Why are the transfer characteristics of a differentiator and an integrator a circle when the input is a sine wave?

Answer 1

It has to do with the fact that what's being plotted isn't a simple function, y = f(x), on an x-y grid, but rather a set of parametric equations plotted on a f(t)-g(t) grid where:

x = f(t) is your input, and

y = g(t) is your output.

Specifically, for the differentiator:

x = f(t)

y = df/dt.

For the integrator:

x = f(t)

y = ∫ f(t) dt.

So, if your input is x = f(t) = sin(t), then your two outputs will be df/dt = cos(t) and ∫ sin(t) dt = -cos(t). That means your coordinate axes for the differentiator and the integrator will be the parametric equations x = sin(t), y = cos(t) and x = sin(t), y = -cos(t), respectively. I don't know if you've ever worked with parametric equations before, but suffice it to say, both of those sets of equations are parametric equations for a circle. I'll add a link to parametric equations below if you want to learn more.