In a perfect sound system, all loudspeakers are identical, and they are all able to play a full frequency range at any listening level. However, most often, this is not a feasible option, either due to space or cost considerations (or both…). Luckily, it is possible to play some tricks to avoid having to install a large-scale sound system to listen to music or watch movies.
Humans have an amazing ability to localise sound sources. With your eyes closed, you are able to point towards the direction sounds are coming from with an incredible accuracy. However, this ability gets increasingly worse as we go lower in frequency, particularly in closed rooms.
In a sound system, we can use this inability to our advantage. Since you are unable to localise the point of origin of very low frequencies indoors, it should not matter where the loudspeaker that’s producing them is positioned in your listening room. Consequently, many simple systems remove the bass from the “main” loudspeakers and send them to a single large loudspeaker whose role it is to reproduce the bass for the entire system. This loudspeaker is called a “subwoofer”, since it is used to produce frequency bands below those played by the woofers in the main loudspeakers.
The process of removing the bass from the main channels and re-routing them to a subwoofer is called bass management.
It’s important to remember that, although many bass management systems assume the presence of at least one subwoofer, that output should not be confused with an (Low-Frequency Effects) LFE (Low-Frequency Effects) or a “.1” input channel. However, in most cases, the LFE channel from your media (for example, a Blu-ray disc or video streaming device) will be combined with the low-frequency output of the bass management system and the total result routed to the subwoofer. A simple example of this for a 5.1-channel system is shown below in Figure 1.
Figure 1: A simple example of a bass management system for a 5.1 channel system.
Of course, there are many other ways to do this. One simplification that’s usually used is to put a single Low Pass Filter (LPF) on the output of the summing buss after the signals are added together. That way, you only need to have the processing for one LPF instead of 5 or 6. On the other hand, you might not want to apply a LPF to an LFE input, so you may want to add the main channels, apply the LPF, and then add the LFE, for example. Other systems such as Bang & Olufsen televisions use a 2-channel bass management system so that you can have two subwoofers (or two larger main loudspeakers) and still maintain left/right differences in the low frequency content all the way out to the loudspeakers.
However, the one thing that older bass management systems have in common is that they typically route the low frequency content to a subset of the total number of loudspeakers. For example, a single subwoofer, or the two main front loudspeakers in a larger multichannel system.
In Bang & Olufsen televisions starting with the Beoplay V1 and going through to the Beovision Harmony, it is possible to change this behaviour in the setup menus, and to use the “Re-direction Level” to route the low frequency output to any of the loudspeakers in the current Speaker Group. So, for example, you could choose to send the bass to all loudspeakers instead of just one subwoofer.
There are advantages and disadvantages to doing this.
The first advantage is that, by sending the low frequency content to all loudspeakers, they all work together as a single “subwoofer”, and thus you might be able to get more total output from your entire system.
The second advantage is that, since the loudspeakers are (probably) placed in different locations around your listening room, then they can work together to better control the room’s resonances (a.k.a. room modes).
One possible disadvantage is that, if you have different loudspeakers in your system (say, for example, Beolab 3s, which have slave drivers, and Beolab 17s, which use a sealed cabinet design) then your loudspeakers may have different frequency-dependent phase responses. This can mean in some situations that, by sending the same signal to more loudspeakers, you get a lower total acoustic output in the room because the loudspeakers will cancel each other rather than adding together.
Another disadvantage is that different loudspeakers have different maximum output levels. So, although they may all have the same output level at a lower listening level, as you turn the volume up, that balance will change depending on the signal level (which is also dependent on frequency content). For example, if you own Beolab 17s (which are small-ish) and Beolab 50s (which are not) and if you’re listening to a battle scene with lots of explosions, at lower volume levels, the 17s can play as much bass as the 50s, but as you turn up the volume, the 17s reach their maximum limit and start protecting themselves long before the 50s do – so the balance of bass in the room changes.
Beosound Theatre
Beosound Theatre uses a new Bass Management system that is an optimised version of the one described above, with safeguards built-in to address the disadvantages. To start, the two low-frequency output channels from the bass management system are distributed to all loudspeakers in the system that are currently being used.
However, in order to ensure that the loudspeakers don’t cancel each other, the Beosound Theatre has Phase Compensation filters that are applied to each individual output channel (up to a maximum of 7 internal outputs and 16 external loudspeakers) to ensure that they work together instead against each other when reproducing the bass content. This is possible because we have measured the frequency-dependent phase responses of all B&O loudspeakers going as far back in time as the Beolab Penta, and created a custom filter for each model. The appropriate filters are chosen and applied to each individual outputs accordingly.
Secondly, we also know the maximum bass capability of each loudspeaker. Consequently, when you choose the loudspeakers in the setup menus of the Beosound Theatre, the appropriate Bass Management Re-direction levels are calculated to ensure that, for bass-heavy signals at high listening levels, all loudspeakers reach their maximum possible output simultaneously. This means that the overall balance of the entire system, both spatially and timbrally, does not change.
The total result is that, when you have external loudspeakers connected to the Beosound Theatre, you are ensured the maximum possible performance from your system, not only in terms of total output level, but also temporal control of your listening room.
There’s one last thing that I alluded to in a previous part of this series that now needs discussing before I wrap up the topic. Up to now, we’ve looked at how a filter behaves, both in time and magnitude vs. frequency. What we haven’t really dealt with is the question “why are you using a filter in the first place?”
Originally, equalisers were called that because they were used to equalise the high frequency levels that were lost on long-distance telephone transmissions. The kilometres of wire acted as a low-pass filter, and so a circuit had to be used to make the levels of the frequency bands equal again.
Nowadays we use filters and equalisers for all sorts of things – you can use them to add bass or treble because you like it. A loudspeaker developer can use them to correct linear response problems caused by the construction or visual design of the device. They can be used to compensate for the acoustical behaviour of a listening room. Or they can be used to compensate for things like hearing loss. These are just a few examples, but you’ll notice that three of the four of them are used as compensation – just like the original telephone equalisers.
Let’s focus on this application. You have an issue, and you want to fix it with a filter.
IF the problem that you’re trying to fix has a minimum phase characteristic, then a minimum phase filter (implemented either as an analogue circuit or in a DSP) can be used to “fix” the problem not only in the frequency domain – but also in the time domain. IF, however, you use a linear phase filter to fix a minimum phase problem, you might be able to take care of things on a magnitude vs. frequency analysis, but you will NOT fix the problem in the time domain.
This is why you need to know the time-domain behaviour of the problem to choose the correct filter to fix it.
For example, if you’re building a room compensation algorithm, you probably start by doing a measurement of the loudspeaker in a “reference” room / location / environment. This is your target.
You then take the loudspeaker to a different room and measure it again, and you can see the difference between the two.
In order to “undo” this difference with a filter (assuming that this is possible) one strategy is to start by analysing the difference in the two measurements by decomposing it into minimum phase and non-minimum phase components. You can then choose different filters for different tasks. A minimum phase filter can be used to compensate a resonance at a single frequency caused by a room mode. However, the cancellation at a frequency caused by a reflection is not minimum phase, so you can’t just use a filter to boost at that frequency. An octave-smoothed or 1/3-octave smoothed measurement done with pink noise might look like you fixed the problem – but you’ve probably screwed up the time domain.
