Electronic – Square wave / Sine wave is more audible

The Fourier transform is linear, which means that if you sum two signals, then also their spectrum will be the sum.

$$ \mathscr{F} [x(t) + y(t)] = \mathscr{F}[x(t)] + \mathscr{F}[y(t)] $$

If you consider the power, you can have two cases:

1. The signals have different frequency
   Then their power just sums up, because: $$ P_{tot} = P_1 cos (\omega _1 t + \phi _1) + P_2 cos (\omega _2 t + \phi _2) $$

2. The signals have the same frequency
   Then if they are in phase you have: $$ P_{tot} = P_1 cos (\omega _0 t + \phi _0) + P_2 cos (\omega _0 t + \phi _0) = (P_1 + P_2) \cdot cos(\omega _0 t + \phi_0) $$ If they're out of phase, the resulting power will be lower and precisely: $$ P_{tot} = P_1 cos (\omega _0 t + \phi _1) + P_2 cos (\omega _0 t + \phi _2) $$ and setting arbitrarily \$\phi_1 = 0\$ we obtain: $$ P_{tot} = P_1 cos (\omega _0 t) + P_2 cos (\omega _0 t + \Delta \phi) $$ which for \$\Delta \phi = \pi\$ gives the subtraction of the signals.

The Fourier series for a triangle wave is Odd harmonics shift in phase by 90 deg from a square wave with 12 dB/ odd octave more attenuation.

and the square wave

Both have only odd harmonics but differ in the slope of the peak value for each harmonic. However triangle harmonics are much smaller. As n increases, the amplitude reduces by 1/n², whereas a square wave reduces by 1/n. Triangle waves harmonics also alternate phase (+/-sin) with increasing n.

To simplify my explanation, the capacitive load on a 50Ω gen. gives a frequency response or Transfer Function of;

, which you know gives the exponential time response to a medium square wave where f is near 1/RC .

But for high frequency where RCs>>1 so the transfer fcn reduces to an integrator Hc(s)=1/RCs transfer function.

Applying this filter to the Fourier series of the square wave, its 1/n harmonic attenuation becomes 1/n² slope on harmonics of the triangle wave. Similarily the triangle wave source when filtered, its harmonic attenuation slope of 1/n² becomes 1/n³.
On a scope all you would see is a sine wave, but on a spectrum analyzer log scale you would see the 1/n³ slope of all harmonics ( i.e. 3rd order slope )

side comments added I believe there is value in the time you spend in the lab to match theory with practise. When it does not match, look at for a better equivalent circuit then verify your assumptions. If you have Java you can play with this programmable signal generator . Have fun ! Spend more time in the lab validating what you learn and bring the lab to your desktop then expand your horizons.

http://www.falstad.com/fourier/ <

use the mouse pointer and left or right click...drag and adjust

• Change the phase of the Fourier series and see the effect on a triangle wave
• Add a spurious resonance in the amplitude of one harmonics, see the waveform
• add a glitch, change the shape anyway you want. (arbitrary waveform)
• add a LPF filter, change the frequency, slide the number of terms in the spectrum, see effect

When Integrate a step input and Dump, you get a Sawtooth waveform Here the mouse is hovering over the fundamental of the Fourier spectrum and the fundamental sinewave amplitude and phase are shown in yellow. Meanwhile I boosted a harmonic to similate a resonance on the sawtooth.

The combinations are "only limited by your imagination"