5.01 Ideal Filters, Practical Specifications & Tolerance Curves | 5.03 Chebyshev, Elliptic & Bessel Filter Approximations

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5.02 Butterworth Filter Approximation

Core Idea

The Butterworth filter is an all-pole continuous-time analog filter designed to have a maximally flat magnitude response in the passband [5.2.2]. It decays monotonically in both the transition band and stopband, making it the mathematical baseline for frequency-selective approximations [5.2.2]. By utilizing stable left-half s-plane poles distributed symmetrically on a semicircle, it provides a physically realizable transfer function [5.2.2].

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1. The Maximally Flat Magnitude Response Criterion

The magnitude-squared frequency response of an $N$-th order continuous-time analog low-pass Butterworth filter is mathematically defined as: |H_a(j\Omega)|^2 = \frac{1}{1 + \left(\frac{\Omega}{\Omega_c}\right)^{2N}} \tag{1} where:

• ${\Omega }_c$ is the 3-dB cutoff frequency (the frequency at which the power drops to half, or the gain drops by $-3\text{ dB}$) [5.2.2].
• $N$ is the integer filter order, which dictates the sharpness of the transition band roll-off [5.2.2].

1.1 Mathematical Proof of Maximally Flat Behavior

An approximation is defined as maximally flat at a point if as many of its derivatives as possible are zero at that point. For a Butterworth filter, we prove that the first $2N-1$ derivatives of the magnitude-squared response with respect to $\Omega$ are exactly zero at the origin ($\Omega =0$) [5.2.2].

Let us define a normalized variable $x=(\Omega /{\Omega }_c{)}^2$. The magnitude-squared response can be expressed as: $f(x)=\frac{1}{1+x^N}$

Using the Binomial/Taylor series expansion of $f(x)$ about $x=0$ (valid for $|x|<1$): $f(x)=(1+x^N{)}^{-1}=1-x^N+x^{2N}-x^{3N}+\dots$

Substituting $x=(\Omega /{\Omega }_c{)}^2$ back into the expansion yields: |H_a(j\Omega)|^2 = 1 - \left(\frac{\Omega}{\Omega_c}\right)^{2N} + \left(\frac{\Omega}{\Omega_c}\right)^{4N} - \left(\frac{\Omega}{\Omega_c}\right)^{6N} + \dots \tag{2}

Let us analyze the derivatives of $|H_a(j\Omega ){|}^2$ with respect to $\Omega$ at $\Omega =0$:

• The terms of the series only exist for powers of $\Omega$ that are integer multiples of $2N$.
• Therefore, any derivative of order $k$ where $1\le k\le 2N-1$ will differentiate ${\Omega }^{2N}$ to a term containing ${\Omega }^{2N-k}$.
• When evaluated at $\Omega =0$, all these derivative terms vanish: ${\frac{d^k|H_a(j\Omega ){|}^2}{d{\Omega }^k}|}_{\Omega =0}=0\text{for }k=1,2,\dots ,2N-1$

This mathematically proves the maximally flat criterion [5.2.2]. The response is exceptionally flat at low frequencies, with the approximation error only starting to grow as $\Omega$ approaches ${\Omega }_c$.

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2. Derivation of Filter Order ($N$) and Cutoff Frequency (${\Omega }_c$)

In practical design scenarios, you are given four primary specification parameters [5.2.2]:

1. ${\Omega }_p$: Passband edge frequency (rad/s) [5.2.1].
2. ${\Omega }_s$: Stopband edge frequency (rad/s) [5.2.1].
3. ${\alpha }_p$: Maximum allowable passband ripple/attenuation (dB) [5.2.1].
4. ${\alpha }_s$: Minimum required stopband attenuation (dB) [5.2.1].

Our goal is to derive the minimum integer order $N$ and the 3-dB cutoff frequency ${\Omega }_c$ to satisfy these boundaries [5.2.2].

2.1 Step-by-Step Order ($N$) Derivation

The attenuation (or loss) of a Butterworth filter in decibels is modeled by: \alpha(\Omega) = 10 \log_{10} \left[ 1 + \left(\frac{\Omega}{\Omega_c}\right)^{2N} \right] \tag{3}

Applying this formula to our passband and stopband edge boundaries yields two coupled equations [5.2.2]: \alpha_p = 10 \log_{10} \left[ 1 + \left(\frac{\Omega_p}{\Omega_c}\right)^{2N} \right] \tag{4a} \alpha_s = 10 \log_{10} \left[ 1 + \left(\frac{\Omega_s}{\Omega_c}\right)^{2N} \right] \tag{4b}

