What is the transfer function for this 4th order elliptic filter circuit?

[pnjbi]

Hi,

I'm trying to find the transfer function to this circuit (see attached image). I put together a MATLAB script file which I will paste the code below. I found the transfer function to be:
((s*L4+1)/s/C3/(s*L4+1+1/s/C3)+L2/C2/(s*L2+1/C2/s))/s/C1/((s*L4+1)/s/C3/(s*L4+1+1/s/C3)+L2/C2/(s*L2+1/C2/s)+1/s/C1)/(((s*L4+1)/s/C3/(s*L4+1+1/s/C3)+L2/C2/(s*L2+1/C2/s))/s/C1/((s*L4+1)/s/C3/(s*L4+1+1/s/C3)+L2/C2/(s*L2+1/C2/s)+1/s/C1)+1)

Can someone verify this is correct?

Matlab Code:
% Convert to S-domain
R1 = 1
C1 = sym('1/(s*C1)')
C2 = sym('1/(s*C2)')
L2 = sym('s*L2')
C3 = sym('1/(s*C3)')
L4 = sym('s*L4')
RL = 1

% L4 in series with RL
Zeq1 = L4 + RL

% Zeq1 in parallel with C3
Zeq2 = (Zeq1*C3)/(Zeq1 + C3)
simple(Zeq2)

% L2 and C2 in parallel
Zeq3 = (L2*C2)/(L2 + C2)
simple(Zeq3)

% Zeq2 in series with Zeq3
Zeq4 = Zeq2 + Zeq3
simple(Zeq4)

% Zeq4 in parallel with C1
ZL = (Zeq4*C1)/(Zeq4 + C1)
simple(ZL)

% Voltage division to get Transfer Function
T = ZL/(ZL+R1)
simple(T)
pause



Thanks!

 

[The Electrician]

What you have done so far looks good, but you stopped too early. What you have called ZL is the driving point impedance at node V1. So when you calculated T=ZL/(ZL+R1), that is only the transfer function from Vin to V1.

Let T1 be what you have already calculated, and then using T1*Vin as the voltage at node V1, we need the voltage at node V2 = T2*V1 and the voltage at node Vout = T3*V2. So continue to calculate T2 = Zeq2/(Zeq2+Zeq3) and T3 = RL/(Zeq1). Then the overall transfer function should be T1*T2*T3.

But, the calculations you have already done don't seem to have given a correct result, even though the steps in your MATLAB script file seem correct. I don't know matlab, so I've done it in Mathematica and attached the result as a .gif file. I typed in your expression and simplified it, and it didn't equal what I got.

 

[The Electrician]

I'm going to try posting the image in two parts. Hopefully it will be more readable.