For my opinion, the simplest solution makes use of the classical feedback formula from H. Black:
$$\frac{V_2}{V_1}=\frac{H(s)}{1-LG}$$
with:
- \$H(s)=H_1(s)H_2(s)\$=Forward transfer function for an open loop (in our case: \$H_1=V_3/V_1\$ for \$R_2\$>>infinite and \$H_2=V_2/V_3\$.)
- Loop gain \$LG\$=Product of all three transfer functions within the loop (with \$V_1=0\$ or \$R_1\$>>infinite).
Note that \$H(s)\$ is positive and the loop gain \$LG\$ must be negative (three inverting stages in series). The transfer functions of the three blocks are basic (inverting lowpass, inverting integrator, inverting amplifier).
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.
Best Answer
As jippie says, there is no single answer to this. But, think about the transfer function and how cascaded stages combine in (Laplace style) transfer functions. In this case, the transfer function is the product of 3 elements, gain, p1, and p2. Response of cascaded OpAmp stages will combine as the product of the stages. So, you could start by designing a 3 stage circuit with each stage carrying out one of the transfer function elements (gain, p1, and p2).
Look at how poles are formed by R and C combinations (Series or parallel) on the input of the OpAmp, and how they are formed by R and C combinations in the feedback path (between the OpAmp output and the negative input). Then you could look at ways to combine stages to end up with fewer stages (probably just one).