Why is my cutting equalizer increasing output level?

Question

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 applied

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

Answer 1

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.

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.

Answer 2

It's because it's a summed audio track. There are peaks and valleys now that have completely changed. think of three separate waveforms with totally different wavelengths. Now, you imagine the peaks and valleys getting summed together. Suppose that you remove the longest ( lowest ) wave. Now you will have parts that are quieter and parts that are louder because of the new sum.

Apply the EQ PRIOR to the final compressor/limiter, you'll appreciate it.

Answer 3

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.