Another, less intuitive example is when you’re building a loudspeaker, and you want to use a filter to fix a resonance that you can hear. It’s quite possible that the resonance (ringing in the time domain) is actually associated with a dip in the magnitude response (as we saw earlier). This means that, although intuition says “I can hear the resonant frequency sticking out, so I’ll put a dip there with a filter” – in order to correct it properly, you might need to boost it instead. The reason you can hear it is that it’s ringing in the time domain – not because it’s louder. So, a dip makes the problem less audible, but actually worse. In this case, you’re actually just attenuating the symptom, not fixing the problem – like taking an Asprin because you have a broken leg. Your leg is still broken, you just can’t feel it.
So far, we have looked at minimum phase and linear phase examples of one basic kind of filter, but everything I’ve shown you are just examples. There are other filters (I could be showing you shelving filters instead of peaking filters – which would be a through-put plus a low- or high-pass instead of a bandpass) and other implementations (for example, there are other ways to make a linear phase filter).
We won’t go through these other versions because we’re not here to learn how to make filters, we’re here to learn why, when someone asks you “which is better, minimum-phase or linear phase?” or “aren’t you worried about the unnatural effects of pre-ringing?” you can answer “it depends” (which is almost always the correct answer for any question related to audio).
At this point you should know that
a filter that changes the frequency response also changes the time response. This is unavoidable.
some filters will also ‘pre-ring’ ahead of the sound
a filter may or may not have an effect on the phase of the signal
If you’re designing (or choosing) a filter, nothing comes for free. (For example, if you want linear phase, the price is latency. There are other prices attached to other design decisions.)
The question that we have not yet addressed is “so what?”
This is the point in the discussion where things are going to get a little fuzzy… Hang on!
You probably read somewhere that human hearing extends from 20 Hz to 20 kHz, therefore anything outside that range is not audible. This is not true. You might have read a little farther where they added an extra detail saying something about your age: the older you are, the lower that top number gets. That might be true.
You can also read that the quietest sound you can hear has a sound pressure level of 20 µPa, which is equivalent to 0 dB SPL, and that the loudest sound you can “hear” (over the sound of your screams of pain) is about 120 dB SPL or so. You might have also read a little farther where they added an extra detail that points out that this is only at 1 kHz. But this is probably also not true.
There are many reasons why all of those numbers are basically meaningless – but the main one is that they’re numbers based on averages. Imagine if an optometrist only carried one strength of glasses, which happened to be the average of the prescriptions required by all of his or her patients. We’d all be tripping over out own feet, and getting blinding headaches caused by wearing the wrong glasses.
Actually, a similar point was once proven by the American Air Force. They wanted to design the perfect airplane cockpit, so they measured all their pilots, averaged all the numbers, and built a seat for the result. Of course, the result was that the seat fit no one, since there was no single person that matched the average.
There are other examples that prove that averages are useless information. For example, I have more than the average number of legs. Also, since one in every three mammals on the earth is a bat, if I have meeting with two other people, one of us must be Batman…
Of course, the message is that we’re all different. For example. I don’t taste things very well. I don’t understand people who talk about the various elements in the taste of wine. All I know is that it’s drinkable, or it’s good for putting on fish & chips. When I eat food, I tend to put on pepper and chill so that I can at least taste something…
Hearing is the same: so all of the stuff I’m about to say is based on averages, which may or may not apply to you. You might be more or less sensitive than the average. Don’t email me to argue about the numbers I’m using here. You’re different. I know… We all are…
Psychoacoustic Masking
Let’s go to an AC/DC concert together. Halfway through You Shook Me (All Night Long), I’ll whisper a secret code word into your left ear. Then, you whisper it back to me.
This exercise will not work. You won’t hear me, which is strange because the changes in air pressure that I’m making by whispering are exactly the same as when AC/DC is not playing – and normally, you’d be able to hear that. Also, your eardrum is wiggling in exactly the same way as a result of that whispering – it also happens to be wiggling to the AC/DC more. So, why can’t you hear me?
The answer lies in your brain. It decides that I’m not as important as AC/DC, so the signal is thrown out as being irrelevant, and so the sound my my whispering is ‘psychoacoustically masked’ by AC/DC. Notice the ‘psycho-‘ part of that, which is the indication that the acoustic masking is happening in your brain, not as a result of a mechanical or physical issue.
This probably doesn’t come as a surprise – at some point in your life, you’ve probably been somewhere where you’ve asked someone to speak up because you can’t hear them over the noise. You’ve been the victim of the limitations of your own brain.
Temporal Masking
What might come as a surprise is that this effect also works when the loud sound (AC/DC) and the quiet sound (me whispering) don’t happen simultaneously.
For example, let’s say that you are getting your photo taken by someone with a flash camera. You’re looking right at the camera, and the flash goes off. For a short while after that, you can’t see anything but the leftover spot in the middle of your vision. If, right after the flash, someone were to hold up some number of fingers and ask “how many fingers?”, you’d have to guess. (This is only an analogous example; the spot in your eyes is not happening in your brain, but the effect is similar.)
Similarly, if I were to fire a gun (not at you… don’t worry), and quickly whisper a word immediately afterwards, you wouldn’t hear the whisper. The gunshot and the whisper didn’t happen simultaneously, but there is temporal masking that causes you to not hear the quieter sound for a little which after the loud sound has happened.
Even weirder, there is an effect called pre-masking. If I were REALLY quick and VERY well-timed, I could whisper the word right BEFORE the gunshot and you also wouldn’t hear it. The loud sound (the gun shot) not only masks quiet sounds that come after it, but also sounds that come before it.
Fig 1. Generalised representation of temporal masking. The gray block is a loud noise. Anything in the pink area won’t be heard by most people most of the time. This plot is intentionally vague. The point is the effect, not the actual values.
So, if I were to play a loud noise (a gunshot, a blast of pink noise, a portion of an AC/DC song) and play a quiet sound before, during, or afterwards, and ask what the quiet sound was (to test if you heard it) your behaviour will match something like the plot in Figure 1. The gray block represents the loud sound. Anything in the pink shape is “stuff you can’t hear”. If the quiet sound occurs much earlier or much later than the loud sound, then it will have to be really quiet for you to not hear it. The closer in time the quiet sound occurs to the loud sound, the louder it has to be for you to hear it.
Of course, this is very general plot, so don’t use it for arguments while you’re drinking beer with your friends. For example, one element that I have not mentioned is frequency content. If the loud sound is the low frequency effects of a recording of distant thunder, and the quiet sound is a kitten mewing, then these two sounds are too far apart in frequency to have any influence on each other, and the graph is just plain wrong.
Signal Envelope
Unless you, like me, spend a lot of time listening to sinusoidal tones, you’ll notice that everything you listen to varies in level over time. This happens on different time scales. The sound pressure level (SPL) in the car while you’re driving to work or at the daycare picking up the kids is much louder than the SPL in your bedroom while you’re dealing with the free-floating existential anxiety that comes to visit at 2:00 in the morning. This is the ‘slow’ time scale. On the ‘fast’ end of the time scale, you have the extreme, and short SPL cause by closing a car door, or the very short, but not very loud sound of a high heel impacting a hardwood or tiled floor. Speech is somewhere in between these two. There are short, spikes caused by “t-” and “k-” sounds, and long-ish portions produced by vowels.
Note that we’re not really talking about how loud or how quiet things are. We’re talking about the change in level from quieter to louder and back again.
If we plot that change over time, we have a view of the signal’s envelope. For example, a single note on a piano has a fast attack (from quieter to louder) and a slow decay (from louder to quieter). A car horn honked in an open area (not a city intersection, where there are buildings to reflect the sound) has almost identical attack and decay envelopes.
