Heavy Metal Analogue

In order to explain the significance of the following story, some prequels are required.

Prequel #1: I’m one of those people who enjoys an addiction to collecting what other people call “junk” – things you find in flea markets, estate sales, and the like. Normally I only come home with old fountain pens that need to be restored, however, occasionally, I stumble across other things.

Prequel #2: Many people have vinyl records lying around, but not many people know how they’re made. The LP that you put on your turntable was pressed from a glob of molten polyvinyl-chloride (PVC), pressed between two circular metal plates called “stampers” that had ridges in them instead of grooves. Easy of those stampers was made by depositing layers of (probably) nickel on another plate called a “metal mother” which is essentially a metal version of your LP. That metal mother was made by putting layers on a “metal master” (also with ridges instead of grooves) which was probably a lamination of tin, silver, and nickel that was deposited in layers on an acetate lacquer disc, which is the original, cut on a lathe. (Yes, there are variations on this process, I know…) The thing to remember in this process is

  • there are three “playable” versions of the disc in this manufacturing process: your LP, the metal mother, and the original acetate that was cut on the lathe
  • there are two other non-playable versions that are the mirror images of the disc: the metal master and the stamper(s).

(If you’d like to watch this process, check out this video.)

Prequel #3: One of my recurring tasks in my day-job at Bang & Olufsen is to do the final measurements and approvals for the Beogram 4000c turntables. These are individually restored by hand. It’s not a production-line – it really is a restoration process. Each turntable has different issues that need to be addressed and fixed. The measurements that I do include:

  • verification of the gain and response of the two channels in the newly-built RIAA preamplifier
    (this is done electrically, by connecting the output of my sound card into the input of the RIAA instead of using a signal from the pickup)
  • checking the sensitivity and response of the two channels from vinyl to output
  • checking the wow and flutter of the drive mechanism
  • checking the channel crosstalk as well as the rumble

The last three of these are done by playing specific test tracks off an LP with signals on it, specifically designed for this purpose. There are sine wave sweeps, sine waves at different signal levels, a long-term sine wave at a high-ish frequency (for W&F measurements), and tracks with silence. (In addition, each turntable is actually tested twice for Wow and Flutter, since I test the platter and bearing before it’s assembled in the turntable itself…)

Prequel #4: Once-upon-a-time, Bang & Olufsen made their own pickup cartridges (actually, it goes back to steel needles). Initially the SP series, and then the MMC series of cartridges. Those were made in the same building that I work in every day – about 50 m from where I’m sitting right now. B&O doesn’t make the cartridges any more – but back when they did, each one was tested using a special LP with those same test tracks that I mentioned above. In fact, the album that they used once-upon-a-time is the same album that I use today for testing the Beogram 4000c. The analysis equipment has changed (I wrote my own Matlab code to do this rather than to dust off the old B&K measurement gear and the B&O Wow and Flutter meter…)

If you’ve read those four pieces of information, you’ll understand why I was recently excited to stumble across a stamper of the Bang & Olufsen test LP, with a date on the sleeve reading 21 March, 1974. It’s funny that, although the sleeve only says that it’s a Bang & Olufsen disc, I recognise it because of the pattern in the grooves (which should give you an indication of how many times I’ve tested the turntables) – even if they’re the mirror image of the vinyl disc.

Below, you can see my latest treasure, pictured with an example of the B&O test disc that I use. It hasn’t “come home” – but at least it’s moved in next-door.

P.S. Since a couple of people have already asked, the short answer is “no”. The long answers are:

  • No, the test disc is no longer available – it never was outside of the B&O production area. However, if you can find a copy of the Brüel and Kjær QR 2010 disc, it’s exactly the same. I suspect that the two companies got together to produce the test disc in the 70s. However, there were also some publicly-available discs by B&O that included some test tones. These weren’t as comprehensive as the “real” test discs like the ones accompanying the DIN standards, or the ones from CBS and JVC.
  • No, the metal master is no longer in good enough shape to use to make a new set of metal mothers and stampers. Too bad… :-(

P.P.S. If you’re interested in the details of how the tests are done on the Beogram 4000c turntables, I’ve explained it in the Technical Sound Guide, which can be downloaded using the link at the bottom of this page. That document also has a comprehensive reading list if you’re REALLY interested or REALLY having trouble sleeping.

“High-Res” Audio: Part 13: Wrapping up

As I’ve stated a couple of times through this series, my reason for writing this stuff was not to prove that high res audio is better or worse than normal res audio (whatever that is…). My reason was to highlight some of the advantages and disadvantages associated with LPCM audio at different bit depths and sampling rates. Just as a bullet-point summary of things-to-remember/consider (with some loose grouping):

  • “High resolution audio” could mean
    • “more than 16 bits per sample”
      or
    • “a sampling rate higher than 44.1 kHz”
      or
    • both.
  • These two dimensions of the specifications have different implications on the signal

  • Doubling the sampling rate only increases your audio bandwidth by 1 octave.
    Yes, it’s twice as much information, but that’s only one octave. If you add an extra octave on top of a piano, you don’t get twice as many notes.
  • Just because you have more bits per sample doesn’t mean that you are actually getting more resolution.
    There are examples out there where a “24-bit recording” is just a 16-bit recording with 8 zeros stuck on the end.
  • Just because you have a higher sampling rate doesn’t mean that you are actually getting a recording that was done at that sampling rate.
    There are examples out there where, if you do a spectral analysis of a “high-res” recording, you’ll see the cutoff filter of the original 44.1 kHz recording.
  • Just because you have a recording done at a higher sampling rate doesn’t mean that the extra information you get is actually useful.



