In Part 1 of this series, I talked about three different options for converting from a fixed-point representation to another fixed-point representation with a bigger bit depth.
This happens occasionally. The simplest case is when you send a 16-bit signal to a 24-bit DAC. Another good example is when you send a 16-bit LPCM signal to a 24- or 32-bit fixed point digital signal processor.
However, these days it’s more likely that the incoming fixed-point signal (incoming signals are almost always in a fixed-point representation) is converted to floating point for signal processing. (I covered the differences between fixed- and floating-point representations in another posting.)
If you’re converting from fixed point to floating point, you divide the sample’s value by 2^(nBits-1). In other words, if you’re converting a 5-bit signal to floating point, you divide each sample’s value by 2^4, as shown below.
Figure 1. Converting a 5-bit fixed point signal to floating point
The reason for this is that there are 2^(nBits-1) quantisation levels for the negative portions of the signal. The positive-going portions have one fewer levels due to the two’s complement representation (the 00000 had to come from somewhere…).
So, you want the most-negative value to correspond to -1.0000 in the floating point world, and then everything else looks after itself.
Of course, this means that you will never hit +1.0. You’ll have a maximum signal level of 1 – 1/2^(nBits-1), which is very close. Close enough.
The nice thing about doing this conversation is that by entering into a floating point world, you immediately gain resolution to attenuate and headroom to increase the gain of the signal – which is exactly what we do when we start processing things.
Of course, this also means that, when you’re done processing, you’ll need to feed the signal out to a fixed-point world again (for example, to a DAC or to an S/PDIF output). That conversion is the topic of Part 4.
In Part 1, I talked about different options for converting a quantised LPCM audio signal, encoded with some number of bits into an encoding with more bits. In this posting, we’ll look at a trick that can be used when you combine these options.
To start, made two signals:
“Signal 1” is a sinusoidal tone with a frequency of 100 Hz. It has an amplitude of ±1, but then encoded it as a quantised 8-bit signal, so in Figure 1, it looks like it has an amplitude of ±127 (which is 2^(nBits-1)-1)
“Signal 2” is a sinusoidal tone with a frequency of 1 kHz and the same amplitude as Signal 1.
Both of these two signals are plotted on the left side of Figure 1, below. On the right, you can see the frequency content of the two signals as well. Notice that there is plenty of “garbage” at the bottom of those two plots. This is because I just quantised the signals without dither, so what you’re seeing there is the frequency-domain artefacts of quantisation error.
Figure 1. Two sinusoidal waveforms with different frequencies. Both are 8-bit quantised without dither.
If I look at the actual sample values of “Signal 1” for the first 10 samples, they look like the table below. I’ve listed them in both decimal values and their binary representations. The reason for this will be obvious later.
Sample number
Sample value (decimal)
Sample Value (binary)
1
0
00000000
2
2
00000010
3
3
00000011
4
5
00000101
5
7
00000111
6
8
00001000
7
10
00001010
8
12
00001100
9
13
00001101
10
15
00001111
Let’s also look at the first 10 sample values for “Signal 2”
Sample number
Sample value (decimal)
Sample Value (binary)
1
0
00000000
2
17
00010001
3
33
00100001
4
49
00110001
5
63
00111111
6
77
01001101
7
90
01011010
8
101
01100101
9
110
01101110
10
117
01110101
The signals I plotted above have a sampling rate of 48 kHz, so there are a LOT more samples after the 10th one… however, for the purposes of this posting, the ten values listed in the tables above are plenty.
At the end of the Part 1, I talked about the Most and the Least Significant Bits (MSBs and LSBs) in a binary number. In the context of that posting, we were talking about whether the bit values in the original signal became the MSBs (for Option 1) or the LSBs (for Option 3) in the new representation.
In this posting, we’re doing something different.
Both of the signals above are encoded as 8-bit signals. What happens if we combine them by just slamming their two values together to make 16-bit numbers?
For example, if we look at sample #10 from both of the tables above:
Signal 1, Sample #10 = 00001111
Signal 2, Sample #10 = 01110101
If I put those two binary numbers together, making Signal 1 the 8 MSBs and Signal 2 the 8 LSBs then I get
0000111101110101
Note that I formatted them with bold and italics just to make it easier to see them. I could have just written 0000111101110101 and let you figure it out.
Just to keep things adequately geeky, you should know that “slamming their values together” is not the correct term for what I’ve done here. It’s called binary concatenation.
Another way to think about what I’ve done is to say that I converted Signal 1 from an 8-bit to a 16-bit number by zero-padding, and then I added Signal 2 to the result.
Yet another way to think of it is to say that I added about 48 dB of gain to Signal 1 (20*log10(2^8) = about 48.164799306236993 dB of gain to be more precise…) and then added Signal 2 to the result. (NB. This is not really correct, as is explained below.)
However, when you’re working with the numbers inside the computer’s code, it’s easier to just concatenate the two binary numbers to get the same result.
If you do this, what do you get? The result is shown in Figure 2, below.
Figure 2. The binary concatenated result of Signal 1 and Signal 2
As you can see there, the numbers on the y-axis are MUCH bigger. This is because of the bit-shifting done to Signal 1. The MSBs of a 16-bit number are 256 times bigger in decimal world than those of an 8-bit number (because 2^8 = 256).
In other words, the maximum value in either Signal 1 or Signal 2 is 127 (or 2^(8-1)-1) whereas the maximum value in the combined signal is 32767 (or 2^(16-1)-1).
