When and how can multiple EQ filters be merged into one?

As I understand it, it should be possible to merge similar filters, e.g. if I have two shelving filters with the same Q and frequency but different gains, I can just merge them into one by adding the gains, and it will have the same effect as two. Is this correct?

In what other cases is this possible?

• IIR of FIR? Series or parallel combinations? – hotpaw2 Sep 29 '16 at 17:25
• @hotpaw2 I don't know, I'm using vDSP_deq22() from Apple's Accelerate framework without much understanding to be honest. – mojuba Sep 29 '16 at 19:46
• vDSP_deq22 is a 2-pole IIR filter. You can combine them in series by using a vDSP_biquad or vDSP_biquadm. You can combine 2 in parallel by using a mixer or vector add. – hotpaw2 Sep 29 '16 at 20:20

This concept generalizes to any series cascade of linear filters. If I have a collection of linear filters with frequency responses $H_1(f), H_2(f), H_3(f), \ldots$ that are applied in series (i.e. an input signal is applied to $H_1$, its output is applied as the input to $H_2$, and so on), that series cascade could be replaced by a single filter whose overall frequency response is equal to $H(f) = H_1(f)H_2(f)H_3(f) \ldots$. This follows directly from the properties of LTI systems.

With that said, I'm not too familiar with some of the audio EQ terminology that you're using, so I'm not sure. If you have two filters with the same frequency response apart from a constant gain factor, like:

$$H_1(f) = k_1 S(f)$$

$$H_2(f) = k_2 S(f)$$

where $S(f)$ defines the shape of the frequency response, then the frequency response of the cascade of the two filters is:

$$H(f) = H_1(f) H_2(f) = k_1 k_2 S^2(f)$$

Note the squaring of $S(f)$; this is an important distinction. Applying two filters in series with the same frequency response shape (apart from a constant) will have a different effect than if you just created one similar filter with a different gain. Specifically, due to the squaring of $S(f)$, the passband will experience twice the ripple and the stopband will experience twice the attenuation than you would observe had you just used one of the two filters.