= Clipped waveform area
{title2=$\int\operatorname{clip}(Gv_i)\,dt$}
A \b[clipped waveform area] is the integral of an amplitude-limited trace over the specified interval. It can differ from the unclipped area by a different proportion than the change in peak height. The <amplifier clipping> article gives the instantaneous limiter model and a rectangular-pulse example; real saturated integrators can also have recovery dynamics.<ACF2101 Low Noise, Dual Switched Integrator datasheet>[p10, Output Overload]{ref}
An original triangular example takes $v(t)=A\max(1-|t|/T,0)$ with $A>0$. Its true area is $AT$. Clip its top at $0<L<A$, with the zero baseline unchanged. The missing region is a triangle of height $A-L$ and base $2T(1-L/A)$, so
$$
\int v_{\rm clipped}(t)\,dt
=AT-\frac{T(A-L)^2}{A}
=T\left(2L-\frac{L^2}{A}\right).
$$
For $A=4\,\mathrm V$, $L=2\,\mathrm V$ and $T=1\,\mathrm{ms}$, the area falls from $4$ to $3\,\mathrm{V\,ms}$. The peak is halved but the area loses only $25\%$. The <clipping loss for a triangular pulse> therefore cannot be inferred merely by applying the peak-loss percentage to its integral.
\Image[/triangular-pulse-clipping-area.png]
{title=Ideal triangular clipping halves the peak while removing one quarter of the waveform area.}
Record input and output together while changing gain or attenuation to test whether the trace is clipped. Recovering area by fitting an assumed triangle is a model-based correction, not recovery of an arbitrary waveform from its plateau. Baseline errors, limited bandwidth and overload recovery require their own measurement models.
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