The table below shows the resulting first 10 values of the combined signal.
Sample number
Sample value (decimal)
Sample Value (binary)
1
0
0000000000000000
2
529
0000001000010001
3
801
0000001100100001
4
1329
0000010100110001
5
1855
0000011100111111
6
2125
0000100001001101
7
2650
0000101001011010
8
3173
0000110001100101
9
3438
0000110101101110
10
3957
0000111101110101
Why is this useful? Well, up to now, it’s not. But, we have one trick left up our sleeve… We can split them apart again, taking that column of numbers on the right side of the table above, cut each one into two 8-bit values, and ta-da! We get out the two signals that we started with!
Just to make sure that I’m not lying, I actually did all of that and plotted the output in Figure 3. If you look carefully at the quantisation error artefacts in the frequency-domain plots, you’ll see that they’re identical to those in Figure 1. (Although, if they weren’t, then this would mean that I made a mistake in my Matlab code…)
Figure 3. The two signals after they’ve been separated once again.
So what?
Okay, this might seem like a dumb trick. But it’s not. This is a really useful trick in some specific cases: transmitting audio signals is one of the first ones to come to mind.
Let’s say, for example, that you wanted to send audio over an S/PDIF digital audio connection. The S/PDIF protocol is designed to transmit two channels of audio with up to 24-bit LPCM resolution. Yes, you can do different things by sending non-LPCM data (like DSD over PCM (DoP) or Dolby Digital-encoded signals, for example) but we won’t talk about those.
If you use this binary concatenation and splitting technique, you could, for example, send two completely different audio signals in each of the audio channels on the S/PDIF. For example, you could send one 16-bit signal (as the 16 MSBs) and a different 8-bit signal (as the LSBs), resulting in a total of 24 bits.
On the receiving end, you split the 24-bit values into the 16-bit and 8-bit constituents, and you get back what you put in.
(Or, if you wanted to get really funky, you could put the two 8-bit leftovers together to make a 16-bit signal, thus transmitting three lossless LPCM 16-bit channels over a stream designed for two 24-bit signals.)
However, if you DON’T split them, and you just play the 24-bit signal into a system, then that 8-bit signal is so low in level that it’s probably inaudible (since it’s at least 93 dB below the peak of the “main” signal). So, no noticeable harm done!
Hopefully, now you can see that there are lots of potential uses for this. For example, it could be a sneaky way for a record label to put watermarking into an audio signal, for example. Or you could use it to send secret messages across enemy lines, buried under a recording of the Alvin and the Chipmunk’s cover of “Achy Breaky Heart”. Or you could use it for squeezing more than two channels out of an S/PDIF cable for multichannel audio playback.
One small issue…
Just to be clear, I actually used Matlab and did all the stuff I said above to make those plots. I didn’t fake it. I promise!
But if you’re looking carefully, you might notice two things that I also noticed when I was writing this.
I said above that, by bit-shifting Signal 1 over by 8 bits in the combined signal, this makes it 48 dB louder than Signal 2. However, if you look at the frequency domain plot in Figure 2, you’ll notice that the 1 kHz tone is about 60 dB lower than the 100 Hz tone. You’ll also notice that there are distortion artefacts on the 1 kHz signal at 3 kHz, 5 kHz and so on – but they’re not there in the extracted signal in Figure 3. So, what’s going on?
To be honest, when I saw this, I had no idea, but I’m lucky enough to work with some smart people who figured it out.
If you go back to the figures in Part 1, you can see that the MSB of a sample value in binary representation is used as the “sign” of the value. In other words, if that first bit is 0, then it’s a positive value. If it’s a 1 then it’s a negative value. This is known as a “two’s complement” representation of the signal.
When we do the concatenation of the two sample values as I showed in the example above, the “sign” bit of the signal that becomes the LSBs of the combined signal no longer behaves as a +/- sign. So, the truth is that, although I said above that it’s like adding the two signals – it’s really not exactly the same.
If we take the signal combined through concatenation and subtract ONLY the bit-shifted version of Signal 1, the result looks like this:
Figure 4. The difference between the combined signals shown in Figure 3 and Signal 1, after it’s been bit-shifted (or zero-padded) by 8 LSBs.
Notice that the difference signal has a period of 1 ms, therefore its fundamental is 1 kHz, which makes sense because it’s a weirdly distorted version of Signal 2, which is a 1 kHz sine tone.
However, that fundamental frequency has a lower level than the original sine tone (notice that it shows up at about -60 dB instead of -48 dB in Figure 2). In addition, it has a DC offset (no negative values) and it’s got to have some serious THD to be that weird looking. Since it’s a symmetrical waveform, its distortion artefacts consist of only odd multiples of the fundamental.
Therefore, when I stated above that you’re “just” adding the two signals together, so there’s no harm done if you don’t separate them at the receiving end. This was a lie. But, if your signal with the MSBs has enough bits, then you’ll get away with it, since this pushes the second signal further down in level.
This is the first of a series of postings about strategies for converting from one bit depth to another, including conversion back and forth between fixed point and floating point encoding. It’ll be focusing on a purely practical perspective, with examples of why you need to worry about these things when you’re doing something like testing audio devices or transmission systems.
As we go through this, it might be necessary to do a little review, which means going back and reading some other postings I’ve done in the past if some of the concepts are new. I’ll link back to these as we need them, rather than hitting you with them all at once.
