= Discrete-jump model of Barkhausen noise
{c}
{title2=$M(t)=M_0+rt+\sum_i\Delta M_i\Theta(t-t_i)$}
An explicit constructed model separates slow <magnetization> drift from discrete collective jumps:
$$
M(t)=M_0+rt+\sum_i\Delta M_i\Theta(t-t_i),\qquad
\frac{dM}{dt}=r+\sum_i\Delta M_i\delta(t-t_i).
$$
Here $\Theta$ is the <Heaviside step function>, $\delta$ the <Dirac delta function>, $r$ a smooth drift rate, and $t_i$ the jump times. The ideal discontinuities stand for events shorter than the resolved timescale; actual <magnetic domain walls> move over finite times. This equation stipulates jump sizes rather than deriving them from atomic <spin>. With $K=\mu_0NA_s$ in the preceding geometry, subtracting the smooth background leaves $e_{\rm jump}=-K\sum_i\Delta M_i\delta(t-t_i)$.
Let a stable <linear filter> have causal <impulse response> $h(t)$, normalized by $\int h(t)\,dt=1$, and let $G$ be a constant dimensionless <voltage gain>. The output is a <convolution>:
$$
v_{\rm out}(t)=G(h*e_{\rm jump})(t)
=-GK\sum_i\Delta M_i h(t-t_i).
$$
A simple <low-pass filter> has $h(t)=\Theta(t)e^{-t/\tau}/\tau$, with <time constant> $\tau>0$. Its output pulse has peak proportional to $\Delta M_i/\tau$ and signed area $-GK\Delta M_i$. Integrating the <convolution> proves area preservation because $\int h=1$; a wider filter response lowers peaks and lengthens tails. This ideal filter models resolution, rather than claiming the cited instrument had exactly this response.
\Image[/barkhausen-discrete-jumps-filtered-voltage.png]
{title=Constructed <magnetization> jumps and filtered <Barkhausen noise> voltage}
{description=Calculated model, not measured traces. The top panel stipulates three jumps of $1000$, $500$ and $500\,\mathrm{A/m}$ at $1.0$, $3.0$ and $3.2\,\mathrm{ms}$, with no smooth drift. The middle panel applies unit-gain exponential <low-pass filters> with $\tau=0.10$ and $0.50\,\mathrm{ms}$ and $K=4\pi\times10^{-11}\,\mathrm{V\,s\,m/A}$. The bottom panel shows cumulative signed <integrals> for the broader filter, with and without an additional zero-DC stage of <time constant> $1.0\,\mathrm{ms}$.}
The nearby positive jumps in the figure have additive negative pulse areas even when their tails overlap. For $\tau=0.50\,\mathrm{ms}$ the first of the pair still has $e^{-0.2/0.5}=e^{-0.4}\approx0.67032$ of its initial height when the second arrives. Thresholding can merge those responses into one detected event; limited <signal bandwidth> also obscures their separation. The total signed area still records the sum if the complete response is integrated, but cannot identify how that sum was partitioned into events. Equal simultaneous jumps are exactly indistinguishable from a single jump of their summed size in this linear measurement. Oppositely signed simultaneous changes can cancel.
Area preservation requires the stated electronics. A zero-DC stage after the <low-pass filter> has response
$$
q(t)=\delta(t)-\frac{\Theta(t)}{T}e^{-t/T},\qquad
\int q(t)\,dt=0.
$$
Consequently $\int G(q*h*e_{\rm jump})(t)\,dt=0$ over the complete response: a main pulse is followed by a compensating tail. Integrating only a threshold-selected lobe, clipping the record, or saturating the <amplifier> breaks the simple pulse-area inference. Recovering a <magnetization> change from such a readout needs the calibrated <frequency response> and an appropriate reconstruction. Neither the audible crackle nor the displayed voltage is a direct staircase of <magnetization>.
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