Category: audio

It’s impossible to build a good loudspeaker. Part 1: Crossovers

So, you want to build a loudspeaker…

One of the first things you’ll find out is that, if you’re building a loudspeaker with moving coil drivers, and unless you want a loudspeaker with very limited capabilities, you’ll probably need to use more than one driver. Starting small, you’ll at least need a bigger driver to produce the lower frequencies and a smaller driver to produce the higher ones. No surprise so far – many people lead meaningful lives with just a tweeter and a woofer.

However, you’ll probably need to ensure that the tweeter doesn’t get too much signal at low frequencies, and the woofer doesn’t get too many highs. In order to do this, you’ll need a crossover. Still no surprises. Most people who build a loudspeaker already know that they’ll need a crossover to keep their drivers happier.

Now for some new stuff – at least for some people. When you make a crossover, you must remember to keep the driver’s characteristics in mind. You can’t just slap a high pass filter on the tweeter and a low pass filter on the woofer and expect things to work. The tweeter is already behaving as a high pass filter all by itself. If the characteristics of the tweeter’s inherent high pass are what you want, then you don’t want to duplicate that filter in the electronics. So, design your filters wisely. I will probably come back to some examples of this some time in a future posting.

However, that is not the topic for today. For today, we will assume that we are building a loudspeaker using two very special drivers. They are:

  1. infinitely small
  2. have bandwidths that go from DC to infinity
  3. have “perfect” impulse responses
  4. and therefore have completely flat phase responses

In other words, we will pretend that each of our drivers is a perfect point source. We’ll also assume that they are not mounted on a baffle (a fancy way of saying “on the front of a box” – usually…). Instead, they’re just floating in space, arbitrarily 25 cm apart (one directly above the other). We’ll arbitrarily make the crossover frequency 500 Hz. Finally, let’s say that we’re arbitrarily 2 m away from the loudspeaker.

The reason for all of these assumptions is that, for the purposes of this posting, we’re only interested in the effects of the crossover on the signal, so I’m making everything else in the system either perfect or non-existent. Of course, this has nothing to do with the real world, but I don’t really care today.

So, if you’ve done a little research, you’ll know that there are a plethora of options to chose from when it comes to crossovers. I’ll assume that we’re building an active loudspeaker with a DSP so we can do whatever we want.

Linkwitz Riley, 4th Order

Let’s start with Old Faithful: a 4th-order Linkwitz-Riley crossover. This is implemented by putting two 12 dB/oct Butterworth filters in series, each with a cutoff frequency equal to the intended crossover frequency. (If you’re using biquads, set your Q to 1/sqrt(2) on each filter). The total low pass section will have a gain of -6.02 dB at the crossover frequency (so will the total high pass section). Since the two sections are 360° out of phase with each other at all frequencies, they’ll add up to give you a total of 0 dB when they sum together at any frequency. However, you must remember that the filter sections used in the crossover have an effect on the phase response of the re-combined signal. As a result, when the two are added back together (at the listening position) the total will also have a modified phase response – even when you are on-axis to the loudspeakers, (and equidistant to the two loudspeaker drivers).

It is also important to remember that the phase relationship of the two sections (coming from the tweeter and the woofer) is only correct when those two drivers are the same distance from the listener. If the tweeter is a little closer to you (say, because the tweeter is on top and you stood up) then its signal will arrive too early relative to the woofer’s and the phase relationship of the two signals will be screwed up, resulting in an incorrect summing of the two signals.

How much the total is screwed up depends on a bunch of factors including

  1. the relative phase responses of the filters in the crossover
  2. the phase responses of the drivers (we’re assuming for this posting that this is not an issue, remember?)
  3. the deviation in those phase responses caused by the mis-alignment of distances to the drivers

The result of this is a deviation in the vertical off-axis response of the loudspeaker. How bad is this? Let’s look!

This figure shows 4 plots. The top one shows the magnitude responses of the two individual sections. As you can see, the crossover frequency is 1 kHz, and both sections are 6 dB down at that frequency.

The second plot shows the total magnitude responses at 5 different vertical angles of incidence to the loudspeaker: -30°, -15°, 0°, 15°, and 30°. So, we’re going from below the loudspeaker to above the loudspeaker. It’s not obvious which plot is for which angle because, for the purposes of this discussion, it doesn’t matter. I’m only interested in talking about how different the loudspeaker sounds at different angles – not the specifics of how it sounds different.

