# transfer function derivation

Discussion in 'Homework Help' started by suzuki, Mar 6, 2012.

1. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
Hi,

I have the following circuit

VIN o--------L1-----C1--------Vout
-
-
L2
-
-
GND

i want to verify that i am getting the transfer function correctly. What i did was add L1 and C1 together, and then did a voltage division between L2 and (L1+C1)

my resulting TF was

H(s) = -ω$\scriptsize(2)$(L2*C1)
----------------------------------
$\scriptsize(2)$(L1+L2)+1

My problem with this equation is that when you evaluate it, it is purely real. Since there are three reactive components, i expected at least a 90 degrees phase angle.

Anyone know where im going wrong?

tia

2. ### Vahe Member

Mar 3, 2011
75
9
I am not sure that I understand your circuit correctly. You have L1 in series with C1 between Vin and Vout and then you have L2 from Vout to ground. Is this correct?

If this is correct, your transfer function should be
$
H(s) = \frac{s L_2}{s L_1 + s L_2 + 1/(s C_1)} = \frac{s^2 L_2 C_1}{s^2 (L_1 + L_2) C_1 + 1}
$

Then,
$
H(j \omega) = \frac{-\omega^2 L_2 C_1}{-\omega^2 (L_1 + L_2) C_1 + 1}
$

3. ### suzuki Thread Starter Member

Aug 10, 2011
119
0

I agree with the transfer function that you come up with, but I am still confused about the meaning. If you were to evaluate H(jw), you would have only a real value, meaning that there is no phase change between Vin and Vout. I just want to verify that this would actually make sense or not.

Since there are an odd number of reactive components, I would have expected that there would be at least a 90 degree phase shift between Vin and Vout. Or does L2 not contribute any phase in this case?

Hope my question is more clear this time.

TIA

4. ### Vahe Member

Mar 3, 2011
75
9
Due to the negative sign, you might actually get 180 degree phase shift. I think you should put the circuit into a simulator and plot the frequency response (magnitude and phase) to get a better idea.