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Easier to Maintain Than to Retake

Easier to Maintain Than to Retake. Alma 59:9

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Easier to Maintain Than to Retake

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  1. Easier to Maintain Than to Retake Alma 59:9   9 And now as Moroni had supposed that there should be men sent to the city of Nephihah, to the assistance of the people to maintain that city, and knowing that it was easier to keep the city from falling into the hands of the Lamanites than to retake it from them, he supposed that they would easily maintain that city. Discussion #16 – Frequency Response

  2. Lecture 16 – Frequency Response Back to AC Circuits and Phasors Frequency Response Filters Discussion #16 – Frequency Response

  3. Frequency Response Frequency Response H(jω): a measure of how the voltage/current/impedance of a load responds to the voltage/current of a source Discussion #16 – Frequency Response

  4. Frequency Response VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω) Discussion #16 – Frequency Response

  5. Frequency Response VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω) Amplitude scaled Phase shifted Discussion #16 – Frequency Response

  6. R1 + RL – C vs(t) + – Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF Discussion #16 – Frequency Response

  7. R1 + RL – C vs(t) + – Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Note frequencies of AC sources Only one AC source so frequency response HV(jω) will be the function of a single frequency Discussion #16 – Frequency Response

  8. Z1 = R1 R1 ZC=1/jωC + RL – Vs ZLD=RL C vs(t) + – ~ + – Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Note frequencies of AC sources • Convert to phasor domain Discussion #16 – Frequency Response

  9. Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Note frequencies of AC sources • Convert to phasor domain • Solve using network analysis • Thévenin equivalent Z1 = R1 ZC=1/jωC Discussion #16 – Frequency Response

  10. Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Note frequencies of AC sources • Convert to phasor domain • Solve using network analysis • Thévenin equivalent Z1 = R1 ZC=1/jωC Vs + VT – + – ~ Discussion #16 – Frequency Response

  11. ZT VT + VL – ZLD + – ~ Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Note frequencies of AC sources • Convert to phasor domain • Solve using network analysis • Thévenin equivalent • Find an expression for the load voltage Discussion #16 – Frequency Response

  12. ZT VT + VL – ZLD + – ~ Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Find an expression for the frequency response Discussion #16 – Frequency Response

  13. ZT VT + VL – ZLD + – ~ Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Find an expression for the frequency response Discussion #16 – Frequency Response

  14. ZT VT + VL – ZLD + – ~ Frequency Response Example 1: compute the frequency response HV(jω) R1 = 1kΩ, RL = 10kΩ, C = 10uF • Find an expression for the frequency response • Look at response for low frequencies (ω = 10) and high frequencies (ω = 10000) ω = 10000 ω = 10 Lowpass filter Discussion #16 – Frequency Response

  15. Frequency Response Lowpass filter

  16. L + RL – R1 is(t) Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH Discussion #16 – Frequency Response

  17. L + RL – R1 is(t) Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH • Note frequencies of AC sources Only one AC source so frequency response HZ(jω) will be the function of a single frequency Discussion #16 – Frequency Response

  18. ZL = jωL is(t) + VL – Is(jω) Z1=R1 ZLD = RL Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH • Note frequencies of AC sources • Convert to phasor domain L + RL – R1 Discussion #16 – Frequency Response

  19. ZL = jωL IL(jω) + VL – Is(jω) Z1=R1 ZLD = RL Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH • Note frequencies of AC sources • Convert to phasor domain • Solve using network analysis • Current divider & Ohm’s Law Discussion #16 – Frequency Response

  20. ZL = jωL IL(jω) + VL – Is(jω) Z1=R1 ZLD = RL Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH • Find an expression for the frequency response Discussion #16 – Frequency Response

  21. Is(jω) Frequency Response Example 2: compute the frequency response HZ(jω) R1 = 1kΩ, RL = 4kΩ, L = 2mH • Find an expression for the frequency response • Look at response for low frequencies (ω = 10) and high frequencies (ω = 10000) ZL = jωL IL(jω) + VL – Z1=R1 ZLD = RL ω = 10000 ω = 10 Lowpass filter Discussion #16 – Frequency Response

  22. 1st and 2nd Order RLC Filters Graphing in Frequency Domain Filter Orders Resonant Frequencies Basic Filters Discussion #16 – Frequency Response

  23. π π -ω ω Frequency Domain Graphing in the frequency domain: helpful in order to understand filters cos(ω0t) π[δ(ω – ω0) + δ(ω – ω0)] Fourier Transform Time domain Frequency domain Discussion #16 – Frequency Response

  24. W/π 1 π/W -W W Frequency Domain Graphing in the frequency domain: helpful in order to understand filters sinc(ω0t) Fourier Transform Time domain Frequency domain Discussion #16 – Frequency Response

  25. Low-pass High-pass Band-pass Band-stop Basic Filters Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies ω ω ω ω Discussion #16 – Frequency Response

  26. Filter Orders Higher filter orders provide a higher quality filter Discussion #16 – Frequency Response

  27. Im ωL ZL π/2 Re R + – + R – Vs(jω) ~ ZR -π/2 + C – + L – ZC I(jω) -1/ωC + VZ(jω) – Z Impedance Impedance of resistors, inductors, and capacitors Phasor domain Discussion #13 – Phasors

