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.
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
If you’ve read the three introductory parts of this series, linked above; and if you’re still awake, then we are ready to start putting things together and jumping to incorrect conclusions…
Let’s say that you’ve been hired to specify a digital audio system for some reason (we’ll assume that it’s an LPCM system – nothing exotic). Using the information I’ve told you so far, you can make two decisions in your specification:
You select a bit depth to be enough to give you the dynamic range you desire. In this case, “dynamic range” means the “distance” in level between the loudest sound you can record / store / transmit (I isn’t say what the “digital audio system” was going to be used for) and the inherent noise floor of the system. If you’re recording the background noise on an airplane while it’s in flight, you don’t need a big dynamic range, because it’s always loud, and never changes. However, if you’re recording a Japanese Taiko Drummer group, you’ll need a huge usable dynamic range because the loud parts of the performance are a LOT louder than the quietest parts.
As we saw in Part 3, an LPCM digital audio system cannot record any audio that has a frequency higher than 1/2 the sampling rate. So, you select a sampling rate that is at least 2x the highest frequency you’re interested in. For example, if you believe the books that say you can hear from 20 Hz to 20,000 Hz, then you might decide that your sampling rate has to be at least 40,000 Hz. On the other hand, if you’re making a subwoofer that you know will never be fed a signal above 120 Hz, then you don’t need a sampling rate higher than 240 Hz.
Don’t get angry yet. I’m just keeping these numbers simple to make the math easy. Later on, I’ll explain why what I just said might not be correct.
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.
Mistake #2
We assumed above that we can define the system’s dynamic range based on the dynamic range of the audio signal itself. However, this makes the assumption that the noise floor of the digital system and the noise floor of your audio signal are identical, which is probably not true. As we saw in Part 2, the noise generated by TPDF dither is white – it has the same probability of having a given amount of energy per Hertz. Since we hear sound logarithmically (meaning that, to us, octaves are equal widths. Equal spacings in Hz are not.) This means that the noise sound “bright” to us – because there’s just as much energy in the top octave (say, 10 kHz to 20 kHz, if you believe the books) as there is in all other frequencies combined from 0 Hz up to 10 kHz.
If, however, the noise floor in your concert hall where the taiko drummers are playing is caused by the air conditioning system, then this noise will be a lot louder in the low frequencies than the the highs – which is not the same.
Therefore it’s too simplistic to say “the noise floor of the digital system” and the “noise floor of the signal” – since these two noise floors are different. (As Steven Wright said: “It doesn’t matter what temperature the room is, it’s always room temperature.”)
Mistake #3
As we’ll see later, if you’re going to do anything to the signal while it’s in the “digital domain”, then you need to take that into consideration when you’re deciding on your sampling rate. It’s not enough to say “useful audio bandwidth times 2” because there are some side effects that need to be remembered…
However, counter-intuitively, it could be that, in order to improve your system, you’ll want to make the sampling rate LOWER instead of HIGHER – so this is not a simple case of “more is better”.
We’ll get to that topic later. For now, I’ll leave you in suspense.
Some details
One thing we saw in Part 3 was that, if we have an audio signal with energy at a frequency higher than 1/2 the sampling rate, and if that signal gets into the analogue-to-digital converter (ADC), then the output of the ADC will contain an error. We’ll get out energy at frequencies that were not in the original, due to the effect called “aliasing“.
Once that’s in the digital audio signal, there’s no removing it, so we need to make sure that the too-high-frequency signals don’t get into the ADC’s input in the first place. This is done using a low-pass filter that (in theory) removes all energy in the signal above the Nyquist frequency (which is equal to 1/2 the sampling rate). Since that low-pass filter prevents aliasing, we call it an anti-aliasing filter. Normally, these days, that antialiasing filter is built into the ADC itself.
As we also saw in Part 3, the digital-to-analogue converter (DAC) has to smooth out the digital signal to convert it from a “staircase” wave to a smoother one. That’s also done with a low-pass filter that eliminates all the harmonics that would be required to make the staircase have sharp corners. Since this is done to re-construct the analogue signal, it’s called a “reconstruction filter“.
