Most digital filters that are applied to audio signals use a “basic” building block called a “biquadratic filter” or “biquad” which consists of 2 feed-forward delays and 2 feed-back delays, each with its own output gain and a delay time of 1 sample. I’ve already talked a little about biquads in this posting, where I showed a couple of different ways to implement it. One of the standard ways is shown below in Figure 1.
Figure 1: A biquad implemented using the “Direct Form 1” method.
The signal flow that I drew for Figure 1 is a little more modular than the way it’s normally shown, but that’s to keep things separate for the purposes of this discussion.
The two feed-forward delays add to the input signal (via gains b0, b1, and b2) and the result shows up at the red arrow. Remember from Part 1 that this portion of the biquad can only make a magnitude response that has (in an extreme case) infinitely deep, sharp dips, and smooth rounded peaks.
The signal from the red arrow onwards goes into the feed-back portion of the filter with two feed-back delays adding through gains -a1 and -a2. Again, remember from Part 1 that this portion of the biquad can make a magnitude response that has infinitely deep, sharp peaks, and smooth rounded dips.
Let’s say that we wanted to make a simple filter – let’s make it a low pass filter – using this biquad. How do we do it?
The simplest way is to cheat and go straight to the answer.
Cheating Option 1: You go to this page at www.earlevel.com and put in the parameters you’re interested in (Filter Type, sampling rate, Fc, Q, etc…) and copy-and-paste the resulting five gains (we’ll call them “coefficients” from now on).
Cheating Option 2: We search on the Interweb for the words “RBJ Audio Cookbook” and then spend some time copying, pasting, and porting the equations that Robert Bristow-Johnson bestowed upon us many years ago* into your processor. You then say “I want a low pass filter at 1000 Hz with a Q of 0.5, please” and the equations spit out the five coefficients that you seek.
However, if you cheat, you’ll never really get a grasp of how those coefficients work and what they’re really doing – and that’s where we’re headed in this little series of articles. So, you might decide to go through this series, and then cheat afterwards (that’s what I would recommend…)
Now, before you go any further, I’ll warn you – the whole purpose of this series is to give you an intuitive understanding. This means that there are things I’m going to (intentionally) skip over, merely mention in passing, or omit completely. So, if you already know what I’m talking about, there’s no point in reading what I’m writing – and there’s certainly no need to email me to remind me that I didn’t mention some aspect of this that you think is important, but I’ve decided is not. If you feel strongly about this, write your own blog.
Back in this posting, I made the following statement:
Generally speaking, digital filters work by taking an audio signal, delaying it, changing the level, and adding the result back to the signal itself.
I then showed an example of a simple digital filter like this:
Figure 1: Simple digital filter
If we use the filter in Figure 1, set the delay to 0, and set the gain to 1, then the output is just the input signal added to itself, so it’s two times the amplitude or about 6 dB louder.
If we leave the gain at 1, and set the delay to something else – let’s say, 1 ms, for example – then, in the very low frequencies (say, 1 Hz) the phase difference caused by the 1 ms delay is almost nothing – therefore the output will be +6 dB. At 500 Hz, however, the 1 ms delay is equal to a 180º phase shift, so the output of the delay will always be opposite in polarity with the non-delayed signal. Therefore, at 500 Hz, this filter will have no output. At 1 kHz, the output will be +6 dB again, because 1 ms = 360º at 1 kHz. At 1.5 kHz, there’s no output (540º phase shift), at 2 kHz, we’re back to +6 dB, and so on all the way up. The result is a magnitude response that looks like this:
Figure 2: The magnitude response of the filter shown in Figure 1, if Delay = 1 ms and Gain = 1. Note that these are two identical plots, the only difference is the scaling of the X-axes.
If I used the same delay time on the same filter structure, but set the gain to something between 0 and 1 – let’s make it 0.75, for example, then the overall shape would be the same, but the effect would be less, as shown in Figure 3.
Figure 3: The magnitude response of the filter shown in Figure 1, if Delay = 1 ms and Gain = 0.75.
