Quiz: How Well Do You Know Your Synthesis?

[Christopher Winkels]

Interesting... the only one I couldn't answer is #2. I'd never heard that "quality" is what Q stands for; I always assumed it was some obscure engineering thing like putting a delta, sigma, or other symbol in.

All the others I'm proud to say I could answer. Yay me.

[Bross]

1. It's the steepness of the filter slope.

More specifically, the amplitude of the output decreases by that many decibals for each octave that you move away from the cutoff frequency. And by definition, the cutoff frequency is the point at which the amplitude is 3 dB below the average amplitude within the passband.

Don't forget the -24dB/dec is an asymptotic expression for the filter response. The actual filter response isn't completely flat below the cutoff frequency and it isn't exactly -24dB/dec above the cutoff. Also the cutoff frequency being the 3dB point is only true for a single pole filter, and especially not a filter with variable Q.

I only used the spell checker to correct "asymptotic"

[HideawayStudio]

1. It's the steepness of the filter slope.

More specifically, the amplitude of the output decreases by that many decibals for each octave that you move away from the cutoff frequency. And by definition, the cutoff frequency is the point at which the amplitude is 3 dB below the average amplitude within the passband.

Don't forget the -24dB/dec is an asymptotic expression for the filter response. The actual filter response isn't completely flat below the cutoff frequency and it isn't exactly -24dB/dec above the cutoff. Also the cutoff frequency being the 3dB point is only true for a single pole filter, and especially not a filter with variable Q.

I only used the spell checker to correct "asymptotic"

Please note that -24dB / decade is not the same thing as -24dB / octave. In electronic music most filters are stated in dB / oct where a single pole filter typically has a -6dB / oct cutoff response and a Moog filter, which is 4 pole, has a -24dB / oct response.

-20dB / dec is actually -6dB / octave

ie. There are 3.32 octaves in a decade, so dB/decade = 3.32 * dB/octave

[OriginalJambo]

BZZZZZT! Wrong! Halve the length and you double the frequency. Frequency is inversely proportional to pipe's lenght. 16' is one octave LOWER than 8'.

Indeed good sir, I stand corrected. The shorter the pipe the higher the pitch. Can't believe I missed that.

[cornutt]

Right; by "inharmonic", we mean that the sounds contain overtones whose frequencies are not integer multiples of the fundemental frequency. Bells do that because they are three-dimensional objects, and can develop vibratory modes across all three axes.

All vibrating bodies including strings and wood panels, are three dimensional. I don't know why bells produce harmonics that are equal or greater in level to the fundamental, but that's not the reason.

I defy you to measure the vibratory modes of a guitar string along its Y and Z axes (Those would be compression waves, which are not significant in a guitar string.) What you say about strings is technically true, but for practical purposes, a string behaves like a one-dimensional object.

As for the piece of wood, it behaves just as a bell does -- it has multiple vibratory modes, and the sound you get from striking it contains inharmonic partials. The main difference, other than that it doesn't sustain well, is that the piece of wood isn't tuned like a bell is. But you can tune a piece of wood and get a bell-ish sound from it. After all, that's what a xylophone is.

[madtheory]

Ah OK, thanks! So, a bell or a piece of wood has stronger vibrations along the Y and Z axes than a stretched string, and that's where the powerful harmonics come from. And that's why a xylophone sounds like that! Cool .
Might that also explain why modelled strings sound a bit weird- because the vibrations in the third dimension are not modelled?

[cornutt]

Might that also explain why modelled strings sound a bit weird- because the vibrations in the third dimension are not modelled?

Could be, but I suspect that whatever vibrations exist in those modes are well above the audio range and get damped out very quickly. More likely, it is because the models treat the string as a "perfect" string, meaning that they do their computations assuming that the string's elasticity is exactly the same along the entire length of the string. Real strings are held stationary at the ends, and that makes the string stiffer at the ends than it is in the middle. Higher harmonics try to bend the string more than the fundemantal does; since the string won't bend well a the ends, the effect is that, for the higher harmonics, the string acts like it's a little shorter than it actually is. This makes the higher harmonics a bit sharp relative to the fundemental. And the effect is worse for thicker strings. That's why pianos are tuned to "stretch" tuning; if they weren't; the high notes would sound flat compared to the bass notes, because the upper harmonics of the bass notes are sharp.

Real strings will also have variations in stiffness along the length of the string, due to variations in manufacturing, impurities in the metal, and so forth. This gets worse as the string is played; picking, bowing, and fretting all cause the strings to develop spots that are stiff or weak. If you've ever heard false harmonics from an old guitar string, that's caused by stiff and weak points causing the string to act as if it were fretted or restrained at some arbitrary point along the length. The basic Karplus-Strong algorithm that a lot of digital synths use to model strings can't model this kind of stuff.

[Soundwave]

OK Here are 15 questions to set the men from the boys....

Here is 15 questions from a boy wanting to be a man!