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I'm using logic studio latest version

On one of my channel strips I'm having an ultrabeat drumkit with an overdrive and a compressor plugin. The track is only playing single snare hits. The channel level peak is at -5dB. So, I wanted to apply an equalizer to cut som frequencies. I usually start by cutting in the very high and low regions of the frequency spectrum. So apply the low cut at 48Hz, gain/slope at 24db/oct and a Q-value of 0.71. I can't hear much difference in my headphones, but the level meter is now showing peaks at 0.4dB. I didn't really expect a low cut to increase output level. Can somebody explain what's going on here?

Image shows output level along with equalizer before and after equalizer is appliedImage shows output level along with equalizer before and after equalizer is applied

Extra information: Adding a high cut as well, increases output level even more, up to 0.7dB.

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Does your equalizer have a makeup gain function? –  Warrior Bob May 23 '13 at 17:05
    
It has a master gain which applies gain on the whole frequency spectrum. Might not be the same thing? However, it is set to zero. –  Mattias Nordqvist May 23 '13 at 17:54
    
What about the "chain-q" (the lit up "link" icon)? –  horatio May 23 '13 at 19:28
    
I believe the chain-q is just a convenience tool, that changes q in relation to changes in gain and vice versa. I think it is supposed to keep the perceived level constant. –  Mattias Nordqvist May 26 '13 at 6:03
    
Could be relevant for this question: gearslutz.com/board/rap-hip-hop-engineering-production/… –  Eugene S May 28 '13 at 8:28

2 Answers 2

up vote 5 down vote accepted

Filtering a signal to remove certain parts of the spectrum on the face of it (and intuitively) should reduce the perceived sound level. This is what common-sense would tell us. However, when it comes to the reshaping of sound with filtering, lowering the sound level doesn't always equate to lowering the peak level. Yes, the perceived sound level may reduce but on many occasions the instantaneous peaks will increase. This does not mean your ears or meters are lying to you.

Take a very simple repetitive waveform like a square wave. If you filtered it to enhance just the fundamental frequency you'd end up with a sinewave (pure tone) that is 27.3% higher in amplitude. This is a peak level increase of about 2.1dB.

When you hear the squarewave you hear the fundamental frequency and all it's harmonics because that is, mathematically how a square wave is constructed. Here is a techy (but good) article that explains how a squarewave consists of a fundamental and harmonics.

If you scroll to the lower half of the page you can see how they've constructed a squarewave from sinewave harmonics. Also notice in the formula (near page top), the 4/Pi - this, if you worked it out on your calculator = 1.273 and is the 27.3% higher in amplitude bit I referred to earlier.

enter image description here

So, in the case of a composite signal of a snare drum, many individual parts of the spectrum may be playing together and various filters may cause the peaks to increase and coincide. Remember the peak value on the meter only has to occur once in a few seconds to give the indication that the resulting filtered signal is bigger.

EDIT Here is a link to a quote from SOS magazine. The quote is an answer to a question related to this subject: -

Fundamentally, the filtering process changes the shape of the waveform, so although there may be less total energy in the signal, the peak amplitude may well increase.

END OF EDIT

There may be other reasons why it appears bigger but I hopefully have demonstrated that there is an absolutely rigourous mathematical reason too.

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Your distillation is a little confusing to me: it is obvious that at certain points in the time series, the amplitude is indeed larger than the original waveform, but if you look at the graphs, the peak amplitude of the "composite wave" is never larger than the peak of the original waveform. Is this because of the time alignment of the waveforms in their illustrations? –  horatio May 24 '13 at 14:40
    
@Horatio - are you referring to the link? If so look at the graph that shows the sum of the 1st, 3rd, 5th, 7th and 9th harmonics - remember that the square wave will be 1V peak with more harmonics introduced - now look at the size of the lowest frequency sinewave - it's peak value is over 1Vpeak (1.273Vpeak). Now if you imagine have a decent squarewave and removing all the harmonics just to leave the fundamental sinwave, you are converting a 1V peak squarewave into a 1.273V peak sinewave and this has a higher amplitude by about 2dB. Does this make more sense? –  Andy aka May 24 '13 at 16:12
    
Any ideas how I can paste an image into my answer? –  Andy aka May 24 '13 at 16:24
    
In that graph the peak value of the original is higher (or equal to the square waveform). Yes you are dealing with a multiplier of 1.2-ish but the harmonics are additive with negative values. I do understand what you are saying, I was only asking (without running a simulation myself) if the fact that the square wave peak never actually is higher than the original sine is due to the all the harmonics used being aligned in time (i.e. is the graph a special case) –  horatio May 24 '13 at 16:27
    
Save the image to disk (or note the url), then edit the post, click the icon that looks sort of like a landscape painting. –  horatio May 24 '13 at 16:31

Depending on the order of processing, it is possible that your cut is resulting in the compressor behaving less aggressively. This in turn might result in the VU meter measuring the signal as stronger depending on how the compressor and VU meter are doing their processing.

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The Equalizer is placed last in the signal chain. Would your answer still apply? –  Mattias Nordqvist May 23 '13 at 7:35
    
Unfortunately no, this wouldn't work if the EQ is last in the chain. I have no idea why then, might just be a bad software implementation... –  AJ Henderson May 23 '13 at 16:53

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