Isolate the frequency ratio terms by dividing by 10 and taking the base-10 exponent [5.2.2]: \left(\frac{\Omega_p}{\Omega_c}\right)^{2N} = 10^{0.1\alpha_p} - 1 \tag{5a} \left(\frac{\Omega_s}{\Omega_c}\right)^{2N} = 10^{0.1\alpha_s} - 1 \tag{5b}

To eliminate the unknown parameter ${\Omega }_c$, we divide Equation (5b) by Equation (5a) [5.2.2]: $\frac{{\left(\frac{{\Omega }_s}{{\Omega }_c}\right)}^{2N}}{{\left(\frac{{\Omega }_p}{{\Omega }_c}\right)}^{2N}}=\frac{10^{0.1{\alpha }_s}-1}{10^{0.1{\alpha }_p}-1}$ \left(\frac{\Omega_s}{\Omega_p}\right)^{2N} = \frac{10^{0.1\alpha_s} - 1}{10^{0.1\alpha_p} - 1} \tag{6}

Take the logarithm (base 10) of both sides of Equation (6) [5.2.2]: $2N{\log }_{10}\left(\frac{{\Omega }_s}{{\Omega }_p}\right)={\log }_{10}\left(\frac{10^{0.1{\alpha }_s}-1}{10^{0.1{\alpha }_p}-1}\right)$

Isolate the filter order $N$ [5.2.2]: N \ge \frac{\log_{10} \left( \frac{10^{0.1\alpha_s} - 1}{10^{0.1\alpha_p} - 1} \right)}{2 \log_{10} \left(\frac{\Omega_s}{\Omega_p}\right)} \tag{7}

The Integer Rule

Because the filter order represents a physical cascade of active components, $N$ must be an integer. You must always round up the calculated value of $N$ to the next higher integer: $N=\lceil N_{\text{calc}}\rceil$ [5.2.2].

2.2 Deriving the 3-dB Cutoff Frequency (${\Omega }_c$)

Once the rounded integer order $N$ is determined, substituting it back into Equations (5a) and (5b) yields two separate values for ${\Omega }_c$ [5.2.2]:

1. From Passband specs: \Omega_{c, \text{pass}} = \frac{\Omega_p}{\left(10^{0.1\alpha_p} - 1\right)^{1/2N}} \tag{8a}
2. From Stopband specs: \Omega_{c, \text{stop}} = \frac{\Omega_s}{\left(10^{0.1\alpha_s} - 1\right)^{1/2N}} \tag{8b}

The Examiner's Convention

In continuous-time design, we always select ${\Omega }_{c,\text{stop}}$ (Equation 8b) [5.2.2]. This guarantees that the stopband attenuation boundary is met exactly (with 0 dB margin), while the passband specification is exceeded with a safe margin (providing less ripple than specified) [5.2.2].

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3. Derivation of Pole Locations on the s-Plane

To construct the physical transfer function $H_a(s)$, we must transition from the frequency domain to the complex $s$-domain by mapping $s=j\Omega ⟹\Omega =s/j$ [5.2.2].

Substituting $\Omega =s/j$ into $|H_a(j\Omega ){|}^2$ yields the system product $H_a(s)H_a(-s)$ [5.2.2]: H_a(s)H_a(-s) = \frac{1}{1 + \left(\frac{s/j}{\Omega_c}\right)^{2N}} = \frac{1}{1 + (-1)^N \left(\frac{s}{\Omega_c}\right)^{2N}} \tag{9}

The poles of this system are the roots of the denominator polynomial [5.2.2]: 1 + (-1)^N \left(\frac{s}{\Omega_c}\right)^{2N} = 0 \implies \left(\frac{s}{\Omega_c}\right)^{2N} = -(-1)^{-N} = (-1)^{N-1} \tag{10}

Solving this equation yields $2N$ poles symmetrically distributed on a circle of radius ${\Omega }_c$ in the $s$-plane [5.2.2].

3.1 The Left-Half s-Plane (LHP) Stability Selection

For the analog system to be causal and stable, all the poles of our transfer function $H_a(s)$ must lie strictly in the left-half of the $s$-plane ($\Re e(p_k)<0$) [4.8.2, 5.2.2]. We discard the $N$ poles lying in the right-half plane (RHP) and keep only the $N$ LHP poles [5.2.2].

These $N$ stable left-half plane poles can be generated directly using the following polar coordinate formula: p_k = \Omega_c e^{j\theta_k} \quad \text{where } \theta_k = \frac{\pi}{2} + \frac{(2k-1)\pi}{2N} \quad \text{for } k = 1, 2, \dots, N \tag{11}

                Im(s)
                  ^
                  |    * (RHP poles discarded)
             x    |
         p1 /     |
           /      |
    x ----o-------+-------> Re(s)
    p2     \      |
            \     |
             x    |
         p3       |    * (RHP poles discarded)
                  |

Figure 1: Symmetrical distribution of Butterworth poles on a left-half s-plane circle of radius ${\Omega }_c$ shown for $N=3$.