The third plot shows the total phase response of the system, at a position that is on axis to the loudspeaker (and therefore equidistant to both drivers). As you can see there, a perfect 2-way loudspeaker with a 4th order Linkwitz-Riley crossover behaves as a 4th-order allpass filter. In other words, at low frequencies, the output is in phase with the input. At the crossover frequency, the output is 180° out of phase with the input. At high frequencies, the output is 360° out of phase with the input.

The fourth plot shows the step response of the total system, at a position that is on axis to the loudspeaker (and therefore equidistant to both drivers). As you can see there, a perfect 2-way loudspeaker with a 4th order Linkwitz-Riley crossover does not give you a “perfect” step response – it can’t, since it acts an allpass filter. The weird shape you see there is cause by the fact that the high frequencies are not “in phase” with the low frequencies. (I know, I know… different frequencies cannot be “in phase”.) Since different frequencies are delayed differently by the total system, they do not add up correctly in the time domain. Thus, although the total output in terms of magnitude is flat (hence the flat on-axis frrequency response) the time response will be weird.

Looking in detail at the step response plot, you can see that it takes about 1.5 ms for the total output to settle to a value of 1. The actual time that it takes is dependent on the crossover frequency. The lower the frequency, the longer it will take. It’s the shape of the step response that’s determined by the crossover’s phase response. What can be seen from the shape is that the high-frequency spike hits first (as we would expect), then the step response drops back to a negative value before heading upwards. It overshoots, peaking at a value of 1.0558 before coming back down, undershooting slightly (to a value of 0.9976)  and finally settling at a value of 1. Note that these values won’t change with changes in crossover frequency – they’ll just happen at a different time. The higher the frequency, the faster the response.

Whether or not this modified time response is worth worrying about (i.e. can you hear it) is also outside of the scope of today’s discussion. All we’re going to say for today is that this temporal distortion exists, and it is different for different crossover strategies as we’ll see below.

Linkwitz-Riley 2nd Order

A second possible crossover strategy is to use a 2nd-order Linkwitz Riley. This is similar to a 4th-order, except that instead of putting two 12 dB/octave Butterworth filters in series to make each section, you put two 6 dB/octave Butterworth filters in series.

Since the total filters applied to make the high pass and low pass sections of this crossover are each made with only two first-order filters (instead of two second-order filters), the high pass and low pass sections are only 180 degrees out of phase with each other (at all frequencies). Consequently, in order to get them to add back together without cancelling completely at the crossover frequency, you have to invert the polarity of one of the sections. (We’ll do this to the high pass section, just in case you can hear your woofers pulling when they ought to push when a kick drum hits). On the plus side, since they’re 180 degrees out of phase at all frequencies, if you DO flip the polarity of your high pass section, they’ll add back together (on axis) to give you a flat magnitude response.

As you can see in the above plots, the slopes of the high pass and low pass sections in this crossover type are more gentle than in the 4th order Linkwitz Riley. This should be obvious, since they have a lower order. In the second plot, you can see that, on-axis, the magnitude response is flat, just as we would expect. However, there are implications on the off-axis response. The deviation from “flat” is greater with the 2nd-order LR than it is with the 4th-order version. Not by much, admittedly, but it is greater. So, if you’re concerned about deviations in your off-axis response in the vertical plane, you might prefer the 4th-order LR over the 2nd-order variant.

If, however, you lay awake at night worring about phase response (you know who you are – yes – I’m talking you YOU) then you might prefer the 2nd-order Linkwitz Riley, since, as you can see in the third plot, the total output is only 180° out of phase with its input in the worst case – only half that of the 4th-order variant. On the other hand, since it’s 180° out of phase, that means that a high voltage going into the system (at high frequencies) will come out as a low pressure. So, if you’re the kind of person who lays awake at night worrying about “absolute phase” (you know who you are – yes – I’m talking to YOU) then this might not be your first choice.

Finally, take a look at the step response in the final plot. You’ll notice immediately that the high frequencies are 180° out iof phase, since the initial transient of the step goes down instead of up. You’ll also notice that the step “recovers” to a value of 1 a little faster than the 4th order Linkwitz Riley. Note that, a 2nd order LR, doesn’t have the overshoot that we saw in the 4th order version.