  • There are many cases where you want equipment that has higher specifications than your audio signal.
  • If you have a volume control after the conversion to analogue, then 93 dB of dynamic range (16 bits, TPDF dithered) might be enough – especially if you listen to music with a limited dynamic range. However, if your volume control is in the digital domain, and you have a speaker that can play loudly, then you’ll probably want more dynamic range, and therefore more bits per sample hitting the DAC.

Like I said, I’m not here to tell you that one thing is better or worse than another thing.

As I said, my intention in writing all of this is to help you to never fall into the trap of assuming that “high resolution audio” is better than “normal resolution audio” in all respects.

More is not necessarily better, sometimes, it’s not even more. Don’t fall victim to misleading advertising.

“High-Res” Audio: Part 12: Outputs

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8a
Part 8b
Part 9
Part 10
Part 11

This series has flipped back and forth between talking about high resolution audio files & sources and the processing that happens in the equipment when you play it. For this posting, we’re going to deal exclusively with the playback side – regardless of the source content.

I work for a company that makes loudspeakers (among other things). All of the loudspeakers we make use digital signal processing instead of resistors, capacitors, and inductors because that’s the best way to do things these days…

Point 1: This means that our volume control is a gain (a multiplier) that’s applied to the digital signal.

We also make surround processors (most of our customers call them “televisions”) that take a multichannel audio input (these days, this is under the flag of “spatial audio”, but that’s just a new name on an old idea) and distribute the signals to multiple loudspeakers. Consequently, all of our loudspeakers have the same “sensitivity”. This is a measurement of how loud the output is for a given input.

Let’s take one loudspeaker model, Beolab 90, as an example. The sensitivity of this loudspeaker is set to be the same as all other Bang & Olufsen loudspeakers. Originally, this was based on an analogue signal, but has since been converted to digital.

Point 2: Specifically, if you send a 0 dB FS signal into a Beolab 90 set to maximum volume, then it will produce a little over 122 dB SPL at 1 m in a free field (theoretically).

Let’s combine points 1 and 2, with a consideration of bit depth on the audio signal.

If you have a DSP-based loudspeaker with a maximum output of 122 dB SPL, and you play a 16-bit audio signal with nothing but TPDF dither, then the noise floor caused by that dither will be 122 – 93 = 29 dB SPL which is pretty loud. Certainly loud enough for a customer to complain about the noise coming from their loudspeaker.

Now, you might say “but no one would play a CD at maximum volume on that loudspeaker” to which I say two things:

  1. I do.
    The “Banditen Galop” track from Telarc’s disc called “Ein Straussfest” has enough dynamic range that this is not dangerous. You just get very loud, but very short spikes when the gunshots happen.
  2. That’s not the point I’m trying to make anyway…

The point I’m trying to make is that, if Beolab 90 (or any other Bang & Olufsen loudspeaker) used 16-bit DACs, then the noise floor would be 29 dB SPL, regardless of the input signal’s bit depth or dynamic range.

So, the only way to ensure that the DAC (or the bit depth of the signal feeding the DAC) isn’t the source of the noise floor from the loudspeaker is to use more than 16 bits at that point in the signal flow. So, we use a 24-bit DAC, which gives us a (theoretical) noise floor of 122 – 141 = -19 dB SPL. Of course, this is just a theoretical number, since there are no DACs with a 141 dB dynamic range (not without doing some very creative cheating, but this wouldn’t be worth it, since we don’t really need 141 dB of dynamic range anyway).

So, there are many cases where a 24-bit DAC is a REALLY good idea, even though you’re only playing 16-bit recordings.

Similarly, you want the processing itself to be running at a higher resolution than your DAC, so that you can control its (the DAC’s) signal (for example, you want to create the dither in the DSP – not hope that the DAC does it for you. This is why you’ll often see digital signal processing running at floating point (typically 32-bit floating point) or fixed point with a wider bit depth than the DAC.

“High-Res” Audio: Part 10: The myth of temporal resolution

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8a
Part 8b
Part 9

If you read about high resolution audio – or you talk to some proponents of it, occasionally you’ll hear someone talk about “temporal resolution” or “micro details” or some such nonsense… This posting is just my attempt to convince the world that this belief is a load of horse manure – and that anyone using timing resolution as a reason to use higher sampling rates has no idea what they’re talking about.

Now that I’ve gotten that off my chest, let’s look at why these people could be so misguided in their belief systems…

Many people use the analogy of film to explain sampling. Even I do this – it’s how I introduced aliasing in Part 3 of this series. This is a nice analogy because it uses a known concept (converting movement into a series of still “samples”, frame by frame) to explain a new one. It also has some of the same artefacts, like aliasing, so it’s good for this as well.

The problem is that this is just an analogy – digital audio conversion is NOT the same as film. This is because of the details when you zoom in on a time scale.

Film runs at 24 frames per second (let’s say that’s true, because it’s true enough). This means that the time between on frame of film being shot and the next frame being shot is 1/24th of a second. However, the shutter speed – the time the shutter is open to make each individual photograph is less than 1/24th of a second – possibly much less. Let’s say, for the purposes of this discussion, that it’s 1/100th of a second. This means that, at the start of the frame, the shutter opens, then closes 1/100th of a second later. Then, for about 317/10,000ths of a second, the shutter is closed (1/24 – 1/100 ≈ 317/10,000). Then the process starts again.

