Digitizing Hysteresis.md (3018B)
1 --- 2 author: Jatin Chowdhury 3 title: Continuous Time Equations for Analog Tape Modeling 4 date: 2/1/2019 5 --- 6 7 ## Hysteresis 8 9 The magnetostatic field recorded to magnetic tape can be described using a hysteresis loop. A circuit simulation of a hysteresis loop by Martin Holters and Udo Zolzer, using the Jiles-Atherton magnetisation model can be found at http://dafx16.vutbr.cz/dafxpapers/08-DAFx-16_paper_10-PN.pdf. They use the following differential equation to describe magnetisation 'M' as a function of magnetic field 'H': 10 11 $$ \frac{dM}{dH} = \frac{(1-c) \delta_M (M_{an} - M)}{(1-c) \delta k - \alpha (M_{an} - M)} + c \frac{dM_{an}}{dH} $$ 12 13 where $M_{an}$ is the anisotropic magnetisation given by: 14 15 $$ M_{an} = M_s L \Big( \frac{H + \alpha M}{a} \Big) $$ 16 17 where $M_s$ is the magnetisation saturation, and $L$ is the Langevin function: 18 19 $$ L(x) = \coth (x) - \frac{1}{x} $$ 20 $$ L'(x) = \frac{1}{x^2} - \coth^2(x) + 1 $$ 21 $$ L''(x) = 2 \coth(x) \cdot (\coth^2(x) - 1) - \frac{2}{x^3} $$ 22 23 Let $Q(t) = \frac{H + \alpha M}{a}$ 24 25 Differentiating, we get: 26 27 $$ \frac{dM}{dt} = \frac{(1-c) \delta_M (M_sL(Q) - M)}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{dH}{dt} + c \frac{M_s}{a} \Big(\frac{dH}{dt} + \alpha \frac{dM}{dt} \Big) L'(Q) $$ 28 29 (wrong see paper) 30 $$ 31 \frac{d^2 M}{dt^2} = \frac{(1-c) \delta_M (M_sL(Q) - M)}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{d^2H}{dt^2} + 32 \frac{(1-c) \delta_M (M_sL'(Q) - \dot{M})}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{dH}{dt} + 33 \frac{(1-c) \delta_M (M_sL(Q) - M)(-\alpha (M_sL'(Q) - \dot{M}))}{((1-c) \delta k - \alpha (M_sL(Q) - M))^2} \frac{dH}{dt} + 34 c \frac{M_s}{a} \Big(\frac{dH}{dt} c \frac{M_s}{a} (\frac{dH}{dt} + \alpha \frac{dM}{dt}) L''(Q) + \frac{d^2H}{dt^2} L'(Q) + \alpha \frac{dM}{dt} c \frac{M_s}{a} (\frac{dH}{dt} + \alpha \frac{dM}{dt}) L''(Q) + \alpha \frac{d^2M}{dt^2} L'(Q) \Big) 35 $$ 36 37 Simplified (wrong see paper): 38 $$ \frac{d^2 M}{dt^2} = \frac{ 39 \frac{(1-c) \delta_M (M_sL(Q) - M)}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{d^2H}{dt^2} + 40 \frac{(1-c) \delta_M (M_sL'(Q) - \dot{M})}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{dH}{dt} + 41 \frac{(1-c) \delta_M (M_sL(Q) - M)(-\alpha (M_sL'(Q) - \dot{M}))}{((1-c) \delta k - \alpha (M_sL(Q) - M))^2} \frac{dH}{dt} + 42 c \frac{M_s}{a} \Big(\frac{d^2H}{dt^2} L'(Q) + c \frac{M_s}{a} L''(Q) (\dot{H} + \alpha \dot{M})^2 \Big)} 43 {1 - c \alpha \frac{M_s}{a} L'(Q)} 44 $$ 45 46 47 $$ \frac{dM}{dt} = \frac{\frac{(1-c) \delta_M (M_sL(Q) - M)}{(1-c) \delta k - \alpha (M_sL(Q) - M)} \frac{dH}{dt} + c \frac{M_s}{a} \frac{dH}{dt} L'(Q)}{1 - c \alpha \frac{M_s}{a} L'(Q)} = f(t, M, \vec{u}) $$ 48 49 where $\vec{u} = \begin{bmatrix} 50 H \\ 51 \dot{H} \\ 52 \ddot{H} 53 \end{bmatrix}$ 54 55 Using trapezoidal rule: 56 57 $$ \dot{\hat{H}}(n) = 2 \frac{\hat{H}(n) - \hat{H}(n-1)}{T} - \dot{\hat{H}}(n-1) $$ 58 59 and similar for $\ddot{\hat{H}}$. Now, using the semi-implicit trapezoidal rule [Yeh]: 60 61 $$ \hat{M}(n) = \hat{M}(n-1) + \frac{T}{2} \frac{f[n, \hat{M}(n-1), \vec{u}(n)] + f[n-1, \hat{M}(n-1), \vec{u}(n-1)]}{1 - \frac{T}{2}\ddot{\hat{M}}(n-1)} $$