Transfer Expansion Curves Between Multibands ???

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mark4man

mark4man

MoonMix Studios
Question:

I want to duplicate curves from one multiband expander to another, for the sake of comparison...& was wondering if someone here could help me w/ the spec transfer.

One of WaveLab's on-board mastering plug-ins is a Spectral Designs (now Cube-Tec) multiband compressor called MultiCompressor (now called MultiComp.)

Every curve seems to be of the expansion type; & the plot display shows negative voltage compression characteristics (where the input & output scales run right to left & top to bottom, moving from FS down [as opposed to a typical (+ voltage) input/output plot.]

This means that curves *above* a linear diagonal plot are 'expansion' curves (& those below are compression.)

[of course...you guys being mastering engineers, you would know that.]

There are no ratio & threshold settings on the MultiComp plug-in. Instead, at the hard knee of the curve (which Spectral Designs calls the 'breakpoint'), 3 specs are displayed on mouseover...input, output & gain. The input & output are displayed as negative values.

On one of my favorite curves, for the LF band, reads as follows:

input -16.6dB
output -1.2dB
gain 15.4dB.

Since I'm not familiar w/ expansion, these numbers are throwing me...but I want to be able to plug them into another mastering multiband...which has expansion capability...but which has the typical ratio & threshold controls.

Could someone who knows how to do the math please tell me how I can use the above expansion type numbers to figure out my ratio & threshold for the 2nd multiband, in order to perform a comparison?

Thanks,

mark4man
 
Why don't you try it and report back (?)

I will be in eager and breathless anticipation of your findings. :D

.
 
chessrock...

just one question...why even bother w/ a response like that? Are you trying to make me feel bad or something...having a rough day...what?

if I could do the math do you think I would ask?
 
mark4man said:
Question:On one of my favorite curves, for the LF band, reads as follows:

input -16.6dB
output -1.2dB
gain 15.4dB.

Since I'm not familiar w/ expansion, these numbers are throwing me...but I want to be able to plug them into another mastering multiband...which has expansion capability...but which has the typical ratio & threshold controls.
Click to expand...
The easy one is the threshold; that will be the input reading at the joint (or breakpoint) of the hard knee.

The compression ratio could be calculated by calculating the difference between the input value and the threshold, and then dividing that by the difference between the output value and the threshold .

For example: If you have a an input value of -4dBFS and a threshold of -8dBFS, that's a difference of 4dB before compression. If the output after compression is -6dBFS, the difference between that and the threshold is 2dB. So, if you take 4/2, you wind up with 2, therefore the compression ratio is 2:1.

When expanding, if the threshold were the same -8dBFS, but the output and input were reversed; i.e. if the input were expanded by 2dB from -6dBFS to -4dBFS at the output, the same formula yields 2/4, or. 0.5. This is a compression ratio of 0.5:1, or in more conventional terms, a ratio of 1:2 (signifying an expansion).

HTH,

G.
 
mark4man said:
input -16.6dB
output -1.2dB
gain 15.4dB.
Click to expand...


Mark ... I'm not trying to be an ass here.

And I'm not going to spew a bunch of Southside babble, to try and make myself look smart.

I'm just trying to help. Seriously, Mark. Look at those numbers again. Stare at those 3 values for a while, until you see a pattern.

Do you find it interesting, at all, that your input plus the gain is equal to your output ? ... :D Think about it for a while and get back with us.

.
 
chessrock said:
Mark ... I'm not trying to be an ass here.

And I'm not going to spew a bunch of Southside babble, to try and make myself look smart.

I'm just trying to help. Seriously, Mark. Look at those numbers again. Stare at those 3 values for a while, until you see a pattern.

Do you find it interesting, at all, that your input plus the gain is equal to your output ? ... :D Think about it for a while and get back with us.

