If you’ve read through the first four parts of this series, then you’re already at a point where you can intuitively understand what’s going on. We just have a couple of details to take care of before finishing off.
Firstly, the plots showing the zeros and poles in the figures you’ve been looking at plots of the “Z-plane” or “Complex-plane“. As I said at the start, we’re only trying to get to an intuitive understanding of these plots – so I’m not going to get into complex numbers, or even much math (apart from what you’ll see below… which isn’t very complicated, and avoids complex numbers).
When I’m developing a new DSP algorithm, I use an application called Max from cycling74.com. Figure 1 shows a screenshot from Max, where I’m using an object to calculate the biquad coefficients to make a low pass filter, as you can see. I’ve then connected the output of that object (it looks like a magnitude response) to a Z-plan representation that shows me the same thing in a different way.
Figure 1: The top plot shows the magnitude response of the filter. The bottom plot shows the Z-plane representation of the same filter.
You may notice that this plot has two poles, one at (0, 0.408) and the other at (0, -0.408). In fact there are two zeros there as well, but they’re situated in the same place, on “on top” of the other, at (-1, 0). This is always true for a biquad – there are always two zeros and two poles. Sometimes, they’re located in the same place, sometimes not, sometimes they’re placed symmetrically, sometimes not, depending on the filter, as we’ll see below.
Let’s look at that Z-plane representation in 3-dimensions:
Figure 2: A 3D view of the Z-plane representation of the filter shown in Figure 1.Figure 3: The same plot as shown in Figure 2, rotated to show the back of the plot.
So, as you would now expect, the poles pull up the edge of the circle, and the zeros (both in the same place) pull down, giving the red line the height that it has.
Now, think back to this Figure from earlier in the series:
Figure 4: Think of the edge of the circle as the frequency
If you therefore look at Figure 3, which is like looking at Figure 4 from the top, you’ll notice that the height of the red line (the edge of the circle is high on the left (in the low frequencies) and drops as you go to the right (the high frequencies). This is the magnitude response that’s shown on the top of Figure 1. The only difference is that it’s on a linear scale instead of a logarithmic scale, so the shape looks a little weird.
Let’s do another one:
Figure 5: A reciprocal peak/dip filter.
Figure 6: A 3D view of the Z-plane representation of the filter shown in Figure 5.Figure 7: The same plot as shown in Figure 6, rotated to show the back of the plot.
Hopefully, now you are able to look at a Z-plane representation of a filter and think about the effect of the poles and zeros on the edge of the circle, and therefore get a rough idea of the magnitude response of the filter…
If not, I apologize for wasting your time. On the other hand, if you’re in a life-threatening situation, this knowledge probably wouldn’t help you anyway… Very few people have gotten a critical injury in a biquad accident.
How I did it
If you want to make these plots for yourself, the math is pretty simple.
Figure 8: How to do the math.
Start by choosing the frequency, which will be a point on the circle. You then find the four distances from the zeros and poles to that point (I’ve indicated those distances in Figure 8 with the variables z1, z2, p1, and p2.) This can be done using the Pythagorean theorem.
To find the gain of the filter at the frequency, you divide the sum of the zeros’ distances by the sum of the poles’ distances. In other words:
(z1 + z2) / (p1 + p2)
That will give you the result as a linear value. If you then want to convert it to decibels, like I’ve done, you do a little extra math like this:
20 * log10 ( (z1 + z2) / (p1 + p2) )
That’s it! You just need to do repeat that math for each frequency that you’re interested in, and you’re done!
I ended Part 3 by saying that DSP engineers think of the frequency scale as a circle rather than as a straight line. The questions are “Why do they think like this? What’s wrong with them?” Although I can’t answer the second question, the answer to the first question is fundamentally simple.
A DSP engineer is not really interested in what a filter does. She or he is interested in how it filters. A normal magnitude response plot shows us mortals the result of what’s happened to the audio after it’s gone through a filter (or a processor in general). Someone making that filter (or system) needs to know how it’s working instead.
So far, in this series, we’ve seen the following:
Digital audio filters are made with feed-forward and feed-back delays with different gains.
Feed-forward delays make narrow dips (and wide peaks) in the magnitude response
Feed-back delays make narrow peaks (and wide dips) in the magnitude response
DSP engineers think of frequency on a circle instead of a straight line
DSP engineers also want to see plots of how a filter works instead of its result on the audio signal.
Let’s put all of this together.
