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I know the importance of the behaviour of a circuit when a sinusoidal signal is applied: by using Fourier Transform, I can see a generic signal as a (continuous) sum of sinusoidal signals with different frequencies. The question is: By applying Laplace Transform to a circuit, I can find the transient (if circuit is stable) response to this sum of sinusoidal signals. If the generic signal is an audio signal, does the transient influence what I hear? I mean, when I listen to a song, I should also hear a "transient" audio (which should be annoying), but I don't listen to this transient. Then: How is transient response dealt in audio circuits?

Example: if vin is a generic audio signal and by using Fourier I get vin= Asin(wt) + A'sin(w't) + A''sin(w''t) + ... and if the circuit is a simple amplifier with a gain equal to k, then vout = kvin In this case I have the same original audio signal, just amplified by a factor k. But we know that every input sinusoidal signal gives raise to transient (that will extinguish if the circuit is stable), then the output is vout = kvin + transient and thus the output audio signal is no more equal to the input audio signal (and i should hear some "modification" to the song I'm listening to...)

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  • \$\begingroup\$ Define a transient that you can hear? THD? \$\endgroup\$ – Sparky256 May 22 '18 at 23:30
  • \$\begingroup\$ To add on Sparky, an audio signal is the sum of any number of sinusoidal signal with different amplitudes. At any moment, you will get a single set of sines. Therefore, audio is by definition always a transient signal that never gets a chance to become steady state. Your system (for the sake of argument let's call it an amplifier) will always try to catch up, but will never have a chance to converge. It will always try to adapt to the input signal thus giving you the output modulation. \$\endgroup\$ – Simon Marcoux May 22 '18 at 23:51
  • \$\begingroup\$ I mean, when a sinusoidal input is given to a circuit, there is a transient that will extinguish if the circuit is stable (and I can campute it by using Laplace), but the "real" output signal is a sinusoidal signal as well (and I can compute it by just using phasors method). Now, let's assume that the generic input audio signal is given (Fourier) by the sum of some sinusoidal signals; then the "real" audio signal output should be the sum of (permanent) sinusoidal signals. The transient corresponding to each sinusoidal input should be some kind of noise, isn't it? Thanks \$\endgroup\$ – Stefanino May 22 '18 at 23:53
  • \$\begingroup\$ Example: if vin is a generic audio signal and by using Fourier I get vin= Asin(wt) + A'sin(w't) + A''sin(w''t) + ... and if the circuit is a simple amplifier with a gain equal to k, then vout = kvin In this case I have the same original audio signal, just amplified by a factor k. But we know that every input sinusoidal signal gives raise to transient (that will extinguish if circuit is stable), then the output is vout = kvin + transient \$\endgroup\$ – Stefanino May 23 '18 at 0:03
  • \$\begingroup\$ in that context your transient will be several magnitude lower than the kvin and will be impossible to actually hear. You will simply have a constant sound on the other side. Depending on the frequencies that are pumped in, there is a lot of chance that the output will not be pleasant (think of a piano that you smash randomly with your fists). This is not the transient sound tho, it is merely a bad choice of frequencies and harmonics resulting in a poor sounding signal. \$\endgroup\$ – Simon Marcoux May 23 '18 at 0:08
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This is why a step/impulse input is used: it determines the dynamics of the system to a (theoretical) infinite sum of sines. The response can be then extrapolated to any input. Keep in mind that you have to separate the theory from the practise, because you'll also have the bandwidth of the system to help, meaning for (e.g.) an audio amplifier, that sudden piano will be well whithin the bandwidth, and so will be most of the signal, so whatever peak response to step input the system has, it will be much less for the practical case, but it will be there.

Here's a minor example with two 4th order filters: a Chebyshev with fc=20kHz, 1dB ripple, frequency normalized to -3dB, and a similar Bessel (no ripple though...). This is the step response:

step

The Bessel has very little overshoot, while the Chebyshev has lots and more group delay. And here's an actual audio sample:

audio

If you look, for example, from ~0.4ms to ~0.6ms, the input (black) is one shape, Bessel follows closely the same shape, while Chebyshev distorts it. Also look at the immediately following double peak. The audio sample (.wav file, hence the sharpness) is a kick drum + snare + cymbal, so enough bandwidth.

Conclusion: you're overthinking this a bit. There are more factors to account for, but the most dominant is the linearity of the phase, which is what brings most distortion. I say mostly, because, if it's an analog filter, you have the sensititvity of the elements, opamp, the supply, etc, while for digital you have the dynamic range, while for both cases the noise (internal or external) is inherent. So, the system will repond to whatever input you have based on its own dynamics: if it distorts less, you'll hear less distortion, or none, and vice-versa.

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