LC Circuit Input Admittance

jegues · Feb 6, 2012

I'm confused as to how I'm supposed to solve for an LC circuit having the following admittance in the s domain,

$Y_{in}(s) = \frac{0.5s}{s^{2}+1} + \frac{2s}{s^{2}+2}$

Can anyone explain how?

t_n_k · Feb 6, 2012

Consider a series LC branch which has an impedance

$Z(s)=Ls+\frac{1}{Cs}=\frac{L(s^2+\frac{1}{LC})}{s}$

The admittance of such a branch is

$Y(s)=\frac{s}{L(s^2+\frac{1}{LC})}$

Now consider two such series LC branches with different L,C values connected in parallel.

Remember: parallel branch admittances are additive

jegues · Feb 6, 2012

t_n_k said: ↑

Consider a series LC branch which has an impedance

$Z(s)=Ls+\frac{1}{Cs}=\frac{L(s^2+\frac{1}{LC})}{s}$

The admittance of such a branch is

$Y(s)=\frac{s}{L(s^2+\frac{1}{LC})}$

Now consider two such series LC branches with different L,C values connected in parallel.

Remember: parallel branch admittances are additive
Click to expand...

So I can start the design with 2 different C values and 2 different L values, (i.e. two L and C series branches in parallel with one another) which I can then reduce to one L and one C?

t_n_k · Feb 6, 2012

The resulting 4 element network may be reducible but presumably not to a simple series LC equivalent.

LC Circuit Input Admittance

jegues Thread Starter Well-Known Member

t_n_k AAC Fanatic!

jegues Thread Starter Well-Known Member

t_n_k AAC Fanatic!

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