Fig 1. Three different representations of an audio signal in time.
Figure 1 shows an 8-second slice of a recording of the Alleluia Chorus from Handel’s “Messiah”, chosen because it’s easy for me to load that into Matlab using the “load handel” command, and I’m very lazy.
The top plot shows the signal in the way we’re used to looking at sound files. The x-axis is time, in ms. The Y-axis shows the linear value of each of the samples. Oddly, this is the way we normally look at sound files, but it represents how we hear sound very poorly, because we don’t hear amplitude linearly.
So, in the middle plot, I’ve taken the same data and, sample-by-sample, plotted each value on a decibel scale (using the equation DisplayOutput = 20*log10(abs(signal)). The absolute value is there because calculating the log of a negative number gives you strange results)
The third plot is the one we’re really interested in. That’s created by connecting the peaks in the middle plot, which results in a running plot of the signal’s level over time. This is its envelope. As you can see in that particular musical example, the attacks (the changes from quieter to louder) are steeper (and therefore faster) than the decays. This is not surprising. It’s hard to get an orchestra and choir to all stop instantaneously…
Fig 2. The same treatment to a different audio signal. In this case, it’s female speech recorded in an anechoic chamber.
Figure 2 shows the same three ways of plotting an audio signal, but in this case, the signal is female speech recorded in an anechoic chamber. Notice that, partly because it’s only one sound source and partly because there is no reverberation in the signal, the decays are almost as fast as the attacks. However, this is a very strange recording. Most people don’t listen to anything in an anechoic chamber, and we don’t typically go to the middle of a football field or a frozen lake to have a conversation.
Back to the “so what?”
Let’s assemble the three collections of information that we’ve been throwing around.
Firstly, focusing on filter response:
We know that filters can ring.
We also know that some filters can pre-ring.
We also know that, unless the Q is really high, that ringing decays pretty quickly.
Finally, we know that, if the frequency that’s ringing in the filter is not present in the signal, it won’t ring because there’s nothing there to ring. (Similarly if you turn up the low bass while listening to a solo piccolo recording, you won’t hear a difference because there’s no bass to boost.)
Secondly, we looked at our own response to quiet and loud signals in time
A loud sound will simultaneously mask a quiet sound that occurs at the same time (fancy-talk for “drown it out so I can’t hear it)
The loud sound will also post-mask a quiet sound that occurs after within a short time window (on the order of 100 – 200 ms)
The loud sound will also pre-mask a quiet sound that occurs before it within a short time window (on the order of 10 – 20 ms)
Finally, we looked at signal envelopes – the change in level of an audio signal over time
Sounds never start instantaneously. In order to do so, they would have to have an infinite frequency range measured at the receiver (e.g. your eardrum, which is most certainly band-limited as well). Even very fast attacks take milliseconds to ramp up.
Sounds in real life have longer decay times
I know, I know, you can name sounds that have very fast attacks (e.g. the pluck of a harpsichord plectrum on a high string, or a single xylophone bar struck in an anechoic environment) and very fast decays (ummm…. good luck finding something that decays more than 100 dB in less than 100 ms…)
The question is: if you have a filter that rings or pre-rings, and you apply it to a sound,
Is that ringing the thing that defines the envelope of the resulting sound? In other words, does the signal’s attack and/or decay envelope change significantly as a result of the time response of the filter?
Is that change in the envelope outside the limits of your ability to detect it due to pre-masking and post-masking?
There is no single answer to this question. If the audio signal is someone hitting a rim shot, and the filter is a peak filter with 12 dB of gain at 500 Hz with a Q of 100, then you’ll hear it ringing. In fact, it will sound like a rim shot with a sine wave generator. If the audio signal is a single bowed note on a ‘cello, and the filter is a dip filter with -3 dB of gain at 100 Hz with a Q of 0.707, then you won’t.
However, (for example) you CANNOT automatically jump to a conclusion that “pre-ringing sounds unnatural” because that starts with the assumption that you can hear it, and therefore it sounds “like” anything.
Whether or not you can hear the effects of a filter applied to an audio signal is dependent not only on the very specific characteristics of the filter, but its interaction with the signal. Change the filter OR change the signal, and the result will change.
This means that you CANNOT say things like “linear phase filters are better (or worse) than minimum phase filters” or “the pre-ringing of a linear phase filter sounds unnatural” because both of those statements start with the (possibly incorrect) assumption that you can hear the effect of the filter.
Now, don’t mis-interpret what I’ve said to mean “you can’t hear a filter ringing” – I didn’t say that. What I said was “just because a filter can be measured to be ringing doesn’t necessarily mean that you can hear it with a given audio signal”.
In this part, we’re going to do something a little weird.
We’ll start by making a minimum phase peaking filter that has Q of 2 and a gain of +6 dB. The response of that filter is shown in Figure 1.
Fig 1. A minimum phase implementation of a peaking filter with a centre frequency of 1 kHz, a gain of 6 dB and a Q of 2.
Almost everything in the plots above should look familiar. The one weird thing is that I’ve left out the data in the time response before T = 0 ms. That’s because the future doesn’t matter, since we’re talking about a non-causal filter, right?
What if I (for some reason) wanted to make a filter that had a desired magnitude response, but a flat phase response? Let’s say that you’re allergic to phase shifts or you belong to an ancient religious cult that thinks that phase shifts are against the order of nature. How would we create that filter?
One way to do it is to create a filter that has the same magnitude response as the one above, but with an opposite phase response so that the two cancel each other. But, how do we create a filter with the opposite phase response?
One trick for doing this is to ignore what I said earlier. Remember when I was being pedantic about how filters advance signals in phase BUT NOT TIME? What if you were to ignore that for a brief moment? Could we use that ignorance to an advantage?
For example, what would happen if we took the time response shown above and reversed it? All of the component frequencies are still there, at all the same relative amplitudes. So the Magnitude Response won’t change. But what happens to the phase response? The answer is this:
Fig 2. The same impulse response as shown in Figure 1, reversed in time.
Now we have a weird filter that can only have an output based on future inputs (therefore it’s non-causal, and therefore not minimum-phase), with the same magnitude response as the one shown in Figure 1. But check out that phase response. It’s the inverse of the first filter.
So reversing time has the effect of flipping the polarity of the phase (not the polarity of the signal!) without modifying the magnitude response. What would have been a 45º phase shift (earlier) becomes a -45º phase shift (later).
So, if I were (conceptually – not really…) to put these two filters in series, feeding the output of one into the input of the other then their total magnitude responses would add (so I’d get a 12 dB boost instead of a 6 dB boost at 1 kHz) and their phase responses would cancel each other out. The end result would look like this:
Fig 3. The combination of the filters in Figure 1 and Figure 2. Notice that it has a gain of 12 dB at 1 kHz (because 6 + 6 = 12)
By now, you’ve probably figured out that what we’re looking at here is a linear phase filter, since can be used to change the magnitude response of a signal without mucking up its phase response.
The catch with this way of implementing a linear phase filter is that it has to see into the future. Of course, in real life this is difficult, but there is a trick you can use to fake it.
If you look at the impulse response plot in Figure 1, you can see that the peak in the response is at Time = 0 ms. As you get later, moving away from that moment in time, the signal is quieter and quieter until it dies away to (almost) nothing. The problem is that ‘almost nothing’ is not the same as ‘nothing’. In fact, if we’re being pedantic, the ringing keeps going forever, which is why it’s called a filter with an Infinite Impulse Response – an IIR filter.
