ECEN 301Discussion #13 – Phasors1 Imaginary JS-H 1: 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.

ECEN 301Discussion #13 – Phasors2 Lecture 13 – Network Analysis with Capacitors and Inductors Phasors A “real” phasor is NOT the same thing used in Star Trek

ECEN 301Discussion #13 – Phasors3 Euler’s Identity Appendix A reviews complex numbers Complex exponential ( e jθ ) is a point on the complex plane 1 1 j -j Re Im sinθ cosθ θ ejθejθ Euler’s equation

ECEN 301Discussion #13 – Phasors4 Phasors Rewrite the expression for a general sinusoid signal: Angle (or argument)magnitude Complex phasor notation for the simplification: NB: The e jwt term is implicit (it is there but not written)

ECEN 301Discussion #16 – Frequency Response5 Frequency Domain Graphing in the frequency domain: helpful in order to understand Phasors -ω-ωω ππ cos(ω 0 t) π[δ(ω – ω 0 ) + δ(ω – ω 0 )] Time domainFrequency domain

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ECEN 301Discussion #13 – Phasors7 Phasors 1.Any sinusoidal signal can be represented by either: Time-domain form: v(t) = Acos(ωt+θ) Frequency-domain form: V(jω) = Ae jθ = A θ 2.Phasor: a complex number expressed in polar form consisting of: Magnitude (A) Phase angle (θ) 3.Phasors do not explicitly include the sinusoidal frequency (ω) but this information is still important

ECEN 301Discussion #13 – Phasors8 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~

ECEN 301Discussion #13 – Phasors9 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation

ECEN 301Discussion #13 – Phasors10 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A)

ECEN 301Discussion #13 – Phasors11 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages

ECEN 301Discussion #13 – Phasors12 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages 4.Convert rectangular back to polar

ECEN 301Discussion #13 – Phasors13 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages 4.Convert rectangular back to polar 5.Convert from phasor to time domain Bring ωt back NB: the answer is NOT simply the addition of the amplitudes of v 1 (t) and v 2 (t) (i.e ), and the addition of their phases (i.e. π/4 + π/12)

ECEN 301Discussion #13 – Phasors14 Phasors Example 1: compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ Re Im π/ Vs(jω)

ECEN 301Discussion #13 – Phasors15 Phasors of Different Frequencies Superposition of AC signals: when signals do not have the same frequency (ω) the e jωt term in the phasors can no longer be implicit i 1 (t) Load +v–+v– I i 2 (t) NB: e jωt can no longer be implicit

ECEN 301Discussion #13 – Phasors16 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 R1R1 R 2 i 1 (t)v s (t) +–+–

ECEN 301Discussion #13 – Phasors17 Phasors of Different Frequencies +R1–+R1– + R 2 – i 1 (t)v s (t) +–+– Example 2: compute the resistor voltages i s (t) = 0.5cos[2π(100t)] A v s (t) = 20cos[2π(1000t)] V R 1 = 150Ω, R2 = 50 Ω

ECEN 301Discussion #13 – Phasors18 Phasors of Different Frequencies +R1–+R1– + R 2 – i 1 (t) 1.Since the sources have different frequencies (ω 1 = 2π*100) and (ω 2 = 2π*1000) use superposition first consider the (ω 1 = 2π*100) part of the circuit When v s (t) = 0 – short circuit Example 2: compute the resistor voltages i s (t) = 0.5cos[2π(100t)] A v s (t) = 20cos[2π(1000t)] V R 1 = 150Ω, R2 = 50 Ω +R1–+R1– + R 2 – i 1 (t)v s (t) +–+–

ECEN 301Discussion #13 – Phasors19 Phasors of Different Frequencies Example 2: compute the resistor voltages i s (t) = 0.5cos[2π(100t)] A v s (t) = 20cos[2π(1000t)] V R 1 = 150Ω, R 2 = 50 Ω + R 1 || R 2 – i 1 (t) 1.Since the sources have different frequencies (ω 1 = 2π*100) and (ω 2 = 2π*1000) use superposition first consider the (ω 1 = 2π*100) part of the circuit

ECEN 301Discussion #13 – Phasors20 Phasors of Different Frequencies Example 2: compute the resistor voltages i s (t) = 0.5cos[2π(100t)] A v s (t) = 20cos[2π(1000t)] V R 1 = 150Ω, R 2 = 50 Ω 1.Since the sources have different frequencies (ω 1 = 2π*100) and (ω 2 = 2π*1000) use superposition first consider the (ω 1 = 2π*100) part of the circuit Next consider the (ω 2 = 2π*1000) part of the circuit +R1–+R1– + R 2 – v s (t) +–+– +R1–+R1– + R 2 – i 1 (t) v s (t) +–+–

