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Imaginary

Imaginary. JS-H 1: 16

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Imaginary

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  1. Imaginary JS-H 1: 16   16 But, exerting all my powers to call upon God to deliver me out of the power of this enemy which had seized upon me, and at the very moment when I was ready to sink into despair and abandon myself to destruction—not to an imaginary ruin, but to the power of some actual being from the unseen world, who had such marvelous power as I had never before felt in any being—just at this moment of great alarm, I saw a pillar of light exactly over my head, above the brightness of the sun, which descended gradually until it fell upon me. Discussion #13 – Phasors

  2. Lecture 13 – Network Analysis with Capacitors and Inductors Phasors A “real” phasor is NOT the same thing used in Star Trek Discussion #13 – Phasors

  3. Im j ejθ 1 sinθ θ Re -1 1 cosθ -j Euler’s Identity • Appendix A reviews complex numbers Complex exponential (ejθ) is a point on the complex plane Euler’s equation Discussion #13 – Phasors

  4. Phasors • Rewrite the expression for a general sinusoid signal: magnitude Angle (or argument) Complex phasor notation for the simplification: NB: The ejwt term is implicit (it is there but not written) Discussion #13 – Phasors

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

  6. 3000 km 0.3 mm 0.3 mm 30 mm 300 km 30 mm 300 m 0.03Å 300Å 0.3 m 3 mm 30 km 3 mm 30 m 0.3Å 3 km 30Å 3 m 3Å … … 1017 Hz 1016 Hz 1012 Hz 1015 Hz 1011 Hz 1010 Hz 1014 Hz 1018 Hz 1013 Hz 1020 Hz 1019 Hz 105 Hz 104 Hz 109 Hz 108 Hz 107 Hz 106 Hz 103 Hz 102 Hz The Electromagnetic Spectrum Gamma rays Radio waves Ultraviolet Infrared Xray Microwaves wavelength Visible light

  7. Phasors • Any sinusoidal signal can be represented by either: • Time-domain form: v(t) = Acos(ωt+θ) • Frequency-domain form: V(jω) = Aejθ = A θ • Phasor: a complex number expressed in polar form consisting of: • Magnitude (A) • Phase angle (θ) • Phasors do not explicitly include the sinusoidal frequency (ω) but this information is still important Discussion #13 – Phasors

  8. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) Discussion #13 – Phasors

  9. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation Discussion #13 – Phasors

  10. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) Discussion #13 – Phasors

  11. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages Discussion #13 – Phasors

  12. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages • Convert rectangular back to polar Discussion #13 – Phasors

  13. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages • Convert rectangular back to polar • Convert from phasor to time domain NB: the answer is NOT simply the addition of the amplitudes of v1(t) and v2(t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12) Bring ωt back Discussion #13 – Phasors

  14. Im + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ π/6 Vs(jω) 14.49 Re 25.10 Phasors • Example 1: compute the phasor voltage for the equivalent voltage vs(t) • v1(t) = 15cos(377t+π/4) • v2(t) = 15cos(377t+π/12) Discussion #13 – Phasors

  15. Load i1(t) Phasors of Different Frequencies Superposition of AC signals: when signals do not have the same frequency (ω) the ejωtterm in the phasors can no longer be implicit I + v – i2(t) NB: ejωt can no longer be implicit Discussion #13 – Phasors

  16. R2 R1 + – vs(t) i1(t) Phasors of Different Frequencies Superposition of AC signals: when signals do not have the same frequency (ω) solve the circuit separately for each different frequency (ω) – then add the individual results Discussion #13 – Phasors

  17. + R2 – + R1 – + – vs(t) i1(t) Phasors of Different Frequencies • Example 2: compute the resistor voltages • is(t) = 0.5cos[2π(100t)] A • vs(t) = 20cos[2π(1000t)] V • R1 = 150Ω, R2 = 50 Ω Discussion #13 – Phasors

  18. + R2 – + R1 – + – vs(t) i1(t) i1(t) Phasors of Different Frequencies • Example 2: compute the resistor voltages • is(t) = 0.5cos[2π(100t)] A • vs(t) = 20cos[2π(1000t)] V • R1 = 150Ω, R2 = 50 Ω • Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of the circuit • When vs(t) = 0 – short circuit + R2 – + R1 – Discussion #13 – Phasors

  19. i1(t) Phasors of Different Frequencies • Example 2: compute the resistor voltages • is(t) = 0.5cos[2π(100t)] A • vs(t) = 20cos[2π(1000t)] V • R1 = 150Ω, R2 = 50 Ω • Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of the circuit + R1|| R2 – Discussion #13 – Phasors