To start, if you’re not familiar with the concept of quantisation and bit depth in an LPCM audio signal, I suggest that you read this posting.
Now that you’re back, you know that if you’re just converting a continuous audio signal to a quantised LPCM version of it, the number of bits in the encoded sample values can be thought of as a measure of the system’s resolution. The more bits you have, the more quantisation steps, and therefore the better the signal to noise ratio.
However, this assumes that you’re using as many of the quantisation steps as possible – in other words, it assumes that you have aligned levels so that the highest point in the audio signal hits the highest possible quantisation step. If your audio signal is 6 dB lower than this, then you’re only using half of your available quantisation values. In other words, if you have a 16-bit ADC, and your audio signal has a maximum peak of -6 dB FS Peak, then you’ve done a 15-bit recording.
But let’s say that you already have an LPCM signal, and you want to convert it to a larger bit depth. A very normal real-world example of this is that you have a 16-bit signal that you’ve ripped from a CD, and you export it as a 24-bit wave file. Where do those 8 extra bits come from and where do they go?
Generally speaking, you have 3 options when you do this “conversion”, and the first option I’ll describe below is, by far the most common method.
Option 1: Zero padding
Let’s simplify the bit depths down to human-readable sizes. We’ll convert a 3-bit LPCM audio signal (therefore it has 2^3 = 8 quantisation steps) into a 5-bit representation (2^5 = 32 quantisation steps), instead of 16-bit to 24-bit. That way, I don’t have to type as many numbers into my drawings. The basic concepts are identical, I’ll just need fewer digits in this version.
The simplest method to do is to throw some extra zeros on the right side of our original values, and save them in the new format. A graphic version of this is shown in Figure 1.
Figure 1. Zero-padding to convert to a higher bit depth.
There are a number of reasons why this is a smart method to use (which also explains why this is the most common method).
The first is that there is no change in signal level. If you have a 0 dB FS Peak signal in the 3-bit world, then we assume that it hits the most-negative value of 100. If you zero-pad this, then the value becomes 10000, which is also the most-negative value in the 5-bit world. If you’re testing with symmetrical signals (like a sinusoidal tone) then you never hit the most-negative value, since this would mean that it would clip on the positive side. This might result in a test that’s worth talking about, since sinusoidal tone that hits 011 and is then converted to 01100. In the 5-bit world, you could make a tone that is a little higher in level (by 3 quantisation levels – those top three dotted lines on the right side of Figure 1), but that difference is very small in real life, so we ignore it. The biggest reason for ignoring this is that this extra “headroom” that you gain is actually fictitious – it’s an artefact of the fact that you typically test signal levels like this with sine tones, which are symmetrical.
The second reason is that this method gives you extra resolution to attenuate the signal. For example, if you wanted to make a volume knob that only attenuated the audio signal, then this conversion method is a good way to do it. (For example, you send a 16-bit digital signal into the input of a loudspeaker with a volume controller. You zero-pad the signal to 24-bit and you now have the ability to reduce the signal level by 141 dB instead of just 93 dB (assuming that you’re using dither…). This is good if the analogue dynamic range of the rest of the system “downstream” is more than 93 dB.) The extra resolution you get is equivalent to 6 dB * each extra bit. So, in the system above:
(5 bits – 3 bits) = 2 extra bits 2 extra bits * 6 dB = 12 dB extra resolution
There is one thing to remember when doing it this way, that you may consider to be a disadvantage. This is the fact that you can’t increase the gain without clipping. So, let’s say that you’re building a digital equaliser or a mixer in a fixed-point system, then you can’t just zero-pad the incoming signal and think that you can boost signals or add them. If you do this, you’ll clip. So, you would have to zero-pad at the input, then attenuate the signal to “buy” yourself enough headroom to increase it again with the EQ or by mixing.
Option 2
The second option is, in essence, the same as the trick I just explained in the previous paragraph. With this method, you don’t ONLY pad the right side of the values with zeros, you pad the values on the left as well with either a 1 or a 0, depending on whether the signals are positive or negative. This means that your “old” value is inserted into the middle of the new value, as shown below in Figure 2. (In this 3- to 5-bit example, this is identical to using option 1, and then dropping the signal level by 6 dB (1 of the 2 bits)).
If your conversion to the bigger bit depth is done inside a system where you know what you’ve done, and if you need room to scale the level of the signal up and down, this is a clever and simple way to do things. There are some systems that did this in the past, but since it’s a process that’s done internally, and we normal people sit outside the system, there’s no real way for us to know that they did it this way.
(For example, I once heard through the grapevine that there was a DAW that imported 24-bits into a 48-bit fixed point processing system, where they padded the incoming files with 12 bits on either side to give room to drop levels on the faders and also be able to mix a lot of full-scale signals without clipping the output.)
Option 3
I only add the third option as a point of completion. This is an unusual way to do a conversion, and I only personally know of one instance where it’s been used. This only means that it’s not a common way to do things – not that NO ONE does it.
In this method, all the padding is done on the left side of the binary values, as shown below in Figure 3.
If we’re thinking along the same lines as in Options 1 and 2, then you could say that this system does not add resolution to attenuate signals, but it does give you the option to make them louder.
However, as we’ll see in Part 2 of this series, there is another advantage to doing it this way…
Nota Bene
I’ve written everything above this line, intentionally avoiding a couple of common terms, but we’ll need those terms in our vocabulary before moving on to Part 2.