This means that, if we pull apart some of the components in the signal chain I showed in Figure 1, it really looks more like this:
I’ve been debating writing a series of postings about “high resolution” audio for a long time – years. Lately, (probably because of some hype generated by some recent press releases) I’ve been getting lots of question (no, that’s not a typo) about it, so it appears the time has come…
To start: the question that I get (a lot) is “If I can’t hear above 20 kHz, then what’s the use of high-res?” As I’ll explain as we go through, this is only one, rather small aspect to consider in this topic. In fact, it might be the least important issue to consider.
However, before I write too much, I’ll say that I’m not going to argue for or against higher resolutions in digital audio systems. I’m only going to go through a bunch of issues that can be used to argue either for or against them. So, there’s not going to be a big reveal at the end of this series telling you that high-res is either better, worse, or no different than whatever you’re using now. It’s merely going to be a discussion of a number of issues that need to be weighed. The problem is that this entire topic is complicated – and there’s no single “right” answer, as I’ll argue as we go along.
To start, let’s get down to basics and look (once again, from the perspectives of this website) at what sound is, and how it’s converted from an analogue electrical signal into a digital representation. The good thing is that I’ve written this introduction before in a different series of postings. So, I’m going to be extremely lazy and just copy-and-paste that information here. I’m not just referring you to another page because I’m intentionally leaving some things out because we’re headed into having a different discussion this time.
A quick introduction to sound
At the simplest level, sound can be described as a small change in air pressure (or barometric pressure) over short periods of time. If you’d like to have a better and more edu-tain-y version of this statement with animations and pretty colours, you could take 10 minutes to watch this video, for example.
That change in pressure can be “captured” by using a microphone, that is (at the simplest level) a device that has a change in air pressure at its input and a change in electrical voltage at its output. Ignoring a lot of details, we could say that if you were to plot a measurement of the air pressure (at the input of the microphone) over time, and you were to compare it to a plot of the measurement of the voltage (at the output of the microphone) over time, you would see the same curve on the two graphs. This means that the change in voltage is analogous to the change in air pressure.
Fig 1. Notice that (in theory, and ignoring a lot of things…) the change in air pressure over time at the input of the microphone is identical to the change in voltage over time at its output. Of course, this is not true in real life – microphones lie like a cheap rug…
At this point in the conversation, I’ll make a point to say that, in theory, we could “zoom in” on either of those two curves shown in Figure 1 and see more and more details. This is like looking at a map of Canada – it has lots of crinkly, jagged lines. If you zoom in and look at the map of Newfoundland and Labrador, you’ll see that it has finer, crinkly, jagged lines. If you zoom in further, and stand where the water meets the shore in Trepassey and take a photo of your feet, you could copy it to draw a map of the line of where the water comes in around the rocks – and your toes – and you would wind up with even finer, crinkly, jagged lines… You could take this even further and get down to a microscopic or molecular level – but you get the idea… The point is that, in theory, both of the plots in Figure 1 have infinite resolution, both in time and in air pressure or voltage.
Now, let’s say that you wanted to take that microphone’s output and transmit it through a bunch of devices and wires that, in theory, all do nothing to the signal. Let’s say, for example, that you take the mic’s output, send it through a wire to a box that makes the signal twice as loud. Then take the output of that box and send it through a wire to another box that makes it half as loud. You take the output of that box and send it through a wire to a measuring device. What will you see? Unfortunately, none of the wires or boxes in the chain can be perfect, so you’ll probably see the signal plus something else which we’ll call the “error” in the system’s output. We can call it the error because, if we measure the input voltage and the output voltage at any one instant, we’ll probably see that they’re not identical. Since they should be identical, then the system must be making a mistake in transmitting the signal – so it makes errors…
Fig 2. If you send an audio signal through some wires and devices that (in theory) do nothing to the signal, you’ll find out that they add some extra stuff that you don’t want.
