Generalised 2nd Order Active Filter Prototype_B
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Generalised 2nd Order Active Filter Prototype_B
- 1. GENERALISED 2ND ORDER ACTIVE FILTER PROTOTYPE Student: A Mostert Supervisor: P OβConnor
- 2. PROJECT OVERVIEW β’ Motivation β’ Key Objectives β’ Circuit Research and Analysis β’ Mathematics β’ Simulation β’ Problems β’ Outcome
- 3. VOLTAGE TRANSFER FUNCTION: GENERALISED TOPOLOGY πππ’π‘ πππ = π» = βπ2 π4 π2 π3 + π1 π3 + π1 π4 + π1 π2
- 4. VOLTAGE TRANSFER FUNCTION: BANDPASS πππ’π‘ πππ = π» = βπ2 π 4 π2 π3 + π 1 π3 + π 1 π 4 + π 1 π2
- 5. VOLTAGE TRANSFER FUNCTION: LOWPASS πππ’π‘ πππ = π» = βπ2 π4 π2 π 3 + π 1 π 3 + π 1 π4 + π 1 π2
- 6. VOLTAGE TRANSFER FUNCTION: HIGHPASS πππ’π‘ πππ = π» = βπ 2 π 4 π 2 π3 + π1 π3 + π1 π 4 + π1 π 2
- 7. VOLTAGE TRANSFER FUNCTION: BANDSTOP πππ’π‘ πππ = π» = βπ2 π 4 π2 π3 + π 1 π3 + π 1 π 4 + π 1 π2
- 8. KEY EQUATIONS πππ’π‘ πππ = π» = βπ2 π 4 π2 π3 + π 1 π3 + π 1 π 4 + π 1 π2 π» ππ = β π 4 πππ3 1 β π2 π 1 π 4 π2 π3) + πππ 1 π2 + π3 π» ππ = ππ 4 π3 1 β π2 π 1 π 4 π2 π3 2 + ππ 1 π2 + π3 2 Eq 1. Eq 2. Eq 3.
- 9. PROVING OUT β’ π» ππ = ππ 4 π3 1 βπ2 π 1 π 4 π2 π3 2 + ππ 1 π2+π3 2 Eq.3 β’ Test function with large and small values of π β’ π» ππ = β π 4 πππ3 1 βπ2 π 1 π 4 π2 π3)+πππ 1 π2+π3 Eq.2 β’ π0 found using real term made equal to zero: 1.13106kHz
- 10. β’ π» ππ = ππ 4 π3 1 βπ2 π 1 π 4 π2 π3 2 + ππ 1 π2+π3 2 Eq.3 β’ Find Gain using eq 3. and value found for π0 : β’ Gain : 2.696 = 8.614dB β’ Cutoff bandwidths at 5.614dB: β’ Make Eq. 3 = 1.90853 and rearrange as quadratic π4 π 1 π 2 π1 π3 2 + π2 π 1 π1 + π3 2 β 0.275 π 2 π3 2 β 2 π 1 π 2 π1 π3 + 1 β’ Use MATLAB to find roots of quadratic
- 11. β’ R1=120 %Resistor R1 β’ R4=330 %Resistor R4 β’ Z2=100*10^(-9) %Capacitor Z2 β’ Z3=5*10^(-6) %Capacitor Z3 β’ A=(R1*R4*Z2*Z3)^2 %coefficient of w^4 β’ B=(R1*R1*(Z2+Z3)^2)-(2*R1*R4*Z2*Z3)-(0.275*R4*R4*Z3*Z3) β’ C=1 %coefficient of w^0 β’ Quad=[A B C] β’ X1=roots(Quad) %array form of quadratic β’ X2=sqrt(X1) %square root of w^2 β’ X3=X2/(2*pi) %frequencies fc1 and fc MATLAB CODE
- 12. MATLAB OUTPUT X3 = 1.0e+03 * 5.1644 0.2477 ππ1 = 5.1644 kHz ππ2 = 247 Hz
- 13. PHASE RESPONSE π» ππ = ππ 4 π3 1 βπ2 π 1 π 4 π2 π3 2 + ππ 1 π2+π3 2 Eq.3 π» ππ = π = 0 β tanβ1 ππ 1 π2+π3 1 βπ2 π 1 π 4 π2 π3 Eq.4 β tanβ1 ππ 1 π2 + π3 1 β π2 π 1 π 4 π2 π3 = β tanβ1 0.6922 1 β 0.0253 = β0.6175
- 14. BANDPASS PHASE SHIFT & FREQUENCY RESPONSE
- 15. ISSUES AND LEARNING CURVE β’ Practical vs Ideal Op-Amp Integrator β’ Positive feedback instability β’ Varying SPICE models β’ Mathematically challenging β’ Technical documentation
- 16. SCHEDULE
- 17. CONCLUSION β’ Most Equations worked out β’ Appreciation for Mathematical software β’ Develop project for 4th year Digital Signal processing