If you look at the figures above, and start at the red 0 line, as you go upwards, the increase in signal can be seen as an increase in the left-most bits in each quantisation value. Reading from left-to-right, the first bit tells us whether the value is positive (0) or negative (1), but after this (for the positive values) the more 1s you have on the left, the higher the level. This is why we call them the Most Significant Bits or MSBs. Of course, this means that the last bit on the right is the Least Significant Bit or LSB.
This means that I could have explained Option 1 above by saying:
The three bits of the original signal become the three MSBs of the new signal.
… which also tells us that the signal level will not drop when converted to the 5-bit system.
Or I could have explained Option 3 by saying:
The three bits of the original signal become the three LSBs of the new signal.
.. which also tells us that the signal level will drop when converted to the 5-bit system.
Being able to think in terms of LSBs and MSBs will be helpful later.
Finally… yes, we will talk about Floating Point representations. Later. But if you can’t wait, read this in the meantime.
I just stumbled across this paper and it struck me as a brilliant idea – detecting symptoms of Parkinson’s disease by analysing frequency modulation of speech.
In Part 1, I talked about how any measurement of an audio device tells you something about how it behaves, but you need to know a LOT more than what you can learn from one measurements. This is especially true for a loudspeaker where you have the extra dimensions of physical space to consider.
Thought experiment: Fridges vs. Mosquitos
Consider a situation where you’re sitting at your kitchen table, and you can hear the compressor in your fridge humming/buzzing over on the other side of the room. If you make a small movement in your chair, the hum from the fridge sounds the same to you. This is partly because the distance from the fridge to you is much bigger than the changes in that distance that result from you shifting your butt.
Now think about the times you’ve been trying to sleep on a summer night, and there’s a mosquito that is flying near your ear. Very small changes in the location of that mosquito result in VERY big changes in how it sounds to you. This is because, relative to the distance to the mosquito, the changes in distance are big.
In other words, in the case of the fridge (that’s say, 3 m away) by moving 10 cm in your chair, you were changing the distance by about 3%, but the mosquito was changing its distance by more than 100% by moving just from 1 cm to 2 cm away.
In other words, a small change in distance makes a big change in sound when the distance is small to begin with.
The challenges of measuring headphones
The methods we use for measuring the magnitude response of a pair of headphones is similar to that used for measuring a loudspeaker. We send a measurement signal to the headphones from a computer, that signal comes out and is received by a microphone that sends its output back to the computer. The computer then is used to determine the difference between what it sent out and what came back. Simple, right?
Wrong.
The problems start with the fact that there are some fundamental differences between headphones and loudspeakers. For starters, there’s no “listening room” with headphones, so we don’t put a microphone 3 m away from the headphones: that wouldn’t make any sense. Instead, we put the headphones on some kind of a device that either simulates an ear, or a head, or a head with ears (with or without ear canals), and that device has a microphone (roughly) where your eardrum would be. Simple, right?
Wrong.
The problem in that sentence was the word “simulates”. How do you simulate an ear or a head or a head with ears? My ears are not shaped identically to yours or anyone else’s. My head is a different size than yours. I don’t have any hair, but you might. I wear glasses, but you might not. There are many things that make us different physically, so how can the device that we use to measure the headphones “simulate” us all? The simple answer to this question is “it can’t.”
This problem is compounded with the fact that measurement devices are usually made out of plastic and metal instead of human skin, so the headphones themselves “see” a different “acoustic load” on the measurement device than they do when they’re on a human head. (The people I work with call this your acoustic impedance.)
However, if your day job is to develop or test headphones, you need to use something to measure how they’re behaving. So, we do.
Headphone measurement systems
There are three basic types of devices that are used to measure headphones.
an artificial ear is typically a metal plate with a depression in the middle. At the bottom of the depression is a microphone. In theory, the acoustic impedance of this is similar to a human ear/pinna + the surrounding part of your head. In practice, this is impossible.
a headphone test fixture looks like a big metal can lying on its side (about the size of an old coffee can, for example) on a base. It might have flat metal sides, or it could have rubber pinnae (the fancy word for ears) mounted on it instead. In the centre of each circular end is a microphone.
a dummy head looks like a simplified model of a human head (typically a man’s head). It might have pinnae, but it might not. If it does, those pinnae might look very much like human ears, or they could look like simplified versions instead. There are microphones where you would expect them, and they might be at the bottom of ear canals, but you can also get dummy heads without ear canals where the microphones are flush with the side of the head.
The test system you use is up to you – but you have to know that they will all tell you something different. This is not only because each of them has a different acoustic response, but also because their different shapes and materials make the headphones themselves behave differently.
That last sentence is important to remember, not just for headphone measurement systems but also for you. If your head and my head are different from each other, AND your pinnae and my pinnae are different from each other, THEN, if I lend you my headphones, the headphones themselves will behave differently on your head than they do on my head. It’s not just our opinions of how they sound that are different – they actually sound different at our two sets of eardrums.
General headphone types
If I oversimplify headphone design, we can talk about two basic acoustical type of headphones: They can be closed (where the back of the diaphragm is enclosed in a sealed cabinet, and so the outside of the headphones is typically made of metal or plastic) or open (where the back of the diaphragm is exposed to the outside world, typically through a metal screen). I’d say that some kinds of headphones can be called semi-open, which just means that the screen has smaller (and/or fewer) holes in it, so there’s less acoustical “transparency” to the outside world.