We’ll draw the circle showing the frequency scale, but then rotate the view to see it in three dimensions. For example, I can pretend to make the surface of of the circle out of a rubber sheet that can be pulled upwards (like a tent) or downwards (like a funnel), whilst always maintaining a circular edge.
If I want to pull the tent upwards, I’ll use a “pole” to do it. That pole has an infinite height (we’re going to need some very stretchy rubber). If I want to pull the funnel downwards, I’ll use something I’ll call a “zero“. (I am not going to go into why zeros and poles are called that, so as to avoid doing too much math.)
So, if I were to put a zero in the middle of the circle, its 2D representation would look like Figure 1 (notice the red circle in the middle showing where the zero is placed), and the 3D version would look like Figure 2:
Figure 1: A 2D representation of a “zero” (indicated by a red circle) in the middle of the circle that shows the frequency scale (the red line shows the scale going counter-clockwise from 0 Hz on the right to Fs/2 on the left)Figure 2: A 3D representation of the same information shown in Figure 1.
If I were to put the pole (indicated by a red ‘x’) in the middle of the circle instead, then the result would look like Figures 3 and 4.
Figure 3: A 2D representation of a “pole” (indicated by a red circle) in the middle of the circle that shows the frequency scale (the red line shows the scale going counter-clockwise from 0 Hz on the right to Fs/2 on the left)Figure 4: A 3D representation of the same information shown in Figure 3.
So far so good… If we were to rotate Figures 2 or 4 and look at the red line that I’ve drawn around the edge, we’d see that it’s flat with a height of 0 (on the vertical scale) all the way around. This is because I’ve carefully placed the zero or the pole at the exact middle of the circle, so it’s pulling equally on all points of the edge of the “tent” or the “funnel”.
However, what would happen if I moved the zero or the pole away from the middle? Some examples of this for a zero moved to the location (-0.75, 0) are shown in Figures 5 to 7, below.
Figure 5: A zero placed at the location (-0.75, 0), shown in a 2-dimensional representationFigure 6: A 3-dimensional view of Figure 5.Figure 7: The same plot as the one shown in Figure 6, rotated to show the “back” of the shape.
As you can see in Figure 7, when the zero is moved away from the centre of the circle, it pulls downwards on the closer edge (notice how the red line is lower than the black line which has a constant height of 0). However, it also doesn’t pull downwards as much on the opposite side of the circle (notice how the red line is higher than the black line on the left side).
Of course, if I were to do the same thing with a pole, everything would behave symmetrically, as shown in Figures 8 to 10.
Figure 8: A pole located at the position (-0.75, 0)Figure 9: a 3-dimensional view of Figure 8Figure 10: The same plot as Figure 9, rotated to show the “back”
We’re almost finished… One more posting to go to wrap up.
I wrote an intuitive explanation of aliasing in this posting and dug in a little deeper, looking at the side-effects of aliasing with audio signals specifically in this posting.
One of the more important figures in that second posting is repeated below in Figure 1.
Figure 1: The black line shows the intended frequency of a sine wave created in the digital domain, expressed as a fraction of the sampling rate (Fs). The red line shows the actual frequency that comes out of the system as a result – its “alias”.
Let’s say that we wanted to make a sine wave generator in the digital domain. This is pretty easy to do using some rather simple math, as follows:
Output(n) = sin(2 * π * Fc / Fs * n)
where Fc is the frequency of the sine wave in Hz, Fs is the sampling rate in Hz, and n is the time, expressed as a sample number.
There are no restrictions on Fc – so if you wanted to plug in a value that is higher than Fs/2 (the Nyquist frequency) then you’ll get a value. However, if you used this math to try to make a sine wave where Fc > Fs/2, then the output will be different from what you expect. This is what’s shown in Figure 1. The red curve shows the actual frequency of the output (read off the Y-axis) for an intended frequency (on the X-axis).
This problem of the difference between input and output is identical to what would happen if you rotated a wheel by some angle, and then asked someone to measure the rotation. For example, look at Figure 2.
Figure 2. The Red arrow shows the actual rotation. The Black arrow shows the assumed rotation.
On the left, it shows a wheel that was rotated clockwise by 90º (indicated by the red arrow). Someone measuring the rotation would say that it was rotated by 90º – a perfect match! If you rotated by 180º (the second example), the person measuring would also get the right answer. However, if you rotated by 270º (the third example, in the middle), the person measuring would (correctly) say that you rotated by 90º counterclockwise. A rotation of 360º gets you back where you started, so it would be measured as 0º. A rotation of 450º (the example on the right) would be measured as a rotation of 90º.