Butterworth, 12 db/octave

Possibly the most common passive crossover type (and therefore, possibly the most common crossover  type, period!) is the 12 dB/octave Butterworth crossover. This is made by using a 2nd-order Butterworth filter for each section (the high pass and the low pass).

 

You’ll notice in the top plot that this means that the filter sections are only 3 dB down at the crossover frequency. This has some implications on the on-axis response. Since the two filter sections are 180° out of phase with each other (at all frequencies – just like the 2nd-order LR crossover) then we have to flip the polarity of one of the sections (the high-pass section again, for all the same reasons) to prevent them from cancelling each other at the crossover frequency when they’re added back together. However, now we have a problem. Since the two sections are in-phase (due to the 180° phase shift plus the polarity flip) and since they’re only 3 dB down at the crossover frequency, when they get added back together, you get more out than you put into the system. This can be seen in the second plot, where the total magnitude response has a bump at the crossover frequency – even when on-axis.

Of course, a 3 dB bump in the magnitude response will be audible, at the very least as a change in timbre (3 dB is, after all, twice the power). We can also see that there is a small, but visible change in the overall magnitude response as you change the vertical angle to the listener.

The third plot shows that a 12 dB/Octave Butterworth crossover, when all other issues are ignored, acts as a 2nd-order allpass filter with a worst-case phase distortion of 180°.

Finally, the fourth plot shows that its step response is similar, but not identical to, the 2nd-order LR crossover. The initial transient goes negative because we have inverted the polarity of the high pass section. Unlike the 2nd-order LR (but similar to the 4th-order LR), however, there is an overshoot and undershoot before the response settles at a value of 1. That overshoot reaches a maximum of 1.1340, and the subsequent undershoot goes down to 0.9942.

 

Butterworth, 18 db/oct

Sometimes, you’ll also hear of people using an 18 dB/octave Butterworth crossover instead of the 12 dB/octave version. These are a little more complicated to implement, but not uncommon.
The responses of this crossover type are plotted below.
The top plot shows that, although the order of the Butterworth high pass and low pass sections are higher (and therefore have steeper slopes), they are still only 3 dB down at the crossover frequency. However, since the two sections are now 270° out of phase, they add together to give you a magnitude of 0 dB – the same as the input – but only when you’re on-axis. As can be seen in the second plot, the off-axis response of an 18 dB/Octave Butterworth crossover is MUCH worse than all of the other crossover types we’ve seen so far. So, if you’re the type of person who worries about off-axis response, or the magnitude response of your ceiling and floor reflections, or the power response of your loudspeaker, then you probably wouldn’t choose this crossover over the previous ones.
The phase response of the total output of this crossover seems a bit strange initially, since you have two filters that are 270° apart at all frequencies, but the summed output has the phase response of a 4th-order allpass. However, what is not seen in this plot are the individual phase responses of the two sections. The low pass section has a phase response that starts at 0° in the low end and drops to -270° in the high end. The high pass section’s phase response starts at -90° in the low end and ends at -360° in the high end. So, although the two sections, individually, have phase response curves that have a similar shape to a third-order allpass, their combined outputs result in a 4th order allpass.
Finally, let’s come to the step response. This one is the busiest one yet, since, after the initial transient and drop, it overshoots (to a value of 1.178), then undershoots (0.971), then overshoots again (1.005) and finally undershoots (0.9994) before finally settling at a value of 1.

Constant Voltage, (using a Butterworth, 18 db/oct high pass)

