= Clipping loss for a triangular pulse
{title2=$\Delta A=T(A-L)^2/A$}
The \b[clipping loss for a triangular pulse] is the area removed when a triangular waveform exceeds an amplitude limit. In the ideal model of <clipped waveform area>, $v(t)=A\max(1-|t|/T,0)$ and the upper limit is $0<L<A$. The original area is $AT$; the missing top triangle has area
$$
\Delta A=\frac{T(A-L)^2}{A},\qquad
\frac{\Delta A}{AT}=\left(1-\frac LA\right)^2.
$$
The geometry is derived in <clipped waveform area>, with its original diagram. Here $T$ is the half-width of the full support, not the width of the clipped plateau.
A useful inverse problem asks how much peak truncation an area-error tolerance permits. If the allowable fractional loss is $0<\eta<1$, the ideal triangle requires
$$
\frac LA\geq1-\sqrt\eta.
$$
For an original $A=5\,\mathrm V$, $T=2\,\mathrm{\mu s}$ pulse, a $1\%$ area-loss allowance requires $L\geq4.5\,\mathrm V$. At equality, the peak is reduced by $10\%$ but the area changes from $10$ to $9.9\,\mathrm{V\,\mu s}$. The plateau duration is only $2T(1-L/A)=0.4\,\mathrm{\mu s}$. This is a shape-dependent calculation, not a universal relationship between peak error and integral error.
An attenuated simultaneous trace can test whether the input is approximately triangular and whether its duration stays fixed as amplitude changes. Baseline and bandwidth must be comparable before an area ratio is meaningful. A real saturated <integrating amplifier> can have additional overload recovery, which the instantaneous clipping model omits.<ACF2101 Low Noise, Dual Switched Integrator datasheet>[p10, Output Overload]{ref} The formula supplies a conditional error estimate; it cannot identify an arbitrary missing waveform from its plateau.
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