  28. Im Re -π/2 + C – + C – + C – ZC -1/ωC Impedance Impedance of capacitors Discussion #13 – Phasors

  29. Im ωL ZL π/2 Re + L – + L – + L – Impedance Impedance of inductors Discussion #13 – Phasors

  30. L + vi(t) – + vo(t) – + vi(t) – + vo(t) – C C L Resonant Frequency Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters) Impedances in series Impedances in parallel Discussion #16 – Frequency Response

  31. ZL=jωL ZC=1/jωC + Vi(jω) – + Vo(jω) – Resonant Frequency Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters) Impedances in series Discussion #16 – Frequency Response

  32. Resonant Frequency Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters) Impedances in series + Vi(jω) – + Vo(jω) – ZEQ=0 Discussion #16 – Frequency Response

  33. ZC=1/jωC ZL=jωL + Vi(jω) – + Vo(jω) – Resonant Frequency Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters) Impedances in parallel Discussion #16 – Frequency Response

  34. Resonant Frequency Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters) Impedances in parallel + Vi(jω) – + Vo(jω) – ZEQ=∞ Discussion #16 – Frequency Response

  35. R Low-pass R L + vi(t) – + vo(t) – + vi(t) – + vo(t) – C C ω1 ω0 ω2 ω3 Low-Pass Filters Low-pass Filters: only allow signals under the cutoff frequency (ω0) to pass 1st Order 2nd Order HL(jω) Discussion #16 – Frequency Response

  36. ZR ZR + Vi(jω) – + Vo(jω) – Low-pass ZC=1/jωC + Vi(jω) – + Vo(jω) – ω1 ω0 ω3 ZR + Vo(jω) – + Vi(jω) – Low-Pass Filters 1st Order Low-pass Filters: ω1: ω→ 0 ω3:ω→ ∞ Discussion #16 – Frequency Response

  37. ZR ZL=jωL Low-pass + Vi(jω) – + Vo(jω) – ZC=1/jωC ω2 ω1 ω3 ω0 ZR ZR + Vi(jω) – + Vi(jω) – + Vo(jω) – + Vo(jω) – Low-Pass Filters 2nd Order Low-pass Filters: ω3:ω→ ∞ ω1:ω→ 0 Resonant frequency ω2:ω = ωn ZR + Vi(jω) – + Vo(jω) – Discussion #16 – Frequency Response

  38. 1st Order High-pass + vi(t) – + vo(t) – C R ω2 ω3 ω1 ω0 2nd Order R + vi(t) – + vo(t) – C L High-Pass Filters High-pass Filters: only allow signals above the cutoff frequency (ω0) to pass HH(jω) Discussion #16 – Frequency Response

  39. High-pass ω1 ω0 ω3 + Vo(jω) – + Vi(jω) – ZR High-Pass Filters 1st Order High-pass Filters: ZC=1/jωC + Vi(jω) – + Vo(jω) – ZR=R ω1: ω→ 0 ω3:ω→ ∞ + Vi(jω) – + Vo(jω) – ZR Discussion #16 – Frequency Response

  40. ZR ZC=1/jωC High-pass + Vi(jω) – + Vo(jω) – ZL=jωL ω3 ω2 ω0 ω1 ZR + Vi(jω) – + Vo(jω) – + Vo(jω) – + Vo(jω) – + Vi(jω) – + Vi(jω) – ZR ZR High-Pass Filters 2nd Order High-pass Filters: ω3:ω→ ∞ ω1:ω→ 0 Resonant frequency ω2:ω = ωn Discussion #16 – Frequency Response

  41. 2nd Order L + vi(t) – + vo(t) – C R ωb ω1 ωa ω2 ω3 Band-pass Band-Pass Filters Band-pass Filters: only allow signals between the passband (ωa toωb) to pass HB(jω) Discussion #16 – Frequency Response

  42. ωb ωa ω3 ω2 ω1 Band-pass + Vo(jω) – + Vi(jω) – + Vo(jω) – + Vo(jω) – + Vi(jω) – ZR ZR ZR Band-Pass Filters 2nd Order Band-pass Filters: ZC=1/jωC + Vi(jω) – + Vo(jω) – ZL=jωL ZR ω3:ω→ ∞ ω1:ω→ 0 Resonant frequency ω2:ω = ωn + Vi(jω) – Discussion #16 – Frequency Response

  43. 2nd Order L + vi(t) – + vo(t) – R C ωb ω1 ωa ω2 ω3 Band-stop Band-Stop Filters Band-stop Filters: allow signals except those between the passband (ωa toωb) to pass HN(jω) Discussion #16 – Frequency Response

  44. ZL=jωL + Vo(jω) – + Vi(jω) – ZR ZC=1/jωC ωb ω3 ω2 ω1 ωa Band-stop + Vo(jω) – + Vi(jω) – + Vo(jω) – + Vi(jω) – ZR ZR Band-Stop Filters 2nd Order Band-stop Filters: ω3:ω→ ∞ ω1:ω→ 0 Resonant frequency ω2:ω = ωn + Vo(jω) – + Vi(jω) – ZR Discussion #16 – Frequency Response

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