Pedantic Sidebar: Some people will call that error that the system adds to the signal “noise” – but I’m not going to call it that. This is because “noise” is a specific thing – noise is random – so if it’s not random, it’s not noise. Also, although the signal has been distorted (in that the output of the system is not identical to the input) I won’t call it “distortion” either, since distortion is a name that’s given to something that happens to the signal because the signal is there. (We would probably get at least some of the error out of our system even if we didn’t send any audio into it.) So, we could be slightly geeky and adequately vague and call the extra stuff “Distortion plus noise” but not “THD+N” – which stands for “Total Harmonic Distortion Plus Noise” – because not all kinds of distortion will produce a harmonic of the signal… but I’m getting ahead of myself…
So, we want to transmit (or store) the audio signal – but we want to reduce the noise caused by the transmission (or storage) system. One way to do this is to spend more money on your system. Use wires with better shielding, amplifiers with lower noise floors, bigger power supplies so that you don’t come close to their limits, run your magnetic tape twice as fast, and so on and so on. Or, you could convert the analogue signal (remember that it’s analogous to the change in air pressure over time) to one that is represented (and therefore transmitted or stored) digitally instead.
What does this mean?
Conversion from analogue to digital and back (but skipping important details)
IMPORTANT: If you read this section, then please read the following postings as well. This is because, in order to keep things simple to start, I’m about to leave out some important details that I’ll add afterwards. However, if you don’t add the details, you could (understandably) jump to some incorrect conclusions (that many others before you have concluded…) So, if you don’t have time to read both sections, please don’t read either of them.
In the example above, we made a varying voltage that was analogous to the varying air pressure. If we wanted to store this, we could do it by varying the amount of magnetism on a wire or a coating on a tape, for example. Or we could cut a wiggly groove in a bit of vinyl that has a similar shape to the curve in the plots in Figure 1. Or, we could do something else: we could get a metronome (or a clock) and make a measurement of the voltage every time the metronome clicks, and write down the measurements.
For example, let’s zoom in on the first little bit of the signal in the plots in Figure 1
Fig. 3 The same curve as was shown in Figure 1 – but zoomed in to the very beginning.
We’ll then put on a metronome and make a measurement of the voltage every time we hear the metronome click…
Fig 4. The same curve (in red) measured at regular intervals (in black)
We can then keep the measurements (remembering how often we made them…) and write them down like this:
We can store this series of numbers on a computer’s hard disk, for example. We can then come back tomorrow, and convert the measurements to voltages. First we read the measurements, and create the appropriate voltage…
Fig. 5. The voltages that we stored as measurements
We then make a “staircase” waveform by “holding” those voltages until the next value comes in.
Fig 6. We make a “staircase” curve using the voltages.
All we need to do then is to use a low-pass filter to smooth out the hard edges of the staircase.
Fig 7. When we smooth out the staircase, we get back the original signal (in red).
So, in this example, we’ve gone from an analogue signal (the red curve in Figure 3) to a digital signal (the series of numbers), and back to an analogue signal (the red curve in Figure 7).
In some ways, this is a bit like the way a movie works. When you watch a movie, you see a series of still photographs, probably taken at a rate of 24 pictures (or frames) per second. If you play those photos back at the same rate (24 fps or frames per second), you think you see movement. However, this is because your eyes and brain aren’t fast enough to see 24 individual photos per second – so you are fooled into thinking that things on the screen are moving.
However, digital audio is slightly different from film in two ways:
The sound (equivalent to the movement in the film) is actually happening. It’s not a trick that relies on your ears and brain being too slow.
If, when you were filming the movie, something were to happen between frames (say, the flash of a gunshot, for example) then it would never be caught on film. This is because the photos are discrete moments in time – and what happens between them is lost. However, if something were to make a very, very short sound between two samples (two measurements) in the digital audio signal – it would not be lost. This is because of something that happens at the beginning of the chain that I haven’t described… yet…
However, there are some “artefacts” (a fancy term for “weird errors”) that are present both in film and in digital audio that we should talk about.