Examples
To show that all these combinations are different, I took three pairs of headphones
open headphones
semi-open headphones
closed headphones
and I measured each of them on three test devices
artificial “simplified” ear
text fixture with a flat-plate
dummy head
In addition, to illustrate an additional issue (the “mosquito problem”), I did each of these 9 measurements 5 times, removing and replacing the headphones between each measurement. I was intentionally sloppy when placing the headphones on the devices, but kept my accuracy within ±5 mm of the “correct” location. I also changed the clamping force of the headphones on the test devices (by changing the extension of the headband to a random place each time) since this also has a measurable effect on the measured response.
Do not bother asking which headphones I measured or which test systems I used. I’m not telling, since it doesn’t matter. Not to me, anyway…
The raw results
I did these measurements using a 10-second sinusoidal sweep from 2 Hz to Nyquist, on a system running at 96 kHz. I’m plotting the magnitude responses with a range from 10 Hz to 40 kHz. However, since the sweep starts at 2 Hz, you can’t really trust the results below 20 Hz (a decade below the lowest frequency of interest is a good rule of thumb when using sine sweeps).
Figure 1: The “raw” magnitude responses of the open headphones measured 5 times each on the three systems
Figure 2: The “raw” magnitude responses of the semi-open headphones measured 5 times each on the three systems
Figure 3: The “raw” magnitude responses of the closed headphones measured 5 times each on the three systems
Looking at the results in the plots above, you can come to some very quick conclusions:
All of the measurements are different from each other, even when you’re looking at the same headphones on the same measurement device. This is especially true in the high frequency bands.
Each pair of headphones looks like it has a different response on each measurement system. For example, looking at Figure 3, the response of the headphones looks different when measured on a flat plate than on a dummy head.
The difference in the results of the systems are different with the different headphone types. For example, the three sets of plots for the “semi-open” headphones (Fig. 2) look more similar to each other than the three sets of plots for the “closed” headphones (Fig. 3)
the scale of these differences is big. Notice that we have an 80 dB scale on all plots… We’re not dealing with subtleties here…
In Part 3 of this series, we’ll dig into those raw results a little to compare and contrast them and talk a little about why they are as different as they are.
People who work in the audio industry use all kinds of different measurements to evaluate the performance of equipment. In many cases, the measurements we do are chosen because they’re easy to do (or because they were easy to do in “The Old Days”), and not because they accurately represent how the equipment actually behaves.
Magnitude response
One simple example of this is what most people call a frequency response but what is actually a magnitude response. This is a measure of how the level of an audio signal is changed by the device under test (the “DUT”) as a function of frequency. For example, if you’re measuring a RIAA-spec preamplifier (used for converting a turntable’s pickup’s output to a “line” level signal), then it should have a magnitude response that looks like the red line in the plot in Figure 1.
Figure 1: The red line shows the correct magnitude response for the frequency-dependent filtering in a RIAA phono preamplifier.
This curve shows that, relative to a signal at 1 kHz, the lower the frequency, the more gain is applied to the signal and the higher the frequency, the more attenuation is applied to the signal. Note that this curve is normalised to the level at 1 kHz, which should actually be +40 dB higher if we were to include the frequency-independent gain of the system.
It’s important to remember that this plot shows us only one thing: the change in level caused by the DUT as a function of a change in frequency of the signal. What this plot does NOT show us is much, much more… For example:
We don’t know anything about the behaviour of the system outside the boundaries of this plot.
We don’t know anything about its phase response.
We don’t know anything about how loud the noise of the DUT is.
We don’t know if this plot is true if we were to measure the DUT at a different input level.
We don’t know whether the DUT would have a different behaviour if the device that was feeding it had a different output impedance.
We don’t know whether the DUT would have a different behaviour if the device that it was feeding had a different input impedance.
We don’t know anything about whether the signal has any non-linear distortion artefacts. (Notice that I didn’t say “…whether the signal is distorted” because we know it’s distorted, since the output of the DUT is not the same as the input of the DUT. Any change in the signal is a form of distortion of the signal.)
I’m not saying that a simple magnitude response plot of a DUT is not useful. I’m just saying that it’s not enough information. It’s like asking for the temperature of a cup of coffee. It’s useful information, but it doesn’t tell you enough to know whether you’re going to enjoy drinking it (unless, of course, you hate coffee…)
This problem gets even worse when you’re measuring the acoustic output of a device like a loudspeaker or a pair of headphones, for example. (The acoustic input of a microphone is a similar problem in the opposite direction.)
Let’s start by thinking about a loudspeaker’s output in real life.
You have a device that radiates sound in space in all directions. Let’s look at that space from the loudspeaker’s perspective and say that this means an angle of rotation around the loudspeaker, and an angle of elevation above/below the loudspeaker. That makes two dimensions.
If we’re talking about the loudspeaker’s magnitude response, then we’re looking at its output level (one dimension) as a function of frequency (one more dimension).
That speaker is (usually) in a room, and you’re probably also there too. We can then that this is in three-dimensional space when we talk about the walls, floor, ceiling, and your location inside that space.
Since the surfaces in the room reflect the audio signal, then the time at which the signal arrives at the listening position must also be considered. The “sound” of a loudspeaker at a listening position before the first reflection arrives is different than after a bunch of reflections are coming in and the room has started resonating as well. So, time adds one more dimension to the problem.
