05 Chapter Map - Analog Filter Design

Chapter 5 Overview & Map of Content (MOC)

Butterworth order derivations, Chebyshev polynomials, Sallen-Key low-pass networks, and frequency mappings.

📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
5.00	5.00 Analog Filter Design Compact Review	Analog Filter Compact Review, Filter Design Formula Sheet
5.01	5.01 Ideal Filters Practical Specifications and Tolerance Cu	Filter Specifications, Ideal Filters, Tolerance Curve, Filter Definition
5.02	5.02 Butterworth Filter Approximation	Butterworth Approximation, Maximally Flat Filter, Butterworth Design
5.03	5.03 Chebyshev and Elliptic Filter Approximations	Chebyshev Filters, Elliptic Filters, Cauer Filters, Inverse Chebyshev
5.04	5.04 Active Filter Realization and Sallen-Key RC Networks	Active Filters, Sallen-Key, Sallen-Key Low-Pass
5.05	5.05 Analog Frequency Transformations	Frequency Transformations, Spectral Mapping, Low-Pass to High-Pass, Low-Pass to Band-Pass, Low-Pass to Band-Stop

🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

Course Teaching Plan: ECE 2107 - Signals and Systems
Previous chapter: 04 Chapter Map - The Sampling Theorem & Multi-Rate Processing
Next chapter: 06 Chapter Map - Two-Port Network Theory

Chapter 5: Analog Filter Design - Compact Review

This compact review sheet compresses all analytical and circuit design concepts of continuous-time analog filters into an exam-optimized, high-density formula guide. It covers ideal filtering parities, practical low-pass filter specs, Butterworth/Chebyshev/Elliptic approximations, Sallen-Key realizations, specialized notch designs, and analog-to-analog frequency transformation formulas.

5.01 Ideal Analog Filters & Impulse Responses

*(Target: Theory Descriptive / 5-Mark Definition)*

Filter Definition: An electrical system that selectively passes a desired band of frequency components without distortion while completely blocking all other frequency components [5.28, 5.5].
Sinc Function Definition: $\operatorname{sinc}(x) = rac{\sin(\pi x)}{\pi x}$ [5.30, 5.3]

Ideal Filter Equations & Non-Causality Symmetries

Ideal filters require “brick-wall” transitions, which physically map to infinite-duration, non-causal time-domain impulse responses [5.28, 5.29].

Ideal Low-Pass Filter (LPF): Passes frequencies in the range $- Ω_{c} \leq Ω \leq Ω_{c}$ [5.29, 5.1]. $H_{ ext{LP}}(j\Omega) = egin{cases} 1, & |\Omega| \le \Omega_c \ 0, & |\Omega| > \Omega_c \end{cases}$ [5.29, 5.1]

ight) = rac{\sin(\Omega_c t)}{\pi t} \quad ext{(derivation)}$$ [5.30, 5.2, 5.5]

Ideal High-Pass Filter (HPF): Passes frequencies for $∣Ω∣ > Ω_{c}$ [5.31, 5.6]. $H_{ ext{HP}}(j\Omega) = 1 - H_{ ext{LP}}(j\Omega) = egin{cases} 0, & |\Omega| \le \Omega_c \ 1, & |\Omega| > \Omega_c \end{cases}$ [5.32, 5.7]

ight) \quad ext{(derivation)}$$ [5.33, 5.8]

Ideal Band-Pass Filter (BPF): Passes frequencies between lower cutoff $Ω_{c 1}$ and upper cutoff $Ω_{c 2}$ [5.33, 5.9]. $H_{ ext{BP}}(j\Omega) = egin{cases} 1, & \Omega_{c1} \le |\Omega| \le \Omega_{c2} \ 0, & ext{otherwise} \end{cases}$ [5.33, 5.9]

ight) - rac{\Omega_{c1}}{\pi} \operatorname{sinc}\left(rac{\Omega_{c1} t}{\pi} ight) = rac{\sin(\Omega_{c2} t) - \sin(\Omega_{c1} t)}{\pi t} \quad ext{(derivation)}$$ [5.34, 5.11]

Ideal Band-Stop Filter (BSF): Rejects frequencies between $Ω_{c 1}$ and $Ω_{c 2}$ [5.34, 5.12]. $H_{e x t BS} (j Ω) = H_{e x t L P 1} (j Ω) + H_{e x t H P 2} (j Ω) = 1 - H_{e x t BP} (j Ω)$ [5.34, 5.35, 5.13]

ight) - rac{\Omega_{c2}}{\pi} \operatorname{sinc}\left(rac{\Omega_{c2} t}{\pi} ight) \quad ext{(derivation)}$$ [5.35, 5.14]

WARNING

The Ideal Filter Causality Trap: Because all four ideal impulse responses contain $sinc (\cdot)$ terms, they are non-zero for $t < 0$ (anticausal) and exist from $t = - \infty$ to $\infty$ [5.30, 5.108]. Thus, ideal “brick-wall” filters are physically unrealizable in real-time systems because they require knowledge of future inputs [5.30, 5.108].

5.02 Practical Filter Specifications & Tolerance Curves

*(Target: Theory Descriptive / Spec Layout)*

Practical filter designs replace the impossible “brick-wall” transition with a gradual roll-off transition band separating passband and stopband tolerance limits [5.36, 5.8].

Passband (0 to Ωp)         Transition Band (Ωp to Ωs)      Stopband (Ωs to π)
       1 + δp  _ _ _ _ _ _ _ _ _ _
              |                   |
Gain    1     |     Passband      |               |      Ripple       |   \   Transition
       1 - δp |_ _ _ _ _ _ _ _ _ _|     \    Slope
                                                  δs     _ _ _ _ _ _ _ _ _ _ _ _ _ _ _  \ _ _ _ _ _ _ _ Stopband Floor (Max δs)
                                                     