  20. + R2 – + R1 – + – + – vs(t) vs(t) i1(t) Phasors of Different Frequencies • Example 2: compute the resistor voltages • is(t) = 0.5cos[2π(100t)] A • vs(t) = 20cos[2π(1000t)] V • R1 = 150Ω, R2 = 50 Ω • Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of the circuit • Next consider the (ω2 = 2π*1000) part of the circuit + R2 – + R1 – Discussion #13 – Phasors

  21. + R2 – + R1 – + – vs(t) i1(t) Phasors of Different Frequencies • Example 2: compute the resistor voltages • is(t) = 0.5cos[2π(100t)] A • vs(t) = 20cos[2π(1000t)] V • R1 = 150Ω, R2 = 50 Ω • Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of the circuit • Next consider the (ω2 = 2π*1000) part of the circuit • Add the two together Discussion #13 – Phasors

  22. i(t) i(t) i(t) + vR(t) – + vC(t) – + vL(t) – R C L + – + – + – + – Vs(jω) vs(t) vs(t) vs(t) ~ ~ ~ ~ I(jω) + VZ(jω) – Z Impedance Impedance: complex resistance (has no physical significance) • Allows us to use network analysis methods such as node voltage, mesh current, etc. • Capacitors and inductors act as frequency-dependent resistors Discussion #13 – Phasors

  23. + – Vs(jω) ~ Im I(jω) I + VZ(jω) – V Z Re Impedance – Resistors Impedance of a Resistor: • Consider Ohm’s Law in phasor form: Phasor Phasor domain NB: Ohm’s Law works the same in DC and AC Discussion #13 – Phasors

  24. i(t) + vL(t) – L + – vs(t) ~ Impedance – Inductors Impedance of an Inductor: • First consider voltage and current in the time-domain NB: current is shifted 90° from voltage Discussion #13 – Phasors

  25. i(t) Im + vL(t) – L + – + – Vs(jω) vs(t) V ~ ~ Re I(jω) -π/2 + VZ(jω) – I Z Impedance – Inductors Impedance of an Inductor: • Now consider voltage and current in the phasor-domain Phasor Phasor domain Phasor Discussion #13 – Phasors

  26. i(t) + vC(t) – C + – + – Vs(jω) vs(t) ~ ~ I(jω) + VZ(jω) – Z Impedance – Capacitors Impedance of a capacitor: • First consider voltage and current in the time-domain Phasor Phasor Discussion #13 – Phasors

  27. Im + – Vs(jω) ~ V I π/2 Re I(jω) + VZ(jω) – Z Impedance – Capacitors Impedance of a capacitor: • Next consider voltage and current in the phasor-domain Phasor domain Discussion #13 – Phasors

  28. 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

  29. Im ωL ZL π/2 Re R + – Vs(jω) ~ ZR -π/2 ZC I(jω) -1/ωC + VZ(jω) – Z Impedance Impedance of resistors, inductors, and capacitors Phasor domain Not a phasor but a complex number AC resistance reactance Discussion #13 – Phasors

  30. Im Im + R – + C – R Re ZR Re + C – -π/2 ZC -π/2 ZC -1/ωC -1/ωC Impedance Practical capacitors: in practice capacitors contain a real component (represented by a resistive impedance ZR) • At highfrequencies or high capacitances • ideal capacitor acts like a short circuit • At lowfrequencies or low capacitances • ideal capacitor acts like an open circuit Practical Capacitor Ideal Capacitor NB: the ratio of ZC to ZR is highly frequency dependent Discussion #13 – Phasors

  31. Im Im + L – ωL ωL ZL ZL π/2 Re π/2 R Re + R – + L – ZR Impedance Practical inductors: in practice inductors contain a real component (represented by a resistive impedance ZR) • At low frequencies or low inductancesZR has a strong influence • Ideal inductor acts like a short circuit • At high frequencies or high inductancesZLdominates ZR • Ideal inductor acts like an open circuit • At high frequencies a capacitor is also needed to correctly model a practical inductor Ideal Inductor Practical Inductor NB: the ratio of ZL to ZR is highly frequency dependent Discussion #13 – Phasors

  32. + R – + Z – + C – Impedance • Example 3: impedance of a practical capacitor • Find the impedance • ω = 377 rads/s, C = 1nF, R = 1MΩ Discussion #13 – Phasors

  33. + Z – Impedance • Example 3: impedance of a practical capacitor • Find the impedance • ω = 377 rads/s, C = 1nF, R = 1MΩ + R – + C – Discussion #13 – Phasors

  34. R1 L ZEQ R2 C Impedance • Example 4: find the equivalent impedance (ZEQ) • ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH Discussion #13 – Phasors

  35. R1 L ZEQ R2 C Impedance • Example 4: find the equivalent impedance (ZEQ) • ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH Discussion #13 – Phasors

  36. ZEQ1 ZEQ Impedance • Example 4: find the equivalent impedance (ZEQ) • ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH R1 L NB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive) Discussion #13 – Phasors

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