ETM Learn · 05 / 08

EQ: magnitude, phase and time

Change a frequency region and connect the result to phase, group delay and impulse response.

01 · Understand

An equalizer changes the balance between frequency regions. Start with an audible purpose: reduce a region that is too prominent or raise one you want to emphasize. Then choose where, how broadly and how much to change.

In a bell filter, Frequency selects the centre, Gain sets boost or cut and controls width: higher Q generally means narrower. Other filter types can use controls differently. The magnitude plot shows boost or cut at each frequency. Later, phase and impulse response show more about timing; an impulse is one sample of value 1 surrounded by zeros.

Terms within reach

Dotted terms offer a definition on hover, keyboard focus or tap. Click to keep it open; Escape or an outside click closes it. All definitions are also available here.

Q

A parameter controlling response width or resonance depending on filter type. In a typical bell filter, higher Q gives a narrower region.

FIR

A finite impulse response filter. A symmetric design can provide linear phase, but not every FIR is symmetric.

all-pass

A filter preserving response magnitude at each frequency while changing phase. Mixing it with dry can change the sum's spectrum.

02 · Play & discover

Choose Bell at 1 kHz and +6 dB. Change frequency, gain and width separately. Once you can explain magnitude, select “Show all controls & plots” to compare phase and time, then try all-pass and mixing with dry.

Predict → change one control → compare → explain the result.

Experiment 01

Start with the bell and identify the boosted region. After revealing all controls, try all-pass: output magnitude stays at 1 but phase changes. Mixing with dry can create boosts and notches.

Filter Bench

idle
Magnitude - dB vs log frequency
Phase - degrees, unwrapped
Group delay - ms · local phase slope
Impulse & step response - time domain
this filter reference / 0 line summed with dry input
vol
audio starts on first click
Peak group delay-
at-
Phase at f₀-
Max phase swing-
Ringing decay-
∫ group delay-

Experiment 02

This compares a minimum-phase IIR and a linear-phase with different magnitudes. It is not a phase-only comparison. The FIR is displayed with delay removed; real-time processing cannot respond to an input before it arrives.

Minimum phase vs linear phase - where the energy sits in time

-
minimum phase linear phase, time-aligned by the DAW the transient
Latency-
Pre-ring at −40 dB-
Min-phase energy in 1 ms-
Transition width-

Check your understanding

Can an all-pass change the result when mixed with dry?

03 · Go deeper

Group delay is in samples for radians/sample; divide by fs for seconds. It is a local phase slope, not a universal transient arrival time. Negative local values do not allow prediction of unknown input.

A symmetric N-tap FIR has group delay (N−1)/2 samples away from phase discontinuities at response zeros. It can ring on either side of its centre; after delay compensation, the early part is called pre-ringing. Magnitude shaping still alters transients. To isolate phase in listening, match magnitude, delay and level. A minimum-phase impulse’s largest sample need not be first.

The mathematics in words

Group delay

τg(ω) = −dφ/dω

Measure how quickly phase φ changes with angular frequency ω, then reverse the sign. With frequency in radians per second the result is seconds. This can vary with frequency; it does not promise a single delay for the entire signal.

Symmetric FIR delay in samples

delay = (N−1)/2

Subtract 1 from tap count N and divide by 2. Divide by sample rate to obtain seconds. Example: 1025 taps give 512 samples, about 10.67 ms at 48 kHz. Phase is undefined at response zeros, where this definition does not apply.

Filter plus dry signal

H(f)+1

H(f) describes the filter's magnitude and phase at frequency f; 1 represents unity-gain dry. Add complex values, accounting for phase direction. Equal magnitudes can reinforce or cancel depending on phase. Each path has gain 1 here.

Numerical challenge: try before revealing the solution

01 · A symmetric FIR has 1025 taps and fs = 48 kHz. What is its group delay away from response zeros?

Hint

Compute (N−1)/2 samples, then divide by fs and convert to ms.

Solution & assumptions

512 / 48000 × 1000 = 10.6667 ms. Host buffering and other processing can add latency. Two such filters in series contribute 1024 samples (21.3333 ms) in total; latency compensation shifts the reference but does not remove the filter’s time spread.

02 · At one frequency an all-pass has phase −120° and magnitude 1. Add its output to dry, both at unity gain. What is the sum’s magnitude relative to dry alone?

Hint

Use |1 + e^(−j120°)|, or 2|cos(120°/2)|.

Solution & assumptions

2|cos 60°| = 1, hence 0 dB relative to dry. At 0° it would be 2 (+6.02 dB); at 180° it would be zero. This assumes summation with unity gains; a normalized 50/50 blend has another factor of 0.5.

A causal stable rational digital filter has a causal stable inverse when its poles and zeros are strictly inside the unit circle. Zeros on the circle are a boundary case of this strict definition. All-pass zeros are reciprocal conjugates of its poles: stable poles inside correspond to zeros outside. Analog all-pass circuits also exist.

Further reading: W3C: Audio EQ Cookbook