Generalised 2nd Order Active Filter Prototype_B
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This document summarizes a student project on generalized 2nd order active filter prototypes. It includes: - Deriving voltage transfer functions for bandpass, lowpass, highpass, and bandstop filters using generalized circuit topologies. - Key equations relating the voltage transfer function to component values. - MATLAB simulation to prove the equations and find cutoff frequencies and gain. - Deriving the phase response equation and simulating the bandpass filter's phase shift and frequency response. - Discussing issues encountered like op-amp stability and varying SPICE models.
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Generalised 2nd Order Active Filter Prototype_B
- 1. GENERALISED 2ND ORDER ACTIVE FILTER PROTOTYPE Student: A Mostert Supervisor: P O’Connor
- 2. PROJECT OVERVIEW • Motivation • Key Objectives • Circuit Research and Analysis • Mathematics • Simulation • Problems • Outcome
- 3. VOLTAGE TRANSFER FUNCTION: GENERALISED TOPOLOGY 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑍2 𝑍4 𝑍2 𝑍3 + 𝑍1 𝑍3 + 𝑍1 𝑍4 + 𝑍1 𝑍2
- 4. VOLTAGE TRANSFER FUNCTION: BANDPASS 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑍2 𝑅4 𝑍2 𝑍3 + 𝑅1 𝑍3 + 𝑅1 𝑅4 + 𝑅1 𝑍2
- 5. VOLTAGE TRANSFER FUNCTION: LOWPASS 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑍2 𝑍4 𝑍2 𝑅3 + 𝑅1 𝑅3 + 𝑅1 𝑍4 + 𝑅1 𝑍2
- 6. VOLTAGE TRANSFER FUNCTION: HIGHPASS 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑅2 𝑅4 𝑅2 𝑍3 + 𝑍1 𝑍3 + 𝑍1 𝑅4 + 𝑍1 𝑅2
- 7. VOLTAGE TRANSFER FUNCTION: BANDSTOP 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑍2 𝑅4 𝑍2 𝑍3 + 𝑅1 𝑍3 + 𝑅1 𝑅4 + 𝑅1 𝑍2
- 8. KEY EQUATIONS 𝑉𝑜𝑢𝑡 𝑉𝑖𝑛 = 𝐻 = −𝑍2 𝑅4 𝑍2 𝑍3 + 𝑅1 𝑍3 + 𝑅1 𝑅4 + 𝑅1 𝑍2 𝐻 𝑗𝜔 = − 𝑅4 𝑗𝜔𝑍3 1 − 𝜔2 𝑅1 𝑅4 𝑍2 𝑍3) + 𝑗𝜔𝑅1 𝑍2 + 𝑍3 𝐻 𝑗𝜔 = 𝜔𝑅4 𝑍3 1 − 𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 2 + 𝜔𝑅1 𝑍2 + 𝑍3 2 Eq 1. Eq 2. Eq 3.
- 9. PROVING OUT • 𝐻 𝑗𝜔 = 𝜔𝑅4 𝑍3 1 −𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 2 + 𝜔𝑅1 𝑍2+𝑍3 2 Eq.3 • Test function with large and small values of 𝜔 • 𝐻 𝑗𝜔 = − 𝑅4 𝑗𝜔𝑍3 1 −𝜔2 𝑅1 𝑅4 𝑍2 𝑍3)+𝑗𝜔𝑅1 𝑍2+𝑍3 Eq.2 • 𝑓0 found using real term made equal to zero: 1.13106kHz
- 10. • 𝐻 𝑗𝜔 = 𝜔𝑅4 𝑍3 1 −𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 2 + 𝜔𝑅1 𝑍2+𝑍3 2 Eq.3 • Find Gain using eq 3. and value found for 𝑓0 : • Gain : 2.696 = 8.614dB • Cutoff bandwidths at 5.614dB: • Make Eq. 3 = 1.90853 and rearrange as quadratic 𝜔4 𝑅1 𝑅2 𝑍1 𝑍3 2 + 𝜔2 𝑅1 𝑍1 + 𝑍3 2 − 0.275 𝑅2 𝑍3 2 − 2 𝑅1 𝑅2 𝑍1 𝑍3 + 1 • Use MATLAB to find roots of quadratic
- 11. • R1=120 %Resistor R1 • R4=330 %Resistor R4 • Z2=100*10^(-9) %Capacitor Z2 • Z3=5*10^(-6) %Capacitor Z3 • A=(R1*R4*Z2*Z3)^2 %coefficient of w^4 • B=(R1*R1*(Z2+Z3)^2)-(2*R1*R4*Z2*Z3)-(0.275*R4*R4*Z3*Z3) • C=1 %coefficient of w^0 • Quad=[A B C] • X1=roots(Quad) %array form of quadratic • X2=sqrt(X1) %square root of w^2 • X3=X2/(2*pi) %frequencies fc1 and fc MATLAB CODE
- 12. MATLAB OUTPUT X3 = 1.0e+03 * 5.1644 0.2477 𝑓𝑐1 = 5.1644 kHz 𝑓𝑐2 = 247 Hz
- 13. PHASE RESPONSE 𝐻 𝑗𝜔 = 𝜔𝑅4 𝑍3 1 −𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 2 + 𝜔𝑅1 𝑍2+𝑍3 2 Eq.3 𝐻 𝑗𝜔 = 𝜑 = 0 − tan−1 𝜔𝑅1 𝑍2+𝑍3 1 –𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 Eq.4 − tan−1 𝜔𝑅1 𝑍2 + 𝑍3 1 – 𝜔2 𝑅1 𝑅4 𝑍2 𝑍3 = − tan−1 0.6922 1 – 0.0253 = −0.6175
- 14. BANDPASS PHASE SHIFT & FREQUENCY RESPONSE
- 15. ISSUES AND LEARNING CURVE • Practical vs Ideal Op-Amp Integrator • Positive feedback instability • Varying SPICE models • Mathematically challenging • Technical documentation
- 16. SCHEDULE
- 17. CONCLUSION • Most Equations worked out • Appreciation for Mathematical software • Develop project for 4th year Digital Signal processing