There is a group of persons who believe that the step response (or the shape of a square wave through the system) is the be-all-and-end-all for determining the quality of a system. The logic goes that, if a square wave goes in, and a square wave comes out, then the system is perfect. This is true – if you mean “perfect at reproducing square waves” – which may or may not be important.
Following this logic, the idea is that, if you take your initial input and make a filtered version, then all you need to do is to subtract that filtered version from the input to get the remainder. If you then add the remainder and the filtered section, you get out what you put in, so the system is perfect. At least, that’s the idea. Let’s see how well that works out, shall we?
(By the way, the name for this classification of crossovers is “constant voltage” crossovers, and there are lots of different ways to implement them. Richard Small wrote some good stuff about them in some AES papers once-upon-a-time, if you’re curious.)
So, let’s look at one fairly-common implementation of a constant voltage crossover. We’ll take the input, filter it with an 18 dB / Octave Butterworth filter and use that for the high pass section. The low pass section is created by subtracting the high pass section from the input, and we just take what we get.
As you can see in the top plot, the result of this is that the low pass section is a little weird. It has a rather large bump in its magnitude response around the crossover frequency. In addition, the slope of the high pass roll-off is not very steep. This all might be okay, if your lower driver is able to handle it, but it might not. (Note that, if we had made a Butterworth low pass and used the subtraction trick to get the high pass section, the bump would have appeared in the high pass section, which would result in too much low-frequency energy in the tweeter, thus likely making it unhappy… That’s why we used a Butterworth high pass to start.)
Let’s skip the next plot and look at the third and fourth. As you can see in both of these, the constant voltage crossover is unique in that it has no phase distortion, and the step response is perfect. This is to be expected, since these are the primary criteria behind the design of this type of crossover.
Now, let’s look at the second plot. As you can see there, the off-axis response of a constant voltage crossover is a complete disaster. So, if you’re the kind of person who thinks that off-axis response is important – or at least worth considering, you should probably stay away from this crossover. However, if you have an acoustically absorptive floor and an acoustically absorptive ceiling (so no vertical reflections) and you never stand up, and you’re inside the room’s critical distance with respect to the loudspeaker, then this little problem might not be an issue for you.
Concluding comments
The thing that you have to remember for all of the stuff I’ve said here is that it’s only applicable within the limitations of the parameters I stated at the beginning. If your drivers are imperfect, or if you have a high pass in series with your low pass section because your loudspeaker driver exists in real life (it does), or if you have a symmetrical driver arrangement (sometimes called a D’Appolito design, named after the first person to be smart enough to publish a paper about it), then all of these results will be different.
Also, the other important thing to remember is that I’m making no claims about whether these “problems” are audible. They might be – and they might not be. But don’t just jump to conclusions all willy-nilly and assume that, because you can see a difference in these plots, you’ll be able to hear the difference in your loudspeakers. Then again, that doesn’t mean that I’m saying “you can’t!” What I’m saying is “I don’t know whether you can hear these issues or not – any of them.”
P.S .
Happy New Year.

Never trust a THD+N measurement: v2

In the last post, I talked about why a THD+N measurement is useless if you don’t know about the type of distortion that you’re measuring. Let’s now talk about another reason why it’s useless in isolation.

Once again, let’s assume that we’re doing a THD+N measurement the old-fahsioned way where we put a sine wave into a device, and apply a notch filter to the output at the same frequency of the sine wave and find the ratio of the level of the sine wave to the output of the notch filter.

This time, instead of taking a signal and distorting it, I’ll do some additive synthesis. In other words, I’ll build a final signal that contains four components (although they’re not entirely independent…):

  1. a “signal” consisting of a 100 Hz sine wave which we’ll call “the fundamental”
  2. sine tones at frequencies that are multiples of the fundamental frequency (in other words, they are harmonically related to the fundamental).
  3. sine tones at frequencies that are not multiples of the fundamental frequency (in other words, they are not harmonically related to the fundamental).
  4. wide-band noise
Version 1: No artefacts

Let’s start by listening to the original 100 Hz sine wave with a level of -10 dB FS without any other additional components. If you hear any distortion or noise, then this is a problem in your playback system (unless your system is so good that you can hear the quantisation error caused by the fact that I didn’t dither the signal).

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Version 2: Wide-band noise

Now let’s add noise. I’ve added noise with a white spectrum and a level such that a THD+N measurement will tell us that we have 10% THD+N (relative to the level of the 100 Hz sine tone signal). In other words, I have a sine wave with a level of -10 dB FS and I have added white noise with a long-term RMS level of  -30 dB FS.