This also means that the same is true for the filter in Figure 2, but the ringing extends infinitely backwards in time.
However, our resolution in measuring and storing the amplitude is not infinite – when the signal is quiet enough, we run out of resolution (‘ticks’ on the ruler) – and when the signal gets that quiet, it effectively becomes the same as nothing.
So, if we decide on the level where ‘almost nothing’ is the same as ‘nothing’ (this is more-or-less up to us if we’re the ones designing the filter), then we can look at the filter’s response and decide when that signal level is reached (both forwards and backwards in time).
For example, with the extremely limited resolution of the plots that I’ve made above, we can decide that ‘nothing’ is what’s left when you get ±10 ms from the impulse peak. Of course, the resolution of the pixels on that plot is not even close to the resolution of the audio signal, so ‘nothing’ on our plot and ‘nothing’ for the audio signal are two different things – but the concept is the same.
Back to the Future
Let’s pretend for a moment that, for the impulse response in Figure 3, we decide that ±10 ms is the window of time we need to get to ‘nothing’. This means that we could implement this filter by sending audio into it, letting it pre-ring with an increasing amount until it hits the maximum peak 10 ms after the sound comes into it, then ringing for another 10 ms until it’s out.
This means that the latency (the delay time between the input and the output of the filter) would appear to be 10 ms. Yes, when you feed in a signal, you immediately start getting something out of the filter, but the output is loudest after the signal has been feeding into the filter for 10 ms.
In other words, if you want to have a linear-phase filter that’s implemented using this method, you are going to have to accept that the cost will be latency. In our case (a filter with a Q of 2, with a centre frequency of 1 kHz, and the decisions we’ve made here) this results in a 10 ms latency. Change a parameter, and you change the latency. For example, as we’ve already seen, the lower the centre frequency, the longer in time the filter will ring. So, if we change the centre frequency of this filter to 100 Hz, then it will have 10 times the latency (because 100 Hz is 1/10 of 1 kHz – therefore the periodicity of the ringing is 10x longer). If we increase the Q, then the ringing lasts longer – both forwards and backwards and time.
Generally speaking, this means that, if you’re building a linear phase filter, you need to decide what characteristics the filter has, and then you need to start making decisions about the quality of the filter (e.g. is the response exactly what you want, or is just almost what you want?) and its latency.
Although I’m not going to talk about implementation much, the latency has two primary considerations. The first is obvious: are you willing to wait for the output? If you’re building a filter for a PA system, then you don’t want the output delayed so much that it sounds like an echo. However, the second issue is one for the person building the filter: latency means memory. This was a bigger problem in the ‘old days’ when memory was expensive, but it’s still an issue.
Why? The problem is that the latency is defined by the frequency and Q of the filter, which define the total time the filter takes to get through the entire impulse response. However, the filter’s processing doesn’t think of time in milliseconds, it thinks in samples. If you double the sampling rate, then you double the amount of memory you need to implement the filter.
So, going from 48 kHz to 192 kHz requires 4x the memory.
A little perspective
One thing to notice in the three figures above is the vertical scale of the time response plots. You can see there that the range is ±0.1, which is a zoomed-in view of the amplitude. This may give you a distorted impression of the level of the ringing relative to the signal (the impulse). Figure 4 should clear this up.
Fig 4. Three plots showing exactly the same data in three different ways.
The top plot in Figure 4 is identical to the top plot in Figure 3. The middle plot is also on a linear amplitude scale, but the range of the plot shows the entire range of the signal. You can see there that the impulse at T = 0 hits a value of 1, and so the pre- and post-ringing isn’t as loud as you might have thought based on the first three figures.
The bottom plot shows the same information, plotted on a dB scale instead. The impulse at T = 0 ms hits 0 dB. Relative to that, the pre- and post-ringing has a maximum value of about -24 dB. You can also see there that the ringing is at least 100 dB down by the time we’re 6 ms away from the main impulse, looking both backwards and forwards in time.
Of course, all of these characteristics would be different for filters with different parameters. The point here is to understand the general characteristics, not the specifics.
Additional Comment
You may read on various fora someone who claims that there’s no such thing as “pre-ringing”. “It’s ‘ripple'” They’ll claim.
This is incorrect.
Pre-ringing and ringing are behaviours that occur in the time domain.
Ripple is a wiggle in the response in the frequency domain. (Say, for example, you zoom into the magnitude response, it won’t be completely flat in some cases – and if it’s not, you have ripple.)
I’m going to start this part by doing something I very, very rarely do: to quote Wikipedia.
“In control theory and signal processing, a linear, time-invariant system is said to be minimum-phase if the system and its inverse are causal and stable.”
However, in my defence, one of the references attached to that statement is Julius O. Smith III, so that makes it okay.
Let’s unwrap that sentence and see if we know enough to know what it’s telling us.
We don’t care about control theory. So let’s ignore that part. We’re only interested in signal processing, where our signal is audio; so we move on.
We already know what a ‘linear, time-invariant” system (like our filters) is, and we now know that we can say that that system is ‘minimum-phase’ if:
the system (our peak filter in the previous part, for example)
and its inverse (our dip filter in the previous part, for example)
are causal
and stable
Let’s deal with the ‘stable’ part first. We know that our two filters are stable because we saw that their poles are inside the unit circle in the Z-Plane representation. (We also know it because they both have ringing that decays instead of increases over time.)
We also know that their zeros are also inside the unit circle, since the zeros of each filter are in the same place as the poles of the other filter, which we already said, are inside the unit circle.
So, what does ‘causal’ mean? It’s really just a fancy word that means that the output of our filter is determined by either the past or the present, or some combination of the two. In real life, all filters and systems are causal, since they can’t do something based on what will happen in the future.
However, if you are not working in real time, you can easily create systems and filters that are non-causal and have outputs that are created by events in the future. One simple example of this is to record your voice, reverse the track, add some reverb, and then reverse it back again. Now you have reverb that ramps up to a sound before it starts. This is non-causal.
Do I care?
Not yet. But keep the two conditions in mind:
Both the filter and its inverse must be ‘causal’. The output of a minimum phase filter can only be the result of the present or the past, never the future.
Both the filter and its inverse must be stable. We like stable…
In this part, I’m going to deviate just a little from something I said at the beginning of this series. To be honest, if I hadn’t admitted this, you probably wouldn’t notice – but I would prefer to keep things clean… The deviation is that, for this part, I’m making a slight change to how Q is defined. This is not serious enough to get into the details of exactly how the definition is different .
Using the slightly-different definition of Q, let’s make a peaking filter with a centre frequency of 1 kHz, a boost of 12 dB and a Q of 2. This will have the response shown below in Figure 1.
Fig 1. The response of a peaking filter. Fc = 1 kHz, gain = 12 dB, Q = 2
Using the same modified definition of Q, let’s also look at the response of a dip filter with the same parameter values, but a gain of -12 dB instead.
Fig 2. The response of a dip filter. Fc = 1 kHz, gain = -12 dB, Q = 2
If you look at the magnitude responses of these two filters, you’ll see that it looks like they are mirror images of each other. In fact, they are.
If you look at the phase responses of these two filters, you’ll also see that it looks like they are mirror images of each other. In fact, they are.
If you look at their impulse responses, you’ll see that it would be difficult to see that they are related at all… But never mind this.
