How do different octaves sound different in the presence of a low-pass filter in a synthesizer?

Question

Per my understanding, when playing a certain pitch, say 440 Hz (middle C), the VCOs will produce their corresponding waveforms (say sawtooth) at that frequency. Because of the characteristics of the waveform, some harmonics will be present, e.g sawtooth includes all integer harmonics of the fundamental frequency. These are then passed through a VCF and filtered according to the filter's parameters.

Now say the VCF is a low-pass one with a cutoff of 1000 Hz, then won't it produce the same sound regardless of the fundamental frequency? e.g what is filtered should be the same regardless of the input signal. If we go up on octave to 880 Hz, the only difference is that the odd harmonics of 440 Hz frequency are no longer present (but the even ones are all there because 880 = 2 x 440) but still anything below 1000 Hz should be equally filtered for both pitches, meaning these notes should sound more or less the same. But this is clearly not the case, otherwise the presence of the filter would defeat the purpose of having a full-size keyboard.

What am I missing?

Answer 1

To illustrate what happens when a waveform is passed through a low-pass filter, have a look at this:

The blue line represents the filter's gain vs. frequency. As you can see, it has a gain of 1 (0dB) for all components of the input signal below 440Hz, but it starts to attenuate components at or above that frequency. The higher the frequency, the greater the attentuation, which is typical of such a filter.

The pink bars represent the harmonics of some input waveform, with a fundamental frequency of 44Hz. The second harmonic will be at 88Hz, the third at 132Hz, and so on. This hypothetical input signal has five harmonics, all of equal amplitude, and because they all fall within the pass band of the filter, none of them are attenuated. The signal would emerge at the filter's output completely unaffected.

The orange bars represent harmonics of the same wave shape, but shifted up in frequency by a factor of ten, so that its fundamental frequency is 440Hz (which by the way is A, not C). This would place the second harmonic at 880Hz, the third at 1320Hz, and so on. Here you can see that some of the harmonics fall within the stop-band of the filter. This means that the fundamental harmonic will pass through at full amplitude, but subsequent harmonics will be attenuated, higher harmonics more than lower ones. This will alter the timbre of the tone.

The green bars represent the waveform shifted up in frequency even further, to a fundamental frequency of 4400Hz. As you can see, all harmonics, including the fundamental, will be severely attenuated, and each subsequent harmonic suffers significantly greater attenuation than the previous. It's clear, I hope, that the signal that emerges from the filter will bear little resemblance to the signal that entered, and its timbre will be heavily affected. It would probably sound a lot more like a pure sinusoidal tone, simply due to the fact that most of the higher harmonics have been significantly suppressed compared to the fundamental.

Answer 2

Out signal will not be the same because each next frequency has smaller amplitude. Remember that y axis is in dB, so change like 3 dB is a lot. So your octave doesn't have "the biggest" frequency. Both signals have different distribution of energy at each frequency.

The pitch is different, because lowest frequencies are different (and ones with the most energy). The timbre is different, because rest of spectrum is different.

So whole your musical tone is different. Still you have similarities between them and you hear them.

Answer 3

Now say the VCF is a low-pass one with a cutoff of 1000 Hz, then won't it produce the same sound regardless of the fundamental frequency? e.g what is filtered should be the same regardless of the input signal

Consider an ideal brick wall filter. Then for a fundamental frequency of 600 Hz (and all 1/k harmonics) the output will be 1 V @ 600 Hz, and for a 700 Hz fundamental the output will be 1 V @ 700 Hz (assuming 1 V). All the other harmonics would be filtered, since the next one for the 1st signal would be at 1.2 kHz, and for the 2nd, 1.4 kHz. Would you say these two are "the same, regardless of the fundamental frequency"?

It's possible you are mistaking in thinking that if the filter has a certain shape of the spectrum, everything at the output must have exactly the same shape. The shape of the filter only influences the input's already existent spectrum, because whatever is present at the input has its own identity. The filter only filters that.

So for the same brick wall filter, a pure white noise at the input would have a perfectly flat spectrum, DC to light. Passing through the filter it will mean that line will be multiplied by the brick wall spectrum (convolution in time domain is multiplication in frequency domain), leaving a brick wall output. If the noise was pink, at the output it would have the same -3 dB/octave slope, but hard-limited at 1 kHz, so a cut triangle. The same for any other spectrum. Conversely, if the input would be grounded then the output would not suddenly have a brick wall shape, it will have nothing, shaped by a brick wall.

In short: the filter alters the input spectrum by its own.