The order of filtering and convolution

Question

Assume I have a source signal s(t) to pass through a system. I record the signal at system output place, which is denoted as d(t). The Green's function (impulse response) of this system is denoted as g(t). Thus, the output can be represented by a convolution as d(t) = s(t) * g(t).

If my target frequency band is between f1 and f2, so I should bandpass filter the output signal. I use F[.] to denote the bandpass filtering operator, I'll get

F[d(t)] = F[s(t) * g(t)]

A similar practice is that I let a filtered source signal to pass the system to get the output signal as d'(t). This process can be represented by

d'(t) = F[s(t)] * g(t)

Can we prove

d'(t) = F[d(t)] ??

Answer 1

Instead of using F[.] to denote the bandpass filtering operator, use convolution. The output $y(t)$ of a signal $x(t)$ through a filter with impulse response $h(t)$ is computed via convolution: $$y(t) = (f * h)(t)$$

Keeping your notation, you want to show that $$\big((s * g) * f\big)(t) = \big((s * f ) * g\big)(t)$$

That's called associativity.

Spoiler, convolution is a linear operation, and has that property (assuming $g$ and $f$ are linear, since convolution would not make sense otherwise)!