# LC Circuit Input Admittance

Thread Starter

#### jegues

Joined Sep 13, 2010
733
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

Joined Mar 6, 2009
5,455
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

Thread Starter

#### jegues

Joined Sep 13, 2010
733
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

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

Joined Mar 6, 2009
5,455
The resulting 4 element network may be reducible but presumably not to a simple series LC equivalent.