If we make the gain a negative value, then the overall shape remains the same, but the high points and the low points swap places because the delayed signal is now cancelling where it was adding, and vice versa.
Figure 4: The magnitude response of the filter shown in Figure 1, if Delay = 1 ms and Gain = -0.75. Notice how the boosts and dips have swapped places as a result of using a negative gain.
Let’s think about this intuitively. If my audio signal cannot exceed a value of 1 (which is normally the way we work… a full-scale signal ranges from -1 to 1) and the gain of the delay output in the filter in Figure 1 also ranges from -1 to 1, then the maximum possible output of the filter is 2.
If I had two delay lines and I were summing all three signals (the input and the two delayed signals) and the gains were still limited within the range of -1 to 1, then the maximum possible output would be 3…
However, the minimum possible output level (not the minimum possible output value) would be no signal (as in the case of a 500 Hz input in Figure 2. This is equivalent to an output of -∞ dB.
If I generalise this, then I can say that if your filter is built using ONLY the summed outputs of feed-forward delays, then the maximum possible output can be easily calculated, and the minimum possible output is no signal.
Still generalising: notice that the “bumps” in the above frequency responses are smooth and rounded, and the dips are pointy notches.
What happens if the filter uses feed-back instead of feed-forward?
Figure 5: A simple filter with a feed-back delay line.
Let’s set the delay time to 1 ms again, and set the gain to 0.99. I chose 0.99 instead of 1 because this means that each time the signal re-circulates back, it will get a little bit quieter. If I had set the gain to 1, then the delay would keep “echoing” forever. If I made it greater than 1, then the output would get louder on each re-circulation, and things would get very loud, sooner or later…
So, if Delay = 1 ms and Gain = 0.99 in the filter in Figure 5, the resulting magnitude response looks like Figure 6.
Figure 6: The magnitude response of the filter in Figure 5 when Delay = 1 ms and Gain = 0.99.
There are some things to notice in Figure 6.
Firstly, notice that the overall “shape” of the curve is upside-down relative to the one in Figure 2. The rounded bits are on the bottom and the pointy bits are on the top. This means that instead of having very narrow notches, you have very narrow resonances that are “singing” like a collection of sinusoidal waves, one at the frequency of each peak.
Secondly, notice that the peaks and the dips are in the same places as in Figure 2. In both cases, the Delay = 1 ms and the gain is positive, so the frequencies that are boosted are the same in both cases. For example, both have a peak at 1 kHz, and a dip at 500 Hz.
Thirdly, notice that the overall level is much, much higher. 40 dB is a LOT louder than 6 dB – this is because the sum of all those re-circulated signals echoing over and over in the filter add up to something loud over time.
If I reduce the gain, but keep it positive, then (just like in the case with the feed-forward filter) the shape of the magnitude response stays the same, it’s just reduced in effect.
Figure 7: Delay = 1 ms and Gain = 0.75. Notice that the peaks are smaller because the lower gain causes the signal to “die away” faster.
If I did the same thing, but set the gain to a negative number (say, -0.99) instead, then each time the signal re-circulates, it flips polarity. The resulting magnitude response looks like Figure 8.
Figure 8: The filter in Figure 5 where Delay = 1 ms and Gain = -0.99.
Notice that this is related to the magnitude response in Figure 4 – we have less output in the low end, and now the first peak is at 500 Hz instead of 1 kHz.
If I generalise this one, then I can say that if your filter is built using ONLY the summed outputs of feed-back delays, then the peaks are much higher than with a feed-forward design because they’re resonating.
Still generalising: notice that the “bumps” in the above frequency responses are pointy (because they’re resonances), and the dips are smooth and rounded.