.
Click to expand...
Unfortunately, Keith...or should I say Dr. House without the Vicodin and without the genius...that little bit of first grade pattern matching will not give him the answer to his question as there is no way to derive threshold from those numbers, and without knowing threshold one cannot determine ratio. He knew from the outset that those numbers alone were insufficient, which is why he asked the question. Which puts him already a step ahead of you.

I simply gave him the answer. No babble, no looking smart. Just the answer to his question. If I were you, I'd worry less about other people looking smart and more about you making yourself look dumb.

G.
 
SouthSIDE Glen said:
... there is no way to derive threshold from those numbers ...
Click to expand...

There's no way to derive anything from those numbers, Professor.

That's the whole point. His question was "Can someone ... please tell me how I can use the above expansion type numbers to figure out my ratio & threshold .."

And the answer is no. You can't. They're completely useless. But you try to answer the question anyway, as usual, just so you can throw a bunch of words out and look smart. :D Why don't you just answer the guy's question and say: "Insufficient data ... does not compute" ?

Oh ... I know. Because then you can't impress anyone with your massive audio intellect!

.
 
chessrock said:
And the answer is no. You can't. They're completely useless. But you try to answer the question anyway, as usual, just so you can throw a bunch of words out and look smart. :D Why don't you just answer the guy's question and say: "Insufficient data ... does not compute" ?
Click to expand...
Because, Keith, I could and did answer his question accurately and correctly. it CAN be done with the info available to him.

The very first sentence I gave told him where to find the threshold. It's at the bend of the hard knee. Most dynamic processor plugs (which is what those plugs he's describing are usually called) will indicate *somewhere* on the display the gain at the cursor position. It sounds like if he puts his cursor on the knee joint, that the "Input gain" value will tell him what the threshold is set for. If not, if that means somethig else, then all he has to do is trace from that knee joint over to the input scale to see what the knee - and therefore the threshold value - is.

That was the first piece of information he was missing (but is available to him nonetheless.) The second, which he directly asked for and I directly answered, was for the formula for calculating ratio from those numbers, which I then gave and explained to him.

If that all sails above your head, that's not my problem. If you need to make yourself feel superior when you don't really know the answer by belittling the questioner, that's not my problem either.

But when you get on my case for giving the right answer, and furthermore NOT giving the WRONG answer - and getting the threshold or ratio by noticing the relationship between the input, output and gain numbers, which is what you originally suggested, is a wrong answer, BTW; and "insufficient data" is a wrong answer also, bonehead...you're 0 for 2 in this thread - then it is my problem, and I'm going to call your smarmy salesperson's ass on it okay? And if you don't like that, you're welcome to leave.

G.
 
Actually I do have a correction/refinement to make in my explanation. The knee in the curve is only the threshold if the line below that knee is a 1:1 (45°) line. If the line deviates from 1:1 right from the bottom - the lower left - then the threshold is -inf; i.e. the "knee" is at the very start of the line.

In such a case, one would have to determine the compression/expansion ratio for that initial slope by calculating the slope of the line; i.e. rise over run. This would mean taking the y value at a given point and dividing it by the x value at that same point. This will give you the first number in the ratio (with the second number being 1, of course.)

Then if there is a second "joint" in the line, the new ratio for the line above that joint would have to be recalculated the same way.

G.
 
You could always treat the two compressors as black boxes and twiddle knobs until the output of the second matches the output of the first, no?
 
Glenn...

Thanks a million...very much appreciated.

I'm going to put a screen shot of the characteristic display up on my test server...[not to have you figure it out for me (as I'm good w/ your explaination), but for you to just have a look; & see if that lower segment of the I/O line is a pure diagonal.] But then...I'm not sure that's the end all & be all because...am I understanding you correctly if I'm thinking that you went on to suggest that it must eminate directly from the bottom corner? (in order to have the breakpoint be the true threshold?) The gain line can't be a 45 with any number as a threshold other than 1:1 (& then...there wouldn't be expansion/compression...what would be the point?) All curves of this type eminate from the corner, correct? If it eminated from 'off the scale', it wouldn't be displaying the true compression/expansion characteristic, right?

