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Comments on Low-pass filter after the output DAC in CD players

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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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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)
Closed-loop gain formula
LvW‭ wrote about 2 years ago · edited about 2 years ago

You wrote: Gain = F / (1 + F/G).

When G is the open-loop gain and F the feedback factor , the closed-loop gain is (according to Blacks famous formula): Gain Acl=G/(1+F*G)

Olin Lathrop‭ wrote about 2 years ago · edited about 2 years ago

That's the same thing with F flipped. You can rearrange my equation to get Gain = G / (1+ G/F). I defined F as "the fraction of the output fed back to the negative input". I suspect your reference defined F as the negative feedback gain. That's the reciprocal of what I used. Substitute this alternate F into the formula above, and you get exactly what you wrote.

LvW‭ wrote about 2 years ago · edited about 2 years ago

Sorry, but I have problems to agree. To me the quantity F is the feeedback factor - which is identical to "the fraction of the output fed back to the negative input" (your definition) . That means: F is always smaller than unity. Example: Feedback resistor is R2 and resistor R1 is from the inv. input to ground.

Hence: F=R1/(R1+R2) and the closed-loop gain for the non-inv. amplifier is as expected (assuming very large open loop gain G): Acl=1/F=1+R2/R1.

See what happens with your formula for infinite gain G: Closed-loop gain would be Acl=F (smaller than unity). And this result is not correct.

Olin Lathrop‭ wrote about 2 years ago

@LvW: Reading that over I realize I wrote that poorly. My F is the reciprocal of your F. I have updated the post to hopefully describe that better.