5.2 FULL BASS, MID FREQUENCY AND FULL TREBLE

Multiplying this � by 1.65 boosts the gain of the full treble curve by approximately 2.5dB to reflect the original analog frequency response [fig 7]

Ba = [th*tl*a*1.65, th, (1 - a)]; Aa = [th*tl, (th + tl), 1];

5.2 FULL BASS, MID FREQUENCY AND FULL TREBLE

Full Bass

Midpoint scoop @ 1kHz

2.5dB difference between full bass and full treble

-15dB range

Mid Frequency

Full Treble

Figure 9. Tone Control Full-Bass, Full-Treble and Mid Frequency Response

It makes sense that these values alphas α = [0:0.1:1-0.1788] make for a better approximation of the original frequency response. As we can see, the original frequency response [fig 7] has only nine response curves, whereas evaluating between 0 – 1 yield eleven responses.

Now that the full treble, full bass and mid frequency are in the right place, we try plotting our new range with equally spaced intermediate values: -

alphas = [0:0.1:1-0.1788];

We obtain [fig 10] which is compared below with [fig 11]. The first disparity we observe pertains to the response belonging to α = 0.1. We observe that, in our approximation, the minimum value of the α = 0.1 curve sits approximately 16.5dB below the maximum value of the full bass curve, whereas in the original response [fig 11] the corresponding α = 0.1 response sits 14dB below the maximum value of the full bass. Comparing the two images side-by-side, it seems that the difference between the max values of α = 0.0 and α = 0.1 appears considerable when compared to the spacing of the other responses. We hypothesise that the responses detailed in [fig 11] are not as equally spaced as we originally assumed. We try experimentally increasing α = 0.1 to see if this brings the response closer to the original.

Figure 10. Analog Frequency Response (Difference in dB Between Full Bass and α = 0.1)

Figure 11. Big Muff Pi Tone Control Frequency Response (Difference Between Full Bass

and α = 0.1) [4]

We see in [fig 12] that α = 0.1+0.49 takes us a lot closer to the original response of [fig 11] but we also observe that now the responses α = 0.1+0.49 and α = 0.2 are closer than they appear in the original response.

Figure 12. Analog Frequency Response When α = 0.1 becomes α = 0.1 + 0.49

By looking at the original response we can assume what we want to do here is achieve a value for α = 0.2 that sits equidistant between α = 0.1 + 0.49 and α = 0.3. To achieve this, we take

the mean W2.+45* 8 2.7X= 0.2245, so we replace α = 0.2 with α = 0.2 + 0.0245 and run the

test again.