In film, if something happened while that shutter was closed for those 317 ten-thousandths of a second, whatever it was that happened will never be recorded. As far as the film is concerned, it never happened.

This is not the way that digital audio works. Remember that, in order to convert an analogue signal into a digital representation, you have to band-limit it first. This ensures (at least in theory…) that there is no signal above the Nyquist frequency that will be encoded as an alias (a different frequency) in the digital domain.

When that low-pass filtering happens, it has an effect in the time domain (it must – otherwise it wouldn’t have an effect in the frequency domain). Let’s look at an example of this…

Let’s say that you have an analogue signal that consists of silence and one almost-infinitely short click that is converted to LPCM digital audio. Remember that this click goes through the anti-aliasing low-pass filter, and then gets sampled at some time. Let’s also say that, by some miracle of universal alignment of planets and stars, that click happened at exactly the same time as the sample was measured (we’ll pretend that this is a big deal and I won’t suggest otherwise for the rest of this posting). The result could look like Figure 1.

Figure 1: A click in the analogue domain (red) sampled and encoded as a digital signal (black circles). The analogue output of the digital system will be the black line

If I zoom in on Figure 1 vertically, it looks like the plot in Figure 2.

Figure 2: A vertical zoom of Figure 1.

There are at least three things to notice in these plots.

  • Since the click happened at the same time as a sample, that sample value is high.
  • Since the click happened at the same time as a sample, all other sample values are 0.
  • Once the digital signal is converted back to analogue later (shown as the black line) the maximum point in the signal will happen at exactly the same time as the click

I won’t talk about the fact that the maximum sample value is lower than the original click yet… we’ll deal with that later.

Now, what would happen if the click did not occur at the same time as the sample time? For example, what if the click happened at exactly the half-way point between two samples? This result is shown in Figure 3.

Figure 3: The click happened half-way in time between two samples.

Notice now that almost all samples have some non-zero value, and notice that the two middle samples (8 and 9) are equal. This means that when the signal is converted to analogue (as is shown with the black line) the time of maximum output is half-way between those two samples – at exactly the same time that the click happened.

Let’s try some more:

Figure 4: The click happened sometime between two samples
Figure 5: The click happened some other time between two samples.

I could keep doing this all night, but there’s no point. The message here is, no matter when in time the click happened, the maximum output of the digital signal, after it’s been converted back to analogue, happens at exactly the same time.

But, you ask, what about all that “temporal smearing” – the once-pristine click has been reduced to a long wave that extends in time – both forwards and backwards? Waitaminute… how can the output of the system start a wave before something happened?

Okay, okay…. calm down.

Firstly, I’ve made this example using only one type of anti-aliasing filter, and only one type of reconstruction filter. The waveforms I’ve shown here are valid examples – but so are other examples… This depends on the details of the filters you use. In this case, I’m using “linear phase” filters which are symmetrical in time. I could have used a different kind of filter that would have looked different – but the maximum point of energy would have occurred at the same time as the click. Because of this temporal symmetry, the output appears to be starting to ring before the input – but that’s only because of the way I plotted it. In reality, there is a constant delay that I have removed before doing the plotting. It’s just a filter, not a time machine.

Secondly, the black line is exactly the same signal you would get if you stayed in the analogue domain and just filtered the click using the two filters I just mentioned (because, in this discussion, I’m not including quantisation error or dither – they have already been discussed as a separate topic…) so the fact that the signal was turned into “digital” in between was irrelevant.

Thirdly, you may still be wondering why the level of the black line is so low compared to the red line. This is because the energy is distributed in time – so, in fact, if you were to listen to these two clicks, they’d sound like they’re the same level. Another way to say it is that the black line shows exactly the same as if the red curve was band-limited. The only thing missing is the upper part of the frequency band. (You may notice that I have not said anything about the actual sampling rate in any of this posting, because it doesn’t matter – the overall effect in the time domain is the same.)

Fourthly, hopefully you are able to see now that an auditory event that happens between two samples is not thrown away in the conversion to digital. Its timing information is preserved – only its frequency is band-limited. If you still don’t believe me, go listen to a digital recording (which is almost all recordings today) of a moving source – anything moving more than 7 mm will do*. If you can hear clicks in the sound as the source moves, then I’m wrong, and the arrival time of the sound is quantising to the closest sample. However, you won’t hear clicks (at least not because the source is moving), so I’m not wrong. Similarly, if digital audio quantised audio events to the nearest sample, an interpolated delay wouldn’t work – and since lots of people use “flanger” and “phaser” effects on their guitar solos with their weekend garage band, then I’m still right…

The conclusion

Hopefully, from now on, if you are having an argument about high resolution audio, and the person you’re arguing with says “but what about the timing information!? It’s lost at 44.1 kHz!” The correct response is to state (as calmly as possible) “BullS#!T!!”

Rant over.


* I said “7 mm” because I’m assuming a sampling rate of 48 kHz, and a speed of sound of 344 m/s. This means that the propagation distance in air is 344/48000 = 0.0071666 m per sample. In other words, if you’re running a 48 kHz signal out of a loudspeaker, the amplitude caused by a sample is 7 mm away when the next sample comes out.

Thought another way, if you have a stereo system, and your left loudspeaker is 7 mm further away from you than your right loudspeaker, at 48 kHz, you can delay the right loudspeaker by 1 sample to re-align the times of arrival of the two signals at the listening position.