The summary (for now)
Repeating myself, because this is the take-away information for this posting:
If your filter is built using ONLY the summed outputs of feed-forward delays, then:
the maximum possible output can be easily calculated
the minimum possible output is no signal
the “bumps” in the above frequency responses are smooth and rounded
the dips are pointy notches.
if your filter is built using ONLY the summed outputs of feed-back delays, then:
the peaks are much higher in level than with a feed-forward design because they’re resonating
the “bumps” in the above frequency responses are pointy because they’re resonating
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
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.
If your processing distorts the signal for some reason, it’s better to have a higher sampling rate to keep the aliased distortion artefacts as far away from the audio signal as possible.
If you are a lazy DSP engineer who thinks that filters give you the expected magnitude response, no matter what the centre frequency, you’d better have a higher sampling rate. (Or you could just stop being lazy and compensate.)
If you need a lower noise floor for the same audio bandwidth, it’s more efficient to add bits than to increase the sampling rate.
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.
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 samplesFigure 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.
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.
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:
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.
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.
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…
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).
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…)
In Part 2 of this series, I talked about the relationship between the noise floor of an LPCM signal and the number of bits used to encode it.
Assuming that the signal is correctly dithered using TPDF dither with a peak-to-peak amplitude of ±1 LSB, then this means that you can easily calculate the dynamic range of your system with a very simple equation:
Dynamic Range in dB = 6.02 * NumberOfBits – 3
(Note that the sampling rate is not part of this equation… That will be useful information later.)
Normally, we’re lazy and we say 6 times the number of bits -3 for the dither – but if you’re really lazy, you leave out the -3 as well.
So, this means that, in a 16-bit system, the noise floor is 93 dB below a sine wave at full scale (6 * 16 -3 = 93) and for a 24-bit system, the noise floor is 141 dB below a sine wave at full scale (you do the math as practice).
Also, we can generalise and say that “adding 1 bit halves the level of the noise floor” (because -6 dB is the same as multiplying by 0.5). However, this is only part of the story.
The noise that’s generated by dither has a “white” characteristic. This means that there is an equal probability of getting some energy per bandwidth (or some say “per Hertz”) over a period of time. This sounds a little complicated, so I’ll explain.
Noise is random. This means that you may or may not get energy at, say 1 kHz, in a given short measurement. However, if you measure white noise for long enough, you’ll eventually see that you got something in every frequency band. Also, you’ll see that, if you look back over the entire length of your measurement of white noise, you got the same amount of energy in the band from 100 Hz to 200 Hz as you did in the band from 1000 Hz to 1200 Hz and the band from 10,000 Hz to 10,200 Hz. (Each of those bandwidths is 200 Hz wide).
There are now two things to discuss:
This distribution of energy is not like the way we hear things. We don’t hear the distance between 100 Hz and 200 Hz as the same distance as going from 1,000 Hz to 1,200 Hz. We hear logarithmically, which means that we hear in multiples of frequency, not additions of bandwidth. So, to use 100 Hz – 200 Hz sounds like the same “distance” as 1,000 Hz to 2,000 Hz. This is why white noise sounds like it is “bright” – or it has emphasis on the high frequencies. If you have a system that has a flat response from 0 to 20,000 Hz, and you play white noise through it, you have the same amount of energy in the top octave (10 kHz to 20 kHz) as you do in all of the octaves below – which is why we hear this as “top-heavy”.
If you had two bands of white noise with equal levels, and let’s say that one ranges from 100 Hz to 200 Hz, and the other is 1000 Hz to 1200 Hz, then the output level of the two of them together will be 3 dB louder than the output level of either of them alone. This is because their powers add together instead of their amplitudes (because the two signals are unrelated to each other).
Let’s put all this (and one or two other things) together:
We know from a previous part in this series that an LPCM digital audio system cannot have signals higher than the Nyquist frequency – 1/2 the sampling rate.
TPDF dither is white noise at a total level that is (6.02 * NumberOfBits – 3) dB below full-scale.