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4. Normalized Butterworth Polynomials Table

When ${\Omega }_c=1\text{ rad/s}$, the transfer function is called the normalized transfer function $H_N(s)$ [5.2.2]. The denominators of these normalized systems are tabulated as the Normalized Butterworth Polynomials [5.2.2].

H_N(s) = \frac{1}{B_N(s)} = \frac{1}{\prod_{k=1}^N (s - p_k)} \tag{12}

Order $N$	Normalized Denominator Polynomial $B_N(s)$ [5.2.2]
1	$s+1$
2	$s^2+\sqrt{2}s+1\approx s^2+1.41421s+1$
3	$(s+1)(s^2+s+1)=s^3+2s^2+2s+1$
4	$(s^2+0.76537s+1)(s^2+1.84770s+1)=s^4+2.61313s^3+3.41421s^2+2.61313s+1$
5	$(s+1)(s^2+0.61803s+1)(s^2+1.61803s+1)$
6	$(s^2+0.51764s+1)(s^2+\sqrt{2}s+1)(s^2+1.93185s+1)$
7	$(s+1)(s^2+1.80194s+1)(s^2+1.24700s+1)(s^2+0.44500s+1)$

4.1 Frequency Scaling Substitution

To find the actual transfer function $H_a(s)$ for a non-unity cutoff frequency ${\Omega }_c$, apply frequency scaling by replacing $s$ with $\left(\frac{s}{{\Omega }_c}\right)$ in the normalized transfer function [5.2.2]: H_a(s) = \left. H_N(s) \right|_{s \, \to \, \frac{s}{\Omega_c}} \tag{13}

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5. Comprehensive Worked Example (13-Mark PYQ Classic)

verbatim past year question (2025/2023)

Design a Butterworth analog low-pass filter with a $1\text{ dB}$ passband ripple, passband edge frequency ${\Omega }_p=2000\pi \text{ rad/s}$, stopband edge frequency ${\Omega }_s=10,000\pi \text{ rad/s}$, and a minimum stopband ripple (attenuation) of $40\text{ dB}$ [5.2.2].

Step-by-Step Solution:

Step 1: Identify Given Specs

• Passband attenuation ${\alpha }_p=1\text{ dB}$ [5.2.2]
• Stopband attenuation ${\alpha }_s=40\text{ dB}$ [5.2.2]
• Passband frequency ${\Omega }_p=2000\pi \approx 6283.18\text{ rad/s}$ [5.2.2]
• Stopband frequency ${\Omega }_s=10,000\pi \approx 31415.93\text{ rad/s}$ [5.2.2]

Step 2: Compute the Required Filter Order ($N$)

Substitute the specs into the algebraic order formula [5.2.2]: $N\ge \frac{{\log }_{10}\left(\frac{10^{0.1\times 40}-1}{10^{0.1\times 1}-1}\right)}{2{\log }_{10}\left(\frac{10,000\pi }{2000\pi }\right)}$

Evaluate the numerator term [5.2.2]: $\text{Num}={\log }_{10}\left(\frac{10^4-1}{10^{0.1}-1}\right)={\log }_{10}\left(\frac{9999}{1.2589-1}\right)={\log }_{10}\left(\frac{9999}{0.2589}\right)={\log }_{10}(38617.36)\approx 4.5868$

Evaluate the denominator term [5.2.2]: $\text{Den}=2{\log }_{10}(5)=2\times 0.69897=1.3979$

Calculate $N$ [5.2.2]: $N\ge \frac{4.5868}{1.3979}\approx 3.2811$

Since $N$ must be an integer, round up to the next integer [5.2.2]: $N=4$

Step 3: Determine the 3-dB Cutoff Frequency (${\Omega }_c$)

Use the stopband-matching convention (Equation 8b) to ensure stopband specs are met exactly [5.2.2]: ${\Omega }_c=\frac{{\Omega }_s}{{\left(10^{0.1{\alpha }_s}-1\right)}^{1/2N}}=\frac{10,000\pi }{{\left(10^4-1\right)}^{1/8}}=\frac{31415.93}{(9999{)}^{0.125}}\text{ rad/s}$

Evaluate $(9999{)}^{0.125}$ [5.2.2]: $9999^{0.125}\approx 3.1622$

Solve for ${\Omega }_c$ [5.2.2]: ${\Omega }_c=\frac{31415.93}{3.1622}\approx 9935\text{ rad/s}$