On to Part 11

“High-Res” Audio: Part 9: Filters at high frequencies

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8a
Part 8b

We’ve already seen that nothing can exist in the audio signal above the Nyquist frequency – one half of the sampling rate. But that’s the audio signal, what happens to filters? Basically, it’s the same – the filter can’t modify anything above the Nyquist frequency. However, the problem is that the filter doesn’t behave well to everything up to the Nyquist and then stop, it starts misbehaving long before that…

Let’s make a simple filter: a peaking filter where Fc=1 kHz, Gain = 12 dB, and Q=1. The magnitude response of that filter is shown in Figure 1.

Figure 1: Magnitude response of a filter with Fc=1 kHz, Gain = 12 dB, and Q=1

What happens if we implement this filter with a sampling rate of 48 kHz and 192 kHz, and then look at the difference between the two? This is shown in Figure 2.

Figure 2: TOP: The black curve shows the same filter from Figure 1, implemented using a biquad running at 48 kHz. The red dotted line shows the same filter, implemented using a biquad running at 192 kHz. BOTTOM: the difference between the two up to the Nyquist frequency of the 48 kHz system. (there’s almost no difference)

As you can see in Figure 2, the filter, centred at 1 kHz, is almost identical when running at 48 kHz and 192 kHz. So far so good. Now, let’s move Fc up to 10 kHz, as shown in Figure 3.

Figure 3: TOP: The black curve shows the magnitude response of a filter with Fc=10 kHz, Gain = 12 dB, and Q=1, implemented using a biquad running at 48 kHz. The red dotted line shows the same filter, implemented using a biquad running at 192 kHz. BOTTOM: the difference between the two up to the Nyquist frequency of the 48 kHz system.

Take a look at the black plot on the top of Figure 3. As you can see there, the 48 kHz filter has a gain of 0 dB at 24 kHz – the Nyquist frequency. Looking at the red dotted line, we can see that the actual magnitude of the filter should have been more than +3 dB. Also, looking at the red line in the bottom plot, which shows the difference between the two curves, the 48 kHz filter starts deviating from the expected magnitude down around 1 kHz already.

So, if you want to implement a filter that behaves as you expect in the high frequency region, you’ll get better results easier with a higher sampling rate.

However, do not jump to the conclusion that this also means that you can’t implement a boost in high frequencies. For example, take a look at Figure 4, which shows a high shelving filter where Fc = 1 kHz, Gain = 12 dB and Q = 0.707.

Figure 4: High shelving filter where Fc = 1 kHz, Gain = 12 dB and Q = 0.707.

As you can see in the bottom plot in Figure 4, the two filters in this case (one running at 48 kHz and the other at 192 kHz) have almost identical magnitude responses. (Actually, there is a small difference of about 0.013 dB…) However, if the Fc of the shelving filter moves to 10 kHz instead (keeping the other two parameters the same) then things do get different.

Figure 5: High shelving filter where Fc = 10 kHz, Gain = 12 dB and Q = 0.707.

As can be seen there, there is a little over a 1 dB difference in the two implementations of the same filter.

Some comments:

  • I’m not going to get into exactly why this happens. If you want to learn about it, look up “bilinear transform”. The short version of the issue is that these filters are designed to work in a system with an infinite sampling rate and bandwidth (a.k.a. analogue), but the band-limiting of an LPCM digital system makes things misbehave as you get near the Nyquist frequency.
  • This does not mean that you cannot design a filter to get the response you want up to the Nyquist frequency.
    If you look at the red dotted curve in Figure 3 and call that your “target”, it is possible to build a filter running at 48 kHz that achieves that magnitude response. It’s just a little more complicated that calculating the gain coefficients for the biquad and pretending as if you were normal. However, if you’re a DSP Engineer and your job is making digital filters (say, for correcting tweeter responses in a digitally active loudspeaker, for example) then dealing with this issue is exactly what you’re getting paid for – so you don’t whine about it being a little more complicated.
  • The side-effect of this, however, is that, if you’re a lazy DSP engineer who just copies-and-pastes your biquad coefficient equations from the Internet, and you just plug in the parameters you want, you aren’t necessarily going to get the response that you think. Unfortunately, this is not uncommon, so it’s not unusual to find products where the high-frequency filtering measures a little strangely, probably because someone in development either wasn’t meticulous or didn’t know about this issue in the first place.

On to Part 10

“High-Res” Audio: Part 8b: Filter Resolution

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8a

In the previous posting, we left off with this drawing of a biquad filter:

Figure 1: A biquad broken into an FIR filter with two 1-sample long delays and 3 gains combined with an IIR filter with two delays and 2 gains.

This is not the normal way to draw the signal flow inside a biquad, since it has a little too much information. Normally you see something like this:

Figure 2: A Biquad shown in “Direct Form 1” implementation with the feed forwards coming first.

or this:

Figure 3: A simpler way to show the same thing

In the versions I show above, the feed-forward half of the biquad comes first, and its output feeds the start of the feedback portion. It is also possible to reverse these, putting the feedback portion first, like this:

Figure 4: A Biquad shown in “Direct Form 2” implementation with the feedback loops coming first.

In theory, these different implementations will all result in the same output if you match the gain values. However, in practice, they are not the same, and this difference is where we need to look for this part of the discussion on high res audio.