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 It should be pretty obvious, even with poor playback equipment, that I have added noise to the 100 Hz tone. This should not be surprising, since a 10% THD+N is pretty bad.
Version 3: 2nd harmonic
For this version, I’ll add a 200 Hz sine tone to the 100 Hz tone. The fundamental (100 Hz) has a level of -10 dB FS. The level of its second harmonic (200 Hz) is -30 dB FS. This means that, again, I get a THD+N value of 10%.
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Version 4: 3rd harmonic
For this version, I’ll add a 300 Hz sine tone to the 100 Hz tone. The fundamental (100 Hz) has a level of -10 dB FS. The level of its third harmonic (300 Hz) is -30 dB FS. This means that, again, I get a THD+N value of 10%.
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Version 5: 2nd to 5th harmonics
For this version, I’ll add a four additional sine tones to the 100 Hz tone. The fundamental (100 Hz) has a level of -10 dB FS. I have added tones at 200 Hz, 300 Hz, 400 Hz and 500 Hz (the 2nd through to the 5th harmonics, inclusive) with a spectral pattern where each successive tone is half the amplitude of the previous. In other words, 500 Hz is half the amplitude of 400 Hz which, in turn, is half the amplitude of the 300 Hz tone, which is half the amplitude of the 200 Hz tone.
I have adjusted the overall level of the harmonics so that we get a THD+N value of 10%. In other words, the RMS level of the signal comprised of the 200 Hz to 500 Hz sine tones (inclusive) is -30 dB FS.
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Version 6: 5 kHz
For this version, I’ll add a 5 kHz sine tone to the 100 Hz tone. The fundamental (100 Hz) has a level of -10 dB FS. The level of the 5 kHz tone is -30 dB FS. This means that, again, I get a THD+N value of 10%.
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Version 7: Noise plus five sine tones with random frequencies
For this version, I’ll add a mess to the 100 Hz tone. The fundamental (100 Hz) has a level of -10 dB FS. To this I added a signal that is comprised of wide-band white noise and 5 sine tones at random frequencies between 0 Hz and 20 kHz (no, I don’t know what they are – but it doesn’t matter for the purposes of this discussion). The levels of the noise and 5 sine tones are random.
I have adjusted the overall level of the signal comprised of the noise and 5 random sine tones so that we get a THD+N value of 10%. In other words, the RMS level of the signal comprised of the noise and 5 random sine tone is -30 dB FS.
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The punch line!

Each of the six signals I’ve presented above in Versions 2 through 7 (inclusive) is a “distorted” version of the original 100 Hz sine tone in Version 1. Each of those six signals will have a measurable THD+N of 10%. However, it is quite obvious that they have very different spectral patterns, and therefore they sound quite different.

This isn’t really revolutionary – it’s jut another reminder that a THD value, in the absence of any other information, isn’t terribly useful – or at least, it doesn’t tell you much about how the signal sounds.

Never trust a THD+N measurement: v1

Caveat: This is basically a geek version of a cover tune. The point that I make here was one that I originally heard someone else present at an AES convention years ago. However, since I haven’t heard anyone tell this story since, I’ve written it here. 

 

Let’s build two black boxes, each of which creates a measurable distortion. We’ll call them Box “A” and Box “B”.

Box “A” has a measured THD+N of 20%. Box “B” has a measured THD+N of 2%. We’ll be using the old-fashioned way of measuring THD+N where we put a sine wave into the device, and apply a notch filter to the output at the same frequency of the sine wave and find the ratio of the level of the sine wave to the output of the notch filter.

Let’s put a 500 Hz sine wave into the boxes and listen to the output. The original sine wave sounds like the following:

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The sine wave at the output of Box “A” (with a THD+N of 20%) sounds like the following:

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The sine wave at the output of Box “B” (with a THD+N of 2%) sounds like the following:

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So far so good. There should be no surprises yet.

Now let’s put a recording of something that I listen to all the time (my own voice) into the same black boxes to see what happens.

We’ll start with the original recording (this is just a file that I happened to have on my hard drive for testing imaging – ignore the fact that it talks about coming from the left channel only – your computer will probably play it as a mono file out both channels – this is irrelevant to the discussion):

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Now let’s listen to how that recording sounds at the output of Box “A” (with a measured THD+N of 20%)

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As you’ll hear, there is no audible distortion on the sound file, despite the fact that it has gone through a box that generates a distortion that we measured to be 20%.