Step 4: Formulate the Normalized Transfer Function $H_N(s)$

For $N=4$, retrieve the factored polynomial from Table 1 [5.2.2]: $H_N(s)=\frac{1}{\left(s^2+0.76537s+1\right)\left(s^2+1.84770s+1\right)}$

Step 5: Frequency Scaling to obtain $H_a(s)$

Replace $s$ with $\left(\frac{s}{9935}\right)$ [5.2.2]: $H_a(s)=\frac{1}{\left[{\left(\frac{s}{9935}\right)}^2+0.76537\left(\frac{s}{9935}\right)+1\right]\left[{\left(\frac{s}{9935}\right)}^2+1.84770\left(\frac{s}{9935}\right)+1\right]}$

To clear the denominators from each quadratic term, multiply the numerator and denominator by ${\Omega }_c^4=9935^4$ [5.2.2]: $H_a(s)=\frac{9935^4}{\left(s^2+[0.76537\times 9935]s+9935^2\right)\left(s^2+[1.84770\times 9935]s+9935^2\right)}$

Calculate individual coefficient values [5.2.2]:

• ${\Omega }_c^2=9935^2\approx 9.8704\times 10^7$ [5.2.2]
• ${\Omega }_c^4=9935^4\approx 9.7425\times 10^{15}$ [5.2.2]
• $0.76537\times 9935\approx 7604.10$ [5.2.2]
• $1.84770\times 9935\approx 18357.10$ [5.2.2]

Assemble the final continuous-time transfer function [5.2.2]: H_a(s) = \frac{9.7425 \times 10^{15}}{\left(s^2 + 7604.10s + 9.8704 \times 10^7\right) \left(s^2 + 18357.10s + 9.8704 \times 10^7\right)} \tag{14}

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6. Common Mistakes That Cost Marks

The Circle Pole Sign-Flip Trap

When solving the poles $p_k={\Omega }_c{e}^{j{\theta }_k}$, always verify that your calculated coordinates place the poles in the Left-Half Plane ($\Re e(p_k)<0$) [4.8.2, 5.2.2]. If any pole real part is positive, you have mistakenly chosen a right-half plane pole, which represents an unstable, non-causal system, yielding zero marks [5.2.2].

The Cutoff Frequency (${\Omega }_c$) Specification Trap

Do not confuse the operating frequencies ${\Omega }_p$ or ${\Omega }_s$ with the cutoff frequency ${\Omega }_c$ [5.2.2]. The gain at ${\Omega }_p$ is $-{\alpha }_p\text{ dB}$ (which is $-1\text{ dB}$ in our numerical) [5.2.2]. The gain at ${\Omega }_c$ is always exactly $-3\text{ dB}$ [5.2.2]. Placing ${\Omega }_c={\Omega }_p$ is a fatal conceptual error.

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 PYQ 2025 Section B Question 4c [13 Marks]

Question: Design a Butterworth low pass filter with 1dB passband ripple, passband edge frequency ${\Omega }_p=2000\pi$ rad/Sec, stop band edge frequency ${\Omega }_s=10,000\pi$ rad/Sec, and a minimum stop band ripple of 40dB [5.2.2].

• Answer Plan: Follow the complete step-by-step design derivation in Section 5. Explicitly write out the calculation of $N=3.28\to 4$, evaluate ${\Omega }_c=9935\text{ rad/s}$ using the stopband spec, state the normalized $N=4$ polynomial, apply frequency scaling, and expand the algebra completely to provide the final numerical transfer function.

7.2 PYQ 2023 Section B Question 4c [9 Marks]

Question: Design a Butterworth analog low pass filter with 1 dB passband ripple, passband edge frequency ${\Omega }_p=2000\pi$ rad/sec, stopband edge frequency ${\Omega }_s=10000\pi$ rad/sec, and a minimum stopband ripple of 40 dB [5.2.2].

• Answer Plan: This is the identical question with a slightly compressed mark weight. Follow the identical pipeline to find $N=4,{\Omega }_c=9935\text{ rad/s}$, but you may save time by skipping the long scalar multiplication steps and writing $H_a(s)$ directly in factored form.

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8. Self-Check Before Moving On

• Can you explain why the first $2N-1$ derivatives of the magnitude response of a Butterworth filter are zero at the origin? [5.2.2]
• Do you know how to derive the minimum filter order $N$ given passband and stopband attenuation parameters? [5.2.2]
• Can you locate stable left-half plane poles using the ${\theta }_k$ angle offset formula? [5.2.2]
• Have you memorized the warning to always use the stopband spec to compute ${\Omega }_c$ to guarantee specification margins? [5.2.2]

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Source: (k.Deergha Rao) signals and systems.pdf (Section 5.2.2).