The first is an error that happens when you mess around with the rate at which you take the measurements (called the “sampling rate”) or the photos (called the “frame rate”) – and, more importantly, when you need to worry about this. Let’s say that you make a film at 24 fps. If you play this back at a higher frame rate, then things will move very quickly (like old-fashioned baseball movies…). If you play them back at a lower frame rate, then things move in slow motion. So, for things to look “normal” you have to play the movie at the same rate that it was filmed. However, as long as no one is looking, you can transfer the movie as fast as you like. For example, if you wanted to copy the film, you could set up a movie camera so it was pointing at a movie screen and film the film. As long as the movie on the screen is running in sync with the camera, you can do this at any frame rate you like. But you’ll have to watch the copy at the same frame rate as the original film… (Note that this issue is not something that will come up in this series of postings about high resolution audio)
The second is an easy artefact to recognise. If you see a car accelerating from 0 to something fast on film, you’ll see the wheels of the car start to get faster and faster, then, as the car gets faster, the wheels slow down, stop, and then start going backwards… This does not happen in real life (unless you’re in a place lit with flashing lights like fluorescent bulbs or LED’s). I’ll do a posting explaining why this happens – but the thing to remember here is that the speed of the wheel rotation that you see on the film (the one that’s actually captured by the filming…) is not the real rotational speed of the wheel. However, those two rotational speeds are related to each other (and to the frame rate of the film). If you change the real rotational rate or the frame rate, you’ll change the rotational rate in the film. So, we call this effect “aliasing” because it’s a false version (an alias) of the real thing – but it’s always the same alias (assuming you repeat the conditions…) Digital audio can also suffer from aliasing, but in this case, you put in one frequency (which is actually the same as a rotational speed) and you get out another one. This is not the same as harmonic distortion, since the frequency that you get out is due to a relationship between the original frequency and the sampling rate, so the result is almost never a multiple of the input frequency. (We’re going to dig into this a lot deeper through this series of postings about high resolution audio, so if it doesn’t immediately make sense, don’t worry…)
Some important details that I left out…
One of the things I said above was something like “we measure the voltage and store the results” and the example I gave was a nice series of numbers that only had 4 digits after the decimal point. This statement has some implications that we need to discuss.
Let’s say that I have a thing that I need to measure. For example, Figure 8 shows a piece of metal, and I want to measure its width.
Fig 8. A piece of metal with a width of “approximately 57 mm”.
Using my ruler, I can see that this piece of metal is about 57 mm wide. However, if I were geeky (and I am) I would say that this is not precise enough – and therefore it’s not accurate. The problem is that my ruler is only graduated in millimetres. So, if I try to measure anything that is not exactly an integer number of mm long, I’ll either have to guess (and be wrong) or round the measurement to the nearest millimetre (and be wrong).
So, if I wanted you to make a piece of metal the same width as my piece of metal, and I used the ruler in Figure 8, we would probably wind up with metal pieces of two different widths. In order to make this better, we need a better ruler – like the one in Figure 9.
Fig 9. The same piece of metal being measured with a vernier caliper. This gives us additional precision (down to 0.05 mm) so we can make a more accurate measurement.
Figure 9 shows a vernier caliper (a fancy type of ruler) being used to measure the same piece of metal. The caliper has a resolution of 0.05 mm instead of the 1 mm available on the ruler in Figure 8. So, we can make a much more accurate measurement of the metal because we have a measuring device with a higher precision.
The conversion of a digital audio signal is the same. As I said above, we measure the voltage of the electrical signal, and transmit (or store) the measurement. The question is: how accurate and precise is your measurement? As we saw above, this is (partly) determined by how many digits are in the number that you use when you “write down” the measurement.
Since the voltage measurements in digital audio are recorded in binary rather than decimal (we use 0 and 1 to write down the number instead of 0 up to 9) then we use Binary digITS – or “bits” instead of decimal digits (which are not called “dits”). The number of bits we have in the number that we write down (partly) determines the precision of the measurement of the voltage – and therefore (possibly), our accuracy…
Just like the example of the ruler in Figure 8, above, we have a limited resolution in our measurement. For example, if we had only 4 bits to work with then the waveform in 4 – the one we have to measure – would be measured with the “ruler” shown on the left side of Figure 10, below.
Fig 10: The waveform from Figure 4 as a voltage (notice the Y-axis on the right). We have to measure these values using the ruler with the resolution shown on the Y-axis on the left.
When we do this, we have to round off the value to the nearest “tick” on our ruler, as shown in Figure 11.
Fig 11: The values from figure 10 (shown as the circles) rounded off to the nearest value on our 4-bit ruler (the red staircase).
Using this “ruler” which gives a write-down-able “quantity” to the measurement, we get the following values for the red staircase:
When we “play these back” we get the staircase again, shown in Figure 12.