``` [5.37, 5.8]

*   **$\Omega_p$ (Passband Edge Frequency):** Boundary defining the end of the passband [5.37, 5.8].
*   **$\Omega_s$ (Stopband Edge Frequency):** Boundary defining the start of the stopband [5.37, 5.8].
*   **$lpha_p$ or $A_p$ (Peak Passband Ripple in dB):** Maximum allowed attenuation in the passband [5.37, 5.8].
    $$lpha_p = -20 \log_{10}(1 - \delta_p) = 10 \log_{10}(1 + \epsilon^2) \quad 	ext{(derivation)}$$ [5.37, 5.49, 5.31]
*   **$lpha_s$ or $A_s$ (Minimum Stopband Attenuation in dB):** Minimum required suppression in the stopband [5.37, 5.8].
    $$lpha_s = -20 \log_{10}(\delta_s) \quad 	ext{(derivation)}$$ [5.37, 5.37, 5.8]
*   **Ripple Amplitude Bounds:**
    $$\delta_p = 1 - 10^{-lpha_p/20} \quad 	ext{and} \quad \delta_s = 10^{-lpha_s/20}$$ [5.37, 5.37]

---

## 5.03 Butterworth Filter Approximation
`*(Target: Mathematical Proof & Numerical Solving)*`

*   **Concept:** Designed to have a **maximally flat magnitude response** at $\Omega = 0$ [5.38]. The first $2N-1$ derivatives of the magnitude-squared response at $\Omega = 0$ are equal to zero [5.38]. Gain decays monotonically in both passband and stopband [5.44].
*   **Magnitude Response Equation:**
    $$|H_a(j\Omega)|^2 = rac{1}{1 + \left(rac{\Omega}{\Omega_c}
ight)^{2N}}$$ [5.38, 5.15]
    *(At cutoff frequency $\Omega = \Omega_c$, the gain is exactly $-3	ext{ dB}$, representing $|H_a(j\Omega_c)| = 1/\sqrt{2}$ [5.38]).*

### Step-by-Step Design Formulas (Derivations)
Starting from the attenuation specifications at $\Omega_p$ and $\Omega_s$ [5.39]:
$$lpha_p = 10 \log_{10} \left[ 1 + \left(rac{\Omega_p}{\Omega_c}
ight)^{2N} 
ight] \quad 	ext{and} \quad lpha_s = 10 \log_{10} \left[ 1 + \left(rac{\Omega_s}{\Omega_c}
ight)^{2N} 
ight]$$ [5.39, 5.17, 5.18]

1.  **Filter Order ($N$):**
    $$N \ge rac{\log_{10}\left(rac{10^{0.1lpha_s} - 1}{10^{0.1lpha_p} - 1}
ight)}{2 \log_{10}\left(rac{\Omega_s}{\Omega_p}
ight)} \quad 	ext{(derivation)}$$ [5.40, 5.23]
    *(Note: $N$ must always be rounded up to the next higher integer [5.40]).*
2.  **Cutoff Frequency ($\Omega_c$):**
    $$\Omega_c = rac{\Omega_p}{(10^{0.1lpha_p} - 1)^{1/2N}} = rac{\Omega_s}{(10^{0.1lpha_s} - 1)^{1/2N}} \quad 	ext{(derivation)}$$ [5.40, 5.19, 5.20]
    *(In practice, using the stopband spec formula guarantees the stopband requirements are exactly satisfied, while passband requirements are exceeded with a safe margin [5.40]).*

### S-Domain Pole Locations
The poles $p_k$ of a normalized Butterworth filter lie symmetrically on a left-half s-plane circle of radius $\Omega_c$ [5.41, 5.25].
$$p_k = \Omega_c e^{j 	heta_k} \quad 	ext{where} \quad 	heta_k = rac{\pi}{2} + rac{(2k-1)\pi}{2N}, \quad k = 1, 2, \dots, N \quad 	ext{(derivation)}$$ [5.41, 5.26, 5.27]

| Order $N$ | Normalized Butterworth Denominator Polynomials $H_N(s)$ (for $\Omega_c = 1$) |
| :--- | :--- |
| **$N=1$** | $s + 1$ |
| **$N=2$** | $s^2 + \sqrt{2}s + 1$ |
| **$N=3$** | $(s + 1)(s^2 + s + 1)$ |
| **$N=4$** | $(s^2 + 0.76537s + 1)(s^2 + 1.8477s + 1)$ |
| **$N=5$** | $(s + 1)(s^2 + 0.61803s + 1)(s^2 + 1.61803s + 1)$ |
| **$N=6$** | $(s^2 + 0.51764s + 1)(s^2 + \sqrt{2}s + 1)(s^2 + 1.931855s + 1)$ |

---

## 5.04 Chebyshev Filter Approximations (Type I & II)
`*(Target: Theory Descriptive & Numerical Solving)*`

### Chebyshev Type I: Passband Ripple
*   **Concept:** Exhibits equal-ripple (equiripple) behavior in the passband and monotonic decay in the stopband [5.55].
*   **Magnitude Response Equation:**
    $$|H(j\Omega)|^2 = rac{1}{1 + \epsilon^2 C_N^2\left(rac{\Omega}{\Omega_p}
ight)}$$ [5.48, 5.28]
*   **Ripple Parameter ($\epsilon$):**
    $$\epsilon = \sqrt{10^{0.1lpha_p} - 1}$$ [5.49, 5.32]
*   **Chebyshev Polynomials $C_N(x)$:**
    $$C_N(x) = egin{cases} \cos(N \cos^{-1}(x)), & |x| \le 1 \ \cosh(N \cosh^{-1}(x)), & |x| > 1 \end{cases}$$ [5.48, 5.49, 5.29]
    *   *Recurrence Formula:* $C_N(x) = 2x C_{N-1}(x) - C_{N-2}(x)$ with $C_0(x)=1, C_1(x)=x$ [5.49].
*   **Filter Order ($N$):**
    $$N \ge rac{\cosh^{-1}\left(\sqrt{rac{10^{0.1lpha_s} - 1}{10^{0.1lpha_p} - 1}}
ight)}{\cosh^{-1}\left(rac{\Omega_s}{\Omega_p}
ight)} \quad 	ext{(derivation)}$$ [5.50, 5.35]
    *(Note: Evaluate using $\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$ [5.50]).*
*   **Left-Half S-Plane Pole Locations:** Poles $p_k = x_k + jy_k$ lie on an **ellipse** in the s-plane [5.50, 5.51]:
    $$x_k = -\sinh(\phi) \sin\left(rac{(2k-1)\pi}{2N}
ight) \quad 	ext{and} \quad y_k = \cosh(\phi) \cos\left(rac{(2k-1)\pi}{2N}
ight) \quad 	ext{(derivation)}$$ [5.51, 5.36, 5.37]
    $$	ext{where} \quad \phi = rac{1}{N} \sinh^{-1}\left(rac{1}{\epsilon}
ight)$$ [5.51, 5.36]

### Chebyshev Type II: Inverse Chebyshev (Stopband Ripple)
*   **Concept:** Exhibits a flat (monotonic) response in the passband and equiripple behavior in the stopband [5.59, 5.63].
*   **Magnitude Response Equation:**
    $$|H(j\Omega)|^2 = rac{1}{1 + \epsilon^2 rac{C_N^2(\Omega_s/\Omega_p)}{C_N^2(\Omega_s/\Omega)}}$$ [5.59, 5.40]
*   **Features:** Contains finite transmission **zeros** on the imaginary $j\Omega$-axis [5.59]:
    $$z_k = j rac{\Omega_s}{\cos\left(rac{(2k-1)\pi}{2N}
ight)} \quad 	ext{for } k=1,2,\dots,N$$ [5.59, 5.42a]

---

## 5.05 Elliptic & Bessel Filters
`*(Target: Theory Descriptive / Comparison)*`

### Elliptic (Cauer) Filter
*   **Concept:** Exhibits equiripple behavior in **both** the passband and the stopband [5.69].
*   **Magnitude Response Equation:**
    $$|H(j\Omega)|^2 = rac{1}{1 + \epsilon^2 U_N^2\left(rac{\Omega}{\Omega_p}
ight)}$$ [5.66, 5.43]
    *(where $U_N(x)$ is the Jacobian elliptic function of order $N$ [5.66]).*
*   **Key Advantage:** Provides the **sharpest possible transition band roll-off** for a given order $N$ among all filter types [5.50, 5.77].
*   **Key Disadvantage:** Highly non-linear phase response in the passband [5.77].

### Bessel Filter
*   **Concept:** An all-pole filter optimized to provide a **perfectly linear phase response** (flat group delay) in the passband [5.72].
*   **Governing Equation:** Uses Bessel polynomials in the denominator [5.72]:
    $$H_a(s) = rac{a_0}{\sum_{n=0}^{N} a_n s^n} \quad 	ext{where} \quad a_n = rac{(2N - n)!}{2^{N-n} n! (N - n)!}$$ [5.72, 5.73, 5.49, 5.50]
*   **Trade-off:** Perfectly preserves waveform pulse shapes without phase dispersion, but exhibits a very slow transition band attenuation slope [5.73, 5.77].

---

## 5.06 Grand Filter Comparison Matrix
`*(Target: Theory Descriptive / High-Yield Exam Questions)*`

This comparison matrix summarizes the physical characteristics, ripple distributions, transition sharpness, and phase behaviors of the 5 key analog filter topologies [5.77]:

| Filter Type | Passband Behavior | Stopband Behavior | Transition Band Roll-off (Same Order) | Phase Linearity (Passband) | Poles/Zeros |
| :--- | :--- | :--- | :--- | :--- | :--- |
| **Butterworth** [5.38, 5.77] | Maximally Flat [5.38] | Monotonic [5.44] | Moderate / Slow [5.77] | Fairly Good [5.77] | Poles only (On a Circle) [5.41] |
| **Chebyshev I** [5.55, 5.77] | Equiripple [5.55] | Monotonic [5.55] | Steep / Fast [5.77] | Poor [5.77] | Poles only (On an Ellipse) [5.50] |
| **Chebyshev II** [5.59, 5.77] | Monotonic [5.63] | Equiripple [5.63] | Steep / Fast [5.77] | Fairly Good | Poles & $j\Omega$-axis Zeros [5.59] |
| **Elliptic** [5.69, 5.77] | Equiripple [5.69] | Equiripple [5.69] | **Sharpest / Fastest** [5.77] | Highly Non-Linear [5.77] | Poles & $j\Omega$-axis Zeros [5.69] |
| **Bessel** [5.72, 5.77] | Monotonic [5.73] | Monotonic | **Poorest / Slowest** [5.73] | **Perfectly Linear** [5.72] | Poles only [5.72] |

---

## 5.07 Active Filter Realization & Sallen-Key Topologies
`*(Target: Circuit Analysis / Numerical Solving)*`

      r1            r2

vi o----[ R1 ]---o----[ R2 ]----+-----------+ | | | --- c1 --- c2 | --- --- +| | | | | +--------------|---------| - ____o vo | | / | +---------| - / | |/ / | | | +---------+

 
The **Sallen-Key active filter** realizes a second-order (2-pole) low-pass system using two passive resistors, two capacitors, and a unity-gain operational amplifier buffer [5.67, 5.23].
 
*   **Nodal Circuit Relations:** Let $V_1$ be the node voltage between $R_1$ and $R_2$, and $V_2$ be the non-inverting terminal node voltage [5.23]:
    $$I_{r1} = I_{r2} + I_{c1} \quad \implies \quad rac{V_i - V_1}{R_1} = rac{V_1 - V_o}{R_2} + s C_1 (V_1 - V_o)$$ [5.23]
    $$I_{r2} = I_{c2} \quad \implies \quad rac{V_1 - V_o}{R_2} = s C_2 V_o \quad \implies \quad V_1 = (1 + s R_2 C_2) V_o$$ [5.23]
*   **System Transfer Function:** Substituting $V_1$ into the primary node equation [5.23]:
    $$H(s) = rac{V_o(s)}{V_i(s)} = rac{1}{s^2 \left(R_1 R_2 C_1 C_2
ight) + s C_2\left(R_1 + R_2
ight) + 1} \quad 	ext{(derivation)}$$ [5.24]
*   **Cutoff Frequency ($\Omega_c$) & Quality Factor ($Q$):**
    $$\Omega_c = rac{1}{\sqrt{R_1 R_2 C_1 C_2}} \quad 	ext{and} \quad Q = rac{\sqrt{R_1 R_2 C_1 C_2}}{C_2(R_1 + R_2)}$$
 
---
 
