EE40Lecture 14Venkat AnantharamComplex Numbers (1)Complex Numbers (2)Arithmetic With Complex NumbersAdditionAdditionSubtractionSubtractionMultiplicationMultiplicationDivisionDivisionPhasorsPhasor from rotating Complex VectorComplex ExponentialsI-V Relationship for a CapacitorCapacitor Impedance (1)Capacitor Impedance (2)ExampleComputing the CurrentInductor ImpedanceExampleSlide 1EE40 Spring 2008 Venkat AnantharamEE40Lecture 14Venkat Anantharam2/27/08 Reading: Chap. 5: phasorsSlide 2EE40 Spring 2008 Venkat AnantharamComplex Numbers (1)• x is the real part• y is the imaginary part• z is the magnitude• θ is the phase(1)j=−θzxyreal axisimaginary axis• Rectangular Coordinates Z = x + jy• Polar Coordinates: Z = z ∠θ• Exponential Form: θcoszx=θsinzy=22yxz +=xy1tan−=θ(cos sin )zjθθ=+Zjjezeθθ==ZZ0211 101190jjejeπ==∠°==∠ °Slide 3EE40 Spring 2008 Venkat AnantharamComplex Numbers (2)22cos2sin2cos sincos sin 1jjjjjjeeeejejeθθθθθθθθθθθθ−−+=−==+=+=Euler’s IdentitiesExponential Form of a complex numberjjezezθθθ===∠ZZSlide 4EE40 Spring 2008 Venkat AnantharamArithmetic With Complex Numbers• To compute phasor voltages and currents, we need to be able to perform computations with complex numbers.– Addition– Subtraction– Multiplication– DivisionSlide 5EE40 Spring 2008 Venkat AnantharamAddition• Addition is most easily performed in rectangular coordinates:A = x + jyB = z + jwA + B = (x + z) + j(y + w)Slide 6EE40 Spring 2008 Venkat AnantharamAdditionReal AxisImaginary AxisABA + BSlide 7EE40 Spring 2008 Venkat AnantharamSubtraction• Subtraction is most easily performed in rectangular coordinates:A = x + jyB = z + jwA - B = (x - z) + j(y - w)Slide 8EE40 Spring 2008 Venkat AnantharamSubtractionReal AxisImaginary AxisABA - BSlide 9EE40 Spring 2008 Venkat AnantharamMultiplication• Multiplication is most easily performed in polar coordinates:A = AM∠θB = BM∠φA × B = (AM × BM) ∠ (θ + φ)Slide 10EE40 Spring 2008 Venkat AnantharamMultiplicationImaginary AxisReal AxisAA × BBSlide 11EE40 Spring 2008 Venkat AnantharamDivision• Division is most easily performed in polar coordinates:A = AM∠θB = BM∠φA / B = (AM/ BM) ∠ (θ − φ)Slide 12EE40 Spring 2008 Venkat AnantharamDivisionImaginary AxisABReal AxisA / BSlide 13EE40 Spring 2008 Venkat AnantharamPhasors• Assuming a source voltage is a sinusoidal time-varying functionv(t) = V cos (ωt + θ)• We can write:• Similarly, if the function is v(t) = V sin (ωt + θ)() ()() cos( ) Re Rejt jtjvt V t V e VeDefine Phasor as Ve Vωθ ωθθωθθ++⎡⎤⎡ ⎤=+= =⎣⎦⎣ ⎦=∠()()22( ) sin( ) cos( ) Re2jtvt V t V t VePhasor Vπωθπθπωθ ωθ+−−⎡⎤=+= +−=⎢⎥⎣⎦=∠Slide 14EE40 Spring 2008 Venkat AnantharamPhasor from rotating Complex Vector{})(tjjwtjeeVetVtvωφφωVReRe)cos()( ==+=Imaginary AxisVReal AxisRotates at uniform angular velocity ωtcos(ωt+φ)The head start angle is φ.Slide 15EE40 Spring 2008 Venkat AnantharamComplex Exponentials• We represent a real-valued sinusoid as the real part of a complex exponential after multiplying by .• Complex exponentials – provide the link between time functions and phasors.– Allow derivatives and integrals to be replaced by multiplying or dividing by jω– make solving for AC steady state simple algebra with complex numbers.• Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.tjeωSlide 16EE40 Spring 2008 Venkat AnantharamI-V Relationship for a CapacitorCv(t)+-i(t)dttdvCti)()( =Suppose that v(t) is a sinusoid:v(t) = Re{VMej(ωt+θ)}Find i(t).Slide 17EE40 Spring 2008 Venkat AnantharamCapacitor Impedance (1)Cv(t)+-i(t)dttdvCti)()( =() ()() () () ()() ()() cos( )2()()22sin( ) cos( )22(2jt jtjt jt jt jtjt jtcVvt V t e edv t CV d CVit C e e j e edt dtCVee CVt CVtjVVZCVIωθ ωθωθ ωθ ωθ ωθωθ ωθωθωωπωωθω ωθθθθπωθ+−++−+ +−++−+⎡⎤=+= +⎣⎦⎡⎤⎡⎤== + = −⎣⎦⎣⎦−⎡⎤=−=−+=++⎣⎦∠== = ∠−⎛⎞∠+⎜⎟⎝⎠VI111)()22jCCjCππωωω−= ∠−=− =Slide 18EE40 Spring 2008 Venkat AnantharamCapacitor Impedance (2)Cv(t)+-i(t)dttdvCti)()( =Phasor definition()()()() cos( ) Re()( ) Re Re1()jtjtjtcvt V t Ve Vdv t deit C CV j CVe Idt dtVVZIjCV jCωθωθωθωθ θωθθθθθω ω+++⇒⇒⎡⎤=+= =∠⎣⎦⎡⎤⎡⎤== = =∠⎢⎥⎣⎦⎣⎦∠== = ∠−=∠VIVISlide 19EE40 Spring 2008 Venkat AnantharamExamplev(t) = 120V cos(377t + 30°)C = 2µF•What is V?•What is I?•What is i(t)?Slide 20EE40 Spring 2008 Venkat AnantharamComputing the CurrentNote: The differentiation and integration operations become algebraic operationsωjdt1⇒∫ωjdtd⇒Slide 21EE40 Spring 2008 Venkat AnantharamInductor Impedance V = jωL IL+-i(t)dttdiLtv)()( =v(t)Slide 22EE40 Spring 2008 Venkat AnantharamExamplei(t) = 1µA cos(2π 9.15 107t + 30°)L = 1µH•What is I?•What is V?•What is
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