Now let’s listen to how the original recording sounds at the output of Box “B” (with a measured THD+N of 2%)

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As you will probably hear in that last sound file, the Box “B” – the one with “only” 2% distortion sounds MUCH worse than either the original sound file or the output of Box “A” which should have much more audible distortion.

So, the question is “why?”

 

Let’s look at the waveforms to see what’s going on here.

The original sine wave looks like the following:

Time plot of original sine wave

After that sine wave has gone through Box “A”, the output looks like the following:

The output of Box “A” when fed with the sine wave

As you can see, I’ve created Box “A” to generate its distortion by clipping the signal at a limits of -0.5 and 0.5.

The output of Box “B” when fed with the same sine wave looks like the following:

The output of Box “B” when fed with the sine wave

If we zoom in on that plot, it looks like the following:

The output of Box “A” when fed with the sine wave

So, as you can see, I’ve made Box “B” to generate a zero-crossing distortion – but a pretty small one.

The reason the THD+N of Box “A” is 20% and that of Box “B” is only 2% is not just because the “damage” done to the signal is bigger with Box “A”. It’s also caused by where the damage is done. This might not make sense, so let’s look at the signals  a little differently.

Let’s do a histogram of the original sine wave. This tells us how often the sample values are a given value. This is shown below in the following plot.

Histogram of the original sine wave

This histogram shows that the sample values in the original sine wave are usually near -1 and +1, and rarely around 0.

Now let’s look at a histogram of the output of Box “A” – the distorted sine wave with 20% THD+N. It looks like the following:

A histogram of the output of Box “A” when fed with the sine wave

As can be seen in the plot above, the sample values from the original sine wave that were below -0.5 are now all congregated at -0.5, and the values that were above 0.5 are now congregated at 0.5. This is the result of the clipping applied to the signal.

By comparison, the histogram of the output of Box “B” is shown below:

A histogram of the output of Box “B” when fed with the sine wave

As you can see by comparing these last two plots, the zero crossing distortion of Box “B” results in a histogram that is more similar to the histogram of the original signal than that of the clipping distortion of Box “A”. This is because the zero crossing distortion distorts the signal where the signal rarely is.

 

Now let’s look at the histograms of the speech signal. Below is a histogram of the original speech recording.

A histogram the original speech signal.

As you can see in this plot, the speech signal is unlike the sine wave in that it is usually at 0, and not at the extreme values of -1 and 1. In addition, you can see that very little, if any, of the signal is below -0.5 or above 0.5 which are the clipping values of Box “A”. Consequently, as you can see below, the histogram of the output of Box “A”, when fed with the speech signal, looks almost the same as the histogram of the original signal, above.

A histogram of the output of Box “A” when fed with the speech signal.

However, the output of Box “B” is different. The histogram of that signal is shown below:

A histogram of the output of Box “B” when fed with the speech signal.

So, as you can see here: the zero crossing distortion is affecting the signal where it is most often, whereas the clipping of Box “A” has no effect on the signal.

The moral of the story

The point that I’ve (hopefully) illustrated here is that the value generated by a THD+N measurement is basically irrelevant when it comes to expressing how a device distorts a normal signal. However, the problem is not with the measurement technique, but the signal that is used in the procedure. We use a sine wave to do a THD+N measurement because that used to be the easy way to do a THD+N measurement back in the old days of signal generators, analogue notch filters, and voltmeters. The problem is that the probability distribution function (PDF) of that sine wave is completely unlike the PDF of a music or speech signal. So, if the distortion of the device affects the signals in the wrong place, then the result of the measurement will not reflect the sound of the device.

the problem with windows

Now, before you start sending me hate mail because you think this posting is a Windows vs. Mac lecture, hold your horses. That’s NOT the kind of windows I’m talking about. This one’s about windowing functions and one (possibly unexpected) effect on the results of the analysis of the impulse response of an allpass filter. So, if you want to debate Windows vs. Mac – go somewhere else. If you think that you can get all riled up over a Blackman Harris window function, read on!

Last week I had to do some frequency-domain analysis of a system that had a small problem with noise in its impulse response measurements. The details of where the noise came from are unimportant. There is only one important thing from the back-story that you need to know – and that is that I was measuring the response of an allpass filter implementation.

