# How to find the transfer function of current H(w) = I_out(w)/I_in(w)?

How do you find the transfer function $$H(\omega)=\frac {I_\text{out}(\omega)}{I_\text{in}(\omega)}$$ of a circuit like the below?

simulate this circuit – Schematic created using CircuitLab

I understand how to do circuits that ask for $$H(\omega) = \frac{U_\text{out}(\omega)}{U_\text{in}(\omega)}$$ as they boil down to a voltage divider relationship after doing a reduction. I have a problem of finding transfer function with the current. In this question Z_0 is the source impedance and is infinite so it is an open circuit.

Is it as simple as (R4 + L5) + (1/R1 + 1/C2 + 1/C3)? (in frequency domain ofc), I know that Z is V/I but I dont know how to use that in this case or for other circuits of type $$\H(\omega) = {I_\text{out}(\omega)}/{U_\text{in}(\omega)}\$$.

• There also exists a current division relationship. Look it up.
– Fred
Commented Aug 16 at 1:54
• @Fred Hi, yea I just noticed it between the left side and the right side, if we do the thevenin equivalent then is it (1/R1 + 1/C2 + 1/C3) / ((1/R1 + 1/C2 + 1/C3) + R4 + L5) in frequency domain? Commented Aug 16 at 2:18

We need to apply the current divider formula, for example in the following way: 1) 2)

• Hi! This is was really helpful thank you very much! Commented Aug 16 at 23:52

This transfer function can easily be obtained via a few sketches with the fast analytical circuits techniques or FACTs. Using this approach, you go straight to the well-ordered result without writing a line of algebra:

Then, if the quality factor is low enough, you can cascade two poles:

• Hi! Thanks for the tip, I will make sure to checkout FACTs! Commented Aug 16 at 23:57

There are two ways as below:

1. Convert the capacitances and inductances as impedances. Now, you would see that all impedances which include $$\Z_0\$$ are in parallel. calculate the impedance for these impedances which are in parallel: $$\{Z_0(jω)}\$$, $$\{R_1}\$$, $$\ \frac {1}{(jω)c_2}\$$ ,$$\ \frac {1}{(jω)c_3}\$$ . Use current divider rule to find the current in the load $$\R_4 + jωL_5\$$.

2. Convert the impedances $$\Z_0\$$,$$\ R_1\$$,$$\ \frac {1}{(jω)c_2}\$$,$$\ \frac {1}{(jω)c_3}\$$,$$\R_4 + jωL_5\$$ into admittances as below.

$$Y(jω) = \frac{ 1}{Z_0(jω)} + \frac{1}{R_1} + (jω)c_2 + (jω)c_3 + \frac{ 1} {R_4 + jωL_5}$$

See that the load $$\R_4 + jωL_5\$$ has the same voltage $$\V_{out}\$$ on top as $$\I_{in}\$$ and ground.

$$V_{out}(jω) = \frac {I_{in}(jω)}{Y(jω)}$$

You would get $$\ I_{out} = \frac{V_{out}} {R_4 + jωL_5} \$$.

The method 1 is slightly lengthy. In both methods, $$\ I_{out}(jω)\$$ would be a function of $$\ I_{in}(jω)\$$. With that, find $$\\frac{I_{out}(jω)}{I_{in} (jω)}\$$