Let’s say I want to make a simple filter that reduces bass in a fairly narrow frequency band. I can use a biquad to do this. For example, if I want a peaking filter that reduces 20 Hz by 12 dB, with a Q of 1, then I get a magnitude response that looks like this:

Figure 5: The magnitude response of a peaking filter where F = 20 Hz, G = -12 dB, Q = 1

If I wanted to build this filter using a biquad in a system with a sampling rate of 48 kHz, it would have the following gain coefficients:

b0 = 0.998049357193933
b1 = -1.994783192754608
b2 = 0.996740671594426
a1 = -1.994783192754608
a2 = 0.994790028788359

We’ll also say that my biquad is implemented like the one shown in Figure 1, above… let’s take a look at that signal flow again:

Figure 6: A copy of Figure 1 with one interior point in the signal flow highlighted in red.

I’ve highlighted a point inside the biquad using a red arrow. Let’s talk about the signal right there, in the middle of the processing…

In the last post, we talked about how, when the signal frequency is very low, a single sample delay has almost the same value at its output as its input, because the phase difference is so small for such a small time. So, let’s start with the (incorrect) assumption that, for those two feed-forward delays at the beginning, their outputs ARE equal to their inputs (because we’re starting with a low frequency). What happens when the input has a value of 1? Then the value at the red arrow is just the sum of the feed forward gains (because I multiplied each of them by 1 and added them together…)

In the case of the filter I described above, this value will be 0.000006836, which is a very small number. Also, if the value coming into the input of the biquad is less than 1, the value at the red arrow will be even smaller! This means that, if you come into the biquad with a low-frequency tone with a level of 0 dB FS, the level at that red arrow will be about -103 dB FS, which is very quiet. The feed-back portion of the biquad, after the red arrow, then has a lot of gain in it to bring the signal level back up towards 0 dB FS again.

So, the issue that we have here is that the FF (Feed Forward) portion of the biquad drops the level A LOT. And the FB portion increases the level A LOT, just to do something like a little 12 dB dip at 20 Hz.

The magnitude of the gains downwards and upwards in those two portions of the biquad are dependent on the parameters of the filter that we’re trying to make, however, we can generalise a little and say that:

  • the lower the frequency
    OR
  • the higher the Q,
  • then the bigger the gain down and up.

In other words, if you have a really low frequency dip, with a really high Q, then the level of the signal at that red arrow will be really low. REALLY low.

How low can you go?

How low is “REALLY low”? let’s see:

Figure 7: The level of a 0 dB FS signal at various frequencies listed in the top right corner, measured inside the biquad shown in Figure 6 at the red arrow, for a peaking filter with a gain of -12 dB, and with variable Q and Fc, in a system running at 48 kHz.

Take a look at Figure 7, which shows some values for one example filter (peaking, Gain = -12 dB, variable Q and Fc, and the test frequency = Fc). Notice that when the Fc is 10 kHz, even at earn Q=32, the signal level at the middle of the biquad is about -38 dB FS or so. However, when the Fc is 20 Hz, it’s -140 dB FS… This is very low.

Now let’s try again at a higher sampling rate: 192 kHz.

Figure 8: Identical parameters to that shown in Figure 7, but in a system running at 192 kHz.

Notice that when we do exactly the same thing running at 192 kHz, the signal levels inside the biquad get much lower. Now for a 20 Hz signal and a Q of 32, the level is around -163 dB FS – a drop of more than 20 dB for 4x the sampling rate.

Why does this happen? It’s because the filter doesn’t “know” that the signal is at 20 Hz. It only knows the relationship between the frequency and the sampling rate. So, in its little world, 20 Hz doesn’t exist. In a system running at 48 kHz, what exists is 20 / 48000 = 0.0004167. This is called the “normalised frequency” where the sampling rate is 1, DC is 0, and everything else is in between. (Note that some textbooks and software say that Nyquist = 1 instead of the sampling rate – but you just need to know what the convention is for the thing you’re reading…) This means that if the sampling rate goes up to 192 kHz, then the normalised frequency for 20 Hz is 20 / 192000 = 0.0001042 (1/4 of the value because the sampling rate was multiplied by 4).

So what?

This is important. If you want to make a low-frequency, high-Q peaking filter in a digital system with a cut of 12 dB, you are forcing the signal to a very low level inside your filter, and then bringing it back up to a normal level again on the way out. If your processing is running with a limited resolution, (e.g. 16-bits, for example) then the signal level can approach or even go below the resolution of your system inside the biquad. This means that, when the signal’s level is raised again on the way out, it’s full of quantisation distortion, and you can’t get rid of it… This is bad.

There are different ways to solve this problem.

  • Increase the resolution of your processing internally. For example, even though your input and output might only be running at 16-bits or 24-bits, maybe you need more resolution inside to make the results of the math better – or at least below the limitations of the input and output.
  • Change the way the biquad is implemented. For example, if you use the implementation shown in Figure 4 (with the feedback before the feed-forward) instead of the one we used, then you don’t drop the signal level and raise it again, you do the opposite. This avoids your quantisation error problem. However, depending on the system, it might overload and clip the signal inside the biquad instead, so then you just end up with a different kind of distortion instead.
  • Reduce your sampling rate to make it closer to your filter’s frequency. The problem I showed above is that the centre frequency of the filter is too far away from the sampling rate. If the sampling rate were lower, then this automatically makes the filter’s centre frequency “higher” in a normalised frequency scale, thus reducing the problem.
  • Other, even more clever solutions that I won’t talk about because they’re not as simple.

This means (for example) that if you’re building a subwoofer with digital filtering, and you know for sure that NOTHING will come out of it above, say 1 kHz (just to pick a random number that’s far enough away from the typical 120 Hz that people normally use…) then it would be dumb to do the filtering at 192 kHz. It’s smarter to run its internal sampling rate at 2 kHz (because we only need to go up to 1 kHz; and we’re not considering anything other issues or artefacts in this posting.)