If I connect the output of the first filter to the input of the second filter, and measure the total throughput of the system, it will look like this:
Fig 3. The response of the combination of the boost and the dip filters from Figures 1 and 2.
Just in case you’re suspicious, I didn’t fake this. I actually connected the boost to the dip and sent an impulse through the whole thing and you’re looking at the result. No tricks! (Note that I could have reversed their order with the same total result.)
What you can see here is that the responses of the dip and the boost negate each other. Whatever one does, the other does exactly the opposite.
Generally speaking, we audio geeks use some special words to describe not-very special cases like this.
Often, you’ll hear us talking about a linear system which is a fancy way of saying ‘the effects of this system can be undone’. In this example, the dip filter can ‘undo’ the effect of the boost (and vice versa) therefore both must be linear filters.
Just as often, you’ll hear us talking about time-invariant systems, which just means that they don’t change over time. Because I implemented those two filters using equations done on my computer, if I run the math again tomorrow, I’ll get exactly the same answer. If I test them using an impulse that is quieter or louder, I also get exactly the same responses. (If I had implemented them using resistors and capacitors and transistors or vacuum tubes, I might not get the same answer tomorrow or with a different signal level because of temperature changes, for example. Although now I’m really splitting hairs, just to make a point.)
The reason I said “just as often” is because, normally we use the two terms together as a package deal. So, we ask whether a system (like something as simple as a filter or as complicated as a reverb unit or an upmixing algorithm) is Linear Time-Invariant or LTI. This is an important question because it packs a lot of information in it.
For example, if a reverb unit is LTI, then I can measure it today with an impulse, and I know that it will behave the same tomorrow with lute music or a snare drum. It does the same thing all day, every day, regardless of the input signal or its level. One measurement, and I can go away and analyse it for the rest of the week.
If it’s not LTI, then its characteristics will change for some reason that I don’t necessarily know. Maybe the internal delays are modulating in time, so its response in 10 seconds will be different than it is now. Maybe it has a compressor or a noise gate built in, so it changes its behaviour according to the level of the signal.
If we get back to our (rather simple) peak / dip filter example. We know they’re LTI (because I said so – and you have to trust me). We also know that the dip filter is the opposite of the boost. The question is “how, exactly, did I make this happen?”
The general answer to this question has already been answered – the magnitude and the phase responses are mirror images of each other. Therefore, for any given frequency, one filter boosts by the same amount that the other cuts, and one filter advances in phase by the same amount that the other delays in phase.
The more geeky answer to this question requires that we look at the Z-Plane, which I’ve talked about throughly in another series of postings starting with this one. I’ll repeat myself a little by saying that a Z-Plane representation shows a different way of looking at the ‘ingredients’ in a filter. It contains ‘poles’ that are placed at frequencies that are infinitely boosted, and ‘zeroes’ that are placed at frequencies that are infinitely cut. By carefully placing poles and zeros relative to each other in the Z-Plane, you can decide how the filter will behave for other frequencies between 0 Hz and the Nyquist frequency.
When you design (or analyse) filters this way, there are a couple of basic rules:
The ‘safe zone’ in the Z-Plane is defined by a circle. If you start placing poles outside it, then the filter can become unstable. If a filter is unstable, this means that its ringing can get louder over time instead of decaying.
If you place a pole in exactly the same place as a zero, they cancel each other out, and the total result is as if neither were there.
So, let’s look at our two filters above in their Z-Plane representations.
Fig 4. The same two filters, including their Z-Plane representations (for a system running at 48 kHz)
Admittedly, the resolution of the display in the software that I’m using to show this isn’t great, but if you compare the Z-Plane plots on the left and right, you can see that the zeros (marked with ‘o’) and the poles (‘x’) swap places. Just to make things a little clearer, I moved the centre frequency to 10 kHz and kept the gain and Q values the same. These are shown in Figure 5.
Fig 5. The same two filters but with their centre frequencies moved to 10 kHz, including their Z-Plane representations (for a system running at 48 kHz)
What’s the point of showing you this? The Magnitude and Phase response plots (which, combined, comprise the filters’ Frequency Responses) are ‘just’ descriptions of the behaviour of the filter. They tell you what happens to a signal that goes through them.
The Z-Plane representations show you how the filters are actually implemented.
It’s like the difference between reading a description of how a cake tastes and reading the recipe.
What you can see in the Z-Plane is not only that the responses of the filters negate each other: they’re built to ensure that this is the case. The poles and zeros of one filter cancel the zeros and poles of the other, and vice versa.
There’s one other extra piece of information that you already know. The fact that the poles for any of these filters are inside the circle helps to tell us that they’re stable and therefore LTI. It also tells us something else that we’ll talk about in the next part.
There are lots of people in audio who will make some claims about one kind of filter being better than another kind of filter because of something to do with the time response. They’ll throw around words like “minimum phase” or “linear phase” or “apodising” or other names, which sound impressive, but don’t really mean anything to normal people. In fact, in most cases, they don’t even mean anything to abnormal people (a.k.a. audio engineers). They’ll even make some statements about why one is better than the other, with some psychoacoustic claims to back themselves up.
One thing to remember is that these terms are very general headings that each sit on top of a lot of sub-headings. It’s also important to separate these terms from the incorrectly-interchanged terms ‘FIR’ and ‘IIR’ (which stand for ‘Finite Impulse Response’ and ‘Infinite Impulse Response’) which are different descriptions for the same filters. For example, many people say “FIR” when they mean “linear phase”, forgetting that an FIR can be used to create a non-linear phase filter.
In this posting, we’ll start to look at the difference between ‘minimum phase’ and ‘linear phase’ filters, but this requires a little set-up first.
Up to now in this series of postings, we’ve only looked at the filters’ magnitude responses (the gain of the filter vs. frequency) and time responses (or impulse responses). Let’s shift gears a little and think about the phase response instead.
Remember from Part 1, we looked at how an impulse is the result of adding an infinite number of cosine waves that all started at the beginning of time, and will continue until the end of time. Those waves all cancel each other out at all moments in time (forwards and backwards) except for that one instant (which we call Time = 0, also known as NOW) where they all add up to make a click.
What happens when we shift the time alignment? The intuitive answer is that we get something different than a simple click. The more we shift the frequency components in time, the more different we get from a simple click.
However, when we talk about shifting frequency components in time, it doesn’t make sense to actually measure that shift in time. I know that sounds like a stupid thing to say, so I’ll illustrate what I mean…
We saw that if we add a bunch of cosine waves together they start looking like an impulse, as shown in Figure 1.
Figure 1: Adding the 5 cosine waves with the same amplitude results in the “pulse” shown in the bottom plot.
What happens if I delay all of those individual waves by 0.5 second (or 500 ms)? The result is shown in Figure 2.
Fig 2. The result of adding the same components, each of them delayed by 500 ms.
It should be pretty obvious that the result in Figure 2 is identical to the result in Figure 1. The only difference is that it’s been shifted in time by 500 ms. The shape of the wave has not changed because we shifted all of the waves together, so their relationship to each other has not changed.
So, if we want to change the shape of the total result, we need to shift the components relative to each other, as shown in Figure 3.
Fig 3. The same components, added together with a different relationship in time produces a different summed total.
Figure 3 shows the same components with the same amplitudes, but shifted so that they all cross the T=0 point at the 0 line instead of at the maximum (as in Figure 1). This means that I’ve shifted each component individually by 90º, which is a different amount of time (in seconds) for each one. (In other words, I’m summing sine waves instead of cosine waves.) The summed result is quite different, as you can see in the bottom plot.
