Study Unit 6: Filter Design - (Part 1) First Order Filters

Study Unit 6: Filter Design - (Part 1) First Order Filters

Lecture Outcomes

  • Topics for today’s discussion:
    • General filter responses
    • First order filters
    • Differentiators and integrators
    • Low-pass, high-pass, and band-pass filters
    • Phase shifters

Common Filter Responses

Ideal Low-Pass Response
  • Frequencies below ω\omega are passed.
  • Frequencies above ω\omega are rejected.
  • Notation:
    • Green: Filter response
    • Blue: Output components
    • Red: Rejected components
  • Illustrations:
    • Input signal depicts various frequencies.
    • Filtered signal shows only frequencies below ω\omega.
Ideal High-Pass Response
  • Frequencies above ω\omega are passed.
  • Frequencies below ω\omega are rejected.
  • Notation:
    • Green: Filter response
    • Blue: Output components
    • Red: Rejected components
  • Illustrations:
    • Input signal contains various frequencies.
    • Filtered signal shows only frequencies above ω\omega.
Ideal Band-Pass Response
  • Frequencies between ω<em>L\omega<em>L and ω</em>H\omega</em>H are passed.
  • Frequencies below ωL\omega_L are rejected.
  • Frequencies above ωH\omega_H are rejected.
  • Notation:
    • Green: Filter response
    • Blue: Output components
    • Red: Rejected components
  • Illustrations:
    • Input signal contains a range of frequencies.
    • Filtered signal only contains frequencies within the ω<em>L\omega<em>L and ω</em>H\omega</em>H range.
Ideal Notch Response
  • Frequencies between ω<em>L\omega<em>L and ω</em>H\omega</em>H are rejected.
  • Frequencies below ωL\omega_L are passed.
  • Frequencies above ωH\omega_H are passed.
  • Notation:
    • Green: Filter response
    • Blue: Output components
    • Red: Rejected components
  • Illustrations:
    • Input signal has a spectrum of frequencies.
    • Filtered signal is missing the frequencies between ω<em>L\omega<em>L and ω</em>H\omega</em>H.
Generic Filter Response and its Characteristics
  • The ratio of the output and input is represented as: H(s)=X<em>oX</em>i=V<em>oV</em>iH(s) = \frac{X<em>o}{X</em>i} = \frac{V<em>o}{V</em>i}
  • Polynomial format:
    • H(s)=N(s)D(s)=a<em>msm+a</em>m−1sm−1+⋯+a<em>1s1+a</em>0b<em>nsn+b</em>n−1sn−1+⋯+b<em>1s1+b</em>0H(s) = \frac{N(s)}{D(s)} = \frac{a<em>ms^m + a</em>{m-1}s^{m-1} + \cdots + a<em>1s^1 + a</em>0}{b<em>ns^n + b</em>{n-1}s^{n-1} + \cdots + b<em>1s^1 + b</em>0}
  • Rewritten to show roots (zeros and poles):
    • H(s)=N(s)D(s)=H<em>0(s−z</em>1)(s−z<em>2)⋯(s−z</em>m)(s−p<em>1)(s−p</em>2)⋯(s−pn)H(s) = \frac{N(s)}{D(s)} = H<em>0 \frac{(s - z</em>1)(s - z<em>2)\cdots(s - z</em>m)}{(s - p<em>1)(s - p</em>2)\cdots(s - p_n)}
  • Stability:
    • Revise these concepts from previous modules (like ELI 220).

Philosophy of the Module

  • Comments about the approach:
    • Balancing math with intuition and circuit understanding.
    • Balancing analysis with design.
  • Extremely useful tool for intuitive reasoning: Asymptotic frequency insight:
    • Low frequencies (DC): lim⁡<em>ω→0Z</em>C=∞\lim<em>{\omega \to 0} Z</em>C = \infty (i.e., open circuit)
    • High frequencies: lim⁡<em>ω→∞Z</em>C=0\lim<em>{\omega \to \infty} Z</em>C = 0 (i.e., short circuit)

The Differentiator

Circuit Analysis
  • Step 1: Current equations (KCL)
    • i<em>1=v</em>in−0ZCi<em>1 = \frac{v</em>{in} - 0}{Z_C}
    • i<em>2=0−v</em>outRi<em>2 = \frac{0 - v</em>{out}}{R}
    • v<em>inZ</em>C=−voutR\frac{v<em>{in}}{Z</em>C} = -\frac{v_{out}}{R}
  • Step 2: Finding the transfer function
    • H(s)=v<em>outv</em>in=−RZC=−RCsH(s) = \frac{v<em>{out}}{v</em>{in}} = -\frac{R}{Z_C} = -RCs
  • Meaning:
    • H(jω)=−jωRC=−j(ω/ω<em>0)=(ω/ω</em>0)∠−90∘H(j\omega) = -j\omega RC = -j(\omega/\omega<em>0) = (\omega/\omega</em>0) \angle -90^\circ
    • ω0=1RC\omega_0 = \frac{1}{RC}

