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Dub Studio · Joined: 20 Jun 2006 Posts: 51 Location: Bristol

A client of mine wants to record some looped/locked grooves, but we are having problems nailing the exact frequencies. We want to find frequencies that have no more than 2 decimal places that when multiplied by 1.8 come up with a whole number. My maths is rubbish Confused

Any ideas?

Steve E. · Posted: Wed Jul 09, 2008 5:22 am Post subject:

Are these sampling frequencies or tones? I'm confused.

OK.....1.8, or 1 4/5 or 9/5, is the exact number of seconds in 1 full rotation at 33 1/3. Got that part.

So maybe you are looking for audible frequencies that end right at the "0" at 9/5's of a second.

hmmmm.

Anyone?

Dub Studio · Joined: 20 Jun 2006 Posts: 51 Location: Bristol

Steve E. · Posted: Wed Jul 09, 2008 6:06 am Post subject:

If it has to work out by decimal places, any multiple of five will do. So, for instance:

5 x 1.8 = 9

(that is, the frequency which beats 5 times in one second will beat exactly 9 times in 1.8 seconds)

10 x 1.8 = 18

(that is, the frequency which beats 10 times in one second will beat exactly 18 times in 1.8 seconds)

20015 x 1.8 = 36027

(to pick a random larger example.... the frequency which beats 20015 times in one second will beat exactly 36027 times in 1.8 seconds)

etc. It's all multiples of 5 at 1 second and multiples of 9 at 1.8 seconds.

**** **** **** ****

But there are eight other whole numbers at 1.8 seconds between those multiples of nine.....and each of those is represented as a fraction at 1 second. And each of those is a multiple of 5/9.

Unfortunately, every single one of these is only accurately represented as an infinitely repeating decimal:

first one (or 5/9) = .5555555555555555.... this equals 1 at 1.8 seconds
second one (or 2 x 5/9..... or 10/9) = 1.111111111111111111....this equals 2 at 1.8
third one (or 3 x 5/9 or 15/9) = 1.6666666666666666....this equals 3 at 1.8
fourth one (or 20/9) = 2.222222222222222....this equals 7 at 1.8
seventh one (or 7 x 5/9 or 35/9) = 3.888888888888888....this equals 7 at 1.8

the 36026th one (or 36026 x 5/9 or 180130/9) =20014.444444....this equals 36026 at 1.8 seconds

*** **** **** **** ****

If you are not limited to decimals, but can do fractions instead, this is suddenly very easy. Then, every multiple of 5/9's will give you a whole number at 9/5's of a second:

5/9 at 1 second is 1 at 1.8 seconds

10/9 (or 2 x 5/9) at 1 second = 2 at 1.8 seconds

15/9 (or 3 x 5/9) at 1 second = 3 at 1.8 seconds

25/9 (or 5 x 5/9) at 1 second = 5 at 1.8 seconds

35/9 (or 7 x 5/9) at 1 second = 7 at 1.8 seconds

45/9 (or 9 x 5/9....which equals 5) at 1 second = 9 at 1.8 seconds.

and two use our 2 consecutive random huge examples:

180130/9 (or 36026 x 5/9) at 1 second = 36026 at 1.8 seconds

180135/9 (or 36027 x 5/9) at 1 second.....
....which equals 20015 as shown earlier) = 36027 at 1.8 seconds

Does that help at all? Smile

Dub Studio · Joined: 20 Jun 2006 Posts: 51 Location: Bristol

wow, yeah makes sense all of a sudden. Cheers Smile

Fractions won't work mainly because tone generators seldom work like that, but also because I think there is a finite amount of decimal places you can use when there are a finite amount of samples per second to work with. I guess its a quantum problem!?

Thanks for your help! I will pass the info on.

H

Steve E. · Posted: Wed Jul 09, 2008 6:23 am Post subject:

To use your example, by the way....

If you wanted to get 50 cycles at 1.8 seconds you would do

50 x 5/9 = (50 x 5) divided by 9 = 250/9 = 27.777777777 hz.

Maybe do an experiment and see if 27.77hz or 27.78 hz is close enough that you don't get a pop.

Dub Studio · Joined: 20 Jun 2006 Posts: 51 Location: Bristol

Ye I find with loops there is almost always some sort of sonic artefact on the crossover point, but hopefully the more accurate the better. Will let you know how it goes.

Steve E. · Posted: Wed Jul 09, 2008 6:37 am Post subject:

There's also a non-math solution. I suspect these pops only enter the situation in the digital realm by chopping the samples at exactly 1.8 seconds. If you make your source loops a little longer than 1.8 seconds, won't it sort of even out, physically when you cut the grooves? I don't know because I've never tried this.

Alternately, lap the ends in the digital realm. Make the sample, say, 2 seconds long, then fade the ends over each other over maybe 2/10 of a second so the loop is 1.8 seconds. At worse you'll get a little phasing at the overlap point. I often do stuff like that when I need to create audio loops for flash animation. I like using Steinberg Wavelab for that sort of stuff. (I have version 5.0 from a million years ago).

Dub Studio · Joined: 20 Jun 2006 Posts: 51 Location: Bristol

Yeh the thing is, the method I use for cutting loops is to play the digital loop over and over, and then drop the needle down for one revolution at any point in the cycle. That way you know that you will capture exactly one loop no matter when you drop the needle. This basically means the sample has to be able to loop continuously without a pop. Inevitably there is some sort of overlapping point on the vinyl, but it can't overlap on the digital loop.

Steve E. · Posted: Wed Jul 09, 2008 4:27 pm Post subject:

Right. I have had to do this, too. Here's how I do it:

Make a sample that's 2 seconds long. (for example)

In the soundfile editing progam of your choice, cut the file more or less in half at a zero point (the zero point isn't even really necessary, actually). These are part A and part B.

NOW....put the second half (B) first and the first half (A) second.

Bring the two halves together so that they overlap by precisely .2 seconds. Put a lap dissolve or fade between the two halves AT THE OVERLAP so that B fades to 0 and A fades to 100% during that .2 seconds. Output your mix of the 1.8 seconds file.

You will now have a loop that is exactly 1.8 seconds long with no pop. It may have a slight phasing issue at the overlap part, but no pop. If you shorten the fade without changing the distance of the files, it may be less noticable. And the first sample of part B directly follows the last sample of part A, because that is where you split the files.

Try it! It should work.