I like to know how the EQ works in this signal path. Let’s say there’s 8 db gain in convolution filter for 200HZ and then another 2 db gain in parametric EQ. Is the parametric EQ works on top of the convolution (i.e. 8db x 2db=16db) or simply add 2db to the convolution (i.e. 8db + 2 db=10db)?

Adding a gain expressed in dB terms is the same as multiplying a gain expressed in linear terms.

So, for example, in analogue electronics terms, a gain of 6dB represents a doubling of voltage (or 4 times the linear power expressed in Watts).

If you have two gain elements each of 6dB then each one multiples the voltage by 2 so the net gain in voltage terms is 2x2=4 which, equates to a gain in linear power terms of 16 (assuming a resistive load).

In the digital audio domain, a 6dB gain equates to multiplying the PCM value by 2 so two 6dB elements results in the PCM value being multiplied by 4 which equates to the same quadrupling of voltage (or current) out of the DAC which again equates to a 16 fold increase in linear power.

I forgot to mention that the 6dB equates to doubling of the voltage is an approximation.

The actual dB value is calculated using the expression 10 x LOG10(Power/ReferencePower) which, for resistive loads, can be mathematically proven to be equivalent to 20 x LOG10(Voltage/ReferenceVoltage) or 20 x LOG10 (Current/ReferenceCurrent)…

So, when you double the voltage, you get a dB increase of 20 x LOG10(2) which evaluates to just slightly more than 6.02.