# Homework Help: 4th Order Elliptic Filter

1. Nov 24, 2007

### 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!

#### Attached Files:

• ###### circuit.JPG
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2. Nov 26, 2007

### 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.

#### Attached Files:

• ###### Ellip4th.jpg
File size:
40.6 KB
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3. Nov 26, 2007

### The Electrician

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

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