If you add white noise signals with equal levels but different bandwidths, you get a 3 dB increase over the level of just one of them
This means that,
if I have a 16-bit, TPDF dithered LPCM audio signal with a sampling rate of 48 kHz, it has a noise floor that is 93 dB below full scale, and that noise has a white characteristic with a bandwidth of 24 kHz (the Nyquist frequency). There will be no noise above that frequency coming out of the system.
if I have a 16-bit, TPDF dithered LPCM audio signalwith a sampling rate of 192 kHz, it has a noise floor that is 93 dB below full scale, and that noise has a white characteristic with a bandwidth of 96 kHz (the Nyquist frequency). There will be no noise above that frequency coming out of the system.
So, the two systems have the same noise floor level overall, but with very different bandwidths… What does this mean?
Well, let’s start by looking at the level of the noise floor in the 48 kHz system (so the noise “only” extends to 24 kHz).
If I double the sampling rate (to 96 kHz), I double the bandwidth of the noise without changing its level, so this means that the portion of the noise that “lives” in the 0 Hz – 24 kHz region drops by 3 dB (because I’m ignoring the top half of the signal ranging from 24 kHz to 48 kHz in the 96 kHz system.
Figure 1: The noise floor caused by TPDF dither in a 16-bit LPCM system with a sampling rate of 96 kHz. The total power of the noise drawn in red is the same as the total power of the noise drawn in blue. But notice that this is plotted on a linear scale for frequency.Figure 1: Re-plotted on a logarithmic scale for the frequency. To us, the top octave is not so important, but to a measurement device, it’s half the signal…
If I had multiplied the original sampling rate by 4 (to 192 kHz) I multiply the bandwidth of the noise by 4 as well (to 96 kHz). This means that the overall level of the noise from 0 to 24 kHz is now 6 dB down from the original version.
In other words: if I multiply the sampling rate by two, but I don’t increase the bandwidth of the noise floor that I’m interested in (say I only care about 20 Hz – 20 kHz), then its level drops by 3 dB.
So what?
Well, you could jump to the conclusion that this proves that higher sampling rates are better. However, that would be a bit (ha hah) premature. Consider that, if you want to drop the (band-limited) noise floor by 6 dB, you have to quadruple the sampling rate – and therefore quadrupling the data rate (and therefore the disc storage, the bandwidth of the transmission system, the error rate, and so on…) A 400% increase in the data is not insignificant.
OR, you could just add one more bit – going from 16 bits to 17 bits will give you the same result with a data increase of only 6.25% – a much smarter decision, no?
The Real World
This little analysis above makes a basic (and possibly incorrect) assumption. The assumption is that, by quadrupling the sampling rate, all other components in the system will remain predictably identical. This may not be true. For example, many DACs (especially older ones) exhibit an increase in their own noise floor when you use them at a higher sampling rate. So, it could be that the benefit you get theoretically is negated by the detriment that you actually get. This is just one example of a flaw in the theory – but it’s a very typical one – especially if you’re building a product instead of just using one.
P.S.
You may have looked at Figures 1 or 2 and are wondering why, if the noise floor is at -93 dB FS in a 16-bit system, I plotted it around -120 dB FS (give or take). The reason is related to the explanation I just gave above. I said in the captions that it’s from a 96 kHz system. This means that the noise extends to the Nyquist frequency at 48 kHz, and that total level is at -93 dB FS. We also know that, if I keep the noise the same, but half the bandwidth that I’m looking at, the level drops by 3 dB. Therefore I can either do math or I can make the following table:
Bandwidth of noise measurement in Hz
Level in dB FS
48,000
-93
24,000
-96
12,000
-99
6,000
-102
3,000
-105
1,500
-108
750
-111
375
-114
187.5
-117
93.75
-120
If you look carefully at the figures, you’ll see that there’s a point every 100 Hz. (It’s most easily visible in the low-frequency range of Figure 2.) So, the level of the noise that I see on a magnitude response plot like this is not only dependent on the noise level itself, but the bandwidths of the divisions that I’ve used to slice it up. In my case, the bandwidth per “slice” is about 100 Hz, so the noise level of each of those little contributors is at about -120 dB FS. If I had used slices only 50 Hz wide, it would show up at -123 instead…
Let’s go back to something I said in the last post:
Mistake #1
I just jumped to at least three conclusions (probably more) that are going to haunt me.
