5.04 Active Filter Realization & Sallen-Key RC Networks | 5.00 Chapter Map - Analog Filter Design

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5.05 Analog Frequency Transformations

Core Idea

Analog frequency transformations are algebraic variable substitutions ($sof(\hat{s})$) that map a standardized, normalized analog low-pass prototype filter (with cutoff frequency ${\Omega }_p=1$ rad/s) into practical analog low-pass, high-pass, band-pass, or band-stop filters with arbitrary cutoff boundaries. This technique eliminates the need to design separate filter algorithms for different frequency specifications from scratch.

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1. The Core Philosophy of Frequency Mapping

In analog filter design, we always design a normalized low-pass prototype filter $H_N(s)$ first, where the passband edge is normalized to ${\Omega }_p=1$ rad/s. Once this prototype transfer function is established, we apply a frequency mapping function to convert the complex s-plane variable of the prototype ($s$) to the target frequency variable ($\hat{s}$):

$s=f(\hat{s})$

This algebraic substitution warps and shifts the imaginary axis of the prototype s-plane onto the imaginary axis of the target $\hat{s}$-plane, mapping the passband and stopband behaviors perfectly.

• Continuous Mapping: The imaginary axis of the $s$-plane ($s=j\Omega$) is mapped directly to the imaginary axis of the $\hat{s}$-plane ($\hat{s}=j\hat{\Omega }$).
• Stability Preservation: Active mapping parameters must ensure that any stable pole in the Left-Half s-plane (LHP) maps strictly to a stable pole in the Left-Half $\hat{s}$-plane, preserving causality and absolute stability.

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2. Low-Pass to Low-Pass (LP-to-LP) scaling

To scale a normalized low-pass filter (with passband edge ${\Omega }_p=1$ rad/s) to a practical low-pass filter with an arbitrary passband edge frequency ${\hat{\Omega }}_p$ rad/s, we apply a linear frequency-scaling substitution:

2.1 The Mathematical Mapping

Let the prototype frequency variable be $s=j\Omega$ and the target scaled frequency variable be $\hat{s}=j\hat{\Omega }$. The transformation is a simple scaling:

s = rac{\hat{s}}{\hat{\Omega}_p} \quad \implies \quad \Omega = rac{\hat{\Omega}}{\hat{\Omega}_p}

• When $\hat{\Omega }=0$, the scaled frequency is $\Omega =0$.
• When the target frequency reaches the target cutoff boundary $\hat{\Omega }={\hat{\Omega }}_p$, the scaled prototype frequency is $\Omega =1$ rad/s.
• The LP-to-LP Transfer Function Rule: H_{LP}(\hat{s}) = H_N(s)\Big|_{s = rac{\hat{s}}{\hat{\Omega}_p}}

2.2 Spec Scaling Adjustments

For Chebyshev and Elliptic filters, the normalization frequency is defined at the passband edge ${\Omega }_p=1$. Therefore: \Omega_s = rac{\hat{\Omega}_s}{\hat{\Omega}_p}

For Butterworth filters, because $\Omega =1$ corresponds to the 3-dB cutoff frequency (${\Omega }_c$), the substitution uses the target 3-dB cutoff frequency ${\hat{\Omega }}_c$: s = rac{\hat{s}}{\hat{\Omega}_c} \quad \implies \quad H_{LP}(\hat{s}) = H_N(s)\Big|_{s = rac{\hat{s}}{\hat{\Omega}_c}}

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3. Low-Pass to High-Pass (LP-to-HP) Mapping

To convert a normalized low-pass prototype into a high-pass filter with a passband starting at ${\hat{\Omega }}_p$ rad/s, we invert the frequency variable. This flips low frequencies to high frequencies and vice versa.

