I am struggling with a curve fitting calculation below. the part I am stuck with is arguably the simplest part, which is determining the coefficients $\\K_3$ and $\\K_4$. It seems like it should be simple/straightforward but it's not clear cut at this stage. some clarity would be appreciated on what I am missing in terms of defining the coefficients of the equation assuming that $\\p = 5$ just to keep it simple.

Formulation for calculating saturation curve part 1

I am trying to calculate $\\I_s$ using equation (II.10) assuming that $\\I_h$ = 0 and $\\I_s = I_0$

if $\lambda\ = \lambda\ _m $ $\sin(\omega t)$

$\lambda\ ^5\ = \lambda\ _m^{5} $ $\sin^{5}(\omega t)$

using the trigonometric power formula:

Trigonometric power formula

$\lambda\ ^5\ = \lambda\ _m^{5} $ $\sin^{5}(\omega t)$ = $\dfrac{1}{16}(10\lambda_m^{5}sin^5(\omega t)-5\lambda_m^{5}sin(3wt)+\lambda_m^{5}sin(5wt))$

enter image description here

note: I used wolfram alpha to complete this formula and couldn't figure out how to use subscript, for this equation $\lambda\ = \lambda\ _m $

I could use some help with determining the coefficients $\\K_3$ and $\\K_4$ what am I missing ?

 Formulation for calculating saturation curve part 2

  • $\begingroup$ You can simplify the problem by noting that $\sin\omega t = \text{Im}[\exp(j\omega t)]$. To find $K_3$ and $K_4$ you need two equations that have to determined from one. You can do that by equating powers of $\lambda_m$ after computing the RMS value of $i_s$. $\endgroup$ – Biswajit Banerjee Sep 5 '19 at 22:06
  • $\begingroup$ Thank you for this @Biswajit, so in the case of the function $\lambda\ ^5\ = \lambda\ _m^{5} $ $\sin^{5}(\omega t)$ would be simplified to $\frac{j}{32}\lambda _m^{5}\ (exp(jwt) - exp(jwt))^{5}$ and to be clear, finding $\ K_4$ and $\ K_5$ would need a simultaneous equation/system of equations and that could be created by calculating the RMS $\ i_s$ using the expansion for $\lambda$ that I've just done? and use Euler's representation to simplify the calculations ? $\endgroup$ – Bradley D Sep 6 '19 at 15:03
  • $\begingroup$ correction, the above expression should be $\frac{j}{32}\lambda _m^{5}\ (exp(jwt) - exp(-jwt))^{5}$ $\endgroup$ – Bradley D Sep 6 '19 at 15:35
  • $\begingroup$ If $\lambda = \lambda_m \exp(j\omega t)$ then $\lambda^5 = \lambda_m^5 \exp(5j\omega t)$. You can extract the imaginary part after all the calculations have been completed with the exponential form. $\endgroup$ – Biswajit Banerjee Sep 6 '19 at 22:01
  • $\begingroup$ Silly me, we're saying similar things, so effectively what I need to do is find the RMS of the following expression $\ i_s = A_1\lambda + A_5\lambda^{5} = A_1(\lambda_m\exp(j\omega t))+A_5(\lambda_m^{5}exp(5j\omega t))$ and then solve the simultaneous equation of $\ i_{s(RMS)}$ and $\ I_s $ ? $\endgroup$ – Bradley D Sep 7 '19 at 1:42

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