---
title: "Second order constant delay filter"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:constant-delay-second-order"
source: https://www.bctengineer.com/q/semester_5/filter_design/constant-delay-second-order
site: BCT Engineering
---

# Second order constant delay filter

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2073 Shrawan, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/constant-delay-second-order
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

What is a constant delay filter? Obtain the transfer function of second order constant delay filter.

## Short answer

A constant delay filter (Bessel-Thomson) has linear phase and maximally flat group delay. The second-order transfer function normalized for 1 s delay is T(s) = 3/(s^2 + 3s + 3).

## Answer

#### Constant delay filter

A **constant delay filter** (Bessel-Thomson filter) is a low-pass filter whose group delay $\tau(\omega) = -d\theta/d\omega$ is (nearly) constant over the passband, i.e. its phase is linear, $\theta(\omega) \approx -\omega\tau_0$. The ideal is $T(s) = e^{-s\tau_0}$: every frequency is delayed by the same $\tau_0$, so the output is an undistorted, delayed copy of the input. Since $e^{-s\tau_0}$ is not rational, it is approximated by an all-pole function whose delay is **maximally flat** at $\omega = 0$.

#### Second-order constant delay filter

Take a second-order all-pole low-pass function normalized for a delay of 1 s at DC:

$$
T(s) = \frac{b_0}{s^2 + b_1 s + b_0}
$$

**Phase and delay.** With $s = j\omega$:

$$
\theta(\omega) = -\tan^{-1}\frac{b_1\omega}{b_0 - \omega^2}, \qquad
\tau(\omega) = -\frac{d\theta}{d\omega} = \frac{b_1(b_0 + \omega^2)}{(b_0-\omega^2)^2 + b_1^2\omega^2}
$$

Expanding the denominator:

$$
\tau(\omega) = \frac{b_1 b_0 + b_1\omega^2}{b_0^2 + (b_1^2 - 2b_0)\omega^2 + \omega^4}
$$

**Conditions for maximally flat delay equal to 1.** $\tau(\omega)$ stays at 1 for as many powers of $\omega$ as possible if the numerator matches the denominator term by term:

$$
\begin{aligned}
\omega^0:&\quad b_1 b_0 = b_0^2 \;\Rightarrow\; b_1 = b_0 \\
\omega^2:&\quad b_1 = b_1^2 - 2b_0 \;\Rightarrow\; b_0 = b_0^2 - 2b_0 \;\Rightarrow\; b_0 = 3
\end{aligned}
$$

So $b_0 = b_1 = 3$ and

$$
T(s) = \frac{3}{s^2 + 3s + 3}, \qquad \tau(\omega) = \frac{9 + 3\omega^2}{9 + 3\omega^2 + \omega^4}
$$

The delay is $1$ s at $\omega = 0$ and deviates only by the $\omega^4$ term, i.e. it is maximally flat. The poles are $s = -1.5 \pm j0.866$ ($\omega_0 = \sqrt{3} = 1.732$, $Q = 1/\sqrt{3} = 0.577$). The denominator $s^2+3s+3$ is the second-order **Bessel polynomial**. For a delay of $\tau_0$ seconds, replace $s$ by $s\tau_0$:

$$
T(s) = \frac{3/\tau_0^2}{s^2 + (3/\tau_0)s + 3/\tau_0^2}
$$

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What property defines a constant delay filter?
   A. Flat magnitude in the passband
   B. Group delay -d theta/dw nearly constant (linear phase)
   C. Zero phase at all frequencies
   D. Infinite stopband attenuation
   Answer: B. Linear phase gives every component the same delay, giving an undistorted delayed copy. (Key point: Constant delay meaning)

2. Why is the ideal T(s) = e^(-s tau0) approximated by an all-pole function?
   A. It is not a rational function
   B. It is unstable
   C. It has no phase
   D. It is only valid at DC
   Answer: A. The exponential is not rational, so a rational all-pole function with maximally flat delay approximates it. (Key point: Why all-pole approximation)

3. In the second-order case, what do the omega^0 and omega^2 coefficient conditions give?
   A. b1 = b0 and b0 = 3
   B. b1 = 2 and b0 = 1
   C. b1 = 3 and b0 = 6
   D. b0 = b1 = 1
   Answer: A. b1*b0 = b0^2 gives b1 = b0; b1 = b1^2 - 2b0 then gives b0 = 3. (Key point: Matching coefficients)

4. Which is the second-order constant delay transfer function (1 s delay)?
   A. T(s) = 1/(s^2 + s + 1)
   B. T(s) = 3/(s^2 + 3s + 3)
   C. T(s) = 3/(s^2 + 6s + 3)
   D. T(s) = 9/(s^2 + 3s + 9)
   Answer: B. With b0 = b1 = 3 the poles are -1.5 +/- j0.866 and Q = 0.577. (Key point: Result)

## Close variants set in other exams

- [2nd order constant delay and delay equalization](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-second-order-equalization) — Filter Design, chapter 2, asked 2×
- [Constant delay filter significance; 2nd order TF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-significance-second-order) — Filter Design, chapter 2, asked 3×

## Related in Filter Design

- [All-pass filter and 2nd order constant delay TF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aallpass-second-order-constant-delay) — Filter Design, chapter 2, asked 2×
- [Third order Bessel-Thomson constant delay filter](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-third-order-bessel) — Filter Design, chapter 2, asked 9×
- [Importance of constant delay filter; 3rd order TF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-importance-third-order) — Filter Design, chapter 2, asked 8×
- [Designing a constant delay filter (2nd order example)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-design-second-order-example) — Filter Design, chapter 2
- [Delay in filter and 3rd order Bessel-Thomson](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Adelay-third-order-bessel-thomson) — Filter Design, chapter 2, asked 8×
- [Steps in designing a constant delay filter](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aconstant-delay-design-steps) — Filter Design, chapter 2, asked 2×

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