The first was that my “digital audio system” was something like the following:
Figure 1
As you can see there, I took an analogue audio signal, converted it to digital, and then converted it back to analogue. Maybe I transmitted it or stored it in the part that says “digital audio”.
However, the important, and very probably incorrect assumption here is that I did nothing to the signal. No volume control, no bass and treble adjustments… nothing.
If you consider that signal flow from the position of an end-consumer playing a digital recording, this was pretty easy to accomplish in the “old days” when we were all playing CDs. That’s because (in a theoretical, oversimplified world…)
the output of the mixing/mastering console was analogue
that analogue signal was converted to digital in the mastering studio
the resulting bits were put on a disc
you put that disc in your player which contained a DAC that converted the signal directly to analogue
you then sent the signal to your “processing” (a.k.a. “volume control”, and maybe some bass and treble adjustment.).
So, that flowchart in Figure 1 was quite often true in 1985.
These days, things are probably VERY different… These days, the signal path probably looks something more like this (note that I’ve highlighted “alterations” or changes in the bits in the audio signal in red):
The signal was converted from analogue to digital in the studio (yes, I know… studios often work with digital mixers these days, but at least some of the signals within the mix were analogue to start – unless you are listening to music made exclusively with digital synthesizers)
The file was transferred to a storage device (let’s say “hard drive”) in a large server farm renting out space to your streaming service
The streaming service encodes the file
If the streaming service does not offer an lossless option, then the file is converted to a lossy format like MP3, Ogg Vorbis, AAC, or something else.
If the streaming service offers a lossless option, then the file is compressed using a format like FLAC or ALAC (This is not an alteration, since, with a lossless compression system, you don’t lose anything)
You download the file to your computer (it might look like an audio player – but that means it’s just a computer that you can’t use to check your social media profile)
You press play, and the signal is decoded (either from the lossy CODEC or the compression format) back to LPCM. (Still not an alteration. If it’s a lossy CODEC, then the alteration has already happened.)
That LPCM signal might be sample-rate converted
The streaming service’s player might do some processing like dynamic range compression or gain changes if you’ve asked it to make all the songs have the same level.
All of the user-controlled “processing” like volume controls, bass, and treble, are done to the digital signal.
The signal is sent to the loudspeaker or headphones
If you’re sending the signal wirelessly to a loudspeaker or headphones, then the signal is probably re-encoded as a lossy CODEC like AAC, aptX, or SBC. (Yes, there are exceptions with wireless loudspeakers, but they are exceptions.)
If you’re sending the signal as a digital signal over a wire (like S/PDIF or USB), the you get a bit-for-bit copy at the input of the loudspeaker or headphones.
The loudspeakers or headphones might sample-rate convert the signal
The sound is (finally) converted to analogue – either one stream per channel (e.g. “left”) or one stream per loudspeaker driver (e.g. “tweeter”) depending on the product.
So, as you can see in that rather long and complicated list (it looks complicated, but I’ve actually simplified it a little, believe it or not), there’s not much relation to the system you had in 1985.
Let’s take just one of those blocks and see what happens if things go horribly wrong. I’ll take the “volume control” block and add some distortion to see the result with two LPCM systems that have two different sampling rates, one running at 48 kHz and the other at 194 kHz – four times the rate. Both systems are running at 24 bits, with TPDF dither (I won’t explain what that means here). I’ll start by making a 10 kHz tone, and sending it through the system without any intentional distortion. If we look at those two signals in the time domain, they’ll look like this:
Figure 1: Two 10 kHz tones. The black one is in a 48 kHz, 24 bit LPCM system. The red one is in a 192 kHz, 24 bit LPCM system.
The sine tone in the 48 kHz system may look less like a sine tone than the one in the 192 kHz system, however, in this case, appearances are deceiving. The reconstruction filter in the DAC will filter out all the high frequencies that are necessary to reproduce those corners that you see here, so the resulting output will be a sine wave. Trust me.
