# Transfer function of complicated circuit

If we have a circuit like this: How can we make a good algebraic expression of the transfer function? I tried to go step by step with circuit analysis and then write expressions for each part of the circuit by itself but then I get an extremely long and complicated algebraic expression that is impossible to work with.

How can we get an expression that is written in the standardized form? Like this:

Where $$\H(s)=\frac{v_L}{v_i}\$$

$$\s = j \omega \$$

$$\n_1, ... ,n_m\$$ are zeros and

$$\p_1, ... ,p_n\$$ are poles.

Once you find $\dfrac{v_L}{v_1}$, then just rearranging the terms will give you the standard form. Let me do this example for you.

Step 1: Calculate $\dfrac{v_L}{v_1}$

From the circuit, the following three equations can be derived directly. $$v_L = k_3v_3 \frac{R_L}{R_L + R_{s3}} \tag{1}$$ $$v_3 = -g_{m2}v_2(r_{\pi3}||\dfrac{1}{sc_{\pi3}})\tag{2}$$ $$v_2 = -(g_{m2}v_2 - g_{m1}v_1)(r_{\pi2}||\dfrac{1}{sc_{\pi2}})\tag{3.a}$$ $$\Rightarrow v_2 = \frac{g_{m1}v_1(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}{1+g_{m2}(r_{\pi2}||\dfrac{1}{sc_{\pi2}})}\tag3$$

Step 2: Rearrange the terms

Now from equation (1), (2) and (3), the value of $v_L$ can be expressed in terms of $v_1$. From this $\dfrac{v_L}{v_1}$ can be calculated. The answer will be in the following form:

$$\frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times \frac{r_{\pi2}}{1+sc_{\pi2}r_{\pi2}+g_{m2}r_{\pi2}}\tag4$$ where $$K_1= -g_{m1}g_{m2}k_3v_3 \frac{R_L}{R_L + R_{s3}}$$

Now its just a matter of rearranging the terms to convert it into standard form. Dividing numerator and denominator of equation (4) by $(1+g_{m2}r_{\pi2})$, $$\Rightarrow \frac{v_L}{v_1} = K_1 \times \frac{r_{\pi3}}{1+sc_{\pi3}r_{\pi3}}\times \frac{\frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}$$

$$\Rightarrow \frac{v_L}{v_1} = K \times \frac{1}{1+sc_{\pi3}r_{\pi3}}\times \frac{1}{1+s\frac{c_{\pi2}r_{\pi2}}{(1+g_{m2}r_{\pi2})}}\tag5$$ where $$K = K_1 \times r_{\pi3} \times \frac{r_{\pi2}}{(1+g_{m2}r_{\pi2})}$$

Now equation (5) is in standard form and value of poles can be calculated directly.

• @Data $v_2$ is the voltage across parallel combination of $r_{\pi2}$ and $c_{\pi2}$. So $v_2$ will be impedance offered by this parallel combination $\times$ current through it. From nodal equation, current is the difference between those two current sources ($g_{m2}v_2-g_{m1}v_1$). The $-$ve sign is because of the polarity of $v_2$. The impedance is the remaining part in the expression (3.a). Eqn (3) is obtained by rearranging eqn (3.a) Jan 19, 2015 at 13:07