In any event...I'll have it up there in about 20 minutes; & come back w/ the link.

Thanks very much,

mark4man
 
mark4man said:
I'm going to put a screen shot of the characteristic display up on my test server..
Click to expand...
Cool. I was thinking about popping up some example screen shots myself, but right from your horse's mouth (so to speak) is even better
mark4man said:
But then...I'm not sure that's the end all & be all because...am I understanding you correctly if I'm thinking that you went on to suggest that it must eminate directly from the bottom corner? (in order to have the breakpoint be the true threshold?) The gain line can't be a 45 with any number as a threshold other than 1:1 (& then...there wouldn't be expansion/compression...what would be the point?) All curves of this type eminate from the corner, correct?
Click to expand...
Yep, all curves eminate from the lower left corner, as that indicates -inf input (zero signal). And a zero input signal is always going to be a zero output, no matter how much expansion or gain you dial in ;). So that means x,y of -inf,-inf as a strting point the whole time.

Now maybe, I have not seen this myself, but maybe there may be a DP plug like that that has an input gain control on it, which could theoretically slide the curve up or down the input scale. In which case, the curve might start above -inf on the input scale. That's easy to adjust for; you'd just change the formula from f = x/y to f = (x/y)+g, where g is the input gain in dB.

And, yes, 1:1 is a 45° angle and no other ratio is. But remember, every "joint" in the curve is a seperate threshold setting, basically; with a DP plug, you actually can have multiple thresholds. So it's possible, for example, to have a line that climbs out of the cellar at 1:2 expansion, say, and then half way up the curve takes a dogleg to the right where the ratio changes to 1:1. There you'll have a 45° line representing 1:1, but it will not be on the "center line" of the whole graph because of the expansion at the lower input volumes.
mark4man said:
In any event...I'll have it up there in about 20 minutes; & come back w/ the link.
Click to expand...
Coolness.

.
 
wait a minute...

looking again at the display...the input value (-16.6dB) & output value (-1.2dB) are unified at the node (breakpoint.) In other words...-16.6dB input is at the exact same location on the display as -1.2dB output. Yet your saying that this breakpoint IS the threshold. I can't see how your calculation works in this scenario.

Oh man...my brain is burnin' now (~!@!#$!%^!&*(!)_+]

mark4man

BTW - I put a new screenshot up...this one w/ those values. But the difference between input & threshold is zero...so the math still wont work?

no wait...you DID say the input reading at the node...so the threshold is -16.6dB. But the difference between input & threshold is zero...so the math still wont work (?)
 
OK, what makes this a bit trickier is that the scales are not linear, so I'm not sure you could actually use the numerical values on the dB scales. A 45° line would still be 1:1, but the slope of a 2:1 will be different on a logarithmic scale than they would be on a linear scale. Plus I'm not positive if those scales are exacty strictly logarithmic or not.

This is kind of a cheat, and maybe someone here knows high school trig better than I, but a way around that would be to make your own linear scale on that graph, with the bottom left being 0, 0, and the top right being 10, 10 or 20, 20 or something like that. Then you could ID "threshold A" as 0,0, and "threshold B" as being 1,9 (or whatever it turns out to be). Then you know that "Slope 1" has a slope of 1/9. That would equal an upward compression (not really an expansion, an expansion would be a negative slope) ratio of 1:9.

Then you'd make a similar calculation for "Slope 2", except your starting coordinates are the coordinates for "threshold 2" (e.g. 1,9 instead of 0,0), and the ending coordinates are 10, 10, or whatever the top of your linear numbering scale is at the upper right. Assuming for example that the numbers are correct on that scale, the slope for "Slope 2" would then be 10-1, 10-9, or 9,1. Therefore "Slope 2" has a downward (standard) compression ratio of 9:1.

So, you have upward compression of 1:9 applied to input levels from -inf to -16.6dBFS, and from -16.6dBFS to 0dBFS you have downward compression of 9:1. Again, assuming a linear scale of 0-10, and assuming that the linear coordinates of "threshold B" actually were exactly 1,9 on that scale.

