Yes I do understand that engineering is full of references to sine waves, which is relevant to linear circuit analysis as the math is greatly simplified. In this case they represented a dual diode nonlinearity by a "sinh". Why? Because we can solve that. Is it really the way diodes work? Maybe "close".

When you have a clipper one diode is forward biased and the other is reversed biased.

The forward diode characteristic is

I(V) = I0*(exp(V/(n Vt)) - 1) ~ I0*exp(V/(n Vt)) ;where n = 1 to 2 and Vt is the 25.8mV thermal voltage.

For a clipper you need to handle the case where the voltage is positive or negative.

Ipos(V) = I0*exp(V/(n Vt)); V > 0

Ineg(V) = - I0*exp(-V/(n Vt)); V < 0

IF we plug in a negative voltage into Ipos(V) then we get a small number so instead of having a function which switches forms at V=0 we can add the two to get a total clipper current for a given applied voltage V:

I(V) = Ipos(V) + Ineg(V) = I0*exp(V/(n Vt)) - I0*exp(-V/(n Vt))

The *hyperbolic* sine function is sinh(x) = (exp(x) - exp(-x))/2

which has the same form.

https://www.mathsisfun.com/sets/function-hyperbolic.htmlSo all they are saying is the sinh(x) function is a good approximation for a clipper.

Notice that the *input* argument is the diode voltage and the result is the diode current - opposite to what people are used to thinking.

Start on page 43 of this document. Everything is explained.

So page 42 gives the sinh() form for the clipper. Then all page 43 is doing is solving the currents and voltages in the circuit. We need clipper voltage Vo for a given input voltage Vi, account for the fact the 2k2, 0.47u and 10n components are present. The behaviour of the those linear parts are modeled with the rest of the equations - the picture to the left of page 43 breaks down "connections". The maths effectively works out the voltages and currents in the circuit.

The maths doesn't assume anything about the waveform. The whole idea is wedging a non-linear part like a clipper into a linear model.