If we look at the magnitude responses of these two signals, they look like Figure 2, below.
Figure 2: The magnitude responses of the two signals shown in Figure 1.
You may be wondering about the “skirts” on either side of the 10 kHz spikes. These are not really in the signal, they’re a side-effect (ha ha) of the windowing process used in the DFT (aka FFT). I will not explain this here – but I did a long series of articles on windowing effects with DFTs, so you can search for it if you’re interested in learning more about this.
If you’re attentive, you’ll notice that both plots extend up to 96 kHz. That’s because the 192 kHz system on the bottom has a Nyquist frequency of 96 kHz, and I want both plots to be on the same scale for reasons that will be obvious soon.
Now I want to make some distortion. In order to make things obvious, I’m going to make a LOT of distortion. I’ve made the sine wave try to have an amplitude that is 10 times higher than my two systems will allow. In other words, my amplitude should be +/- 10, but the signal clips at +/- 1, resulting in something looking very much like a square wave, as shown in Figure 3.
Figure 3: Distorted 10 kHz sine waves. The black one is in a 48 kHz, 24 bit LPCM system. The red one is in a 192 kHz, 24 bit LPCM system.
You may already know that if you want to make a square wave by building it up using its constituent harmonics, you need to have the fundamental (which we’ll call Fc. In our case, Fc = 10 kHz) with an amplitude that we’ll say is “A”, you then add the
3rd harmonic (3 times Fc, so 30 kHz in our case) with an amplitude of A/3.
5th harmonic (5 Fc = 50 kHz) with an amplitude of A/5
7 Fc at A/7
and so on up to infinity
Let’s look at the magnitude responses of the two signals above to see if that’s true.
Figure 4: The magnitude responses of the two signals shown in Figure 3.
If we look at the bottom plot first (running at 192 kHz and with a Nyquist limit of 96 kHz) the 10 kHz tone is still there. We can also see the harmonics at 30 kHz, 50 kHz, 70 kHz, and 90 kHz in amongst the mess of other spikes we’ll get to those soon…)
Figure 5. Some labels applied to Figure 4 for clarity, showing the harmonics of the square waves that are captured by the two systems
Looking at the top plot (running at 48 kHz and with a Nyquist limit of 24 kHz), we see the 10 kHz tone, but the 30 kHz harmonic is not there – because it can’t be. Signals can’t exist in our system above the Nyquist limit. So, what happens? Think back to the images of the rotating wheel in Part 3. When the wheel was turning more than 1/2 a turn per frame of the movie, it appears to be going backwards at a different speed that can be calculated by subtracting the actual rotation from 180º (half-a-turn).
The same is true when, inside a digital audio signal flow, we try to make a signal that’s higher than Nyquist. The energy exists in there – it just “folds” to another frequency – its “alias”.
We can look at this generally using Figure 6.
Figure 6: A general plot of aliasing, showing the intended frequency in black and the actual output frequency in red.
Looking at Figure 6: If we make a sine tone that sweeps upward from 0 Hz to the Nyquist frequency at Fs/2 (half the sampling rate or sampling frequency) then the output is the same as the input. However, when the intended frequency goes above Fs/2, the actual frequency that comes out is Fs/2 minus the intended frequency. This creates a “mirror” effect.
If the intended frequency keeps going up above Fs, then the mirroring happens again, and again, and again… This is illustrated in Figure 7.
Figure 7: An extension of Figure 5 to a higher intended frequency.
This plot is shown with linear scales for both the X- and Y-axes to make it easy to understand. If the axes in Figure 7 were scaled to a logarithmic scaling instead (which is how “Frequency Response” are normally shown, since this corresponds to how we hear frequency differences), then it would look like Figure 8.
