Basics On Active-RC Low-Pass Filters

Just some basic understandings on analog filters which is inspired by ‘The Guru’ in our company. To clarify my thoughts, I will write in the format of Q&A. There are four questions to answer:

  1. What do we dream for a low-pass (LP) filter?
  2. Why complex poles are required?
  3. How to generate complex poles without inductor?
  4. Any real-life example?

Q1) What do we dream for a low-pass (LP) filter?

An ideal one, which has a brick-wall response. We only receive what we intend to receive, pure and loss-free. But, in reality…


Fig.1 Brick-wall response (in red) Vs. reality (in bluish)

Q2) Why complex poles are required?

Complex poles help to lift up the magnitude around the cut-off frequency by contributing larger pole quality factor (Q).

If we only have real poles, though higher-order gives better roll-off, the loss of magnitude around cut-off frequency becomes bigger. 


Fig.2 A system from order 1 to 5 which only have real poles

Now we move to a system which has complex poles. Taking the 5th-order Butterworth filter as an example, which has a real pole and two pairs of complex poles, the complex poles with a Q of 1.618 help to compensate the loss of magnitude around cut-off frequency. It tries to approximate the brick-wall response.


Fig.3 Generating 5th-order butterworth lowpass by multiplying (cascading) three transfer functions (1 one-pole + 2 biquads)

Q3) How to generate complex poles without inductor?

The answer is Feedback! R and C only generate real poles. When feedback is applied around a system containing real roots, the closed-loop transfer function may contain complex roots. 

Let’s think of this example: an amplifier with two poles. Its transfer function can be written as:

A(s)=\frac{A_0 \omega_1 \omega_2}{(s+\omega_1)(s+\omega_2)}.

The poles are generated by Rs and Cs in the amplifier and they are real. Now assume a negative feedback of beta is placed around the amplifier. The closed-loop transfer function becomes:

A_{cl}(s)=\frac{A(s)}{1+\beta A(s)}= \frac{A_0 \omega_1 \omega_2}{s^2+(\omega_1+\omega_2)s + \omega_1 \omega_2(1+\beta A_0)}.

We can then calculate the two poles of the closed-loop transfer function:

p_{1,2}=-\frac{\omega_1 + \omega_2}{2} \pm \frac{1}{2} \sqrt{(\omega_1 + \omega_2)^2 - 4\omega_1 \omega_2(1+\beta A_0)}.

By increasing \beta A_0, complex poles can be achieved! 


Fig.4 An illustration example of root-locus of poles when A*beta is increased from 0 to infinity

Q4) Any real-life example?

Of course. Fig.4 shows the Two-Thomas biquad. Without the feedback resistor R2, the open-loop transfer function has two real poles: one pole generated by R3 and C1 and the other pole at origin. With a feedback resistor applied, the two poles will move towards each other, arrive at the same position, and then leave the real axis, becoming complex poles.


Fig.4 The common Two-Thomas biquad filter (Wikipedia)

This entry was posted in Analog Design, Circuit Analysis and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s