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

Question

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)}\$.

Answer 1

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

Answer 2

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:

Answer 3

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)}\$