P.S.

For this discussion, I used the specific example of a peaking filter with a gain of -12 dB, and I was varying the Q and the Fc. I was also measuring the level of the signal using a sine wave with a frequency that was the same as Fc in each case. However, the general lesson here about low frequency and high-Q filtering holds for other filter types and implementations as well.

On to Part 9.

“High-Res” Audio: Part 8a: Filter resolution (The Setup…)

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7

Almost every audio system has filters or equalisers in it for some reason or another. Originally, equalisers were named that because they were put in on long-distance telephone lines to make the balance of the frequency content more equal. Nowadays, we use equalisers to do things like add bass, or to add more bass.

In the “old days” audio filters were made by building circuits with resistors, capacitors, and inductors: if you choose the relationships between the values of these devices correctly, you can affect the magnitude response as you choose. The problem was production tolerances: if you take two resistors out of the package, and both are supposed to have the same resistance – they’ll be close, but they won’t be identical.

One of the great things about audio filters implemented in a digital system is that you don’t need to worry about variations as a result of production differences. Since digital filters are “just math”, you put the same equation in every device, and you get the same answer for the same input every time. (In the same way that, if I have two calculators on my desk, and I put “2 x 3″ into both of them, and press”=”, I’ll get the same answer on both devices.)

So, to start, let’s talk a little about how a digital filter works. Generally speaking, digital filters work by taking an audio signal, delaying it, changing the level, and adding the result back to the signal itself. Let’s take a simple example, shown in Figure 1.

Figure 1: Simple digital filter

Let’s say that, to start, we make the gain in that signal flow = 1, and set the delay to equal 1 sample. In this case:

  • At a very low frequency, the output of the delay has almost exactly the same value as its input (because 1 sample is a phase difference of almost 0º at a low frequency). When you add a signal to (nearly) itself, you get twice the output – a gain of 6 dB.
  • As the frequency of the input goes higher and higher, the delay (of 1 sample) is more and more significant, and therefore its output value gets more and more different from its input value.
  • When the input signal’s frequency is 1/2 of the sampling rate, then a delay of 1 sample is equal to a 90º phase shift. When you add a sine wave to itself with a “delay” (actually a “phase shift”) of 90º, the result is a magnitude that is 3 dB higher than the original.
  • When the input signal’s frequency is 1/3 of the sampling rate, then a delay of 1 sample is equal to a 120º phase shift. When you add a sine wave to itself with a “delay” (actually a “phase shift”) of 120º, there’s no change in the magnitude (the level).
  • When the frequency is 1/2 the sampling rate, then each consecutive sample is 180º out of phase with the previous one, so the sum of the delay and the signal results in complete cancellation, and you get no output at all.

An example of the magnitude response plot of this is shown below in Figure 2.

Figure 2: The magnitude response of the filter shown in Figure 1 when the delay is set to 1 sample, the gain is set to 1 and Fs=48 kHz.

If we reduce the gain to, say 0.5, then the effect of adding the delayed signal is reduced. The overall shape of the magnitude response is the same, it’s just less, as shown in Figure 3.

Figure 3: The magnitude response of the filter shown in Figure 1 when the delay is set to 1 sample, the gain is set to 0.5 and Fs=48 kHz.

Notice in Figure 3 that the boost in the low end is less, and the dip in the high end is also less than in Figure 2. So, by adjusting the gain on the delayed signal that’s added to the original signal, we can adjust how much this filter is affecting the signal.

What happens when we change the delay? If we make it 2 samples instead of 1, then the phase difference between the output and the input of the delay will be bigger for a given frequency. This also means that the delay will be equivalent to a 180º phase shift at 1/4 Fs (instead of 1/2). Also, at 1/2 Fs, the delay will be equivalent to a 360º phase shift, so the signal adds constructively, just like it does in the low frequencies. So, the resulting magnitude response will look like Figure 4.

Figure 4: The magnitude response of the filter shown in Figure 1 when the delay is set to 2 samples, the gain is set to 1 and Fs=48 kHz.

Again, if we reduce the gain, we reduce the effect of the filter, as can be seen in Figure 5.

Figure 5: The magnitude response of the filter shown in Figure 1 when the delay is set to 2 samples, the gain is set to 0.5 and Fs=48 kHz.

Now let’s make things a little more complicated. We can add another delay and another gain to get a little more control of things.

Figure 6: Getting a little more complicated

I’m not going to get very detailed about this – but if each of those delays is just one sample long, and we only play with the gains g1 and g2, we can start getting some nice control over the response of this filter. Below are some examples of the results we can get with just this filter, playing with the gains.

Figure 7: g1 = 0.5, g2 = 0.5
Figure 8: g1 = -0.5, g2 = 0.5
Figure 9: g1 = 0.75, g2 = -0.75
Figure 10: g1 = -1, g2 = -0.1

So, as you can see, all I need to do is to play with those two gains to get some nice control over the magnitude response.

Up to now, everything I’ve done is to add a delayed copy of the input to itself. This is what is known as a “feed-forward” design because (as you can see in Figures 1 and 6) I’m feeding the signal forwards in the flow to be added to itself. However, if there’s a “feed-forward”, it must be because we want to distinguish it from a “feed-back” design.

This is a filter where we delay the output (instead of the input) multiply that by a gain, and add it to the signal, as shown in Figure 11.