You can also shift some components differently (measured in phase) as well. For example, take a look at Figure 4. In that one, the first 4 components with the lowest frequencies are cosine waves, and I’ve shifted the 5th component by 90º. As you can see in the bottom plot, just shifting one component can make a large difference.
Fig 4. Shifting one of the 5 frequency components by 90º also has a significant effect on the total result.
And it probably goes without saying, but I’ll say it anyway, that if you change the relative levels of the components, you’ll also change their total sum, as shown in Figure 5.
Fig 5. Notice that the cosines are all aligned in phase, but the amplitude of the highest frequency is dropped by 50%, resulting in a different summed total.
Let’s turn this around (finally…). In the examples above, I was playing with the components’ amplitudes and relative phases to produce different total summed results, even through the frequencies of the components were the same each time.
If we think of this backwards, we can conclude that, if the time response of a filter is NOT a perfect impulse, then it must have done something to the relative levels and/or the relative phases of the collection of infinite frequency components that went through it. Using math (the same Fourier Transform that I mentioned in Part 2) we can take the impulse response and calculate what happened to the components, both in amplitude (the Magnitude Response) and phase (the Phase Response), which together give us the filter’s Frequency Response.
Let’s look at an example: a bandpass filter with a centre frequency of 1 kHz and a Q of 2, shown in Figure 6.
Fig 6. The top plot is the impulse response where you can see the ringing. The middle plot is the magnitude response where you can see the gain applied by the filter to a given frequency. The bottom plot is the phase response, which I’ll talk about below.
The top and middle plots in Figure 6 should not come as surprises now, so let’s talk about that bottom plot. What is shows us, generally speaking, is that if you send a sinusoidal wave through the bandpass filter at the centre frequency (1 kHz) then the output will have the same phase as the input, since the red line is at 0 degrees at 1 kHz.
Fig 7. The input and output of the bandpass filter from Figure 6 when the signal is a 1 kHz sinusoidal tone. Notice that the output has the same amplitude as the input (hence the gain of 0 dB in the Magnitude Response) and the two signals are in phase (the tops align, for example).
If the sinusoidal wave that you send in is above 1 kHz, then the output will be later in phase than the input. This does NOT necessarily mean that it’s delayed in time. We can’t know this because as soon as I said “sinusoidal wave”, this implied that it has no start or stop time – it’s just a sinusoidal tone that has always been there and will always be there. (In order to start or stop it, you need other frequency components.)
Philosophically, this may be difficult to consider – but think of it the same way you you experience seeing Niagara Falls. You really have no first-hand knowledge of when the water started falling or when it will stop – it’s as if it’s always been doing this and it always will – and you just get to see it for a small slice of time in its “infinitely”-long existence.
Fig 8. The same filter, showing the input and output with a 2 kHz sinusoidal tone. Notice that the output has dropped in level and it appears to be late relative to the input – it’s shifted to the right by a little less than 90º.
It’s really important to remember that what we’re looking at in Figure 8 is a phase shift and NOT a time delay (even though it looks like it). Repeat this sentence until you believe it before looking at the next plot.
Fig 9. The same filter, showing the input and output with a 500 Hz sinusoidal tone. Notice that the output has dropped in level and it appears to be early relative to the input – it’s shifted to the left by a little less than 90º.
Figure 9 shows an example of why you have to believe that we’re not talking about a time delay – just a phase shift. As you can see there, in the case of a bandpass filter, if the signal frequency is below the centre frequency, the phase shift is backwards, which looks like the output is ahead of the input. Of course, this is impossible. Bandpass filters are not time machines.
Now go back and look at the bottom plot in Figure 6. You’ll see that frequencies above the centre frequency of the filter (1 kHz) have a phase shift that is below 0º – they’re negative numbers approaching -90º as the frequency increases. Compare this to Figure 8 and you can make the link that a negative phase shift is “later” (in phase, not in time!).
Conversely, lower frequencies have a positive phase shift in Figure 6, which (as can be seen in Figure 9) correspond to a phase shift that moves “earlier”.
Remember that a peak/dip filter is a combination of a bandpass and a throughput. So now let’s look at the phase shift that results when you use one.
Fig 10. The impulse, magnitude, and phase responses of a peaking filter with a centre frequency of 1 kHz, a gain of 12 dB, and a Q of 2.
Looking at the magnitude response, it should now be fairly easy to see the merging of a throughput (which would be a straight line at 0 dB across all frequencies) and a bandpass (which causes the bump around 1 kHz).
It should be almost as easy to see the merging in the phase response as well. A throughput would have a phase response of 0º at all frequencies – which is why the plot starts at 0º in the very low frequencies and ends at 0º in the very high frequencies (because the bandpass doesn’t have much contribution out there). In the middle, the phase response of the bandpass shows up; so around 1 kHz, the phase responses of Figure 10 and 6 are very similar.
Let’s change the Q and see what happens.
Fig 11. The impulse, magnitude, and phase responses of a peaking filter with a centre frequency of 1 kHz, a gain of 12 dB, and a Q of 10.
Figure 11 shows the same peaking filter with the Q increased to 10. Notice 5 things (not in any obvious order):
The bump in the magnitude response is narrower
The ringing starts at a lower level
The impulse response is ringing for a lot longer in time
The deviation from 0º in the phase response has a narrower bandwidth.
The slope of the phase response at 1 kHz is steeper.
Let’s put some of these together. I’ll take these in a slightly different order, but after reading the paragraphs below, the points above should all interlock.
The bump in the magnitude response is narrower; therefore it has a smaller bandwidth. This should be expected, since Q = Fc/BW, so if we don’t change Fc, then the higher Q goes, the smaller BW gets.
Notice that both the filter in Figure 10 and the filter in Figure 11 have a gain at Fc of 12 dB. However, since the Q is lower in Figure 10, this means that, overall, more frequencies are boosted by more. Consequently, if you have a signal that has all frequencies in it (say, pink noise or Metallica), then the output of Figure 10’s filter will be generally louder than the output of Figure 11’s. Another way to see this is that the level of the start of the ‘tail’ of the impulse response is higher.
There is a direct link between the length of time the filter rings (which you can see in the impulse responses) and the slope of the phase response. The steeper the slope at a given frequency, the longer the filter will ring at that frequency. So, if you only look at the phase response plots, it’s easy to tell which of the two filters will ring for a longer time, and at what frequency. This will come in handy in the next part.
Let’s put together a couple of things that were said in the last postings, which should help to support each other:
A peak or a dip filter is created by adding a bandpass filter to a throughput, as shown in Figure 1.
Fig 1. The individual building blocks of a peak/dip filter
To change from peak to dip, you switch the polarity of the bandpass portion by making the “gain” negative instead of positive. (In other words, you subtract the bandpass from the throughput instead of adding it). To change the gain of the peak/dip filter, you change the gain of the bandpass portion. To change the Q of the peak/dip, you change the Q of the bandpass.
We also saw at the end of Part 3 that changing the gain does not change the rate of the decay.
This should all come together nicely to make sense for the first of the three points. For example, since the bandpass portion is the part that’s ringing, and since changing the gain of the peak (or dip) is just a matter of changing the gain applied to the bandpass portion, then there is no reason why the decay rate of the ringing should change. It will start at a higher or lower level, but its decay slope will be the same.
Q vs Time
We also saw at the end of Part 3 that changing the Q will change the slope of the decay inversely proportionally, but that changing the frequency will change the slope of the decay proportionally.