ECEN 301Discussion #13 – Phasors21 Phasors of Different Frequencies Example 2: compute the resistor voltages i s (t) = 0.5cos[2π(100t)] A v s (t) = 20cos[2π(1000t)] V R 1 = 150Ω, R 2 = 50 Ω 1.Since the sources have different frequencies (ω 1 = 2π*100) and (ω 2 = 2π*1000) use superposition first consider the (ω 1 = 2π*100) part of the circuit Next consider the (ω 2 = 2π*1000) part of the circuit Add the two together +R1–+R1– + R 2 – i 1 (t)v s (t) +–+–

ECEN 301Discussion #13 – Phasors22 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 v s (t) +–+– ~ R + v R (t) – i(t) v s (t) +–+– ~ C + v C (t) – i(t) v s (t) +–+– ~ L + v L (t) – i(t) V s (jω) +–+– ~ + V Z (jω) – I(jω) Z

ECEN 301Discussion #13 – Phasors23 Impedance – Resistors Impedance of a Resistor: Consider Ohm’s Law in phasor form: V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Re Im I V Phasor Phasor domain NB: Ohm’s Law works the same in DC and AC

ECEN 301Discussion #13 – Phasors24 Impedance – Inductors Impedance of an Inductor: First consider voltage and current in the time-domain v s (t) +–+– ~ L + v L (t) – i(t) NB: current is shifted 90° from voltage

ECEN 301Discussion #13 – Phasors25 Impedance – Inductors Impedance of an Inductor: Now consider voltage and current in the phasor-domain v s (t) +–+– ~ L + v L (t) – i(t) V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Phasor Phasor domain Re Im I V -π/2

ECEN 301Discussion #13 – Phasors26 Impedance – Capacitors Impedance of a capacitor: First consider voltage and current in the time-domain v s (t) +–+– ~ C + v C (t) – i(t) V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Phasor

ECEN 301Discussion #13 – Phasors27 Impedance – Capacitors Impedance of a capacitor: Next consider voltage and current in the phasor-domain V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Phasor domain Re Im I V π/2

ECEN 301Discussion #13 – Phasors28 Impedance +L–+L– +C–+C– Re Im -π/2 π/2 R -1/ωC ωLωL ZRZR ZCZC ZLZL Phasor domain +R–+R– V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Impedance of resistors, inductors, and capacitors

ECEN 301Discussion #13 – Phasors29 Impedance Re Im -π/2 π/2 R -1/ωC ωLωL ZRZR ZCZC ZLZL Phasor domain V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Impedance of resistors, inductors, and capacitors AC resistancereactanceNot a phasor but a complex number

ECEN 301Discussion #13 – Phasors30 Impedance Practical capacitors: in practice capacitors contain a real component (represented by a resistive impedance Z R ) At high frequencies or high capacitances ideal capacitor acts like a short circuit At low frequencies or low capacitances ideal capacitor acts like an open circuit +C–+C– +R–+R– +C–+C– Re Im -π/2 -1/ωC ZCZC Re Im -π/2 R -1/ωC ZRZR ZCZC Ideal Capacitor Practical Capacitor NB: the ratio of Z C to Z R is highly frequency dependent

ECEN 301Discussion #13 – Phasors31 Impedance Practical inductors: in practice inductors contain a real component (represented by a resistive impedance Z R ) At low frequencies or low inductances Z R has a strong influence Ideal inductor acts like a short circuit At high frequencies or high inductances Z L dominates Z R Ideal inductor acts like an open circuit At high frequencies a capacitor is also needed to correctly model a practical inductor Re Im π/2 ωLωL ZLZL Re Im π/2 R ωLωL ZRZR ZLZL Ideal Inductor Practical Inductor NB: the ratio of Z L to Z R is highly frequency dependent +L–+L– +R–+R– +L–+L–

ECEN 301Discussion #13 – Phasors32 Impedance Example 3: impedance of a practical capacitor Find the impedance ω = 377 rads/s, C = 1nF, R = 1MΩ +R–+R– +C–+C– +Z–+Z–

ECEN 301Discussion #13 – Phasors33 Impedance Example 3: impedance of a practical capacitor Find the impedance ω = 377 rads/s, C = 1nF, R = 1MΩ +R–+R– +C–+C– +Z–+Z–

ECEN 301Discussion #13 – Phasors34 Impedance Example 4: find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH Z EQ R2R2 C R1R1 L

ECEN 301Discussion #13 – Phasors35 Impedance Example 4: find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH Z EQ R2R2 C R1R1 L

ECEN 301Discussion #13 – Phasors36 Impedance Example 4: find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH Z EQ R1R1 L Z EQ1 NB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive)