, you would get an output component at 100kHz with a 20Vpp amplitude. But the output dynamic range is just 10V and the amp will begin saturating just when the 100kHz component is at half its amplitude (the tiny superimposed 1kHz component is just irrelevant at this point). So for the most part of the 100kHz cycle the amp is in full saturation, so it won't amplify anything.

And even during the most of the rest of the cycle, when the amp is near saturation, it will have a very non-linear response and it will distort heavily, producing a lot of spurious harmonics, and cross-modulation products between the two signals.

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