## 5.08 Specialized Filters: Notch by Pole-Zero Placement
`*(Target: Circuit Design / Pole-Zero Placement)*`
 
*   **Concept:** A narrow band-stop filter designed to completely suppress a single target frequency (e.g., $50	ext{ Hz}$ power line hum) [5.6, 5.106].
*   **Pole-Zero Placement Rule:** 
    *   Place a pair of conjugate **zeros** directly on the imaginary axis at the target notch frequency: $z_{1,2} = \pm j\Omega_0$ [5.106, 5.107].
    *   To make the notch sharp and maintain flat unity gain across all other frequencies, place a pair of conjugate **poles** very close to the zeros, shifted slightly left into the stable s-plane [5.107]: $p_{1,2} = -\Omega_0 \cos 	heta \pm j\Omega_0 \sin 	heta$ [5.107, 5.108].
*   **Transfer Function:**
    $$H(s) = rac{(s - j\Omega_0)(s + j\Omega_0)}{(s + \Omega_0 \cos 	heta - j\Omega_0 \sin 	heta)(s + \Omega_0 \cos 	heta + j\Omega_0 \sin 	heta)}$$ [5.108]
    $$H(s) = rac{s^2 + \Omega_0^2}{s^2 + 2\Omega_0 \cos 	heta \cdot s + \Omega_0^2} \quad 	ext{(derivation)}$$ [5.108]
    *(Note: As $	heta 	o \pi/2$ ($\cos	heta 	o 0$), the notch bandwidth approaches $0$, creating an infinitely sharp spike suppression [5.107]).*
 
---
 
## 5.09 Analog-to-Analog Frequency Transformations
`*(Target: Numerical Solving / System Mapping)*`
 
To design a high-pass, band-pass, or band-stop filter, we first map the desired specs back to a normalized prototype LPF ($\Omega_p = 1	ext{ rad/s}$), design the low-pass transfer function $H_N(s)$, and then replace $s$ using the appropriate algebraic frequency scaling transformation [5.79]:

LPF (Prototype) Desired HPF Desired BPF Desired BSF | | | \ /| |\ \ | | / ---| |— ---| |--- ---| | | |--- ---| | |--- 0 1 0 Ωp Ωp1 0 Ωp2 Ωs1 0 Ωs2

 
*   **Low-Pass to Low-Pass (LPF to LPF):**
    $$s \longrightarrow rac{s}{\hat{\Omega}_p} \quad 	ext{(derivation)}$$ [5.80, 5.81b]
*   **Low-Pass to High-Pass (LPF to HPF):**
    $$s \longrightarrow rac{\hat{\Omega}_p}{s} \quad 	ext{(derivation)}$$ [5.83, 5.84]
*   **Low-Pass to Band-Pass (LPF to BPF):**
    $$s \longrightarrow rac{s^2 + \hat{\Omega}_0^2}{B \cdot s} \quad 	ext{(derivation)}$$ [5.90, 5.55]
    *   *Symmetry Formulas:* bandwidth $B = \hat{\Omega}_{p2} - \hat{\Omega}_{p1}$ and geometric center frequency $\hat{\Omega}_0 = \sqrt{\hat{\Omega}_{p1} \hat{\Omega}_{p2}}$ [5.89, 5.90, 5.54a, 5.54b].
*   **Low-Pass to Band-Stop (LPF to BSF):**
    $$s \longrightarrow rac{B \cdot s}{s^2 + \hat{\Omega}_0^2} \quad 	ext{(derivation)}$$ [5.97, 5.59]
    *   *Symmetry Formulas:* stopband bandwidth $B = \hat{\Omega}_{s2} - \hat{\Omega}_{s1}$ and geometric notch center $\hat{\Omega}_0 = \sqrt{\hat{\Omega}_{s1} \hat{\Omega}_{s2}}$ [5.98, 5.97, 5.60].
 
---
 
## 5.10 Common Mistakes That Cost Marks
 
*   **The Pole Stability Signs Swap:** When writing the normalized Butterworth transfer function $H(s) = rac{1}{\prod (s - p_k)}$, remember that the stable poles $p_k$ lie in the LHP and have a **negative real part** [5.41]. If your denominator factors contain terms like $(s - 1)$, you have selected unstable poles, which will result in a zero grade [5.46].
*   **Filter Order Rounding Error:** When evaluating the order formula for Butterworth or Chebyshev filters, if you obtain $N = 3.12$, you **must round up** to $N = 4$ [5.40, 5.50]. Selecting $N = 3$ fails to satisfy the stopband suppression specifications [5.40].
*   **Frequency Parameter Radian Unit Omission:** Textbook and exam questions often provide frequency values in Hertz (e.g., $f_p = 1000	ext{ Hz}$) [5.153]. You **must convert these to angular frequency** ($\Omega = 2\pi f$) before applying any order or design formulas [5.45].
*   **Chebyshev Type I vs Type II Ripple Placement:** Do not confuse the two types [5.55, 5.59]. Type I has ripples in the passband [5.55]; Type II has ripples in the stopband [5.63].
 
---
 
## 5.11 PYQ Bank — Verbatim Questions & Answer Plans
 
### [KUET 2025, 2023 - 13 Marks]
*   **Question:** *Design a Butterworth analog low-pass filter with $1	ext{ dB}$ passband ripple, passband edge frequency $\Omega_p = 2000\pi	ext{ rad/s}$, stopband edge frequency $\Omega_s = 10000\pi	ext{ rad/s}$, and a minimum stopband ripple of $40	ext{ dB}$.* [5.45]
*   **Answer Plan:**
    1.  Extract specs: $lpha_p = 1	ext{ dB}$, $lpha_s = 40	ext{ dB}$, $\Omega_p = 2000\pi$, $\Omega_s = 10000\pi$ [5.46].
    2.  Calculate order $N$ using the order formula to get $N = 3.28 	o 	ext{round up to } N = 4$ [5.47].
    3.  Select $N=4$ normalized Butterworth polynomial: $H_N(s) = rac{1}{(s^2 + 0.76537s + 1)(s^2 + 1.8477s + 1)}$ [5.47, 5.46].
    4.  Determine cutoff frequency $\Omega_c = rac{\Omega_s}{(10^{4} - 1)^{1/8}} = 9935	ext{ rad/s}$ [5.47].
    5.  Perform low-pass scaling substitute $s 	o s / 9935$ to obtain the final transfer function $H_a(s)$ [5.47, 5.48].
 
### [KUET 2025 - 3 Marks]
*   **Question:** *Design an elliptic LPF with: $f_p = 1	ext{ kHz}, f_s = 1.5	ext{ kHz}, A_p = 1	ext{ dB}, A_s = 40	ext{ dB}$. Determine: (i) order, $N$; (ii) ripple factor $\epsilon$; (iii) Transfer function.*
*   **Answer Plan:**
    1.  Convert frequencies: $\Omega_p = 2000\pi$, $\Omega_s = 3000\pi$, selectivity factor $k = \Omega_p / \Omega_s = 2/3 pprox 0.667$ [5.70, 5.44].
    2.  Calculate ripple factor: $\epsilon = \sqrt{10^{0.1} - 1} pprox 0.5088$ [5.49, 5.32].
    3.  Estimate order $N$ using the elliptic order estimation formulas [5.67, 5.45] to find the minimum integer order $N$.
    4.  Extract the corresponding transfer function from the normalized elliptic lookup tables [5.69, 5.70].
 
### [KUET 2024 - 4 Marks]
*   **Question:** *Draw the specifications for an analog low-pass filter with tolerance curve.*
*   **Answer Plan:** Reconstruct a labeled low-pass filter specifications graph showing the passband, stopband, transition band, passband peak ripple bounds $1 \pm \delta_p$, and the stopband attenuation floor $\delta_s$ [5.37, 5.8].
 
### [KUET 2021, 2017 - 5 Marks]
*   **Question:** *Determine the output response of a low-pass RC network for an input signal $x(t) = e^{-t/RC}u(t)$.*
*   **Answer Plan:** Solve using the time-domain convolution integral $y(t) = x(t) * h(t)$, where the RC network impulse response is $h(t) = rac{1}{RC} e^{-t/RC}u(t)$ [5.8, 8.181]. Alternatively, convert to s-domain: $X(s) = rac{1}{s + 1/RC}$, $H(s) = rac{1/RC}{s + 1/RC}$ and apply inverse Laplace transform to get $y(t) = rac{t}{RC} e^{-t/RC}u(t)$ [8.182].
 
---
 
## 5.12 Self-Check Before Moving On
 
*   [ ] Can you write down the four ideal filter impulse responses from memory? [5.30, 5.33, 5.34, 5.35]
*   [ ] Can you derive the Butterworth filter order $N$ and cutoff frequency $\Omega_c$ formulas from passband/stopband attenuation equations? [5.40]
*   [ ] Do you know how to calculate Chebyshev Type I pole locations on an ellipse? [5.50, 5.51]
*   [ ] Can you derive the Sallen-Key second-order low-pass transfer function using nodal analysis? [5.23, 5.24]
*   [ ] Do you have the low-pass to band-pass and low-pass to band-stop frequency transformations memorized? [5.90, 5.97]
 
---
_Source: ECE 2107 Syllabus, (K. Deergha Rao) Signals and Systems Ch 5, Rabiul Sir class notes_
 
---
 
**Related Concepts:** [[5.02_Butterworth_Filter_Approximation|5.02 Butterworth Filter Approximation]] | [[5.03_Chebyshev_and_Elliptic_Filter_Approximations|5.03 Chebyshev, Elliptic & Bessel Filter Approximations]] | [[5.04_Active_Filter_Realization_and_Sallen-Key_RC_Networks|5.04 Active Filter Realization & Sallen-Key RC Networks]]
 
---
 
# 5.01 Ideal Filters, Practical Specifications & Tolerance Curves
 
> [!abstract] Core Idea
> 
> **Filtering** is a fundamental frequency-domain operation designed to pass a desired band of frequency components without distortion while completely suppressing all other undesired frequency bands. While ideal continuous-time filters have "brick-wall" rectangular frequency profiles, they require non-causal sinc-shaped impulse responses that extend infinitely into negative time, making them physically unrealizable. Practical filter design therefore relies on specifying mathematical tolerance bands (passband ripple, stopband attenuation, and transition widths) that can be approximated by realizable rational transfer functions.
 
---
 
## 1. What is Filtering?
In signal processing, **filtering** is defined as an operation that selectively shapes the spectrum of an input signal. A filter is an LTI system characterized by a frequency response $H(j\Omega)$ that acts as a spectral gate.
 
### 1.1 Major Applications of Filtering
Examiners frequently ask why filtering is critical in engineering. Key modern applications include:
1.  **Noise Suppression:** Stripping high-frequency thermal noise from low-frequency physiological signals (e.g., ECG or EEG).
2.  **Demodulation & Baseband Recovery:** Extracting message signals $m(t)$ from modulated carrier waves $s(t)$ in communication receivers.
3.  **Anti-Aliasing:** Placing a strict low-pass filter before an Analog-to-Digital Converter (ADC) to band-limit the input signal to under $f_s/2$.
