Electronic – How to we find transfer function of this n/w

Let me label your circuit nodes for better understanding of your question

^{simulate this circuit – Schematic created using CircuitLab}

To answer your question I must first clarify some definitions:

• Error gain function: a gain k that multiplies the error e = r-y
• Transfer function: the behavior of the closed-loop system y/r

Now,

$$ y=k\cdot e $$ $$ e=r-y $$ $$ y=k\cdot (r-y) $$

So, the error gain function follows as

$$ k=\frac{y}{r-y} $$

So, you don't even need the transfer function to establish the value of k. You just need y and r.

Anyway the transfer function is

$$ TF=\frac{y}{r}=\frac{k}{k+1} $$

So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as

$$ k=\frac{TF}{1-TF} $$

Note also that, with negative feedback, in stable system conditions, and with finite k:

$$ 0\leq TF<1 $$

A simple (mathematical!) way to compute a transfer function for a circuit is to find the voltage at the output using the impedances of the components.

For a simple \$L\$-\$C\$ circuit (i.e. if you remove \$L_2\$ from the above circuit), we can use voltage divider rule to find the voltage across \$C\$:

$$V_o = V_i \times \frac{Z_C}{Z_C + Z_L} $$

(Where \$Z_C\$ is the impedance of the capacitor (= \$1/j\omega C\$) and \$Z_L\$ is the impedance of the inductor (= \$j\omega L\$))

Which gives you the transfer function as

$$\frac{V_o}{V_i} = \frac{Z_C}{Z_C + Z_L} = \frac{1}{1-\omega^2LC} $$

If we add "\$L_2\$ + a series resistance \$R_L\$" parallel to \$C\$, then we need to consider the combined impedance of "\$C\$ parallel (\$L_2\$ series \$R_L\$)" in the place of \$Z_C\$ in the above equation. After doing all the math, it gives us something big for the final transfer function:

$$\frac{V_o}{V_i} = \frac{R_L}{R_L + j\omega(L_1+L_2) - \omega^2L_1R_LC - j \omega^3L_1C}$$

Now, if you want to see the transfer function without \$R_L\$, just set \$R_L = \infty\$ in the above equation.