Figure 11: A simple digital filter using feedback.

This feedback means that, if the gain is not equal to zero, once a signal gets into the input, the output will last forever. This is why this kind of filter is called an infinite impulse response filter (or IIR filter): because if an impulse (a short spike) gets into it, there will be a signal at the output until the end of time (theoretically, at least…).

And, yes… the output of a filter without a feedback loop will eventually stop, which means it’s a finite impulse response filter (or FIR). Stop the input signal, wait for the last delay to send its signal through, and the output stops.

So what?

Most filters in most digital audio devices are built on a combination of these two types of designs. If you take Figure 6 and you combine it with an extended version of Figure 11 (with two delays instead of just one) you get Figure 12:

Figure 12: An FIR filter with two delays and 3 gains combined with an IIR filter with two delays and 2 gains.

This combination of FIR and IIR filters is a powerful little tool that forms the heart of almost every digital audio filter in the world. (yes, there are exceptions, but they’re definitely exceptions…). There are different ways to implement it (for example, you could put the IIR before the FIR, or you could re-draw it to share the delays) but the result is the same.

This little tool is what we call a “biquadratic filter” or “biquad” for short. (The reason it’s called that is that the effect it has on the signal (its “transfer function”) can be mathematically expressed as the ratio of two quadratic equations – but I will not say anything more about that.) Whenever developers are building a new digital audio device like a loudspeaker or a pair of headphones or a car audio system, it’s common in the early meetings to hear someone ask “how many biquads will we need?” which is a way of asking “how much processing power and memory will we need?” (In the same way that I can measure prices in pizzas, or when I was a kid I would ask “how many more Sesame Streets until we’re there?”)

At this point, you may be asking why I’ve gone through all of this, since I haven’t said anything about high resolution audio. The reason is that, in the next posting, we’ll look at what’s going on inside that biquad when you use it to do filtering – and how that changes, not only with the filters you’re building, but how they relate to the sampling rate and the bit depth…

“High-Res” Audio: Part 7: Conversion

Part 1
Part 2
Part 3
Part 4
Part 5
Part 6

Back in Part 5 of this series, I described an example of a pretty typical / normal signal flow for an audio signal that you’re playing from a streaming service to a “smart-ish” loudspeaker in your house. If you read through that list, you’ll see that I mentioned that the signal might be sampling-rate converted two times (once in your player, and once again in your loudspeaker or headphones).

Let me say something very clearly, before we go any further:

  • There’s no guarantee that this is happening.
    For example, many players don’t sampling-rate convert the signal if the device they’re sending the signal is compatible with the sampling rate of the signal. However, many players do sampling-rate convert the signal – and many devices (like DACs, for example) are not compatible with all sampling rates, so the player is forced to do something about it.
  • Sampling rate conversion is not necessarily a bad thing.
    There are many good sampling rate converters out there in the world. In fact, you can use a high-quality sampling rate converter to reduce problems with jitter coming in from an “upstream” device or transmission path.

However, sampling rate conversion is not necessarily a good thing either… so the more of them you have in your audio signal path, the better you want them to be. In an optimal case, the artefacts caused by the sampling rate converter will not be the “weakest link” in the audio chain.

However, this last statement is very easy to mis-interpret, as I alluded to in Part 6. The problem is that, if I say “I have a sampling rate converter with a THD+N of -100 dB relative to the signal level” this might look pretty good. However, if the signal and the SRC artefacts are in COMPLETELY different frequency bands, and you’re playing the signal out of a loudspeaker that can’t produce the signal (say, because it’s too low in frequency) then 100 dB might not be nearly good enough. In other words, it’s not a mere numbers-game… you have to know how to interpret the data…

A what?

Maybe we should first back up a little and talk about what a sampling rate converter is. As you saw in Part 1, at its most basic level, LPCM digital audio is just a way of describing a signal by storing a long string of measurements that were made at a regular time interval. Each of those measurements is called a “sample” and the rate at which you measure the samples (per second) is called the “sampling rate”. A CD, for example, uses a standard sampling rate of 44,100 samples per second, or 44.1 kHz. Other systems use other rates.

If you want to listen to a CD on a loudspeaker with built-in digital processing, and the loudspeaker happens to have an internal sampling rate that is NOT 44.1 kHz (let’s say that it’s 48 kHz), then you need to somehow convert the sampling rate from 44.1 kHz to 48 kHz to get things to work properly. (This is a little like having a gearbox in a car – your engine does not turn at the same speed as your wheels – you put gears in-between to convert the rotational speed of the engine to the rotational speed of the wheels.)

One sneaky way to do this is to use an analogue connection – you convert the 44.1 kHz digital signal to an analogue one using a DAC, and then re-sample the analogue signal using an ADC running at 48 kHz. This is simple, and (if you choose your DAC and ADC properly) potentially a really good solution. In the “old days” (up to the 1990s) before digital SRCs became really good, this was the best way to do it (assuming you had access to some decent gear).

There are many ways to make a fully-digital SRC. For example:

Let’s say that you have an audio signal that’s been sampled at some sampling rate that we’ll call “Fs1” (for “Sampling Frequency 1”) , as is shown in Figure 1.

Figure 1: A signal recorded at some sampling rate.

You then want to have the same signal, represented at a different sampling rate, which we’ll call Fs2. The old signal (in black) and the new sampling rate (the red dots and the gridlines) can be seen in Figure 2.

Figure 2: The original signal at Fs1 and the new samples that we want to create (in red) at Fs2.