We’ll ignore the non-linear distortion artefacts produced by the loudspeaker and the fact that they radiate in different directions differently, since it’s already complicated enough… However, if we were to add things like changes in the response due to temperature of the voice coil or directionally-dependent distortion artefacts like breakup, this would wind up being a much longer discussion…
So, just looking at the small list of “usual suspects” above, we can see that evaluating the sound of a single loudspeaker in a listening room is at least an 8-dimensional problem. And this doesn’t even take things like 2-channel stereo or 7.1.4 multichannel or whether you’re listening to Aretha Franklin or Stockhausen into account…
In other words, it’s complicated. So, we use reductionism to try to start to get an idea of what’s going on. We put a microphone directly in front of a loudspeaker and measure its magnitude response at one level using one kind of test signal (e.g. a swept sine wave or an MLS) and we remove all the room’s reflections somehow. This reduces our 8-dimensional problem to a 2-dimensional version: we have level as a function of frequency and nothing else, since we’ve chosen to throw away everything else by the way we did the measurement.
Figure 2: The on-axis, free-field magnitude response of a loudspeaker.
For example, take a look at the magnitude response shown in Figure 2, which is a real measurement of a real loudspeaker. This measurement was performed using a swept-sine (a sinusoidal wave with a frequency that changes smoothly over time, typically from low to high) with a microphone on-axis to the loudspeaker at a distance of 3 m. The measurement was time-windowed to remove the room reflections, and therefore can be considered to be a “free field” (a sound field that is free of reflections) measurement. However, the roll-off in the low end is actually a combination of the actual response of the loudspeaker and the artefacts of using a shorter time window. (We would have needed to use a much bigger room to get less influence from the time windowing.)
So, this plot ONLY tells us how the loudspeaker behaves at one point in infinite space, when we’re ONLY asking “how does the level of the loudspeaker’s output vary with changes in frequency and we ONLY play sinusoidal signals at one level.” This is all useful information, but we need to know more – otherwise, we’ll jump to conclusions about whether this loudspeaker sounds “good” or not.
Just like looking at ONLY the temperature of a cup of coffee, this doesn’t give us enough of the story to know how the loudspeaker will “sound” (no matter what a magazine reviewer will try and tell you…).
In other words, if we use reductionism to understand the problem, you simplify the question so much that the problem you wind up understanding is not the same as the thing you’re trying to understand in the first place.
For example, if we measure that same loudspeaker at a different angle (by rotating the loudspeaker and leaving the microphone in place) we’ll see a magnitude response like the one shown in Figure 3.
Figure 3: The free-field magnitude response of the same loudspeaker, measured at 90º off-axis.
This magnitude response is the output of the same loudspeaker at 90º off-axis, which might be what’s heading towards your side-wall. If your side wall is perfectly reflective, then this is therefore the magnitude response of your first reflection, which might be a bad thing if you think that it’s important.
So, when you’re looking at any one measurement of anything, you don’t have enough information to know enough to make a general evaluation. However, unfortunately, many people will run with this information and make the evaluation anyway. It’s data, and data doesn’t lie, so this tells the truth, right?
Wrong. Because it’s only a portion of the total truth.
For example, you can say that “organic food is good for me” but I have an allergy to peanuts. So if I eat organic peanuts, I have about 20 minutes to get to a hospital. Much longer than that and I need a funeral home instead. “Organic” is true, but not enough information for me to know whether or not it’ll be an uneventful meal.
After I posted the last two parts of this series (which I thought wrapped it up…) I received an email asking about whether there’s a similar thing happening if you remove the reconstruction (low-pass) filter in the digital-to-analogue part of the signal path.
The answer to this question turned out to be more interesting than I expected… So I wound up turning it into a “Part 3” in the series.
Let’s take a case where you have a 1 kHz signal in a 48 kHz system. The figure below shows three plots. The top plot shows the individual sample values as black circles on a red line, which is the analogue output of a DAC with a reconstruction filter.
The middle plot shows what the analogue output of the DAC would look like if we implemented a Sample-and-hold on the sample values, and we had an infinite analogue bandwidth (which means that the steps have instantaneous transitions and perfect right angles).
The bottom plot shows what the analogue output of the DAC would look like if we implemented the signal as a pulse wave instead, but if we still we had an infinite analogue bandwidth. (Well… sort of…. Those pulses aren’t infinitely short. But they’re short enough to continue with this story.)
Figure 1. A 1 kHz sine wave
If we calculate the spectra of these three signals , they’ll look like the responses shown in Figure 2.
Figure 2. The spectra of the signals in Figure 1.
Notice that all three have a spike at 1 kHz, as we would expect. The outputs of the stepped wave and the pulsed wave have much higher “noise” floors, as well as artefacts in the high frequencies. I’ve indicated the sampling rate at 48 kHz as a vertical black line to make things easy to see.
We’ll come back to those artefacts below.
Let’s do the same thing for a 5 kHz sine wave, still in a 48 kHz system, seen in Figures 3 and 4.
Figure 3. A 5 kHz sine wave
Figure 4. The spectra of the signals in Figure 3.
Compare the high-frequency artefacts in Figure 4 to those in Figure 2.
Now, we’ll do it again for a 15 kHz sine wave.
Figure 5. A 15 kHz sine wave
Figure 6. The spectra of the signals in Figure 5.
There are three things to notice, comparing Figures 2, 4, and 6.