Make sense?

G.

UPDATE EDIT: As far as the first formula I gave before we got into slopes, that forumla wont work *at* the threshold level itself, because the threshold and the input level are the same. What you need to do is pick a point on the slope above the threshold, and use *those* in/out values minus the threshold value.

It really turns out to be the same thing as the other method, you're still calculating the slope using the threshold as the first point on the slope, but your doing it based upon relative dB values at some point along the line as the second point instead of the upper-right endpoint of the line as the second point.
 

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[posted prior to my reading your update...but still on the beam, I think]:

no...because all numbers here are in the negative; & these are all truly expansion curves. Look at the scales. The input scale runs 0 to -90, right to left. The output scale runs 0 to -90, bottom to top. Just like I mentioned in my first post, any breakpoint above a 1:1 diagonal plot on a reverse scale scale such as this represents expansion.

But I think the ratio equation may be simpler, here (???)...if we look at the characteristics (regardless of whether or not they're in reverse.) In looking at the curve in question...following the plot from the zero points on the scales...one could say that for every 16.6dB's of input, 1.2dB's of expansion occurs. That's the accepted premise, right? Ratio is always relative to *above* the threshold.

[I was also going to ask you if you flipped your original ratio math...since were talking expansion here...& since the threshold is normally relative to output, w/ compression...but that's another story for another day.]

But we could still say that the above premise is true, right...for every 16.6dB of input over the threshold we get 1.2dB of expansion or 15.4dB of output....meaning the ratio; being input over output; is 16.6 over 15.4, or 1:1.077. If a tight ratio such as this were used for compression...we would call it "limiting".

Am I right, or is it the trig thing?

[This also make sense in terms of a conventional expander plug-in, as well. On the unit I'm trying to transfer these values to (UA's Precision Multiband)...when it's switched to expansion, the ratio dial offers 1:1 to 1:4.]

???

mark4man
 
well...

This may prove to be a fruitless venture anyway.

I took the numbers from my favorite 4 band settings on the MultiComp...figured out my thresholds & ratios [based on my above logic...(if you could call it that)]...plugged them into the Precision Multiband; & A/B'd them on one of the tunes from my CD.

First of all...there's something going on w/ WaveLab's MultiComp that is literally indescribable...when I pull off the soft clip function, it pegs the master section meters @ +7.1 dB !!! Maybe there's some type of super gain staging built into the MultiComp's design, but however nice the Precision Multiband sounds w/ the MultiComp's settings plugged in…w/ its output set @ -1dB (to match that of the MultiComp), the meters don't even achieve FS.

So I pull the output gain back on the MultiComp 7.1dB, to try & at least do a sound for sound comparison...& it's not even close...the MultiComp sounds richer & deeper & smoother & brighter & everythinger.

So now I run the output up on the Precision Multiband by 7.1dB...& it's distorting all over creation.

So...my numbers are either incredibly wrong...or the spectral design specs on the MultiComp far exceed anything accomplished by Universal Audio...or something.

I should apply the MultiComp, w/ these settings, to an evenly EQ'd composition...& let you guys hear this thing. It sounds like it rivals anything ever done at Sterling Sound...in a little f_ck_'n software box. Unbelievable.

mark4man
 
I was sort of dissapointed when after getting WL6, they give you tha dinasauor of an MBC!! good luck w/ that; your way ahead of the herd in that your taking a analytical approach instead of just twisting knobs!! ( there a dearth of knobs on that thing!)

Any how , now that I made a terse , unhelpfull comment, I feel so smart!:p
Hey Glen , Remember , you don't like MBC"S anyway!!!

:D
:D:D
:D:D:D
 
Using the Multicompressor picture following the link above,
http://www.moonjams.com/artlinks.htm
Can anyone explain the relationship between the left side and the right side of that picture? Do the four separate lines under characteristic represent the four frequency bands on the left?
 
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