1EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu1Announcements HW #2 due on Tuesday at 6pm in Cory. Midterm #1 on 7/12 12:00-1:30. Location TBD.EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu2Lecture #6OUTLINE Chap 5 Phasors Complex ImpedancesReadingChap 5.1-5.42EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu3Complex Numbers x is the real part y is the imaginary part z is the magnitude θ is the phaseθzxyreal axisimaginary axisEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu4More Complex Numbers Polar Coordinates: A = z ∠θ Rectangular Coordinates: A = x + jyθcoszx =θsinzy=22yxz +=xy1tan−=θ3EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu5Summary of Phasors Phasor (frequency domain) is a complex number:X = z ∠θ= x + jy Sinusoid is a time function:x(t) = z cos (ωt + θ)EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu6ExamplesFind the time domain representations ofX = -1 + j2 V = 104V - j60V A = -1mA - j3mA4EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu7Arithmetic With Complex Numbers To compute phasor voltages and currents, we need to be able to perform computation with complex numbers.AdditionSubtractionMultiplicationDivisionEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu8Addition Addition is most easily performed in rectangular coordinates:A = x + jyB = z + jwA + B = (x + z) + j(y + w)5EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu9AdditionReal AxisImaginary AxisABA + BEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu10Subtraction Subtraction is most easily performed in rectangular coordinates:A = x + jyB = z + jwA - B = (x - z) + j(y - w)6EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu11SubtractionReal AxisImaginary AxisABA - BEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu12Multiplication Multiplication is most easily performed in polar coordinates:A = AM∠θB = BM∠φA × B = (AM × BM) ∠ (θ + φ)7EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu13MultiplicationReal AxisImaginary AxisABA × BEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu14Division Division is most easily performed in polar coordinates:A = AM∠θB = BM∠φA / B = (AM/ BM) ∠ (θ − φ)8EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu15DivisionReal AxisImaginary AxisABA / BEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu16Complex Exponentials We represent a real-valued sinusoid as the real part of a complex exponential. Complex exponentials provide the link between time functions and phasors. Complex exponentials make solving for AC steady state an algebraic problem.9EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu17Complex Exponentials A complex number A = z ∠θcan be represented asA = z ∠θ= z ejθ= z cos θ + j z sin θ A complex exponential isejωt= cos ωt + j sin ωt What do you get when you multiply A and ejωtand find the real part?EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu18Complex ExponentialsAejωt= z ejθejωt= z ej(ωt+θ)z ej(ωt+θ) = z cos (ωt+θ) + j z sin (ωt+θ) Re[Aejωt] = z cos (ωt+θ)10EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu19Sinusoids, Complex Exponentials, and Phasors Sinusoid:z cos (ωt+θ) Complex exponential:Aejωt= z ej(ωt+θ) Phasor:V = z ∠θEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu20Phasor Relationships for Circuit Elements Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor. A complex exponential is the mathematical tool needed to obtain this relationship.11EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu21I-V Relationship for a CapacitorSuppose that v(t) is a sinusoid:v(t) = VMej(ωt+θ)Find i(t).Cv(t)+-i(t)dttdvCti)()( =EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu22Computing the CurrentdtedVCdttdvCtijtjMθω+==)()()()( tCvjeCVjtijtjMωωθω==+12EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu23Phasor Relationship Represent v(t) and i(t) as phasors:V = VM∠θI = jωC V The derivative in the relationship between v(t) and i(t) becomes a multiplication by jωin the relationship between V and I.EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu24Examplev(t) = 120V cos(377t + 30°)C = 2μF What is V? What is I? What is i(t)?13EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu25I-V Relationship for an InductorV = jωL ILv(t)+-i(t)dttdiLtv)()( =EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu26Examplei(t) = 1μA cos(2π 9.15 107t + 30°)L = 1μH What is I? What is V? What is v(t)?14EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu27Resistor I-V relationshipvR= iRR ………….VR= IRR where R is the resistance in ohms,VR= phasor voltage, IR= phasor current(boldface indicates complex quantity)Capacitor I-V relationshipiC= CdvC/dt ...............Phasor current IC= phasor voltage VC/capacitive impedance ZC: Æ IC= VC/ZC where ZC = 1/jωC , j = (-1)1/2and boldface indicates complex quantityInductor I-V relationshipvL= LdiL/dt ...............Phasor voltage VL= phasor current IL/inductive impedance ZL Æ VL= ILZLwhere ZL= jωL, j = (-1)1/2 and boldface indicates complex quantityEE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu28Phasor Diagrams A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes). A phasor diagram helps to visualize the relationships between currents and voltages.15EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu29Impedance AC steady-state analysis using phasorsallows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:V = IZ Z is called impedance.EE40 Summer 2005: Lecture 6 Instructor: Octavian Florescu30Some Thoughts on Impedance Impedance depends on the frequency ω. Impedance is (often) a complex number. Impedance allows us to use the same solution techniques for AC steady state as
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