# Explanation of how to create circuit from transfer function

So I'm in this lab course that must be taken concurrently with the main upper div circuits class, but we were assigned a lab that our professor did not explain and that the lab manual does not explain.

The task is to design a circuit with transfer function characteristics. I understand that the poles will dictate the roots of the denominator and the zeros are self explanatory, but I'm not sure how this actually plays into actual design.

I would really appreciate if someone could walk me through an example of how to do this, on say part a or part b.

Thank you! ## 1 Answer

I won't solve your professor's problem, but the key to understanding this is that capacitors and inductors have "resistance" with imaginary coefficients. The idea is to combine the resistance and the reactance as a complex sum $$Z = R + jX,$$ that we call the impedance (and $j$ is the imaginary number $\sqrt{-1}$).

What is reactance $X$? It is the opposition of a circuit element to a change in current or voltage, due to that element's inductance or capacitance. By change, we mean how it reacts to an AC current (as opposed to a DC current). It can be further split into capacitive reactance $X_C$ and inductive reactance $X_L$. $$Z = R + j(X_C + X_L)$$

Looking at an ideal capacitor, for instance, we know that $$i = C \frac{dv}{dt},$$ so that if we generate an AC signal $$v(t) = V_0 e^{j\omega t},$$ then $$i(t) = C\frac{d}{dt}\{V_0 e^{j\omega t}\} = C V_0\cdot j\omega\cdot e^{j\omega t}$$ So (keeping in mind $v = iR$, or the complex equivalent $v = iZ$) the impadance $$Z = jX_C = \frac{v(t)}{i(t)} = \frac{1}{j\omega C},$$ or $$X_C = \frac{1}{j^2\omega C} = -\frac{1}{\omega C}$$

Similarily, for an inductor, we have $$v = L \frac{di}{dt},$$ which leads to $$jX_L = \frac{v(t)}{i(t)} = j L\omega,$$ or $$X_L = L\omega.$$

So, now we can use passive components such as resistances, capacitors and inductors as generalized resistors, and we can create voltage transfer functions using simple voltage dividers. As an example, a $1 \mu F$ capacitor followed by a resistor of $1 k\Omega$ will have transfer function (keeping in mind $s = j\omega$):

$$\frac{Z_R}{Z_R + Z_C} = \frac{10^3}{10^3-j\frac{1}{\omega 10^{-6}}} = \frac{10^3\cdot j\omega}{10^3 j\omega + 10^{6}} = \frac{s}{s+0.001}$$

• I understand what a transfer function is and what an impedance is. What I'm asking is: how do I work backwards to realize that there will be an inductor in this location of the circuit, a resistor in this other location, and a capacitor in that location? I know how to do the math, but I'm not sure for 2nd order circuits how to get what I want working backward. – jonnyd42 Feb 6 '16 at 22:00
• It boils down to knowing voltage division and how you get zeros and poles from it. Also, if you know poles, e.g. $s_1$ and $s_2$, then you know that the function is $\frac{P(s)}{Q(s)(s-s_1)(s-s_2)}$. – Pål-Kristian Engstad Feb 6 '16 at 22:02
• Perhaps it would help you if you tried to find out how many topologies you could make using 3 elements only? For instance, L || (R+C), (L+R)||C, etc. – Pål-Kristian Engstad Feb 6 '16 at 22:08