Electronic – Designing a circuit from a transfer function

Here is the vital part of your question: -

This tells me that you'll need two cascaded 2nd order high-pass-filters. Cascading the filters is the same as the multiply in the middle of the bottom equation.

Here's a sallen-key high-pass filter (remember you'll need two cascaded): -

The transfer function for it is: -

Now you need to convert your individual 2nd order equations into a form that suits the sallen-key formulas. From experience (and with a little help from google and wiki) your formulas are of the form: -

And this means that for the left-hand part of your equation, 0.765 = Wo/Q AND 1 = (Wo)^2

By my reckoning, this means Q = 1/0.765 and Wo = 1. Equate these values to the sallen-key formulas for Wo and Q to get the resistor and capacitor values for the left hand stage. Then repeat for the right hand stage of your formula. This isn't as easy as it sounds and a little trial and error will be needed. Assume both capacitors are the same value and that R1 is half of R2 - try and get values that match Q and Wo - if Q is too low make R1 a bit smaller and repeat/iterate.

Alternatively, use a website where you can enter F (Wo/2Pi) and Q. Here is one that looks suitable. It gave the following result for the first part of your transfer function: -

Note that there is a little tiny discrepency in the numbers due to the suggested website using standard resistor and capacitor sizes. Maybe you can find one that doesn't default to using standard values.

Then it is just a simple matter of cascading the output from the left-hand sallen-key filter into the input of the right hand sallen-key filter and you have your answer.

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.