4.  **Frequency-Division Multiplexing (FDM):** Demultiplexing multiple adjacent channels sharing a single medium by isolating individual carriers.
5.  **Structural Vibration Analysis:** Isolating mechanical vibration frequencies of a bridge or high-rise building caused specifically by wind forces rather than passing vehicles.
 
---
 
## 2. Mathematical Models of Ideal Analog Filters
An **ideal analog filter** possesses a frequency response with flat, unity-gain passbands and zero-gain stopbands, separated by discontinuous vertical cuts called **brick-wall boundaries**.
 
### 2.1 The Ideal Low-Pass Filter (LPF)
An ideal low-pass filter passes all frequency components below a specified **cutoff frequency** $\Omega_c$ and completely blocks all components above it:
$$H_{\text{LP}}(j\Omega) = \begin{cases} 1, & |\Omega| \le \Omega_c \\ 0, & |\Omega| > \Omega_c \end{cases}$$
 
#### Step-by-Step Impulse Response Derivation:
The impulse response $h_{\text{LP}}(t)$ is the inverse Fourier transform of $H_{\text{LP}}(j\Omega)$:
$$h_{\text{LP}}(t) = \mathcal{L}^{-1}\{H_{\text{LP}}(j\Omega)\} = \frac{1}{2\pi} \int_{-\infty}^{\infty} H_{\text{LP}}(j\Omega) e^{j\Omega t} d\Omega$$
Substitute the limits of the passband:
$$h_{\text{LP}}(t) = \frac{1}{2\pi} \int_{-\Omega_c}^{\Omega_c} (1) e^{j\Omega t} d\Omega = \frac{1}{2\pi} \left[ \frac{e^{j\Omega t}}{jt} \right]_{-\Omega_c}^{\Omega_c}$$
$$h_{\text{LP}}(t) = \frac{e^{j\Omega_c t} - e^{-j\Omega_c t}}{2\pi jt} = \frac{\sin(\Omega_c t)}{\pi t}$$
Using the normalized sinc function definition where $\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$:
$$h_{\text{LP}}(t) = \frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right)$$
 
---
 
### 2.2 The Ideal High-Pass Filter (HPF)
An ideal high-pass filter blocks all frequencies below $\Omega_c$ and passes all frequencies above it:
$$H_{\text{HP}}(j\Omega) = \begin{cases} 0, & |\Omega| \le \Omega_c \\ 1, & |\Omega| > \Omega_c \end{cases}$$
We can express the ideal high-pass response as a parallel subtraction of the low-pass block from an all-pass channel:
$$H_{\text{HP}}(j\Omega) = 1 - H_{\text{LP}}(j\Omega)$$
Taking the inverse Fourier transform of both sides yields the impulse response:
$$h_{\text{HP}}(t) = \delta(t) - \frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right)$$
 
---
 
### 2.3 The Ideal Band-Pass Filter (BPF)
An ideal band-pass filter passes frequencies situated within a restricted band between cutoff frequencies $\Omega_{c1}$ and $\Omega_{c2}$ (where $\Omega_{c1} < \Omega_{c2}$):
$$H_{\text{BP}}(j\Omega) = \begin{cases} 1, & \Omega_{c1} \le |\Omega| \le \Omega_{c2} \\ 0, & \text{otherwise} \end{cases}$$
Its impulse response is derived by taking individual passband integrals:
$$h_{\text{BP}}(t) = \frac{1}{2\pi} \int_{-\Omega_{c2}}^{-\Omega_{c1}} e^{j\Omega t} d\Omega + \frac{1}{2\pi} \int_{\Omega_{c1}}^{\Omega_{c2}} e^{j\Omega t} d\Omega$$
$$h_{\text{BP}}(t) = \frac{\sin(\Omega_{c2} t) - \sin(\Omega_{c1} t)}{\pi t} = \frac{\Omega_{c2}}{\pi} \text{sinc}\left(\frac{\Omega_{c2} t}{\pi}\right) - \frac{\Omega_{c1}}{\pi} \text{sinc}\left(\frac{\Omega_{c1} t}{\pi}\right)$$
 
---
 
### 2.4 The Ideal Band-Stop Filter (BSF)
An ideal band-stop (or band-rejection) filter blocks all frequency components lying within the band from $\Omega_{c1}$ to $\Omega_{c2}$ and passes all other frequencies:
$$H_{\text{BS}}(j\Omega) = \begin{cases} 1, & |\Omega| \le \Omega_{c1} \text{ and } |\Omega| \ge \Omega_{c2} \\ 0, & \Omega_{c1} < |\Omega| < \Omega_{c2} \end{cases}$$
Representing the band-stop filter as a parallel sum of an LPF and an HPF:
$$H_{\text{BS}}(j\Omega) = H_{\text{LP}}(j\Omega) + H_{\text{HP}}(j\Omega)$$
Taking the inverse Fourier transform yields the impulse response:
$$h_{\text{BS}}(t) = \delta(t) + \frac{\Omega_{c1}}{\pi} \text{sinc}\left(\frac{\Omega_{c1} t}{\pi}\right) - \frac{\Omega_{c2}}{\pi} \text{sinc}\left(\frac{\Omega_{c2} t}{\pi}\right)$$
 
---
 
## 3. The Non-Causality & Unrealizability Proof
A critical theoretical question in ECE 2107 examinations is: **"Prove that ideal filters are physically unrealizable."**
 
### Proof of Non-Causality:
1.  **Definition of Causality:** For an LTI system to be physically causal, its impulse response must satisfy:
    $$h(t) = 0 \quad \text{for } t < 0$$
2.  **The Ideal LPF Impulse Response:** As derived in Section 2.1, the impulse response of an ideal LPF is:
    $$h_{\text{LP}}(t) = \frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right) = \frac{\sin(\Omega_c t)}{\pi t}$$
3.  **Evaluating for Negative Time ($t < 0$):**
    For any $t < 0$ (excluding the discrete zero-crossings where $\Omega_c t = -k\pi$):
    $$h_{\text{LP}}(t) \neq 0$$
    *{Because the Sinc function envelope decays as $1/t$, it has non-zero oscillatory tails extending all the way back to $t = -\infty$}*.
4.  **Physical Meaning:** This implies that the system produces an output response *before* the input impulse is applied at $t = 0$. This requires anticipation of future inputs, which is physically impossible.
5.  **Conclusion:** Therefore, the ideal filter is **non-causal** and cannot be physically built using real-world resistors, capacitors, and active components.
 
---
 
## 4. Practical Analog Filter Specifications & Tolerance Curves
Since ideal "brick-wall" filters are physically impossible, practical filters must tolerate some deviation from unity gain in the passband and some non-zero leakage in the stopband. The transition from the passband to the stopband cannot be instantaneous; it must occur over a finite **transition band**.
 
### 4.1 Parameter Definitions
To design a realizable rational filter, we define six mandatory parameters:
*   $\Omega_p$ (**Passband Edge Frequency**): The frequency bounding the end of the passband.
*   $\Omega_s$ (**Stopband Edge Frequency**): The frequency defining the start of the stopband.
*   $\delta_p$ (**Passband Peak Ripple**): The maximum allowed amplitude deviation from unity gain in the passband.
*   $\delta_s$ (**Stopband Peak Leakage**): The maximum allowed amplitude gain in the stopband.
*   $\alpha_p$ (**Peak Passband Ripple in dB** or **Passband Loss**): 
    $$\alpha_p = -20 \log_{10}(1 - \delta_p) \text{ dB}$$
*   $\alpha_s$ (**Minimum Stopband Attenuation in dB** or **Stopband Loss**):
    $$\alpha_s = -20 \log_{10}(\delta_s) \text{ dB}$$
 
### 4.2 Practical Filter Specifications Tolerance Graph
 
The following graph maps the tolerance boundaries that any practical low-pass filter's magnitude response $|H(j\Omega)|$ must respect:

Amplitude |H(jΩ)|
   ^

1+δp |-------------+
1 | ~ ~ ~ ~ | \ Passband Ripple Band 1-δp |-------------+--------------------------------- | |
| | \ Transition Band | |
| |
δs | | +------------------------ ⇐ Stopband Floor 0 +-------------+---------+----------------------⇒ 0 Ωp Ωs Ω (Frequency)

> *Placeholder Source: Figure 5.8 (Specifications of a low-pass analog filter) — (k.Deergha Rao) signals and systems.pdf (Ch 5)*.

---

## 5. Summary of Ideal Filter Responses

| Filter Type | Frequency Domain Profile $H(j\Omega)$ | Impulse Response $h(t)$ |
| :--- | :--- | :--- |
| **Low-Pass (LPF)** | $1 \text{ for } |\Omega| \le \Omega_c$; else $0$ | $\frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right)$ |
| **High-Pass (HPF)** | $1 \text{ for } |\Omega| > \Omega_c$; else $0$ | $\delta(t) - \frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right)$ |
| **Band-Pass (BPF)** | $1 \text{ for } \Omega_{c1} \le |\Omega| \le \Omega_{c2}$; else $0$ | $\frac{\Omega_{c2}}{\pi}\text{sinc}\left(\frac{\Omega_{c2}t}{\pi}\right) - \frac{\Omega_{c1}}{\pi}\text{sinc}\left(\frac{\Omega_{c1}t}{\pi}\right)$ |
| **Band-Stop (BSF)** | $0 \text{ for } \Omega_{c1} < |\Omega| < \Omega_{c2}$; else $1$ | $\delta(t) + \frac{\Omega_{c1}}{\pi}\text{sinc}\left(\frac{\Omega_{c1}t}{\pi}\right) - \frac{\Omega_{c2}}{\pi}\text{sinc}\left(\frac{\Omega_{c2}t}{\pi}\right)$ |

---

## 6. Common Mistakes That Cost Marks

> [!danger] **The Sinc Definition Discrepancy**
> 
> Standard mathematical software (like MATLAB) utilizes the **normalized** sinc function: $\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$. However, many engineering textbooks define it as the **unnormalized** function: $\text{Sa}(x) = \frac{\sin(x)}{x}$. In examinations, always state your definition clearly before writing out step-responses. If utilizing the normalized sinc function, the LPF impulse response **must** include the scaling factors: $h(t) = \frac{\Omega_c}{\pi} \text{sinc}\left(\frac{\Omega_c t}{\pi}\right)$.

> [!warning] **The Inverse Log Formula Sign Trap**
> 
> When solving for the peak ripple parameters $\delta_p$ and $\delta_s$ from dB specifications, remember the signs:
> $$\delta_p = 1 - 10^{-\alpha_p/20} \quad \text{and} \quad \delta_s = 10^{-\alpha_s/20}$$
> Writing positive exponents (e.g., $10^{\alpha_s/20}$) results in astronomical values that make no physical sense for fractional gain limits.

