Voltage Divider Rule using Admittance

I have a question about the Voltage Divider rule. I am looking at a example in a textbook to calculate a transfer function for this circuit.

----inductor------|------------| | | | | C R VIN a E VOUT p S | | | | ------------------|------------|

Transfer Function is

Vout(s) (1/sL)

------ = -------------------------------- Vin(s) (1/sL) + [sC + (1/R)]

looking at the transfer function it appears that the voltage divider rule for admittances is like.

------ g1 ----------|-------| | | Vin g g Vout 2 3 | |

--------------------|-------|

g1 Vout = Vin * ( 1/(g1 + g2 + g3) ) * g1 ) = Vin *

--------------------- g1 + g2 + g3

Assume above are resistors where : g1 = .5 mhos, g2 = .25 mhos, and g3 = .1666 mhos or r1=2 ohms r2=4 ohms r3=6 ohms

then using admittances and my theory above:

.5 Vout = Vin * ------------ = 0.545 Vin .5 + .25 + .1666

Using Voltage Division for Resistance:

Vout = Vin * (r2 || r3) / ( r1 + ( r2 || r3) ) = 2.4 / 4.4 = .545 * Vin

Is the approach I took for using admittances correct for all cases. It seems to be working for the example at the end.

I was looking at page 11 of RC Active Filter Design Handbook by F.W. Stephenson.

It makes the equation much easier using admittances over impedences.

I can not find admittance example in my textbooks.

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golson
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