The first thing is that artefacts for the stepped and pulsed waves have the same frequency components.
The second thing is that those artefacts are related to the signal frequency and the sampling rate. For example, the two spikes immediately adjacent to the sampling rate are Fs ± Fc where Fs is the sampling rate and Fc is the frequency of the sine wave. The higher-frequency artefacts are mirrors around multiples of the sampling rate. So, we can generalise to say that the artefacts will appear at
n * Fs ± Fc
where n is an integer value.
This is interesting because it’s aliasing, but it’s aliasing around the sampling rate instead of the Nyquist Frequency, which is what happens at the ADC and inside the digital domain before the DAC.
The third thing is a minor issue. This is the fact that the level of the fundamental frequency in the pulsed wave is lower than it is for the stepped wave. This should not be a surprise, since there’s inherently less energy in that wave (since, most of the time, it’s sitting at 0). However, the artefacts have roughly the same levels; the higher-frequency ones have even higher levels than in the case of the stepped wave. So, the “signal to THD+N” of the pulsed wave is lower than for the stepped wave.
In Part 1, we looked at what happens when you try to record a signal whose frequency is higher than 1/2 the sampling rate (which, from now on, I’ll call the Nyquist Frequency, named after Harry Nyquist who was one of the people that first realised that this limit existed). You record a signal, but it winds up having a different frequency at the output than it had at the input. In addition, that frequency is related to the signal’s frequency and the sampling rate itself.
In order to prevent this from happening, digital recording systems use a low-pass filter that hypothetically prevents any signals above the Nyquist frequency from getting into the analogue-to-digital conversion process. This filter is called an anti-aliasing filter because it prevents any signals that would produce an alias frequency from getting into the system. (In practice, these filters aren’t perfect, and so it’s typical that some energy above the Nyquist frequency leaks into the converter.)
So, this means that if you put a signal that contains high frequency components into the analogue input of an analogue-to-digital converter (or ADC), it will be filtered. An example of this is shown in Figure 1, below. The top plot is a square wave before filtering. The bottom plot is the result of low-pass filtering the square wave, thus heavily attenuating its higher harmonics. This results in a reduction in the slope when the wave transitions between low and high states.
Figure 1: A square wave before and after low-pass filtering.
This means that, if I have an analogue square wave and I record it digitally, the signal that I actually record will be something like the bottom plot rather than the top one, depending on many things like the frequency of the square wave, the characteristics of the anti-aliasing filter, the sampling rate, and so on. Don’t go jumping to conclusions here. The plot above uses an aggressively exaggerated filter to make it obvious that we do something to prevent aliasing in the recorded signal. Do NOT use the plots as proof that “analogue is better than digital” because that’s a one-dimensional and therefore very silly thing to claim.
However…
… just because we keep signals with frequency content above the Nyquist frequency out of the input of the system doesn’t mean that they can’t exist inside the system. In other words, it’s possible to create a signal that produces aliasing after the ADC. You can either do this by
creating signals from scratch (for example, generating a sine tone with a frequency above Nyquist) or
by producing artefacts because of some processing applied to the signal (like clipping, for example).
Let’s take a sine wave and clip it after it’s been converted to a digital signal with a 48 kHz sampling rate, as is shown in Figure 2.
Figure 2: The red curve is a clipped version of the black curve.
When we clip a signal, we generate high-frequency harmonics. For example, the signal in Figure 2 is a 1 kHz sine wave that I clipped at ±0.5. If I analyse the magnitude response of that, it will look something like Figure 3:
Figure 3: The magnitude response of Figure 2, showing the upper harmonics that I created by clipping.
The red curve in Figure 2 is not a ‘perfect’ square wave, so the harmonics seen in Figure 3 won’t follow the pattern that you would expect for such a thing. But that’s not the only reason this plot will be weird…
Figure 3 is actually hiding something from you… I clipped a 1 kHz sine wave, which makes it square-ish. This means that I’ve generated harmonics at 3 kHz, 5 kHz, 7 kHz, and so on, up to ∞ Hz..
Notice there that I didn’t say “up to the Nyquist frequency”, which, in this example with a sampling rate of 48 kHz, would be 24 kHz.
Those harmonics above the Nyquist frequency were generated, but then stored as their aliases. So, although there’s a new harmonic at 25 kHz, the system records it as being at 48 kHz – 25 kHz = 23 kHz, which is right on top of the harmonic just below it.
In other words, when you look at all the spikes in the graph in Figure 3, you’re actually seeing at least two spikes sitting on top of each other. One of them is the “real” harmonic, and the other is an alias (there are actually more, but we’ll get to that…). However, since I clipped a 1 kHz sine wave in a 48 kHz world, this lines up all the aliases to be sitting on top of the lower harmonics.
So, what happens if I clip a sine wave with a frequency that isn’t nicely related to the sampling rate, like 900 Hz in a 48 kHz system, for example? Then the result will look more like Figure 4, which is a LOT messier.
Figure 4: The magnitude response of a 900 Hz square wave, plotted with a logarithmic frequency axis in the top axis and a linear axis in the bottom.
A 900 Hz square wave will have harmonics at odd multiples of the fundamental, therefore at 2.7 kHz, 4.5 kHz, and so on up to 22.5 kHz (900 Hz * 25).
