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5.00 Analog Filter Design Compact Review

Nakib Noor14 min read

📋 Note Properties
COURSE[[ECE 2107 - Signals and Systems]]
CHAPTER MAP[[05 Chapter Map - Analog Filter Design]]
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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 [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 [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 and upper cutoff [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 and [5.34, 5.12]. [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 terms, they are non-zero for (anticausal) and exist from to [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_

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