---

## 7. PYQ Bank — Verbatim Questions & Answer Plans

### 7.1 PYQ 2024 Question 4a [08 Marks]
**Question:** Draw the specifications for analog low pass filter with tolerance curve.
*   **Answer Plan:**
    1.  Sketch the exact specifications curve shown in **Section 4.2**, clearly labeling the vertical axis ($1+\delta_p$, $1-\delta_p$, and $\delta_s$) and horizontal axis ($\Omega_p$, $\Omega_s$, and the transition band).
    2.  Write out the mathematical definitions of passband ripple $\alpha_p$ (dB) and minimum stopband attenuation $\alpha_s$ (dB).
    3.  Define the terms: Passband Edge, Stopband Edge, Passband Ripple, and Stopband Attenuation.

### 7.2 PYQ 2023 Question 4a [04 Marks]
**Question:** What is filtering? What are the applications of it?
*   **Answer Plan:**
    1.  Define filtering as a frequency-domain operation designed to select certain bands while blocking others.
    2.  State at least four practical applications with keywords (Noise suppression, ECG signal cleanup, FDM Demux, demodulation, anti-aliasing).

---

## 8. Self-Check Before Moving On

- [ ] Can you derive the impulse response of an ideal LPF from its frequency domain rect function?
- [ ] Do you know why ideal filters are mathematically non-causal?
- [ ] Can you write down the algebraic formulas linking decibel parameters ($\alpha_p, \alpha_s$) with peak ripple values ($\delta_p, \delta_s$)?
- [ ] Can you draw the practical low-pass tolerance curve from memory?

---

*Source: (k.Deergha Rao) signals and systems.pdf (Chapter 5), Azmat Sir-2309008.pdf (Lecture Slides), signal checklist.md.*

---

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

---

# 5.02 Butterworth Filter Approximation

> [!abstract] 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].

---

## 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:
$$\left. \frac{d^k |H_a(j\Omega)|^2}{d\Omega^k} \right|_{\Omega=0} = 0 \quad \text{for } k = 1, 2, \dots, 2N-1$$

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 $\Omega_c$.

---

## 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]:
$$N \ge \frac{\log_{10} \left( \frac{10^{0.1\alpha_s} - 1}{10^{0.1\alpha_p} - 1} \right)}{2 \log_{10} \left(\frac{\Omega_s}{\Omega_p}\right)} \tag{7}$$

> [!important] **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**:
    $$\Omega_{c, \text{stop}} = \frac{\Omega_s}{\left(10^{0.1\alpha_s} - 1\right)^{1/2N}} \tag{8b}$$

> [!tip] **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].

---

## 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 $s = j\Omega \implies \Omega = s/j$ [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 $N=3$.*

---

## 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]:
$$H_a(s) = \left. H_N(s) \right|_{s \, \to \, \frac{s}{\Omega_c}} \tag{13}$$

---

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

> [!note] **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]:
$$H_a(s) = \frac{9.7425 \times 10^{15}}{\left(s^2 + 7604.10s + 9.8704 \times 10^7\right) \left(s^2 + 18357.10s + 9.8704 \times 10^7\right)} \tag{14}$$

---

## 6. Common Mistakes That Cost Marks

> [!danger] **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].

> [!warning] **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.

---

## 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.

---

## 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]

---

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

---

[[5.02_Butterworth_Filter_Approximation|5.02 Butterworth Filter Approximation]] | [[5.04_Active_Filter_Realization_and_Sallen-Key_RC_Networks|5.04 Active Filter Realization & Sallen-Key RC Networks]]

---

# 5.03 Chebyshev & Elliptic Filter Approximations

> [!abstract] Core Idea
> 
> While Butterworth filters offer a perfectly flat magnitude response at the cost of a wide transition band, **Chebyshev** and **Elliptic** filters trade flatness in the passband and/or stopband to achieve dramatically narrower transition widths {sharper roll-offs} for the same filter order [5.2.6]. By distributing mathematical ripples across the operating bands, these approximations provide highly optimized hardware realizations [5.2.3, 5.2.4].

---

## 1. The Philosophical Shift: Ripple vs. Transition Width

When designing practical filters, we are constrained by a fundamental trade-off: **passband flatness versus transition roll-off steepness** [5.2.6]. 
*   **Butterworth:** Maximally flat passband response, but requires a very high order $N$ to achieve steep attenuation [5.2.2].
*   **Chebyshev Type I:** Permits controlled **ripples in the passband** to secure a much faster roll-off in the transition region and a monotonic stopband [5.2.3].
*   **Chebyshev Type II (Inverse Chebyshev):** Maintains a perfectly flat (monotonic) passband, but introduces **ripples in the stopband** to maximize transition steepness [5.2.3].
*   **Elliptic (Cauer):** Allows **ripples in both the passband and stopband**, yielding the absolute steepest transition band possible for any given filter order $N$ [5.2.4, 5.2.6].

---

## 2. Chebyshev Type I Approximation (Passband Ripple)

The magnitude-square response of an $N$-th order Chebyshev Type I filter is mathematically defined by [5.2.3]:
$$|H(j\Omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\Omega}{\Omega_p}\right)}$$

Where:
*   $\epsilon$ is the **ripple factor** {controls the depth of passband ripples} [5.2.3].
*   $\Omega_p$ is the **passband edge frequency** [5.2.1].
*   $T_N(x)$ is the **$N$-th order Chebyshev polynomial** [5.2.3].

### 2.1 The Chebyshev Polynomials
The Chebyshev polynomials are defined piecewise as [5.2.3]:
$$T_N(x) = \begin{cases} 
\cos(N \cos^{-1}(x)), & |x| \le 1 \quad \text{(Passband)} \\ 
\cosh(N \cosh^{-1}(x)), & |x| > 1 \quad \text{(Stopband)} 
\end{cases}$$

These polynomials can be generated recursively using the recurrence relation [5.03]:
$$T_N(x) = 2x T_{N-1}(x) - T_{N-2}(x) \quad \text{with} \quad T_0(x) = 1, \ T_1(x) = x$$

#### First Five Chebyshev Polynomials:
*   $T_0(x) = 1$
*   $T_1(x) = x$
*   $T_2(x) = 2x^2 - 1$
*   $T_3(x) = 4x^3 - 3x$
*   $T_4(x) = 8x^4 - 8x^2 + 1$
*   $T_5(x) = 16x^5 - 20x^3 + 5x$

### 2.2 Deriving Ripple Factor ($\epsilon$) and Order ($N$)
1.  **Ripple Factor ($\epsilon$):** At the passband edge ($\Omega = \Omega_p$), $T_N(1) = 1$. The passband attenuation $\alpha_p$ (in dB) is [5.2.3]:
    $$\alpha_p = 10 \log_{10}(1 + \epsilon^2) \implies \epsilon = \sqrt{10^{0.1 \alpha_p} - 1}$$
2.  **Filter Order ($N$):** At the stopband edge ($\Omega = \Omega_s$), the attenuation must satisfy $\alpha_s$ [5.2.3]:
    $$\alpha_s = 10 \log_{10}\left(1 + \epsilon^2 \cosh^2\left(N \cosh^{-1}\left(\frac{\Omega_s}{\Omega_p}\right)\right)\right)$$
    Solving for $N$ yields the critical design formula [5.2.3]:
    $$N \ge \frac{\cosh^{-1}\left(\frac{\sqrt{10^{0.1 \alpha_s} - 1}}{\epsilon}\right)}{\cosh^{-1}\left(\frac{\Omega_s}{\Omega_p}\right)}$$
    Where $\cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right)$ is evaluated algebraically [5.2.3].

### 2.3 Pole Locations on the s-Plane Semicircle
Unlike Butterworth poles (which lie on a circle), Chebyshev Type I poles lie symmetrically on a **s-plane ellipse** [5.2.3]. The stable Left-Half s-Plane (LHP) poles are given by [5.2.3]:
$$p_k = \sigma_k + j \omega_k, \quad k = 1, 2, \dots, N$$
Where [5.2.3]:
$$\sigma_k = -\sinh(\beta) \sin\left(\frac{(2k-1)\pi}{2N}\right)$$
$$\omega_k = \cosh(\beta) \cos\left(\frac{(2k-1)\pi}{2N}\right)$$
$$\beta = \frac{1}{N} \sinh^{-1}\left(\frac{1}{\epsilon}\right) = \frac{1}{N} \ln\left(\frac{1}{\epsilon} + \sqrt{\frac{1}{\epsilon^2} + 1}\right)$$

### 2.4 Normalized Chebyshev Type I Denominator Polynomials (Ripple = 1 dB)
For $\alpha_p = 1\text{ dB} \implies \epsilon = 0.5088$, the normalized s-domain transfer functions are [5.2.3]:

| $N$ | Denominator Polynomial $D(s)$ | Scaling Constant $H_0$ |
| :--- | :--- | :--- |
| **1** | $s + 1.9652$ | $1.9652$ |
| **2** | $s^2 + 1.0977s + 1.1025$ | $0.98261$ |
| **3** | $s^3 + 0.9883s^2 + 1.2384s + 0.4913$ | $0.49131$ |
| **4** | $s^4 + 0.9528s^3 + 1.4539s^2 + 0.7426s + 0.2756$ | $0.24565$ |
| **5** | $s^5 + 0.9368s^4 + 1.6888s^3 + 0.9744s^2 + 0.5805s + 0.1228$ | $0.12283$ |

> [!important] The $H_0$ Scaling Rule
> 
> To ensure correct passband scaling at DC ($s=0$), the numerator constant is calculated as [5.2.3]:
> *   For **odd $N$**: $H_0 = \prod |p_k| = D(0) \implies H_s(0) = 1$
> *   For **even $N$**: $H_0 = \frac{\prod |p_k|}{\sqrt{1+\epsilon^2}} = \frac{D(0)}{\sqrt{1+\epsilon^2}} \implies H_s(0) = \frac{1}{\sqrt{1+\epsilon^2}}$

---

## 3. Chebyshev Type II (Inverse Chebyshev) Approximation

Chebyshev Type II filters keep a perfectly flat passband (like Butterworth) but introduce equal-amplitude ripples in the stopband to maintain sharp transition properties [5.2.3, 5.2.6].

The magnitude-square response is given by [5.2.3]:
$$|H(j\Omega)|^2 = \frac{1}{1 + \epsilon^2 \frac{T_N^2(\Omega_s/\Omega_p)}{T_N^2(\Omega_s/\Omega)}} = \frac{T_N^2(\Omega_s/\Omega)}{T_N^2(\Omega_s/\Omega) + \delta_s^2 T_N^2(\Omega_s/\Omega_p)}$$

### 3.1 Zeros and Poles Formulation
1.  **Transmission Zeros ($z_k$):** Because the response decays to finite ripples instead of zero at high frequencies, the transfer function possesses zeros on the imaginary axis [5.2.3]:
    $$z_k = j \frac{\Omega_s}{\cos\left(\frac{(2k-1)\pi}{2N}\right)}, \quad k = 1, 2, \dots, N$$
    *(Note: For odd $N$, the zero at $k = \frac{N+1}{2}$ lies at infinity).*
2.  **Poles ($p_k$):** The poles of a Type II filter are the **reciprocals of a prototype Type I filter's poles** [5.2.3]:
    $$p_k = \frac{-\sigma_k \Omega_s}{\sigma_k^2 + \Omega_k^2} + j \frac{\omega_k \Omega_s}{\sigma_k^2 + \Omega_k^2}$$
    Where $\sigma_k$ and $\omega_k$ are the prototype elliptical coordinates computed with stopband ripple parameter $\delta_s = \frac{1}{\sqrt{10^{0.1 \alpha_s} - 1}}$ [5.2.3].

---

## 4. Elliptic (Cauer) Filter Approximation

The **Elliptic Filter** is the most mathematically efficient analog filter approximation [5.2.4, 5.2.6]. By distributing equiripples across **both** the passband and the stopband, it yields the sharpest possible roll-off for a given order $N$ [5.2.4].

The magnitude-square response is defined as [5.2.4]:
$$|H(j\Omega)|^2 = \frac{1}{1 + \epsilon^2 U_N^2\left(\frac{\Omega}{\Omega_p}\right)}$$
Where $U_N(x)$ is the **Jacobian elliptic function** of order $N$ [5.2.4].

### 4.1 Elliptic Parameter Design Rules
The sharp transition boundary is characterized by the **selectivity factor ($k$)** [5.2.4]:
$$k = \frac{\Omega_p}{\Omega_s} \quad \text{(where } 0 < k < 1\text{)}$$

To estimate the required order $N$, we apply the following Jacobian coordinate algorithm [5.2.4]:
1.  Compute the complementary selectivity parameter $k'$ [5.2.4]:
    $$k' = \sqrt{1 - k^2}$$
2.  Evaluate the initial coordinate parameter $\rho_0$ [5.2.4]:
    $$\rho_0 = \frac{1 - \sqrt{k'}}{2(1 + \sqrt{k'})}$$
3.  Calculate the modular constant $\rho$ [5.2.4]:
    $$\rho = \rho_0 + 2\rho_0^5 + 15\rho_0^9 + 150\rho_0^{13}$$
4.  Solve for the minimum required integer order $N$ [5.2.4]:
    $$N \ge \frac{\log_{10}\left(16 \frac{10^{0.1 \alpha_s} - 1}{10^{0.1 \alpha_p} - 1}\right)}{\log_{10}(1/\rho)}$$

---

## 5. High-Yield Comparative Analysis

To help you secure maximum marks on qualitative questions, here is the official ECE 2107 comparison of the five primary analog filter families [5.2.6, 5.2.7]:

### 5.1 The 5-Way Filter Property Comparison Matrix

| Parameter | Butterworth | Chebyshev I | Chebyshev II | Elliptic | Bessel |
| :--- | :--- | :--- | :--- | :--- | :--- |
| **Passband Shape** | Monotonic (Flat) | Equiripple | Monotonic (Flat) | Equiripple | Monotonic |
| **Stopband Shape** | Monotonic | Monotonic | Equiripple | Equiripple | Monotonic |
| **Transition Width** | Extremely Wide | Medium-Narrow | Medium-Narrow | Steepest (Ultra-Narrow) | Extremely Wide |
| **Phase Linearity** | Moderately Linear | Non-linear | Moderately Linear | Highly Non-linear | **Perfectly Linear** |
| **Transient Response** | Moderate overshoot | Poor (Ringing) | Moderate overshoot | Severely degraded ringing | **Zero Overshoot** |

### 5.2 Order Requirements for Identical Specifications [Table 5.5]
*   **Target Specifications:** $f_p = 450\text{ Hz}$, $f_s = 550\text{ Hz}$, $\alpha_p = 1\text{ dB}$, $\alpha_s = 35\text{ dB}$ [5.2.6]:

[ORDER STEEPNESS COMPARISON] Butterworth (N = 24) ======================== Chebyshev I (N = 9) ========= Chebyshev II (N = 9) ========= Elliptic (N = 5) =====


---

## 6. Comprehensive Worked Examples (The Exam Killers)

### 6.1 Elliptic Parameter Design [PYQ 2025 - 13 Marks]
**Question:** Design an Elliptic Low-Pass Filter with $f_p = 1\text{ kHz}$, $f_s = 1.5\text{ kHz}$, $\alpha_p = 1\text{ dB}$, and $\alpha_s = 40\text{ dB}$. Determine:
1.  The filter order $N$.
2.  The ripple factor $\epsilon$.
3.  The normalized s-domain transfer function $H_N(s)$.

#### Step-by-Step Solution:

1.  **Calculate Selectivity Factors:**
    $$\Omega_p = 2\pi(1000) = 2000\pi \text{ rad/s}, \quad \Omega_s = 2\pi(1500) = 3000\pi \text{ rad/s}$$
    $$k = \frac{\Omega_p}{\Omega_s} = \frac{2000\pi}{3000\pi} = \frac{2}{3} \approx 0.6667$$
    $$k' = \sqrt{1 - k^2} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \frac{\sqrt{5}}{3} \approx 0.745356$$
2.  **Calculate Jacobian Coordinates:**
    $$\rho_0 = \frac{1 - \sqrt{0.745356}}{2(1 + \sqrt{0.745356})} = \frac{1 - 0.86334}{2(1 + 0.86334)} = \frac{0.13666}{3.72668} \approx 0.03667$$
    Since $\rho_0$ is small, the higher order terms are negligible:
    $$\rho \approx \rho_0 = 0.03667$$
3.  **Evaluate Filter Order $N$:**
    $$N \ge \frac{\log_{10}\left(16 \frac{10^{0.1(40)} - 1}{10^{0.1(1)} - 1}\right)}{\log_{10}(1/\rho)} = \frac{\log_{10}\left(16 \frac{9999}{0.258925}\right)}{\log_{10}(27.269)}$$
    $$N \ge \frac{\log_{10}(617875.2)}{1.43567} = \frac{5.7909}{1.43567} \approx 4.03$$
    Rounding up to the next integer yields:
    $$\mathbf{N = 5}$$
4.  **Calculate Ripple Factor $\epsilon$:**
    $$\epsilon = \sqrt{10^{0.1 \alpha_p} - 1} = \sqrt{10^{0.1(1)} - 1} = \sqrt{0.258925} \approx \mathbf{0.5088}$$
5.  **Formulate Normalized Transfer Function:**
    From the normalized Elliptic reference table for $\alpha_p = 1\text{ dB}, \alpha_s = 40\text{ dB}$, and $N = 5$ [5.2.4]:
    $$H_N(s) = \frac{0.0470 s^4 + 0.2201 s^2 + 0.2299}{s^5 + 0.9234 s^4 + 1.8471 s^3 + 1.1292 s^2 + 0.7881 s + 0.2299}$$

---

## 7. Common Mistakes That Cost Marks

> [!danger] **The Even-Order DC Scaling Trap**
> 
> In Chebyshev Type I designs, when the order $N$ is **even**, the magnitude response at $s=0$ does not start at $1.0$. It starts at $\frac{1}{\sqrt{1+\epsilon^2}}$ (due to even symmetry ending on a ripple trough) [5.2.3]. Failing to apply this scaling factor to $H_0$ will result in a **loss of 3–5 marks** on your transfer function.

> [!warning] **The Inverse Chebyshev Ripple Location Mix-Up**
> 
> Do not mix up the ripple regions of Type I and Type II Chebyshev filters. Type I has ripples in the passband and is monotonic in the stopband [5.2.3]. Type II has ripples in the stopband and is monotonic in the passband [5.2.3]. Examiners love testing this distinction via 4-mark short questions.

---

## 8. PYQ Bank — Verbatim Questions & Answer Plans

### 8.1 PYQ 2025 Question 5c [13 Marks]
**Question:** Design an Elliptic Low-Pass Filter with $f_p = 1\text{ kHz}$, $f_s = 1.5\text{ kHz}$, $\alpha_p = 1\text{ dB}$, and $\alpha_s = 40\text{ dB}$. Determine: (i) order, n; (ii) ripple factor, $\epsilon$; (iii) Transfer function.
*   **Answer Plan:** Follow the step-by-step math layout shown in **Section 6.1**. Write out the selectivity variables, use the logarithm formula for order approximation, define the ripple factor, and write the final normalized transfer function to secure all 13 marks.

---

## 9. Self-Check Before Moving On

- [ ] Can you recursively generate Chebyshev polynomials up to $T_5(x)$? [5.03]
- [ ] Do you know how to calculate the stable s-plane poles for a Type I Chebyshev filter? [5.2.3]
- [ ] Do you understand why Chebyshev Type II filters have zeros on the $j\Omega$ axis? [5.2.3]
- [ ] Can you evaluate the Jacobian coordinate algorithm ($\rho_0, \rho$) to calculate Elliptic orders? [5.2.4]

---

*Source: (k.Deergha Rao) signals and systems.pdf (Chapter 5), Signals, Systems, and Networks MOC Roadmap (Chapter 5).*

---

[[5.03_Chebyshev_and_Elliptic_Filter_Approximations|5.03 Chebyshev & Elliptic Filter Approximations]] | [[5.05_Analog_Frequency_Transformations|5.05 Analog Frequency Transformations]]

---

# 5.04 Active Filter Realization & Sallen-Key RC Networks

> [!abstract] Core Idea
> 
> Passive filter topologies containing resistors, capacitors, and inductors ($RLC$) become highly bulky, expensive, and lossy at low frequencies (e.g., audio or medical ranges) because low-frequency inductors require large physical iron cores. **Active filter realization** eliminates inductors entirely by using operational amplifiers (op-amps) alongside resistors and capacitors ($RC$) [5.1]. The **Sallen-Key topology** is the classic second-order active filter structure used to implement complex conjugate poles without inductors, providing excellent frequency selectivity, high input impedance, and zero loading effects [1.298, 4.46].

---

## 1. Passive vs. Active Filter Realizations

Before deriving the active filter equations, it is critical for examinations to understand **why** modern signal processing relies heavily on active realizations rather than classical passive $RLC$ circuits [1.41].

### 1.1 Limitations of Passive $RLC$ Filters
1. **Bulky and Heavy Inductors:** At low frequencies ($f < 100\text{ kHz}$), the required inductance values are on the order of millihenries or henries. Achieving these values with passive coils requires large winding counts and heavy ferromagnetic cores, which are physically impractical [1.41].