3.1 The Mathematical Mapping

The algebraic substitution is an inversion mapping:

s = rac{\hat{\Omega}_p}{\hat{s}} \quad \implies \quad \Omega = -rac{\hat{\Omega}_p}{\hat{\Omega}}

• Time-Frequency Boundary Warping:
  • At DC ($\hat{\Omega }=0$), the prototype frequency is $\Omega =\infty$ (maps passband stop behaviors to zero frequency).
  • At the target passband edge ($\hat{\Omega }={\hat{\Omega }}_p$), the prototype frequency is $\Omega =-1$ rad/s (the negative sign disappears under magnitude squaring).
  • At infinite frequency ($\hat{\Omega }=\infty$), the prototype frequency is $\Omega =0$ (maps prototype DC passband behaviors to high frequencies).
• The LP-to-HP Transfer Function Rule: H_{HP}(\hat{s}) = H_N(s)\Big|_{s = rac{\hat{\Omega}_p}{\hat{s}}}

3.2 High-Pass Spec Warping Equations

To find the required normalized prototype low-pass stopband edge frequency ${\Omega }_s$ to design a target HPF with specs ${\hat{\Omega }}_p$ and ${\hat{\Omega }}_s$: \Omega_s = rac{\hat{\Omega}_p}{\hat{\Omega}_s}

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3.3 Complete Step-by-Step Solved Numerical [Example 5.5 / PYQ Classic]

Question: Design a third-order Butterworth analog high-pass filter with the following specifications:

• Passband edge frequency: $f_p=30.777$ Hz, Passband ripple: lpha_p = 1 dB.
• Stopband edge frequency: $f_s=10$ Hz, Stopband ripple: lpha_s = 20 dB.

Step 1: Convert Cyclic Frequencies to Radian Frequencies

\hat{\Omega}_p = 2\pi f_p = 2\pi(30.777) pprox 193.38 ext{ rad/s} \hat{\Omega}_s = 2\pi f_s = 2\pi(10) pprox 62.83 ext{ rad/s}

Step 2: Warp Specifications to Normalized Low-Pass Prototype

Using the high-pass frequency warping relation: ${\Omega }_p=1extrad/s$ \Omega_s = rac{\hat{\Omega}_p}{\hat{\Omega}_s} = rac{30.777}{10} = 3.0777 ext{ rad/s} \quad ext{[114]}

Step 3: Find the Filter Order $N$

Using the Butterworth order estimation equation:

ight)}{2 \log_{10}(\Omega_s/\Omega_p)} = rac{\log_{10}\left(rac{10^2 - 1}{10^{0.1} - 1} ight)}{2 \log_{10}(3.0777)} = rac{\log_{10}(99 / 0.2589)}{2 imes 0.4882} = rac{2.5829}{0.9764} pprox 2.6447$$ Rounding to the next higher integer yields: $$N = 3 \quad ext{[114]}$$ #### Step 4: Write the Normalized 3rd-Order Butterworth Low-Pass Prototype From the standard normalized polynomial table ($N=3$, Table 5.1): $$H_N(s) = rac{1}{(s+1)(s^2 + s + 1)} = rac{1}{s^3 + 2s^2 + 2s + 1}$$ #### Step 5: Find the Prototype LPF Cutoff Frequency $\Omega_c$ Since we want the stopband specification to be met exactly at $\Omega_s = 3.0777$: $$\Omega_c = rac{\Omega_s}{(10^{0.1lpha_s} - 1)^{1/2N}} = rac{3.0777}{(99)^{1/6}} pprox 1.4309 ext{ rad/s} \quad ext{[115]}$$ #### Step 6: Scale the Prototype to Actual Low-Pass Transfer Function $H_{LP}(s)$ We substitute $s o rac{s}{\Omega_c} = rac{s}{1.4309}$ in $H_N(s)$: $$H_{LP}(s) = rac{1}{\left(rac{s}{1.4309} + 1 ight)\left(\left(rac{s}{1.4309} ight)^2 + rac{s}{1.4309} + 1 ight)} = rac{1.4309^3}{(s + 1.4309)(s^2 + 1.4309s + 2.0475)}$$ $$H_{LP}(s) = rac{2.93}{s^3 + 2.8619s^2 + 4.0952s + 2.93} \quad ext{[115]}$$ #### Step 7: Apply the LP-to-HP Transformation We substitute $s o rac{\hat{\Omega}_p}{\hat{s}} = rac{3.0777}{\hat{s}}$ into the low-pass transfer function $H_{LP}(s)$ to get the final target high-pass transfer function $H_{HP}(\hat{s})$: $$H_{HP}(\hat{s}) = rac{2.93}{\left(rac{3.0777}{\hat{s}} ight)^3 + 2.8619\left(rac{3.0777}{\hat{s}} ight)^2 + 4.0952\left(rac{3.0777}{\hat{s}} ight) + 2.93}$$ Multiply numerator and denominator by $\hat{s}^3$ to clear the fractions: $$H_{HP}(\hat{s}) = rac{2.93 \hat{s}^3}{2.93 \hat{s}^3 + 4.0952(3.0777) \hat{s}^2 + 2.8619(3.0777^2) \hat{s} + 3.0777^3}$$ $$H_{HP}(\hat{s}) = rac{2.93 \hat{s}^3}{2.93 \hat{s}^3 + 12.6038 \hat{s}^2 + 27.1086 \hat{s} + 29.1528}$$ Divide through by $2.93$ to normalize the leading coefficient: $$H_{HP}(\hat{s}) = rac{\hat{s}^3}{\hat{s}^3 + 4.3017\hat{s}^2 + 9.2521\hat{s} + 9.9499} \quad ext{[116]}$$ --- ## 4. Low-Pass to Band-Pass (LP-to-BP) Mapping To transform a normalized low-pass filter into a band-pass filter centered at geometric mean frequency $\hat{\Omega}_0$ and possessing a passband bandwidth of $B$ rad/s, we apply a second-order frequency mapping. ### 4.1 The Mathematical Mapping The algebraic substitution is a quadratic-ratio mapping: $$s = rac{\hat{s}^2 + \hat{\Omega}_m^2}{B \hat{s}} \quad \implies \quad \Omega = rac{\hat{\Omega}_m^2 - \hat{\Omega}^2}{B \hat{\Omega}}$$ where: * $\hat{\Omega}_{p1}$ = Lower passband edge frequency of target filter. * $\hat{\Omega}_{p2}$ = Upper passband edge frequency of target filter. * $B = \hat{\Omega}_{p2} - \hat{\Omega}_{p1}$ is the **Passband Bandwidth**. * $\hat{\Omega}_m = \sqrt{\hat{\Omega}_{p1}\hat{\Omega}_{p2}}$ is the **Geometric Mean Center Frequency**. * **The LP-to-BP Transfer Function Rule:** $$H_{BP}(\hat{s}) = H_N(s)\Big|_{s = rac{\hat{s}^2 + \hat{\Omega}_m^2}{B \hat{s}}}$$ ### 4.2 Nodal Edge Warping Matrix The target stopband specifications ($\hat{\Omega}_{s1}$ and $\hat{\Omega}_{s2}$) map to two prototype stopband edges, $A_1$ and $A_2$: $$A_1 = rac{\hat{\Omega}_{s1}^2 - \hat{\Omega}_{p1}\hat{\Omega}_{p2}}{(\hat{\Omega}_{p2} - \hat{\Omega}_{p1})\hat{\Omega}_{s1}} \quad ext{[119]}$$ $$A_2 = rac{\hat{\Omega}_{s2}^2 - \hat{\Omega}_{p1}\hat{\Omega}_{p2}}{(\hat{\Omega}_{p2} - \hat{\Omega}_{p1})\hat{\Omega}_{s2}} \quad ext{[119]}$$ To satisfy the stopband requirements symmetrically, we select the tighter constraint: $$\Omega_s = \min(|A_1|, |A_2|)$$ --- ### 4.3 Complete Step-by-Step Solved Numerical [Example 5.7 / PYQ Classic] **Question:** Design a first-order Butterworth analog band-pass filter with the following specifications: * Lower passband edge frequency: $f_{p1} = 41.4$ Hz, Upper passband edge frequency: $f_{p2} = 50.95$ Hz. * Passband bandwidth: $B = f_{p2} - f_{p1} = 9.55$ Hz. * Lower stopband edge frequency: $f_{s1} = 7.87$ Hz. * Normalized prototype specifications derived as: $\Omega_p = 1$, $\Omega_s = 8.26$, passband ripple: $lpha_p = 2$ dB, stopband ripple: $lpha_s = 10$ dB. #### Step 1: Calculate Geometric Mean Center Frequency $\hat{\Omega}_m$ Using the geometric mean formula on cyclic frequencies: $$f_m^2 = f_{p1} f_{p2} = 41.4 imes 50.95 = 2109.3 ext{ Hz}^2$$ #### Step 2: Establish the Normalized LPF Prototype Since the order is given as $N=1$, the normalized Butterworth LPF is: $$H_N(s) = rac{1}{s + 1} \quad ext{[121]}$$ Calculate the prototype 3-dB cutoff frequency $\Omega_c$: $$\Omega_c = rac{\Omega_s}{(10^{0.1lpha_s} - 1)^{1/2N}} = rac{8.26}{(10^1 - 1)^{1/2}} = rac{8.26}{3} pprox 2.7533 ext{ rad/s}$$ Scale the normalized prototype to build actual LPF transfer function $H_{LP}(s)$: $$H_{LP}(s) = H_N(s/\Omega_c) = rac{2.7533}{s + 2.7533} \quad ext{[122]}$$ #### Step 3: Apply LP-to-BP Transformation We substitute $s o rac{\hat{s}^2 + \hat{\Omega}_m^2}{B \hat{s}} = rac{\hat{s}^2 + 2109.3}{9.55 \hat{s}}$ into the low-pass transfer function $H_{LP}(s)$: $$H_{BP}(\hat{s}) = rac{2.7533}{rac{\hat{s}^2 + 2109.3}{9.55 \hat{s}} + 2.7533}$$ Multiply numerator and denominator by $9.55 \hat{s}$: $$H_{BP}(\hat{s}) = rac{2.7533 imes 9.55 \hat{s}}{\hat{s}^2 + 2.7533(9.55) \hat{s} + 2109.3}$$ $$H_{BP}(\hat{s}) = rac{26.2943 \hat{s}}{\hat{s}^2 + 26.2943 \hat{s} + 2109.3} \quad ext{[122, 123]}$$ --- ## 5. Low-Pass to Band-Stop (LP-to-BS) Mapping To transform a normalized low-pass prototype into a band-stop (notch) filter, we invert the band-pass substitution, routing DC and infinite frequencies to the output while completely blocking a middle band. ### 5.1 The Mathematical Mapping The algebraic substitution is an inverted quadratic mapping: $$s = rac{B \hat{s}}{\hat{s}^2 + \hat{\Omega}_m^2} \quad \implies \quad \Omega = rac{B \hat{\Omega}}{\hat{\Omega}_m^2 - \hat{\Omega}^2}$$ where: * $B = \hat{\Omega}_{s2} - \hat{\Omega}_{s1}$ is the **Stopband Bandwidth**. * $\hat{\Omega}_m = \sqrt{\hat{\Omega}_{s1}\hat{\Omega}_{s2}}$ is the **Geometric Mean Notch Frequency**. * **The LP-to-BS Transfer Function Rule:** $$H_{BS}(\hat{s}) = H_N(s)\Big|_{s = rac{B \hat{s}}{\hat{s}^2 + \hat{\Omega}_m^2}}$$ ### 5.2 Passband Edge Warping Equations The passband limits ($\hat{\Omega}_{p1}$ and $\hat{\Omega}_{p2}$) map to two prototype stopband variables, $A_1$ and $A_2$: $$A_1 = rac{\hat{\Omega}_{s1}\hat{\Omega}_{s2} - \hat{\Omega}_{p1}^2}{(\hat{\Omega}_{s2} - \hat{\Omega}_{s1})\hat{\Omega}_{p1}} \quad ext{[127]}$$ $$A_2 = rac{\hat{\Omega}_{p2}^2 - \hat{\Omega}_{s1}\hat{\Omega}_{s2}}{(\hat{\Omega}_{s2} - \hat{\Omega}_{s1})\hat{\Omega}_{p2}} \quad ext{[127]}$$ We select the tighter constraint: $$\Omega_s = \min(|A_1|, |A_2|)$$ --- ### 5.3 Complete Step-by-Step Solved Numerical [Example 5.9 / PYQ Classic] **Question:** Design an analog band-stop Butterworth filter with the following specifications: * Lower passband edge: $f_{p1} = 22.35$ Hz, Upper passband edge: $f_{p2} = 447.37$ Hz. * Lower stopband edge: $f_{s1} = 72.65$ Hz, Upper stopband edge: $f_{s2} = 137.64$ Hz. * Passband ripple: $lpha_p = 3$ dB, Stopband ripple: $lpha_s = 15$ dB. #### Step 1: Calculate Geometric Mean Notch Frequency $\hat{\Omega}_m^2$ and Bandwidth $B_s$ $$\hat{\Omega}_m^2 = f_{s1} f_{s2} = 72.65 imes 137.64 = 10000 ext{ Hz}^2 \quad ext{[131]}$$ $$B_s = f_{s2} - f_{s1} = 137.64 - 72.65 = 64.99 ext{ Hz} \quad ext{[131]}$$ #### Step 2: Establish the Normalized LPF Prototype Given order $N=1$, the normalized LPF is $H_N(s) = rac{1}{s+1}$. The prototype stopband edge is $\Omega_s = \min(|A_1|, |A_2|) pprox 6.5397$. Calculate the prototype 3-dB cutoff frequency $\Omega_c$: $$\Omega_c = rac{\Omega_s}{(10^{0.1lpha_s} - 1)^{1/2N}} = rac{6.5397}{(10^{1.5} - 1)^{1/2}} = rac{6.5397}{\sqrt{30.6228}} pprox 1.1818 ext{ rad/s} \quad ext{[130]}}$$ Scale the normalized prototype to build actual LPF transfer function $H_{LP}(s)$: $$H_{LP}(s) = H_N(s/\Omega_c) = rac{1.1818}{s + 1.1818} \quad ext{[130]}$$ #### Step 3: Apply LP-to-BS Transformation Using the band-stop scaling coefficient: $$k = B_s \Omega_s = 64.99 imes 6.5397 pprox 425$$ Substitute $s o rac{k\hat{s}}{\hat{s}^2 + \hat{\Omega}_m^2} = rac{425 \hat{s}}{\hat{s}^2 + 10000}$ into $H_{LP}(s)$: $$H_{BS}(\hat{s}) = rac{1.1818}{rac{425 \hat{s}}{\hat{s}^2 + 10000} + 1.1818}$$ Multiply numerator and denominator by $(\hat{s}^2 + 10000)$: $$H_{BS}(\hat{s}) = rac{1.1818(\hat{s}^2 + 10000)}{1.1818\hat{s}^2 + 425\hat{s} + 1.1818(10000)}$$ Divide through by $1.1818$ to normalize the leading quadratic coefficient: $$H_{BS}(\hat{s}) = rac{\hat{s}^2 + 10000}{\hat{s}^2 + rac{425}{1.1818}\hat{s} + 10000} = rac{\hat{s}^2 + 10000}{\hat{s}^2 + 360\hat{s} + 10000} \quad ext{[131, 132]}$$ --- ## 6. Unified Frequency Mapping Reference Table | Target Filter Class | Algebraic s-Domain Substitution | Target Frequency Axis Mapping ($\Omega$) | Boundary Warping Equation ($\Omega_s$) | | :--- | :--- | :--- | :--- | | **Low-Pass (LP)** | $$s = rac{\hat{s}}{\hat{\Omega}_p}$$ | $$\Omega = rac{\hat{\Omega}}{\hat{\Omega}_p}$$ | $$\Omega_s = rac{\hat{\Omega}_s}{\hat{\Omega}_p}$$ | | **High-Pass (HP)** | $$s = rac{\hat{\Omega}_p}{\hat{s}}$$ | $$\Omega = -rac{\hat{\Omega}_p}{\hat{\Omega}}$$ | $$\Omega_s = rac{\hat{\Omega}_p}{\hat{\Omega}_s}$$ | | **Band-Pass (BP)** | $$s = rac{\hat{s}^2 + \hat{\Omega}_m^2}{B \hat{s}}$$ | $$\Omega = rac{\hat{\Omega}_m^2 - \hat{\Omega}^2}{B \hat{\Omega}}$$ | $$\Omega_s = \min(|A_1|, |A_2|)$$ where $$A = rac{\hat{\Omega}_s^2 - \hat{\Omega}_m^2}{B \hat{\Omega}_s}$$ | | **Band-Stop (BS)** | $$s = rac{B \hat{s}}{\hat{s}^2 + \hat{\Omega}_m^2}$$ | $$\Omega = rac{B \hat{\Omega}}{\hat{\Omega}_m^2 - \hat{\Omega}^2}$$ | $$\Omega_s = \min(|A_1|, |A_2|)$$ where $$A = rac{B \hat{\Omega}_p}{\hat{\Omega}_m^2 - \hat{\Omega}_p^2}$$ | --- ## 7. Common Mistakes That Cost Marks > [!danger] **The Radian vs. Cyclic Frequency Trap** > > Exam questions often specify frequencies in cyclic Hertz ($f$ in Hz). **Never** plug cyclic Hertz values directly into your s-domain formulas! You must convert them to radian frequency ($\Omega = 2\pi f$) before starting your calculations. > [!warning] **The Bandwidth Radian Scaling Oversight** > > For band-pass and band-stop filters, the bandwidth $B$ must be converted to rad/s ($B_{ ext{rad}} = 2\pi B_{ ext{Hz}}$) if you are substituting directly, or keep all calculations strictly in cyclic Hertz during intermediate prototyping, multiplying the final scaling constant by $2\pi$ when mapping to $s$. Failing to scale bandwidth by $2\pi$ shifts the poles, completely corrupting the filter shape. > [!danger] **The Butterworth Cutoff ($\Omega_c$) vs. Chebyshev Passband ($\Omega_p$) Edge Confusion** > > Remember: Butterworth filters are designed using the 3-dB cutoff frequency ($\Omega_c$), while Chebyshev/Elliptic filters are scaled using the exact passband edge frequency ($\Omega_p$). Substituting Chebyshev parameters into Butterworth formulas results in incorrect attenuation values at the boundaries. --- ## 8. Verbatim Past Year Questions ### 8.1 PYQ 2025 Question 6a / 2022 Question 6a **Question:** Define analog frequency transformation? * **Answer Plan:** Explain that analog frequency transformation is a mathematical mapping technique where the complex frequency variable $s$ of a normalized low-pass filter is replaced by a function $f(\hat{s})$ to map its specifications directly to a target low-pass, high-pass, band-pass, or band-stop filter. Outline the four fundamental substitution equations. ### 8.2 PYQ 2017 Question 5c **Question:** Design a third-order Butterworth analog high-pass filter with a passband edge of $30.777$ Hz and stopband edge of $10$ Hz, under passband attenuation of $1$ dB and stopband attenuation of $20$ dB. * **Answer Plan:** Follow the exact step-by-step mathematical derivation shown in **Section 3.3**. Write out the frequency conversions, prototype specifications, order calculations ($N=3$), cutoff calculations ($\Omega_c = 1.4309$), low-pass scaling, and the final HPF substitution to secure all marks. --- ## 9. Self-Check Before Moving On - [ ] Can you write out the exact s-domain substitutions for LP-to-LP, LP-to-HP, LP-to-BP, and LP-to-BS transformations from memory? - [ ] Do you know how to calculate the geometric mean frequency and radian bandwidth for a band-pass specification? - [ ] Do you remember the rule to convert cyclic Hz to radian rad/s before substituting variables? - [ ] Can you explain why Chebyshev filters are scaled differently than Butterworth filters? --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 5), Azmat Sir-2309008.pdf, Rabiul sir class note.pdf (Class Lecture Slides).*