Transfer function of cascaded filters

[FrankJ777]

1. Problem statement
Find the transfer function , H(s) by two methods. Use Nodal analysis with Cramer's method, and decompose into simpler sections and use the property, H(s) = H1(s) * H2(s).

Homework Equations

H(s) = H1(s) * H2(s)
H(s) = Vout/Vin
Cramer's Rule. X = Dx/D : where D is the determinant and Dx is the determinate with the x column.[/B]

The Attempt at a Solution

I used nodal with Cramer's Method, and decomposed into to sections. The transfer functions were similar but the denominators differ by a term, and I don't know why.

First using nodal with Cramer's method. [/B]

Then decomposing the filter into two filters, finding H(s)1,H(s)2, then H(s)=H(s)1xH(s)2.

Here are the two transfer functions. You can see that they differ by a single term in the denominator, G_L/sL circled in red.

I'm sure they should be the same, but I'm not sure where I'm going wrong. Can anyone point me in the right direction?
Thanks a lot.

 

[Hesch]

Then decomposing the filter into two filters, finding H(s)1,H(s)2, then H(s)=H(s)1xH(s)2.

Separated the two transfer functions, H(s)1 and H(s)2, are correct. But when you connect the filters, the input of (2) will load the output of (1).
Thus the transfer function H(s)1 will depend on L and R_L.

( I don't know Cramers method. )

 

[NascentOxygen]

To treat it as two simple transfer functions and apply H = H₁ x H₂
there must exist a buffer amplifier between the two. Without that buffer stage, the signal output of filter 1 is determined not only by the CR elements in filter 1 but also by the elements in the second filter (the so-called 'loading' effect). This absence of a buffer stage means you cannot separate the network into independent blocks; it must be analysed as a whole.

The only way you may be able to justify analysing the network as a pair of independent filters is if you ensure the loading effect is minimal by setting the impedance of the L-R stage to be >> the impedance of the C-R stage. The result will still be an approximation, and this is rarely acceptable in any filter application.

 

[donpacino]

http://www.learnabout-electronics.org/Amplifiers/images/Zin-Zout.gif

Lets look at two amplifiers, a1 and a2. they are connected in series, with the output of a1 going into a2.
each amplifier has a input resistance R_inx and an output resistance R_outx

the gain of source 1 will be dependant upon the relationship between the output resistance of source 1, and the input resistance of source 2. the relationship is essentially a voltage divider.

Since our source is a perfect voltage source, it has zero output impedance, so the input gain will be 1.

the output of stage one will be
H1=a1*R_in2 /(R_in2+R_out1)

now knowing that, how would you proceed with your system?

 

[rude man]

Post 2 is right-on. Go with it.
@Hesch, cramer's rule is just a way of solving simultaneous linear equations using determinants.
www.purplemath.com/modules/cramers.htm