This a very useful theory in the application of parametric equalizer in audio recording, mixing and mastering. The bandwidth (Δf) is given by a relationship:
Bandwidth =Center Frequency / Q
Definition of Bandwidth
In addition, the bandwidth is also defined as:
Bandwidth = f2 – f1
Where f1 and f2 are -3dB points in the peak amplitude for the center frequency (f0). Supposing a frequency boost using a parametric equalizer can be graphically depicted:
bandwidth q, and the 3dB points
-3dB points (also known as “half power point”) means that the energy or loudness level drops by half. Supposing you will boost the center frequency by 12dB. The -3dB point (point at which the loudness level drops by half) is 12/2= 6dB.
When you cut a frequency level, say for example cutting -8dB on a certain frequency band, the -3dB point is -8dB/2 or -4dB.
In audio mixing, two values are needed to define as EQ setting (like what most EQ suggestions you find in audio recording website). One is the center frequency and the other is the Q. This is because most parametric equalizers need that setting (EQ and Q) to define a frequency adjustment. You might notice that parametric equalizers in Adobe Audition or other DAW do have the same requirements.
The Q is also a measure of how wide or narrow is the frequency adjustment. Large Q means wide frequency adjustment and small Q means small frequency adjustment.
Computing f given Q and the center frequency
It would be interesting to compute f1 and f2 provided that you know the Q and the center frequency.
Given:
f2 = yf1 where y could be any positive real number
And also f0 = √(f2 * f1)
Source: http://www.rane.com/note170.html
Using all the above equations, the y can be derived in terms of Q:
Solving for y in terms of Q
Take note that y can have two values due to the ±. if you use the – sign, then:
f1 = yf2
And if you use the + sign then:
f2 = yf1
The important is to remember that f2 > f1
Approach to solving f1 and f2, given Q and the center frequency:
Step1. Solve y from the above equation by substituting the given value of Q.
Step2. Once the y is solved (two possible y values), substitute y value to
f2 = yf1 for the use of + sign or f1 = yf2 for the use of – sign.
Step3: Now since f0 = √(f2 * f1), then squaring both sides:
(f0)2=f2 * f1
Now supposing you use the + sign in the calculation of y then:
f2 = yf1
Expressing the entire equation in terms of f1:
(f0)2=(yf1) * f1
Step4: Rearranging the above equation to express f1 in terms of f0
(f1)2=(f0)2/y
Therefore: f1 = √((f0)2/y)
If you use the negative sign then you will use the reverse: f1 = yf2 to solve for f2. And now you have solved f1 and f2 given the center frequency and Q.
Applications- Prevent Overlapping of Frequency Range during Boosting in Mixing
It would be exciting to apply what you have just learned in the previous section. In the above equations, it is possible to compute f1 and f2 provided you know the center frequency (f0) and Q:
In audio mixing, you will be using a parametric equalizer to shape the sound of your mix. One important considerations is to prevent overlapping adjustments in your EQ settings. For example, say you are mixing bass and kick drum. And you want the bass guitar to occupy the lower levels in the bass frequency spectrum while the kick drum would occupy the upper bass region. Then you decide to apply the following EQ settings:
Bass Guitar
Boost 50Hz, +3dB Q=1.0
Kick drum:
Boost 100Hz, +3dB Q=1.0
Question: Does the boost settings overlapped? Overlapping is undesirable as it will artificially boost certain frequency range that you are not intending to boost. You can analyze whether the EQ adjustments are overlapping by implementing the theory illustrated previously.
Let’s compute the f1 and f2 of the bass guitar parametric EQ adjustments:
Let’s compute Y first:
y= 0.38
y= 2.62
There are two values but the bigger value (2.62) is the result of the + sign. Let’s use that.
The compute f1 given the center frequency f0 and Y:
f1 = √((f0)2/y)
Solving f1 (use your calculator): 30.9Hz
Then f2 is equal to:
f2 = yf1
f2= (2.62)*(30.9)= 80.9Hz
Now for the kick drum EQ adjustment, solving for Y:
Y= 0.38
Y=2.62
It is the same because the Q is still set to 1.0 for the kick drum. Let’s still use the bigger value that is 2.62. Computing for f1:
f1= 61.78Hz
Now, its obvious that the boost settings overlapped because the lower value of kick drum adjustment that is 61.78Hz is within the f1 and f2 boundary of bass guitar adjustments (30.9Hz to 80.9Hz).
What is the solution in case of overlapping? You might to assign a tighter Q such as 2.0 instead of 1. If you use Q=2.0, then y=1.645 (using the bigger value). Then following are the f1 and f2 of the bass guitar and kick drums:
Bass guitar: f1=39Hz, f2=64Hz
Kick drums: f1=78Hz, f2=128Hz
See? There is no more overlapping since the bass f1 frequency level (78Hz) is now beyond bass guitar f2 (64Hz). Of course, you don’t need to this calculation everytime you are mixing. It is because modern parametric equalizers offer visuals, plots and graphs in addition to the settings so that you can directly observe if overlapping occurs.
Overlapped
Without overlapping, you can be certain that there are no artificial boost on unintended frequency range. This is commonly caused by wide Q adjustment on your parametric equalizer.
Content last updated on August 22, 2012