Fig 12: The output of the measurements. Notice that all values sit exactly on one of the values for the “ruler” on the left Y-axis of the plot.
Of course, this means that, by rounding off the values, we have introduced an error in the system (just like the measurement in Figure 8 has a bigger error than the one in Figure 9). We can calculate this error if we just subtract the original signal from the output signal (in other words, Figure 12 minus Figure 10) to get Figure 13.
Fig 13: The error that we produced due to the rounding off of the signal when we did the measurements. Notice that the error is always less than 0.5 of a “tick” of the ruler on the left Y-axis.
In order to improve our accuracy of the measurement, we have to increase the precision of the values. We can do this by adding an extra digit (or bit) to the number that we use to record the value.
If we were using decimal numbers (0-9) then adding an extra digit to the number would give us 10 times as many possibilities. (For example, if we were using 4 digits after the decimal in the example at the start of this posting, we have a total of 10,000 possible values – 0.0000 to 0.9999. If we add one more digit, we increase the resolution to 100,000 possible values – 0.00000 to 0.99999 ).
In binary, adding one extra digit gives us twice as many “ticks” on the ruler. So, using 4 bits gives us 16 possible values. Increasing to 5 bits gives us 32 possible values.
If you’re listening to a CD, then the individual measurements of each voltage – the “sample values” – are stored with 16 bits, which means that we have 65,536 possible values to pick from.
Remember that this means that we have more “ticks” on our ruler – but we don’t necessarily increase its range. So, for example, we’re still measuring a voltage from -1 V to 1 V – we just have more and more resolution with which we can do that measurement.
As I’ve talked about in a previous posting, when a reciprocal peak/dip filter says “Q”, there’s no knowing what it might mean, because there are at least 7 different definitions of Q (3 for boosts and 4 for dips).
For many people, this doesn’t really matter. If you’re just playing with an EQ to make things sound better right now, then the values on the display really don’t matter: it’s the sound that counts.
If you’re like me, you need to be able to navigate between different pieces of software and hardware, and to get the same EQ response from them, then you’ll also need to know firstly that you can’t trust the display, and secondly, how to “translate” from device to device when necessary.
For example, take a look at Figure 1
Figure 1: The magnitude response of two peaking filters, both with Fc=1 kHz, Gain = +12 dB, Q = 2
This shows two magnitude responses, however, these are the measurements of two equalisers with identical settings: Fc = 1 kHz, Gain = +12 dB, Q = 2.
The black curve shows the response of an equaliser that uses the -3 dB points to define the bandwidth of the filter, and therefore the Q is based on 1/(2 zeta). The red curve shows the response of an equaliser that uses the mid-point (in this case, +6 dB because the Gain is +12 dB) to define the bandwidth of the filter.
The difference between these two plots is shown below in Figure 2.
Figure 2: The difference between the two curves in Figure 1.
We’d have a similar problem if we were cutting instead of boosting, as shown in Figure 3.
Figure 3: The magnitude response of two peaking filters, both with Fc=1 kHz, Gain = -12 dB, Q = 2
You have to think upside down in this case, because the 1/(2 zeta) filter is actually using the 3 dB UP points to measure bandwidth; but we’ll ignore that and move on.
If you need to translate between the two systems shown above, there’s a pretty easy way to do it.
I’ll assume that you are implementing your filter using the mid-point definition of the bandwidth, so you need to convert into that system rather than out of it. (I’m making this assumption because it’s the one that Robert Bristow-Johnson used in his Audio Cookbook, which was freely copy-and-pasteable, which means that you find it everywhere these days.) Get the parameters from the filter you want to copy.
We’ll call these parameters Fc (for centre frequency, in Hz), (Gain in dB), and . I’m calling it because it’s a Q based on 1/(2 zeta) and we’ll need to keep it separate from our other Q, which I’ll call (for Robert Bristow-Johnson).
Convert the gain into linear.
