Hi,
Is it possible to combine or simplify the following transfer functions with Mathematica?
Subscript[Z, 1]=Subscript[R, L]/(1+Subscript[sR, L] Subscript[C, k])
Subscript[Z, 2]=Subscript[Z, 1]+1/Subscript[sC, pt]
Subscript[Z, 3]=(Subscript[Z, 2]*Subscript[sL, h])/(Subscript[Z, 2]+Subscript[sL, h])
Subscript[Z, 4]=Subscript[Z, 3]+Subscript[R, pri]+s Subscript[L, st]
Subscript[Z, 5]=(1/Subscript[Z, 4]+s Subscript[C, p]+s Subscript[C, v])^-1
Subscript[Z, in]=Subscript[Z, 5]+Subscript[R, Ls]+s Subscript[L, s]+1/(s Subscript[C, s])
TraditionalForm[FullSimplify[(Subscript[Z, 1]*Subscript[Z, 3]*Subscript[Z, 5])/(Subscript[Z, 2]*Subscript[Z, 4]*Subscript[Z, in])]]
Subscript[Z, 1] = Subscript[R, L]/(
1 + Subscript[sR, L] Subscript[C, k])
TransferFunctionModel[Subscript[Z, 1], s]
Subscript[Z, 2] = Subscript[Z, 1] + 1/Subscript[sC, pt]
FullSimplify[TransferFunctionModel[Subscript[Z, 2], s]]
Zn = Together[ExpandAll[%]]
pcs1 = Denominator[Zn]
Subscript[Z, 3] = (Subscript[Z, 2]*Subscript[sL, h])/(Subscript[Z,
2] + Subscript[sL, h])
FullSimplify[TransferFunctionModel[Subscript[Z, 3], s]]
Subscript[Z, 4] =
Subscript[Z, 3] + Subscript[R, pri] + s Subscript[L, st]
TraditionalForm[
FullSimplify[TransferFunctionModel[Subscript[Z, 4], s]]]
Subscript[Z, 5] = (1/Subscript[Z, 4] + s Subscript[C, p] +
s Subscript[C, v])^-1
FullSimplify[TransferFunctionModel[Subscript[Z, 5], s]]
Subscript[Z, in] =
Subscript[Z, 5] + Subscript[R, Ls] + s Subscript[L, s] + 1/(
s Subscript[C, s])
TransferFunctionModel[Subscript[Z, in], s]
TraditionalForm[
FullSimplify[(Subscript[Z, 1]*Subscript[Z, 3]*Subscript[Z, 5])/(
Subscript[Z, 2]*Subscript[Z, 4]*Subscript[Z, in])]]
FullSimplify[TransferFunctionModel[%, s]]
for example you get the following for Z, 2
(1 + Subscript[R, L] Subscript[sC, pt] +
Subscript[C, k] Subscript[sR, L])/(Subscript[sC, pt] +
Subscript[C, k] Subscript[sC, pt] Subscript[sR, L])
For the remaining Z,3 & Z,4 & Z,5, Z,in etc. equations, it is even more complicated.
how can I simplify the functions given above?
This is about the transfer function of a circuit.
Thanks for your helps