The next harmonic is 24.3 kHz (900 Hz * 27), which will show up in the plots at 48 kHz – 24.3 kHz = 23.7 kHz. The next one will be 26.1 kHz (900 Hz * 29) which shows up in the plots at 21.9 kHz. This will continue back DOWN in frequency through the plot until you get to 900 Hz * 53 = 47.7 kHz which will show up as a 300 Hz tone, and now we’re on our way back up again… (Take a look at Figure 7, below for another way to think of this.)
The next harmonic will be 900 Hz * 55 = 49.5 kHz which will show up in the plot as a 1.5 kHz tone (49.5 kHz – 48 kHz).
Depending on the relationship between the square wave’s frequency and the sampling rate, you either get a “pretty” plot, like for the 6 kHz square wave in a 48 kHz system, as shown in Figure 5.
Figure 5: the magnitude response of a 6 kHz square wave in a 48 kHz system
Or, it’s messy, like the 7 kHz square wave in a 48 kHz system in Figure 6.
Figure 6: The magnitude response of a 7 kHz square wave in a 48 kHz system.
The moral of the story
There are three things to remember from this little pair of posts:
Some aliased artefacts are negative frequencies, meaning that they appear to be going backwards in time as compared to the original (just like the wheel appearing to rotate backwards in Part 1).
Just because you have an antialiasing filter at the input of your ADC does NOT protect you from aliasing, because it can be generated internally, after the signal has been converted to the digital domain.
Once this aliasing has happened (e.g. because you clipped the signal in the digital domain), then the aliases are in the signal below the Nyquist frequency and therefore will not be removed by the reconstruction low-pass filter in the DAC. Once they’re mixed in there with the signal, you can’t get them out again.
Figure 7: This is the same as Figure 4, but I’ve removed the first set of mirrored alias artefacts and plotted them on the left side as being mirrored in a “negative frequency” alternate universe.
One additional, but smaller problem with all of this is that, when you look at the output of an FFT analysis of a signal (like the top plot in Figure 7, for example), there’s no way for you to know which components are “normal” harmonics, and which are aliased artefacts that are actually above the Nyquist frequency. It’s another case proving that you need to understand what to expect from the output of the FFT in order to understand what you’re actually getting.
One of the best-known things about digital audio is the fact that you cannot record a signal that has a frequency that is higher than 1/2 the sampling rate.
Now, to be fair, that statement is not true. You CAN record a signal that has a frequency that is higher than 1/2 the sampling rate. You just won’t be able to play it back properly, because what comes out of the playback will not be the original frequency, but an alias of it.
If you record a one-spoked wheel with a series of photographs (in the old days, we called this ‘a movie’), the photos (the frames of the movie) might look something like this:
As you can see there, the wheel happens to be turning at a speed that results in it rotating 45º every frame.
The equivalent of this in a digital audio world would be if we were recording a sine wave that rotated (yes…. rotated…) 45º every sample, like this:
Notice that the red lines indicating the sample values are equivalent to the height of the spoke at the wheel rim in the first figure.
If we speed up the wheel’s rotation so that it rotated 90º per frame, it looks like this:
And the audio equivalent would look like this:
Speeding up even more to 135º per frame, we get this:
and this:
Then we get to a magical speed where the wheel rotated 180º per frame. At this speed, it appears when we look at the playback of the film that the wheel has stopped, and it now has two spokes.
In the audio equivalent, it looks like the result is that we have no output, as shown below.
However, this isn’t really true. It’s just an artefact of the fact that I chose to plot a sine wave. If I were to change the phase of this to be a cosine wave (at the same frequency) instead, for example, then it would definitely have an output.
At this point, the frequency of the audio signal is 1/2 the sampling rate.
What happens if the wheel goes even faster (and audio signal’s frequency goes above this)?
Notice that the wheel is now making more than a half-turn per frame. We can still record it. However, when we play it back, it doesn’t look like what happened. It looks like the wheel is going backwards like this:
Similarly, if we record a sine wave that has a frequency that is higher than 1/2 the sampling rate like this:
Then, when we play it back, we get a lower frequency that fits the samples, like this:
Just a little math
There is a simple way to calculate the frequency of the signal that you get out of the system if you know the sampling rate and the frequency of the signal that you tried to record.
Let’s use the following abbreviations to make it easy to state:
Fs = Sampling rate
F_in = frequency of the input signal
F_out = frequency of the output signal
IF F_in < Fs/2 THEN F_out = F_in
IF Fs > F_in > Fs/2 THEN F_out = Fs/2 – (F_in – Fs/2) = Fs – F_in
Some examples:
If your sampling rate is 48 kHz, and you try to record a 25 kHz sine wave, then the signal that you will play back will be: 48000 – 25000 = 23000 Hz
If your sampling rate is 48 kHz, and you try to record a 42 kHz sine wave, then the signal that you will play back will be: 48000 – 42000 = 6000 Hz
So, as you can see there, as the input signal’s frequency goes up, the alias frequency of the signal (the one you hear at the output) will go down.
There’s one more thing…
Go back and look at that last figure showing the playback signal of the sine wave. It looks like the sine wave has an inverted polarity compared to the signal that came into the system (notice that it starts on a downwards-slope whereas the input signal started on an upwards-slope). However, the polarity of the sine wave is NOT inverted. Nor has the phase shifted. The sine wave that you’re hearing at the output is going backwards in time compared to the signal at the input, just like the wheel appears to be rotating backwards when it’s actually going forwards.
In Part 2, we’ll talk about why you don’t need to worry about this in the real world, except when you REALLY need to worry about it.