Figure 8: The same information shown in Figure 7, plotted on a logarithmic scale instead. Note that this example is for a system running at 48 kHz (therefore with a Nyquist frequency of 24 kHz), and an intended input frequency (in black) going up to 3 times 48 kHz = 144 kHz.
Coming back to our missing 30 kHz harmonic in the 48 kHz LPCM system: Since 30 kHz is above the Nyquist limit of 24 kHz in that system, it mirrors down to 24 kHz – (30 kHz – 24 kHz) = 18 kHz. The 50 kHz harmonic shows up as an alias at 2 kHz. (follow the red line in Figure 7: A harmonic on the black line at 48 kHz would actually be at 0 Hz on the red line. Then, going 2000 Hz up to 50 kHz would bring the red line up to 2 kHz.)
Similarly, the 110 kHz harmonic in the 192 kHz system will produce an alias at 96 kHz – (110 kHz – 96 kHz) = 82 kHz.
If I then label the first set of aliases in the two systems, we get Figure 9.
Figure 9: The first set of aliased frequency content in the two systems.
Now we have to stop for a while and think about what’s happened.
We had a digital signal that was originally “valid” – meaning that it did not contain any information above the Nyquist frequency, so nothing was aliasing. We then did something to the signal that distorted it inside the digital audio path. This produced harmonics in both cases, however, some of the harmonics that were produced are harmonically related to the original signal (just as they ought to be) and others are not (because they’re aliases of frequency content that cannot be reproduced by the system.
What we have to remember is that, once this happens, that frequency content is all there, in the signal, below the Nyquist frequency. This means that, when we finally send the signal out of the DAC, the low-pass filtering performed by the reconstruction filter will not take care of this. It’s all part of the signal.
So, the question is: which of these two systems will “sound better” (whatever that means)? (I know, I know, I’m asking “which of these two distortions would you prefer?” which is a bit of a weird question…)
This can be answered in two ways that are inter-related.
The first is to ask “how much of the artefact that we’ve generated is harmonically related to the signal (the original sine tone)?” As we can see in Figure 5, the higher the sampling rate, the more artefacts (harmonics) will be preserved at their original intended frequencies. There’s no question that harmonics that are harmonically related to the fundamental will sound “better” than tones that appear to have no frequency relationship to the fundamental. (If I were using a siren instead of a constant sine tone, then aliased harmonics are equally likely to be going down or up when the fundamental frequency goes up… This sounds weird.)
The second is to look at the levels of the enharmonic artefacts (the ones that are not harmonically related to the fundamental). For example, both the 48 kHz and the 192 kHz system have an aliased artefact at 2 kHz, however, its level in the 48 kHz system is 15 dB below the fundamental whereas, in the 192 kHz system, it’s more than 26 dB below. This is because the 6 kHz artefact in the 48 kHz system is an alias of the 30 kHz harmonic, whereas, in the 192 kHz system, it’s an alias of the 190 kHz harmonic, which is much lower in level.
As I said, these two points are inter-related (you might even consider them to be the same point) however, they can be generalised as follows:
The higher the sampling rate, the more the artefacts caused by distortion generated within the system are harmonically related to the signal.
In other words, it gives a manufacturer more “space” to screw things up before they sound bad. The title of this posting is “Mirrors are bad” but maybe it should be “Mirrors are better when they’re further away” instead.
Of course, the distortion that’s actually generated by processing inside a digital audio system (hopefully) won’t be anything like the clipping that I did to the signal. On the other hand, I’ve measured some systems that exhibit exactly this kind of behaviour. I talked about this in another series about Typical Problems in Digital Audio: Aliasing where I showed this measurement of a real device:
Figure 10: A measurement of a real device showing some kind of distortion and aliased artefacts of a swept sine tone. Half of the aliasing is immediately recognizable as going downwards when the tone is going upwards.
However, I’m not here to talk about what you can or can’t hear – that is dependent on too many variables to make it worth even starting to talk about. The point of this series is not to prove that something is better or worse than something else. It’s only to show the advantages and disadvantages of the options so that you can make an informed choice that best suits your requirements.