Then do the following:
IF
ELSEIF
ELSE your filter isn’t doing anything because
END
Example 1
If you have a -3 dB-based filter with the following parameters:
and you want to implement that using the Bristow-Johnson equations, then you’ll have to use the following parameters:
Example 2
If you have a -3 dB-based filter with the following parameters:
and you want to implement that using the Bristow-Johnson equations, then you’ll have to use the following parameters:
Two Extra Things…
If the filter that you’re translating FROM is based on Andy Moorer’s design (which is based on the gain mid-point if the gain is within the ±6 dB range, but based on the 3 dB points if it’s outside that), then you’ll have to write your own IF/THEN statements.
If you’re implementing a filter that was specified for RBJ’s equations in a system that’s based on 1/(2 zeta), then you’re probably smart enough to figure out how to do the above in reverse.
One additional addendum
IF you don’t like IF/THEN statements for some reason or another (code optimisation, for example)
THEN you could do it this way instead:
What I’ve done there is to fold the decibel-to-linear conversion into the equation. I’ve also converted the gain in dB to an absolute value before converting to linear. That way, it’s always positive, so you always divide.
When it comes to audio, the “signal” is an easy thing to define. It’s what you want to listen to – a song, the dialogue in the movie – whatever it was that you wanted to hear that made you turn on the loudspeaker in the first place.
Let’s say that, normally, we listen to music – so that’s the signal. And, although “music” means different things to different people, most of the time, “music” will contain energy at more than one frequency, and its level will change over time. For example, compare the two plots in Figure 1.
Figure 1: The top plot is a 1-second slice from Jennifer Warnes’s “Bird on a Wire”. The bottom plot is a 1-second slice of a 997 Hz sine tone.
Looking at Figure 1, it seems obvious that the level of “Bird on a Wire” changes over time, but the level of a sine wave doesn’t. However, that’s not as obvious when we zoom into that plot, as is shown in Figure 2, below.
Figure 2: a 10-ms slice out of Figure 1.
From Figure 2, we can easily establish the obvious fact that “Bird on a Wire” and a sine wave are different. However, now it’s not as obvious that the sine wave as a constant level – it repeats itself periodically – which is why we call it “periodic” – but what is its level?
The simplest way to determine the level of a signal is similar to the way yesterday’s share prices are shown in the financial section of the newspaper. In that case, you are told the highest price and the lowest price for the day. In audio, we sometimes talk to the “peak-to-peak” amplitude of a signal. This is the difference between the highest and and the lowest peak (more accurately called a “trough”) of the signal in whatever amount of time you’ve been measuring. For example, take a look at Figure 3.
Figure 3: Two different signals with the same peak-to-peak amplitude.
In Figure 3, I’ve drawn two signals. The top one is a 100 Hz sine wave with a peak-to-peak amplitude of 2 (because the difference between the highest peak (+1) and the lowest peak (-1) is 2). The bottom signal is a 100 Hz sine wave with a peak-to-peak amplitude of 0.1 – but with two clicks – one hitting +1 and the other hitting -1. So, if I just look at the peaks of that second signal, it also has a peak-to-peak amplitude of 2.
So, although it was easy to find the peak-to-peak amplitudes of those two signals, it should be obvious that this does not give a fair indication of how loud they appear to be.
However, if you’re building a piece of audio equipment (like an amplifier or an EQ, for example), this measurement does give you an idea of the “worst case” limits of the signal that might come through the system. So it’s not a useless measurement.
An additional problem with a peak-to-peak measurement of a signal is that it doesn’t tell you anything about asymmetry across the 0 line. (In an analogue world, we’d call that a “DC offset” because there would be a DC voltage that is added to the AC waveform.) For example, both of the signals in Figure 4 have a peak-to-peak amplitude of 1, but they are different…
Figure 4: Two sine waves with the same frequency and the same peak-to-peak amplitude. One of them will be nice to a loudspeaker driver (the top one…) but the other will not (the bottom one)
If you’re lazy, you can do half of a peak-to-peak measurement. This is where you just check the maximum value of either the peak or the trough. We call this a “peak” amplitude measurement.
This has its problems, though. For example, take a look at Figure 5.
Figure 5: Two signals, each with a peak amplitude of 1.
Here, we see two signals. The top one is a sine wave. The bottom one was a sine wave until I squished its negative-going half with a cheap compressor. As you can probably see, the top waveform is symmetrical – the negative half of the signal is the same as the positive half of the signal, just upside-down. It is also easily obvious that the second signal on the lower plot is not symmetrical. Its positive peak is higher than its negative peak.