There is a nice little rule-of-thumb that’s used by electrical engineers for measuring the Q of a filter. Let’s say that you can’t (or couldn’t be bothered to take the time to) measure the frequency or magnitude response, and you want to figure out the Q based on the time response only, you can calculate this by looking at its impulse response.
Fig 2. The time response of an unknown peaking filter. (You can tell it’s peaking because the ringing cosine wave starts above the 0 line, just like the initial impulse.)
For example, Figure 2 shows the initial part of the impulse response of an unknown filter. I’ve highlighted two points that are reasonably close to the tops of two of the cosine wave cycles. I picked the first one (on the left) and then noted its Y value (Y = 0.027). Then I found a top of another wave that was as close to half that value as I could find. You can see there that it’s 2 cycles later, where Y = 0.0149.
So, you multiply the number of cycles it takes to drop by 50% (in this example, 2 cycles) and multiply that by 4.53, which results in a value of about 9. This is a good estimate of the Q of the filter (which is actually 10, if I measure it using the -3 dB points in the magnitude response).
If you’d like to read the long version of this, check out this page.
Note that it doesn’t matter which cycle I chose to get the first value, since the rate of decay is the same through the entire time response of the filter. In other words, if I chose the 3rd cycle to do the first measurement, I would have found that the 5th cycle is about 50% lower because it’s also 2 cycles later.
It also doesn’t matter whether we’re talking about peaks or dips, since, as we already know, from a perspective of the individual building blocks of the filter, these are the same thing.
So what?
Of course, most normal people aren’t measuring the time response of filters to calculate the Q. However, this piece of information is good from the opposite perspective: if you know the Q of the filter, you can figure out how fast it’s decaying. For example, a filter with a Q of 2 will take 2 / 4.53 = 0.44 cycles to decay by 50% (or 6 dB). If you know the frequency, then you can then translate that into a decay rate per seconds, because the period in seconds (the total time of one cycle of the wave) = 1 / Fc.
So, if that filter with a Q of 2 has an Fc of 100 Hz, then the period is 1/100 = 0.01 sec, and therefore it will decay by 6 dB (50%) in 0.44 cycles * 0.01 sec/cycle = 0.0044 sec or 4.4 ms.
If the Fc of the filter is 5 kHz, then the the period is 1/5000 = 0.0002 sec, and therefore it will decay by 5 dB in 0.0002 * 0.44 = 0.000088 sec = 88 µsec. (This is roughly equivalent to 2 samples at 48 kHz.)
Another good thing to remember is that Q = Fc / BW where BW is the bandwidth of the response measured between the two -3 dB points. This means, for example, that if Q = 1, then Fc = BW, therefore the bandwidth is about 1 octave. If Q = 2, then the bandwidth is about 1/2 of an octave, if Q = 12 then the bandwidth is about 1 semitone (1/12th of an octave), and so on.
Now we’ve seen that if we have a filter that results in either a peak or a dip in the magnitude response, we’ll also result in the signal ringing in time. We’ve also seen that the frequency of the ringing is the centre frequency of the filter. Now let’s dig a little deeper into the behaviour of that ringing; or, more specifically its decay characteristics.
We’ll repeat the process from Part 2: measure the impulse response of a peaking filter where Fc = 1 kHz, gain = +12 dB, and Q = 2. However, this time I’ll look at the time response with a different scaling. Instead of plotting the linear value over time, I’ll convert each instantaneous value to dB and plot that. This looks like Figure 1.
Fig 1. The same filter from Part 1, but now I’m plotting the impulse response on an instantaneous decibel scale.
The important thing to notice here is that, when I plot the instantaneous amplitude in decibels (in other words, on a logarithmic scale), the decay is a straight line with a slope.
Let’s get two things out of the way here. This isn’t really decibels, because decibels requires some time averaging. Also, I’m actually plotting the absolute value of the impulse response in a decibel scale, because if I try to calculate the log of a negative number, things get ugly. This means that the math I’m actually using to create the bottom plot is
20 * log10(abs(signal))
If I draw a line across the tops of the bumps in that plot, I can look at the decay of the filter’s ringing as in Figure 2.
Fig 2. The blue line shows the decay rate of the filter’s ringing. In this particular case, the decay is about -1360 dB per second.
For this filter, the decay rate of the ringing is -1360 dB per second (which is very fast). Let’s change some parameters and see what happens.
If I increase the gain of the filter without changing the Fc or the Q, I get the following:
Fig 3. Changing the gain to +20 dB makes the ringing louder overall, but it decays at the same rate: about -1360 dB per second.Fig 4. Fc = 1 kHz, Gain = +12 dB, Q = 4. Now the decay of the ringing is about -680 dB / second.Fig 5. Fc = 2 kHz, Gain = +12 dB, Q = 2. Now the decay of the ringing is about -2720 dB / second.
I could plot lots more of these so that you start to see a pattern, but I’ll jump to the punch lines and you can use the plots above to check that things make sense.
If I have a filter that is using a definition of Q = Fc / BW (where BW is the distance between the -3 dB points down from the maximum), then:
Changing the gain does not change the rate of the decay (all least, as long as it’s a boost, according to what we’ve seen so far…)
Changing the Q will change the slope of the decay inversely proportionally if we’re measuring the slope in dB/sec. For example, if I multiply the Q by 2, the ringing decays twice as slowly. If I multiply the Q by 10, the ringing will take 10 times longer to decay to the same level.
Changing the frequency will change the slope of the decay proportionally if we’re measuring the slope in dB/sec. For example, if I multiply the frequency by 2, the ringing will decay twice as fast.
Let’s talk about the last of these first, since it’s the easiest to understand conceptually. In the plots above, I’m showing the time in seconds. So, the higher the frequency, the more cycles I’m showing in the same plot. However, if I were plotting time in cycles of the cosine wave instead, the slope would be the same regardless of frequency.
In other words, the level of the ringing decays by the same amount per number of cycles of the cosine wave.
This is why, if you count the number of “bumps” in the dB plots in Figure 2 and 5, you’ll see that they are the same number. It takes about 12 cycles to get down to -100 dB, but the shorter the cycles (because the frequency is higher) the faster you get there when measuring in seconds. If the X-axis were not “Time in milliseconds”, but “Time in periods of the centre frequency” instead, then the slopes would be identical in Figures 2 and 5.
If you throw a rock into a pond on a windless day, you’ll see the ripples moving away in an expanding circle from the place where the rock hit. The ripples are places on the water where the water is either higher or lower than where it was before you hit the rock. The water itself only moves up and down, but the waves expand sideways. (You can see this if there is something floating on the water, for example – it bobs up and down as the waves go by.)
A similar thing happens when you pluck a guitar string. The point where your finger plucked is the same as the point where the rock landed in the water, and waves radiate away from that place on the string in two directions (because there are only two directions to travel in on a string: this way and that way). However, when those waves reach the end of the string, they reflect and come back in the opposite direction.
In both cases, the water and the guitar string, the wave has some speed at which it travels. It’s slow enough on the water for you to watch it, but it’s much too fast on a guitar string. In fact, it’s so fast that, when you pluck it, the wave travels to the end of the string, reflects in the opposite direction, hits the other end of the string, reflects again, and gets back to where you plucked it in about 1/82nd of a second if it’s the low E string. Since the wave doesn’t stop there – it keeps going, repeating the back-and-forth journey along the length of the string every 1/82nd of a second, then we hear a note with a fundamental frequency of 82 Hz (82 cycles per second): a low E.