2. **High Losses (Low $Q$):** Physical inductors have significant winding resistance. This internal resistance dissipates power, degrades the sharpness of the filter (lowers the quality factor $Q$), and deviates the circuit from its ideal mathematical model.
3. **Insertion Loss (No Gain):** Passive filters cannot amplify signals. The maximum gain of a passive network is always less than or equal to $1$ ($0\text{ dB}$).
4. **Severe Loading Effects:** Passive filters are highly sensitive to source and load impedances. Connecting a load resistor directly to a passive filter alters its pole-zero locations, distorting the frequency response and necessitating complex impedance-matching networks.

### 1.2 Advantages of Active $RC$ Realizations
* **Inductor Elimination:** Active filters replace bulky inductors with a combination of active op-amps and passive $RC$ elements, significantly reducing weight, cost, and physical size [1.41].
* **High Input & Low Output Impedance:** Op-amps act as physical isolating buffers. They draw negligible current from the source (infinite input impedance) and can drive heavy loads easily (near-zero output impedance), completely eliminating loading effects and allowing simple cascaded designs.
* **Integrated Gain:** Active filters can easily provide voltage or power gain ($A_v \ge 1$) by configuring the feedback path of the op-amp.

---

## 2. Sallen-Key Low-Pass Filter Topology

The **Sallen-Key filter** (introduced by R. P. Sallen and E. L. Key in 1955) is the most popular active filter topology for realizing second-order ($2\text{nd}$-order) filter blocks [1.298]. 

### 2.1 Circuit Schematic
A unity-gain second-order active low-pass Sallen-Key filter consists of two resistors ($R_1, R_2$), two capacitors ($C_1, C_2$), and an op-amp configured as a voltage follower {unity-gain buffer} [4.46]:

                 C1 (Feedback Path)
           +-----------||------------+
           |                         |
           |   R1          R2        |    |\

Vi(s) o-----+---///---v1---///—+—v2---|
| | ____o Vo(s) ----- C2 |- / | ----- | / | | |/ | GND | | | +-------------+ (Unity Gain)


### 2.2 Functional Path Breakdown
* **$R_1$ and $R_2$:** Form the primary resistive path carrying the input signal toward the non-inverting terminal.
* **$C_2$:** Acts as a low-pass element that shunts high-frequency signals directly to ground [4.46].
* **$C_1$:** Provides a positive feedback path from the low-impedance op-amp output back to the intermediate node $v_1$ [4.46]. This feedback boosts the signal near the cutoff frequency, creating the desired sharp "knee" (resonance) in the frequency response.

---

## 3. Step-by-Step Nodal Derivation of the Transfer Function

To secure full marks on analytical exam questions, you must prove the system function of the Sallen-Key circuit from foundational circuit principles using node-voltage analysis [4.46]:

### Step 1: Establish Node Voltages & Op-Amp Constraints
Let the non-inverting input node of the op-amp be $v_2(t)$, and the intermediate node between $R_1$ and $R_2$ be $v_1(t)$. 
Because the op-amp is configured as an ideal unity-gain voltage follower [4.46]:
$$v_2(s) = v_o(s)$$

Since the input terminals of an ideal op-amp draw zero current, the current through capacitor $C_2$ is exactly equal to the current through $R_2$. We write the current through $C_2$ in the $s$-domain as [4.46]:
$$I_{C2}(s) = s C_2 v_2(s) = s C_2 v_o(s)$$

### Step 2: Formulate Node Equation at $v_2$
Applying Kirchhoff's Current Law (KCL) at node $v_2$ [4.46]:
$$\frac{v_1(s) - v_2(s)}{R_2} = I_{C2}(s)$$

Substitute $v_2(s) = v_o(s)$ and $I_{C2}(s) = s C_2 v_o(s)$ into this expression [4.46]:
$$\frac{v_1(s) - v_o(s)}{R_2} = s C_2 v_o(s)$$

Multiply by $R_2$ and isolate $v_1(s)$ [4.46]:
$$v_1(s) - v_o(s) = s R_2 C_2 v_o(s)$$
$$v_1(s) = (1 + s R_2 C_2) v_o(s) \quad \text{--- [Equation 1]}$$

### Step 3: Formulate Node Equation at $v_1$
Applying KCL at the intermediate node $v_1$ (sum of outgoing currents must equal zero) [4.46]:
$$I_{R1}(s) + I_{R2}(s) + I_{C1}(s) = 0$$
$$\frac{v_1(s) - v_i(s)}{R_1} + \frac{v_1(s) - v_o(s)}{R_2} + s C_1 (v_1(s) - v_o(s)) = 0$$

Multiply the entire equation by $R_1 R_2$ to clear the denominators:
$$R_2 (v_1(s) - v_i(s)) + R_1 (v_1(s) - v_o(s)) + s R_1 R_2 C_1 (v_1(s) - v_o(s)) = 0$$

Regroup the terms to isolate the input voltage $v_i(s)$ on one side:
$$R_2 v_i(s) = (R_1 + R_2 + s R_1 R_2 C_1) v_1(s) - (R_1 + s R_1 R_2 C_1) v_o(s) \quad \text{--- [Equation 2]}$$

### Step 4: Substitute and Eliminate $v_1(s)$
Now, substitute the expression for $v_1(s)$ from **Equation 1** into **Equation 2** [4.46]:
$$R_2 v_i(s) = (R_1 + R_2 + s R_1 R_2 C_1) \cdot (1 + s R_2 C_2) v_o(s) - (R_1 + s R_1 R_2 C_1) v_o(s)$$

Divide both sides by $R_2$ to make the algebra easier to trace [4.47]:
$$v_i(s) = \left[1 + \frac{R_1}{R_2} + s R_1 C_1\right] (1 + s R_2 C_2) v_o(s) - \left[\frac{R_1}{R_2} + s R_1 C_1\right] v_o(s)$$

Expand the product of the two binomial terms:
$$\left(1 + \frac{R_1}{R_2} + s R_1 C_1\right) (1 + s R_2 C_2) = 1 + \frac{R_1}{R_2} + s R_1 C_1 + s R_2 C_2 + s R_1 C_2 + s^2 R_1 R_2 C_1 C_2$$

Substitute this expansion back into the input equation [4.47]:
$$v_i(s) = \left[\left(1 + \frac{R_1}{R_2} + s R_1 C_1 + s R_2 C_2 + s R_1 C_2 + s^2 R_1 R_2 C_1 C_2\right) - \left(\frac{R_1}{R_2} + s R_1 C_1\right)\right] v_o(s)$$

Notice how the term $\frac{R_1}{R_2}$ and the term $s R_1 C_1$ subtract out completely! This simplifies the equation to [4.47]:
$$v_i(s) = \left[s^2 R_1 R_2 C_1 C_2 + s (R_1 + R_2) C_2 + 1\right] v_o(s)$$

### Step 5: Formulate the Final System Transfer Function
The continuous-time system function $H(s)$ is the ratio of output to input [4.45, 4.47]:
$$H(s) = \frac{v_o(s)}{v_i(s)} = \frac{1}{s^2 (R_1 R_2 C_1 C_2) + s (R_1 + R_2) C_2 + 1}$$

To write this in standard quadratic second-order form, divide the numerator and denominator by the lead coefficient $R_1 R_2 C_1 C_2$ [4.47]:
$$H(s) = \frac{\frac{1}{R_1 R_2 C_1 C_2}}{s^2 + s \frac{R_1 + R_2}{R_1 R_2 C_1} + \frac{1}{R_1 R_2 C_1 C_2}}$$

---

## 4. Parameter Mapping & Filter Design

For standard active filter design, we map our derived transfer function directly to the standard $2\text{nd}$-order LPF model [1.51, 1.296]:
$$H(s) = \frac{\omega_0^2}{s^2 + 2\zeta \omega_0 s + \omega_0^2} = \frac{\omega_0^2}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2}$$

Where:
* **$\omega_0$** is the undamped natural angular frequency {cutoff frequency} [1.296].
* **$\zeta$** is the damping ratio [1.296].
* **$Q$** is the Quality Factor governing peaking at the cutoff band [1.296].

By comparing the coefficients of the Sallen-Key transfer function, we establish the following design relations:

### 4.1 Natural Cutoff Frequency ($\omega_0$)
$$\omega_0^2 = \frac{1}{R_1 R_2 C_1 C_2} \implies \omega_0 = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}$$

### 4.2 Quality Factor ($Q$)
$$\frac{\omega_0}{Q} = \frac{R_1 + R_2}{R_1 R_2 C_1} \implies Q = \frac{\omega_0 \cdot R_1 R_2 C_1}{R_1 + R_2} = \frac{\sqrt{R_1 R_2 C_1 C_2}}{(R_1 + R_2) C_2} = \frac{\sqrt{\frac{C_1}{C_2}} \cdot \sqrt{R_1 R_2}}{R_1 + R_2}$$

### 4.3 High-Yield Equal-Component Simplification
If we set $R_1 = R_2 = R$ and $C_1 = C_2 = C$:
$$\omega_0 = \frac{1}{\sqrt{R^2 C^2}} = \frac{1}{R C}$$
$$Q = \frac{\sqrt{\frac{C}{C}} \cdot \sqrt{R^2}}{R + R} = \frac{R}{2R} = 0.5$$

> [!warning] **The Equal-Component Limit**
> 
> Designing with $R_1=R_2$ and $C_1=C_2$ limits the Quality Factor to $Q = 0.5$. This represents an **over-damped response** with a very soft roll-off, which cannot approximate a sharp Butterworth response ($Q = 1/\sqrt{2} \approx 0.707$) or Chebyshev peaking. Therefore, practical active filters must be designed with unequal components [1.51].

---

## 5. ECE 2108 Laboratory Connection & MATLAB Codes

In **ECE 2108 (Signal & Systems Laboratory) Experiment 5**, students must implement continuous state-space models and observe their step, magnitude, and phase responses [1.245, 1.260].

### 5.1 Nodal-to-State-Space Conversion
To analyze the Sallen-Key filter in MATLAB, we represent our derived $2\text{nd}$-order system using a companion state-space model [1.260]:
$$\dot{\mathbf{X}}(t) = \mathbf{A}\mathbf{X}(t) + \mathbf{B}\mathbf{U}(t)$$
$$\mathbf{Y}(t) = \mathbf{C}\mathbf{X}(t) + \mathbf{D}\mathbf{U}(t)$$

For the Sallen-Key parameters $b_0 = \frac{1}{R_1 R_2 C_1 C_2}$ and $a_1 = \frac{R_1 + R_2}{R_1 R_2 C_1}$:
$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ -b_0 & -a_1 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad \mathbf{C} = \begin{bmatrix} b_0 & 0 \end{bmatrix}, \quad \mathbf{D} = \begin{bmatrix} 0 \end{bmatrix}$$

### 5.2 Lab Manual MATLAB Script [1.262, 1.263]
Below is the standard, fully documented MATLAB script to model, plot, and verify the Sallen-Key low-pass filter:

```matlab
% ECE 2108 Experiment 5: Sallen-Key Active Low-Pass Filter Analysis
clc; clear all; close all;

% 1. Define Component Values
R1 = 10e3;  % 10 kOhms
R2 = 10e3;  % 10 kOhms
C1 = 22e-9; % 22 nF
C2 = 10e-9; % 10 nF

% 2. Calculate State-Space Coefficients
b0 = 1 / (R1 * R2 * C1 * C2);
a1 = (R1 + R2) / (R1 * R2 * C1);

% 3. Define State Matrices (Control Canonic Form)
A = [0, 1; -b0, -a1];
B = [0; 1];
C = [b0, 0];
D = 0;

% 4. Initialize State-Space System
sys = ss(A, B, C, D);

% 5. Generate Plots
figure(1)
step(sys) % Plots the transient step response of the active circuit
title('Sallen-Key Filter: Step Response');
grid on;

figure(2)
bode(sys) % Plots the frequency magnitude (dB) and phase (degrees) response
title('Sallen-Key Filter: Bode Plot');
grid on;

% 6. System Verification Checks
stb = isstable(sys);
if stb == 1
    disp('System Status: Absolutely Stable');
else
    disp('System Status: Unstable');
end

% 7. Controllability Check
S = ctrb(sys);
if det(S) == 0
    disp('Controllability: Not Controllable');
else
    disp('Controllability: Fully Controllable');
end

% 8. Observability Check
V = obsv(sys);
if det(V) == 0
    disp('Observability: Not Observable');
else
    disp('Observability: Fully Observable');
end

6. Common Mistakes That Cost Marks

The Op-Amp Output Node KCL Violation

When writing nodal equations for the Sallen-Key circuit, never apply KCL at the op-amp output node $v_{o}$ ! An ideal op-amp output terminal acts as a low-impedance voltage source that can supply or sink arbitrary amounts of current to maintain its output voltage. The current leaving the op-amp output is mathematically unknown, meaning KCL at node $v_{o}$ is unsolvable and leads to a zero-mark evaluation [4.46].

The Passive Cascade Confusion

Do not confuse a cascaded passive RC circuit (which is formed by simply connecting two passive RC stages) with the active Sallen-Key circuit [1.298].

Cascading passive RC networks can only yield real, separated poles, which limits the maximum quality factor to $Q \leq 0.5$ and rounds off the frequency response curve.

The active Sallen-Key network uses positive feedback ( $C_{1}$ ) to realize complex conjugate poles, allowing $Q \geq 0.707$ to create highly sharp Butterworth or Chebyshev low-pass transitions.

7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 ECE 2107 Question [10 Marks]

Question: Obtain the system function of the second-order Sallen-Key low-pass filter circuit shown in the diagram and discuss its stability [4.46].

Answer Plan:
1. Draw the annotated schematic labeling node voltages $v_{1}$ , $v_{2}$ , and the output follower relation $v_{2} = v_{o}$ [4.46].
2. Set up the node equation at $v_{2}$ to write $v_{1} (s) = (1 + s R_{2} C_{2}) v_{o} (s)$ [4.46].
3. Apply KCL at node $v_{1}$ , substitute $v_{1} (s)$ to isolate $v_{i} (s)$ and show the step-by-step algebraic cancelation of the $R_{1} / R_{2}$ and $s R_{1} C_{1}$ terms [4.47].
4. Write the final system function $H (s) = \frac{1}{s ^{2} R _{1} R _{2} C _{1} C _{2} + s ( R _{1} + R _{2} ) C _{2} + 1}$ [4.47].
5. Discuss stability: Since all component values ( $R_{1}, R_{2}, C_{1}, C_{2}$ ) are positive, all poles of $H (s)$ lie strictly in the left-half s-plane (LHP). Therefore, the Sallen-Key low-pass filter is absolutely stable [4.47].

8. Self-Check Before Moving On

Can you list the four main limitations of passive $R L C$ circuits at low frequencies? [1.41]
Can you sketch the Sallen-Key unity-gain active low-pass schematic from memory? [4.46]
Do you know how to derive the Sallen-Key transfer function without looking at the node substitutions? [4.46, 4.47]
Why is an equal-component active filter ( $R_{1} = R_{2}, C_{1} = C_{2}$ ) limited to $Q = 0.5$ ? [1.296]
Can you write the state-space matrices of a second-order Sallen-Key transfer function in Control Canonic Form? [1.260]

Source: (k.Deergha Rao) signals and systems.pdf (Chapter 4), ECE 2108 Laboratory Manual (Experiment 5), Azmat Sir-2309008.pdf (Class Slides).

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

5.05 Analog Frequency Transformations

Core Idea

Analog frequency transformations are algebraic variable substitutions ( $so f (\overset{s}{^})$ ) that map a standardized, normalized analog low-pass prototype filter (with cutoff frequency $Ω_{p} = 1$ 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.

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 $Ω_{p} = 1$ 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 ( $\overset{s}{^}$ ):

$s = f (\overset{s}{^})$

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

Continuous Mapping: The imaginary axis of the $s$ -plane ( $s = j Ω$ ) is mapped directly to the imaginary axis of the $\overset{s}{^}$ -plane ( $\overset{s}{^} = j \hat{Ω}$ ).
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 $\overset{s}{^}$ -plane, preserving causality and absolute stability.

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

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

2.1 The Mathematical Mapping

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

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

When $\hat{Ω} = 0$ , the scaled frequency is $Ω = 0$ .
When the target frequency reaches the target cutoff boundary $\hat{Ω} = \hat{Ω}_{p}$ , the scaled prototype frequency is $Ω = 1$ 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 $Ω_{p} = 1$ . Therefore: $\Omega_s = rac{\hat{\Omega}_s}{\hat{\Omega}_p}$

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

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 $\hat{Ω}_{p}$ 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 ( $\hat{Ω} = 0$ ), the prototype frequency is $Ω = \infty$ (maps passband stop behaviors to zero frequency).
- At the target passband edge ( $\hat{Ω} = \hat{Ω}_{p}$ ), the prototype frequency is $Ω = - 1$ rad/s (the negative sign disappears under magnitude squaring).
- At infinite frequency ( $\hat{Ω} = \infty$ ), the prototype frequency is $Ω = 0$ (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 $Ω_{s}$ to design a target HPF with specs $\hat{Ω}_{p}$ and $\hat{Ω}_{s}$ : $\Omega_s = rac{\hat{\Omega}_p}{\hat{\Omega}_s}$

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: $Ω_{p} = 1 e x t r a d / s$ $\Omega_s = rac{\hat{\Omega}_p}{\hat{\Omega}_s} = rac{30.777}{10} = 3.0777 ext{ rad/s} \quad ext{[114]}$

Step 3: Find the Filter Order $N$

Using the Butterworth order estimation equation:

007_index

Chapter 5 - Analog Filter Design

05 Chapter Map - Analog Filter Design

📚 Study Notes Index

🎯 Exam Weight

Chapter 5: Analog Filter Design - Compact Review

5.01 Ideal Analog Filters & Impulse Responses

Ideal Filter Equations & Non-Causality Symmetries

5.02 Practical Filter Specifications & Tolerance Curves

6. Common Mistakes That Cost Marks

7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 ECE 2107 Question [10 Marks]

8. Self-Check Before Moving On

5.05 Analog Frequency Transformations

1. The Core Philosophy of Frequency Mapping

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

2.1 The Mathematical Mapping

2.2 Spec Scaling Adjustments

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

3.1 The Mathematical Mapping

3.2 High-Pass Spec Warping Equations

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

Step 1: Convert Cyclic Frequencies to Radian Frequencies

Step 2: Warp Specifications to Normalized Low-Pass Prototype

Step 3: Find the Filter Order $N$

Graph View

Table of Contents

007_index

Chapter 5 - Analog Filter Design

05 Chapter Map - Analog Filter Design

📚 Study Notes Index

🎯 Exam Weight

🔗 Related Resources

Chapter 5: Analog Filter Design - Compact Review

5.01 Ideal Analog Filters & Impulse Responses

Ideal Filter Equations & Non-Causality Symmetries

5.02 Practical Filter Specifications & Tolerance Curves

6. Common Mistakes That Cost Marks

7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 ECE 2107 Question [10 Marks]

8. Self-Check Before Moving On

5.05 Analog Frequency Transformations

1. The Core Philosophy of Frequency Mapping

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

2.1 The Mathematical Mapping

2.2 Spec Scaling Adjustments

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

3.1 The Mathematical Mapping

3.2 High-Pass Spec Warping Equations

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

Step 1: Convert Cyclic Frequencies to Radian Frequencies

Step 2: Warp Specifications to Normalized Low-Pass Prototype

Step 3: Find the Filter Order N

Graph View

Table of Contents

Step 3: Find the Filter Order $N$