However, both of these signals have a maximum positive peak of 1 – therefore their peak amplitudes are both 1 (but their peak-to-peak amplitudes and their apparent loudnesses are different).
You might think that an easy way around this problem is to look at the absolute value of the signals and find the peaks that way. However, as you can see in Figure 6, in the case of asymmetrical signals, this does not change anything.
Figure 6: The same signals shown in Figure 5 – but the plot shows the absolute values of the signals. Note that their peak amplitudes are still the same here.
Another way to look at the signal is to take an average of the level over time. However, if the signal is symmetrical (like a sine wave, for example) this would not work, since the average will probably be 0. This is because, if the signal is symmetrical, then the average of all of the negative values in the signal (over time) average out to be the negative equal of the average of all of the positive values. So we can’t just use the average of the signal directly… However, with a little extra math, we can do something useful.
I’m going to skip quickly over some old-fashioned math here in order to jump to the punchline which is: “the power in an AC signal (like a sine wave) is proportional to the square of the signal.”
The reason for this can be explained by combining Ohm’s Law and Watt’s Law as follows:
V = IR
where V is electromotive force (or voltage) in volts, I is current in amperes, and R is the resistance in ohms.
P = VI
where P is the power in watts, and V and I are the same as above.
If we fiddle with Ohm’s Law like this:
V = IR
therefore
I = V/R
Then we can replace the “I” in Watt’s Law like this
P = VI
P = V * V/R
P = V2 / R
So, with that last equation, we can see that the Power (in watts) is proportional to the square of the Voltage (in volts). So, if you double the voltage, you get 4 times the power (because 22 = 4).
We could do the same thing for current, as follows:
P = VI
P = IR * I
P = I2 R
So what? Well, one thing this tells us is that, if you want double the power (for example, from a loudspeaker’s output or the heat from a hair dryer) then you’ll need 4 times the amplitude of the signal feeding it (for example, 4 times the voltage at the same current level or 4 times the current with the same voltage).
Now, let’s come back to the problem at hand… What’s the level of the signal? Well, we start by taking our signal and find its equivalent power (by squaring its instantaneous amplitude value over time – so, for example, if it’s a digital signal, we take the value of each sample and multiply it by itself). Part of the effect of this squaring of the signal is that it removes the negative portion of the signal (because a negative number multiplied by a negative number is a positive number).
We then take a slice of time, and average all of the values that we just created by squaring the original values. Now we have the average (or “mean”) power in the signal.
However, we’re not interested in the power of the signal, we’re interested in its “average” amplitude (say, its voltage). So, to get back from power, we take the square root of the average that we just calculated.
By doing all of this, we are finding the Root of the Mean of the Square of the voltage – the RMS level.
If we apply this math to a sine wave, the result will be something like what’s shown in Figure 6.
Figure 6. See the following text.
In Figure 6, the black curve is the original sine wave with a frequency of 100 Hz and a peak amplitude of 1.0 (and no DC offset). The red curve shows the result of squaring all the values in the sine wave (which is why it’s called a “sine squared” wave or sin2 wave). If we find the average of all of the values in the red curve, the result would be 0.5. The square root of 0.5 is approximately 0.707 – which is shown as the blue line in the plot.
So, the RMS value of a sine wave with a peak value of 1 is 0.707. What does this mean? The easiest way to think of this is that if you had an old-fashioned incandescent light bulb and you powered it with a 1Vp (1 Volt Peak) AC voltage sine wave, it would be exactly the same brightness as if you connected it to a 0.707 V DC battery instead. If you wanted to use a battery to power your toaster, and you wanted it to make toast just as quickly as it normally does, then the battery will have to have a voltage that is 0.707 * the peak value of the AC voltage that normally feeds it. (Note that, if you live in North America, then the electrical signal feeding your toaster is 110 V RMS – so you’ll need a 110 V battery. If you live in Europe, then your toaster is fed with 220 V RMS – so you’ll need a 220 V battery. If you live somewhere else, you might need something else… Note that the electrical company has already done the RMS calculation for you…)
So, an RMS measurement of an AC signal tells us what DC value would result in the same power consumption.