That ringing that happens on the guitar string will happen no matter how you start the movement on it. You could hit the string with a chopstick, you could just thump the side of the guitar with your fist, you could even stand next to the guitar and cough loudly. All of these things will “inject” energy into the string, causing it to move, and the wave starts banging back and forth.
The rate of repetition is dependent on two things: the length of the string and the speed of the wave. The speed of the wave is dependent on two things: the mass of the string (e.g. how heavy is 1 m of it?) and the tension (how tightly is it stretched?) Increase the tension, and you increase the speed of the wave. Decrease the mass and you increase the speed of the wave. Increase the speed of the wave, and the repetition takes less time, so you hear a higher note.
That frequency at which the string will naturally ring is called a resonance. A child on a swing will go back and forth at the same rate (number of times per second) no matter how gently or forcefully you push them – apply energy, and the system will resonate.
Now, let’s think about that push of the child, the rock hitting the water, or the pluck of the guitar string. All of those things are a short injection of energy: a kind of impulse, and the way the child, the water, or the string behaves afterwards is its impulse response – how it responds to that impulse.
But here’s a strange thing to consider. This means that the note (the frequency) that you hear from the guitar string was one of the many frequencies in the initial pluck itself.
So, another way to think of this is that, by plucking the string, you inject a signal with all frequencies in it, and all of those frequencies decay (“die away”) very quickly except for one.
Okay, okay, if we’re going to be pedantic, I should be including not only the fundamental frequency but all of the additional harmonics; typically multiples of that frequency. But we don’t need to complicate things with the truth at the moment…
What does this have to do with filters?
From a “big picture” point of view, a guitar string is a filter. I feed in some signal (the pluck) and I get out a modified version of that signal (the note ringing). From the same perspective, a filter in an equaliser is the same: I feed in a signal (music) and I get out a modified version of it (the same music, but slightly louder at 1 kHz, for example). What’s interesting is that the two things basically work the same way.
Let’s take the example of the filter at the end of Part 1: a peaking filter with a boost of 12 dB at 1 kHz, with a Q of 2. If I feed in a sine wave (which only contains energy at 1 frequency) at a very low frequency (say, 100 Hz or lower) then the level of the output will equal that of the input. If I do the same with a very high frequency (say, 10 kHz) then the level of the output will also equal that of the input. However, if I feed in a sine wave at 1 kHz, the output will be 4 times louder than the input (+12 dB = 4 time the amplitude because 20*log10(4) = 12-ish).
Fig 1. The magnitude response of the example filter that we’re working with for now.
At some other frequency around 1 kHz, I’ll get a different answer. However, this is a VERY long and tedious way to measure the magnitude response of the filter. Another option is to measure its impulse response.
If I feed the input of the filter with an impulse (which is a sound that contains all frequencies at the same level, as we saw in Part 1), and look at the filters output in time, it might look like this:
Fig 2. The impulse response of the example filter from Fig 1.
Notice that the impulse looks like an impulse at Time = 0, but then something extra happens afterwards – like a guitar string ringing in time. If I zoom in vertically and look at the same plot, it will look like Figure 3.
Fig 3. The same data shown in Figure 2, but zoomed in vertically.
And if we zoom in horizontally as well, it will look like this.
Fig 4. The same data again, focusing on the initial part of the response
So, as you can see there, it’s almost as if we kept the impulse, and then just added a cosine wave with a period (a repetition time) of 1 ms, starting at Time = 0 and decaying over time. In fact, that’s exactly what the filter does.
Time response to Frequency response
The excuse I gave above for sending an impulse through the filter (instead of sine waves) was that this will be a faster way to measure its response. The time response of the filter is already done. We can see that in the figures above. But how do we see the filter’s frequency response? This is done using a clever bit of math called a Fourier Transform, which lets you take a signal in time, and analyse its content by frequency. I won’t explain that here, but if you’re interested in how it works, you can start by reading this.
If I take the total impulse response (also known as a time response measurement) of the filter: in other words, I send in an impulse, I record the output and don’t stop recording until the ringing has decayed to a level low enough that I no longer care (for the purposes of this discussion, at least). Then, I do a Fourier Transform of the recording, I get something like Figure 5.
Fig 5. In this example, the “portion” of the time response that I’ve used is the entire time response. Bear with me.
There is no new information in Figure 5. It’s just a setup for Figures 6 and 7.
Let’s now start slicing up the time response selectively to see what frequencies are contained in the output of the filter at what time. We’ll start by just taking the first and second samples of the impulse at the output, shown in Figure 6.
Fig 6. The magnitude response is a measurement of ONLY the first 2 samples of the impulse, which are shown in the middle plot.
As you can see in Figure 6, if I remove the ringing that comes after the impulse, then the response of the signal has an almost-flat magnitude response and a gain of about 2 dB or so. This should not come as a surprise, since it’s almost an impulse. The only real difference between the portion that I’ve used and a real impulse is that the second value is not 0. So far so good…
Let’s look at the remainder of the time response. This is shown in Figure 7.
Fig 7. The magnitude response of the remaining portion of the time response, omitting the initial onset of the impulse.
Figure 7 shows something interesting. We see the response of a band-pass filter with a centre frequency of 1 kHz, and a gain of 9 dB, which is the response of the filter after the initial impulse has passed.
What does this all mean!?
If we leave out one important thing for now, this means that a peaking filter that has a boost of 12 dB, an Fc of 1 kHz and a Q of 2 is actually the sum of two things:
a through-put with a little gain (about 1 dB)
a bandpass filter with a gain of about 9 dB
This is, in essence, true. You can create a peaking filter by summing a bandpass filter to a through-put. However, an important point to realise here is that the band pass signal essentially comes after the onset of the signal. In Part 3, we’ll talk about whether this is a problem – or, more accurately, when this might be a problem. For now, however, I’ll throw one more example at you.
Up to now, we’ve only looked at the example of a peaking filter with a boost. What happens when the filter has a cut instead?
Fig 8. The time and magnitude responses of a dip filter where Fc = 1 kHz, Gain = -12 dB and Q = 2.
Notice that a dip filter also rings in time after the initial impulse, but decays much faster than the equivalent boost. (I’ll have to be a bit more careful about my use of the word “equivalent”, actually – but I’ll straighten that out at the end of the series. To be continued…)
Fig. 9: Similar to the boost, the first onset of the impulse has a nearly-flat magnitude response.
Fig 10. The decay of the dip filter is also a slightly-strange-looking band-pass filter, but with an overall gain of about -6 dB.
Okay, what’s going on here? A peaking filter with a boost is a through-put plus a bandpass. A dip filter is ALSO a through-put plus a somewhat quieter (sort-of) bandpass. This doesn’t make any sense.
Actually it doesn’t make any sense because there’s a piece of information that I’m leaving out – the phase of the ringing. Notice that, with the peaking filter, the decay portion starts positive and then goes negative initially. With the dip filter, the decay starts negative and goes positive. So, the previous paragraph should have read: “A peaking filter with a boost is a through-put PLUS a bandpass. A dip filter is ALSO a through-put MINUS a somewhat quieter (sort-of) bandpass.”
The phases of the decays of the bandpass portions are opposite for the two filters. Another way to think of this is that the ringing in the dip filter cancels the energy around 1 kHz in the initial impulse, whereas the ringing in the peak filter adds to it.
However, it’s really important to note for now that both filters – the peak and the dip result in ringing in time.