The (Miller) Integrator

  • Transfer function:
    • H(s)=v<em>outv</em>in=−ZCR=−1RCsH(s) = \frac{v<em>{out}}{v</em>{in}} = -\frac{Z_C}{R} = -\frac{1}{RCs}
  • Meaning:
    • H(jω)=−1jωRC=−1j(ω/ω<em>0)=j(ω/ω</em>0)=1(ω/ω0)∠+90∘H(j\omega) = -\frac{1}{j\omega RC} = -\frac{1}{j(\omega/\omega<em>0)} = \frac{j}{(\omega/\omega</em>0)} = \frac{1}{(\omega/\omega_0)} \angle +90^\circ
    • ω0=1RC\omega_0 = \frac{1}{RC}

The (Deboo) Integrator

  • Transfer function:
    • H(s)=v<em>outv</em>in=1RCsH(s) = \frac{v<em>{out}}{v</em>{in}} = \frac{1}{RCs}

Low-Pass Filter with Gain

  • Transfer function:
    • H(jω)=−R<em>2R</em>11jωR<em>2C+1=H</em>011+j(ω/ω0)H(j\omega) = -\frac{R<em>2}{R</em>1} \frac{1}{j\omega R<em>2 C + 1} = H</em>0 \frac{1}{1 + j(\omega/\omega_0)}
  • Design equations:
    • H<em>0=−R</em>2R1H<em>0 = -\frac{R</em>2}{R_1}
    • ω<em>0=1R</em>2C\omega<em>0 = \frac{1}{R</em>2 C}

High-Pass Filter with Gain

  • Transfer function:
    • H(jω)=−R<em>2R</em>1jωR<em>1CjωR</em>1C+1=H<em>0j(ω/ω</em>0)1+j(ω/ω0)H(j\omega) = -\frac{R<em>2}{R</em>1} \frac{j\omega R<em>1 C}{j\omega R</em>1 C + 1} = H<em>0 \frac{j(\omega/\omega</em>0)}{1 + j(\omega/\omega_0)}
  • Design equations:
    • H<em>0=−R</em>2R1H<em>0 = -\frac{R</em>2}{R_1}
    • ω<em>0=1R</em>1C\omega<em>0 = \frac{1}{R</em>1 C}

Band-Pass Filter with Gain

  • Transfer function:
    • H(jω)=−R<em>2R</em>1jωR<em>1C</em>1jωR<em>1C</em>1+11jωR<em>2C</em>2+1=H<em>0j(ω/ω</em>L)1+j(ω/ω<em>L)11+j(ω/ω</em>H)H(j\omega) = -\frac{R<em>2}{R</em>1} \frac{j\omega R<em>1 C</em>1}{j\omega R<em>1 C</em>1 + 1} \frac{1}{j\omega R<em>2 C</em>2 + 1} = H<em>0 \frac{j(\omega/\omega</em>L)}{1 + j(\omega/\omega<em>L)} \frac{1}{1 + j(\omega/\omega</em>H)}
  • Design equations:
    • H<em>0=−R</em>2R1H<em>0 = -\frac{R</em>2}{R_1}
    • ω<em>L=1R</em>1C1\omega<em>L = \frac{1}{R</em>1 C_1}
    • ω<em>H=1R</em>2C2\omega<em>H = \frac{1}{R</em>2 C_2}

Phase Shifter

  • Transfer function:
    • H(jω)=−jωRC+1jωRC+1H(j\omega) = \frac{-j\omega RC + 1}{j\omega RC + 1}
  • Insight into the circuit:
    • Consider phase at opamp inputs:
      • ∠v+=∠+90∘\angle v_+ = \angle +90^\circ
      • ∠v<em>−=∠v</em>+\angle v<em>- = \angle v</em>+
    • Consider the current through the top branch:
      • ∠i<em>R1∝∠v</em>i−v<em>−R</em>1∝∠0∘−(+90∘)R1∝∠−90∘\angle i<em>{R1} \propto \angle \frac{v</em>i - v<em>-}{R</em>1} \propto \angle \frac{0^\circ - (+90^\circ)}{R_1} \propto \angle -90^\circ