There is just one problem: part of the RMS calculation is the “M” part – we are finding the mean of the values over some period of time. The length of time that we’re going to use is easy to choose if it’s a sine wave – we just make sure that the length of time (we call it a “time constant”) is at least as long as one period of the sine wave itself. If it’s smaller, then the RMS value will bob up and down as the sine wave goes up and down.
However, if we’re going to try to use the RMS method to find the level of a music signal, we’re going to have to make some tough choices… For example, let’s find the RMS value of the “Bird on a Wire” sample, using different time constants, shown below.
Figure 7: The green plots show the same thing: a 200-ms slice out of Bird on a Wire. The red plots show the RMS level of the green signal for 4 different time constants – from top to bottom: 1 ms, 10 ms, 100 ms, and 1 second.
If we convert the plot in Figure 7 to a decibel representation by taking 20*log10 of each sample value, we get the plots in Figure 8. (Note that this is not the same as dB FS, since we are not comparing the result to the RMS value of a full-scale sine wave… but that’s a topic for another posting.)
Figure 8: The decibel equivalents of the plots shown in Figure 7.
There are some things that are evident in Figure 8. The most obvious one is that there is a link between the RMS time constant and the variability of the RMS level when the signal that you’re analysing is not periodic. Looking at this short 200 ms-long example from Bird on a Wire, with the four time constants that I used, the range of results are as shown below in Figure 9.
Figure 9. The ranges of RMS levels in Figure 8
[table]
RMS Time constant, Min (dB), Max (dB), Range (dB)
Original signal, -∞, -9.1, ∞
1 ms, -43.6, -10.8, 32.8
10 ms, -27.2, -15.8, 11.4
100 ms, -24.0, -19.4, 4.6
1 sec, -23.2, -22.8, 0.4
[/table]
Of course, it’s important to remember that if I had picked a different signal or different RMS time constants, I would have gotten different results.
The question to ask here is:
“If I want to know the level of that 200 ms slice of Bird on a Wire, which RMS time constant should I use?”
or
“which of those four plots tells me the signal’s level?”
The answer is that none of these is correct – or all of them are, even though they show different things. The problem is that music has such a wide frequency range – from 20 Hz to 20,000 Hz. Therefore, if you choose a time constant that is long enough to give you a stable measurement at 20 Hz (which will be at least 50 ms – 1/20th of a second or one period of a 20 Hz wave), then it will be the length of 1000 periods of the 20 kHz portion of the signal.
Of course, you could argue that you care more about the 20 Hz part than the 20,000 Hz part – but that’s dependent on what you’re doing. If you’re measuring the signal that’s being sent to a tweeter, then you’re probably not interested in what’s going on at 20 Hz at all…
So what?
We’re heading (in a future posting) towards talking about measuring a system’s (or a devices’s) ability to deliver a wide range of signal levels. We’re going to talk about its “signal to noise ratio” which is a measurement of how much louder the signal (the music) is than the noise that the system itself generates. The idea in the design of all audio systems is that you want to make that ratio as big as possible so that you cannot hear the noise because it’s so much quieter than the signal.
The problem is that we’re going to have to measure how loud the signal can be – and compare that to how loud the signal actually is at any given moment. In order to understand the concepts in that discussion, then it’s necessary to understand the concepts that I introduced above, namely the following:
Peak-peak level
Peak level
RMS level
the relationship between RMS Time constant and the RMS level
(Gain in dB), and 
. I’m calling it 
(for Robert Bristow-Johnson).![\[G_{lin} = 10^\frac{G_{dB}}{20}\]](http://www.tonmeister.ca/wordpress/wp-content/ql-cache/quicklatex.com-5da79ddd7eef7bebd2ff0a8cb915d15f_l3.png "Rendered by QuickLaTeX.com")
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![\[Q_{rbj} = \frac {2} {\sqrt{ 3.9811}} = 1.0024\]](http://www.tonmeister.ca/wordpress/wp-content/ql-cache/quicklatex.com-e54e30d0230089601b1f302f89c444a7_l3.png "Rendered by QuickLaTeX.com")
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