So, I did my MLS measurement of the allpass filter and, because I had noise in the impulse response, I chose to use a windowing function to clean up the impulse response’s tail. Now, I know that, by using a windowing function (or a DFT, for that matter), there are consequences that one needs to be aware of. However, the consequence that I stumbled on was a new one for me – although in retrospect, it should not have been.

Here’s a sterilised version of what happened, just in case it’s of use.

Below is a plot showing a (very clean) impulse response of an allpass filter. To be more specific, it’s a 4th order Linkwitz Riley crossover with a crossover frequency of 100 Hz, where I summed the outputs of the high pass and low pass components together to make an output. (We will not discuss why I did it this way, since that information is outside the scope of this discussion.) In addition, I have plotted three windowing functions, a Hann, a Hamming and a Blackman Harris.

Impulse response of an allpass filter with a centre frequency of 100 Hz and three windowing functions with a length of 65536 samples (Fs = 65536 Hz).

Note that the length of the windowing functions is big – 65536 samples to be exact. As you can see in the plot, the ringing of the allpass filter is negligible in this plot by the time we get to the end of the window. This can also be seen below in the next two plots where I’ve shown the impulse response after it has been windowed by the three (actually four, if we include rectangular as a function), scaled in linear and dB FS. (I know, I know, dB FS is an RMS measurement and I plotted this as instantaneous values – sue me.)

The result of the windowing functions on the impulse response.
The result of the windowing functions on the impulse response plotted in dB FS.

So, if you now take those windowed impulse responses and calculate their magnitude and phase responses, you get the plots shown below.

Magnitude responses for the windowed impulse responses. Window length = 65536 samples, FFT length = 65536, Fs = 65536. Allpass Fc = 100 Hz.
Phase responses (in degrees) for the windowed impulse responses. Window length = 65536 samples, FFT length = 65536, Fs = 65536. Allpass Fc = 100 Hz.

“So what?” I hear you cry. The magnitude responses of the four versions of the windowed impulse response are all identical enough that their plots lie on top of each other. This is also true for their phase responses. “I see what I would expect to see – what are you complaining about?” I hear you cry.

Well, let me tell you. The plots above show the results when you use a 65536-point FFT and a 65536-sample window (okay, okay, DFT – sue me).

Let’s do all that again, but with a 65536-point FFT and a 1024-point window instead (I did this in MATLAB, so it’s zero-padding the impulse responses with the remaining 65536-1024 = 64512 samples.)

Impulse response of an allpass filter with a centre frequency of 100 Hz and three windowing functions with a length of 1024 samples (Fs = 65536 Hz).

Now we can see immediately, that the ringing in the allpass filter’s impulse response hasn’t settled down by the time we get to the end of the window. This can also be seen in the following two plots.

The result of the windowing functions on the impulse response.
The result of the windowing functions on the impulse response plotted in dB FS.

As you can see there, the impulse response itself (aka “Rectangular” windowing) is only about 60 dB below its peak when we reach the end of the window. How does this then affect our magnitude response?

Magnitude responses for the windowed impulse responses. Window length = 1024 samples, FFT length = 65536, Fs = 65536. Allpass Fc = 100 Hz.

As you can see there, the implications on the rectangular window is a ripple in the low end of the calculated magnitude response. As you can also see there, the result of attenuating the tail of the allpass filter’s impulse response before we unceremoniously cut it off is that we lose low-end in the magnitude response. The more we attenuate in the windowing function, the more low end we lose.

Of course, this also has implications on the phase response of the windowed impulse responses, as is shown below.

Phase responses (in degrees) for the windowed impulse responses. Window length = 1024 samples, FFT length = 65536, Fs = 65536. Allpass Fc = 100 Hz.

The moral of this story is not a new one: beware of the effects of a windowing function on your analysis.

In my personal case, it’s a memorable lesson, since I didn’t get to this conclusion immediately. This is because I was measuring the allpass with different Fc’s – and what I saw in my magnitude response was a shelving response (I was using a Blackman Harris window). When I changed the Fc of the allpass, the shelving response that I saw moved appropriately. So, my conclusion was that there was a problem in my filter that I was measuring. It took some time (too much time!) before I figured out (with the help of some more level-headed friends) that my problem was the window length and my windowing function, not the filter that I was measuring. Won’t make that mistake again for a while…