If we were to do this a lot, and plot the results, they’d look like Figure 3.
Figure 3: The results of my little thought experiment shown in Figure 2.
Now compare Figure 3 to Figure 1. Notice how they’re identical? This is important because it’s a graphic example of exactly the way frequencies “wrap” in a digital audio world. This “wrapping” is the result of the fact that a sinusoidal wave (a signal containing only one frequency) is just a 2-dimensional view of a 3-dimensional rotation (I showed this with photos of a Slinky™ in this posting.
When we normal people look at a magnitude response of a device – let’s say, a low-pass filter, we put it on a nice cartesian plot with the frequency displayed on a straight line on the X-axis and the magnitude displayed on a straight line called the Y-axis. This looks something like Figure 4.
Figure 4: A typical, familiar way of viewing a magnitude response of something (in this case, a low-pass filter with Fc=1 kHz).
However, this is only a portion of the truth. The truth extends further than the limits of that plot. I conveniently stopped plotting at Fs/2 (since the filter that I made is running at 48 kHz, this plot goes up to 24 kHz). I also didn’t plot anything below 20 Hz – and I certainly didn’t extend the plot below 0 Hz into the negative frequencies… (“Negative frequencies?” I hear you ask… These are the same as positive frequencies, except that 3-dimensional wheel is rotating in the opposite direction; but since we’re only looking at it on-edge from one location, we can’t tell whether it’s rotating clockwise or counter-clockwise. See this posting if you want to go further.)
Let’s try extending the plot. First, I’ll show Figure 4, but using a linear scale for the frequency instead of a logarithmic scale. This is shown in Figure 5.
Figure 5: The same information shown in Figure 4, but on a linear frequency scale. Notice how 1 kHz is quite low compared to 24 kHz.
If I then were to plot beyond Fs/2, then the magnitude response would be a mirrored version of the one you see in Figure 4. The same would be true if I were to plot below 0 Hz. This is shown in Figure 6.
Figure 6: The same information shown in Figure 5, but showing an extended frequency range.
What does this mean? It means for example that, if I had an LPCM system running at 48 kHz, and I were to digitally generate a sine tone at 48 kHz, then the result would be the same as making a “sine tone” at 0 Hz (or “DC”) because all of the samples would have the same value – neither 0 Hz nor 48 kHz would be a sinusoidal wave in a 48 kHz system. If I then, inside the same system, sent that “48 kHz sine tone” through a low-pass filter with a cutoff frequency of 1 kHz, then it would go through un-impeded (just like a 0 Hz signal would get through a low-pass filter).
Assembling the pieces
Let’s take the illustration I just showed in Figure 6, and consider it, knowing what I showed in the comparison between Figures 3 and 1.
Although we normal people show each other magnitude responses that look like the one in Figure 4, this is not the way people who make digital signal processing (DSP) software think. They see the frequency axis on a circle that goes from 0 Hz up to Fs/2 (the Nyquist frequency), and then wraps back around to 0 Hz (= Fs). This weird way of viewing the world is shown in Figure 7.
Figure 7: DSP engineers think of the frequency axis as the top half of a circle that starts at 0 (Hz) on the right side, and goes (linearly) up to Fs/2 when you get to the opposite side. An example, with a system running at 48 kHz, is shown on the right.
There are some very good reasons why DSP engineers think like this – one of which you already know (the wrapping and aliasing issue). There are some reasons I’m not going to talk about here (but you can read this if you’re interested), and there are some other reasons that I’m headed towards…
However, before we move on to the next chapter in our little saga, it’s best to get really comfortable with the plots in Figure 7. I especially want you to get used to some specific things, in order of importance:
The frequency scale is circle – it’s not a straight line.
The scale starts on the right (at the 3 o’clock position) and goes counter-clockwise to the left (the 9 o’clock position).
The scale is linear, not logarithmic, like you’re used to seeing.
The maximum frequency is the Nyquist frequency, so it’s defined by the sampling rate.
Once the point on the circle goes beyond the Nyquist, we’ve started aliasing, and so we’ve entered a symmetrical world that mirrors the half below the Nyquist. (In other words, when we get a little farther, you’ll see that the top and the bottom of that circle are mirror images of each other – as I’ve already hinted in Figure 6 looking at the frequency range from 0 to 48 kHz.)
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…