How do we do this? One way is to draw straight lines between the original samples, and calculate the values at the point on the line that corresponds with the time of the new samples. This is called “linear interpolation” (because it’s based on drawing straight lines between the original samples), and it’s shown in Figure 3.

Figure 3: An example of linear interpolation for converting to the new sampling rate.

A better way to do this is to use some fancy math to calculate where the signal would be after the reconstruction filter smoothed it back to the original (band-limited) input. There are different ways to do this (in other words, different mathematical strategies) that are outside the scope of this posting, however, I’ve shown an example of a piecewise cubic spline interpolation implementation in Figure 4, below.

Figure 4: An example of piecewise cubic spline interpolation for converting to the new sampling rate.

However, let’s say that:

  • you’ve been given the job of building a sampling rate converter, but
  • you think that the examples I gave above are way to complicated…

What do you do? One possibility is to look at the sample value that you want to output, find the closest sample (in time) in the original signal, and use that. This is a technique commonly called “nearest neighbour” for obvious reasons – and it’s one of the worst-performing SRC strategies you can use. An example of this is shown in Figure 5, below. Notice that the new values (the red circles) are identical to the closest original value

Figure 5: An example of “nearest neighbour” interpolation for converting to the new sampling rate. Note that each new values in red is a copy of the closest value in black.

If we look at these two signals without the sample values, we’ll see some pretty nasty distortion in the time domain, as shown in Figure 6.

Figure 6: The same signals shown in Figure 5 without the circles.

So what?

The plots above show the results of good and bad SRCs in the time domain, but what does this look like in the frequency domain? Let’s take a couple of specific examples.

Figure 7: 500 Hz sine tone at 0 dB FS, Linear interpolation from 44.1 kHz to 48 kHz.

Figure 8: 500 Hz sine tone at 0 dB FS, Piecewise cubic spline interpolation from 44.1 kHz to 48 kHz.

Figures 7 and 8 look almost identical. There are the windowing artefacts of the frequency analysis that I’m doing are larger than most of the artefacts caused by the interpolation implementations. However, you may notice a couple of spikes sticking up between 1 kHz and 10 kHz in Figure 7. These are the most obvious frequency-domain artefacts of the distortion caused by linear interpolation. Notice however, that those artefacts are about 80 dB down from the signal – so that’s pretty good for a cheap implementation.

However, let’s look at the same 500 Hz tone converted using the “nearest neighbour” strategy.

Figure 9: 500 Hz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.

Now you can see that things have really fallen apart The artefacts are almost up to 40 dB below the signal level, and they’re quite far away in frequency, so they’ll be easy to hear. Also remember that the artefacts that are generated here are inside the audio band, so they will not be eliminated later in the chain by a reconstruction filter in a DAC, for example. They’re there to stay.

There’s one more interesting thing to consider here. Let’s try the same nearest neighbour algorithm, converting between the same two sampling rates, but I’ll put in signals at different frequencies.

Figure 10: 50 Hz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.
Figure 11: 5 kHz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.

Figure 10 shows the same system, but the input signal is a 50 Hz sine wave (instead of 500 Hz). Notice that the artefacts are now about 60 dB down (instead of 40 dB).

Figure 11 shows the same system again, but the input signal is a 5 kHz sine wave. Notice that the artefacts are now only about 20 dB down.

So, with this poor implementation of an SRC, the distortion-to-signal ratio is not only dependent on the algorithm itself, but the signal’s frequency content. Why is this?

Think back to the way the “nearest neighbour” strategy works. You’re simply copying-and-pasting the value of the nearest sample. However, the lower the frequency, the less change there is in the signal from sample to sample. So, as your signal’s frequency goes down (more accurately, as it gets further away from the sampling rate), the smaller the error that you create with this system. At 0 Hz, there would be no error, because all of the samples would have the same value.

So, (for example) if your job is to build the SRC in the first place, and you measure it with a 50 Hz tone, you’ll see that the artefacts are 60 dB below the signal and you’ll pat yourself on the back and go to lunch. Then, some weeks later, when the customer complaints start coming in about tweeter distortion, you’ll think it must be someone else’s fault… but it isn’t…

Conclusion

What does this have to do with “High Resolution Audio”? Well, the problem is that most audio gear does not run at crazy-high sampling rates (this is not necessarily a bad thing), so if you play a high-res file, you’re probably sampling rate converting (this is not necessarily a bad thing).

However, if your gear does have a bad SRC in the signal flow (and, yes, this is not uncommon with modern audio gear) then you either need to

  • play the signal with a different (e.g. not-high-res) sampling rate to find out if it’s better,
    OR
  • buy better gear,
    OR
  • at least check for a firmware update.

Note that first recommendation of the three: Because the quality of a sampling rate converter is very dependent upon the relationship between the input and the output sampling rates, it can happen that a “normal” resolution audio signal (say, at 44.1 kHz) will sound better on your particular equipment than a “high” resolution audio signal (say, at 192 kHz) because of this. Of course, the opposite could be true (say, because your gear is running at 48 kHz and it’s easier to get to that from 192 kHz (just multiply by 1/4) than it is to get there from 44.1 kHz (just multiply by 480/441…)

This doesn’t mean that “low-res” is better than “high-res” – it just means that your particular equipment deals with it better. (In the same way that purely from the point of view as a fuel, gasoline might have more energy per litre than diesel fuel, but it’s a terrible choice to put in the tank of a car that’s expecting diesel…)

On to Part 8a.