09/29/2009 Reading: Hambley Chapter 5 and Appendix A
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1 EE40 Lec 10 Complex Numbers and Phasors Prof. Nathan Cheung 09/29/2009 Reading: Hambley Chapter 5 and Appendix A Slide 1
2 OUTLINE Phasors as notation for Sinusoids Arithmetic with Complex Numbers Complex impedances Circuit analysis using complex impedances Dervative/Integration as multiplication/division Phasor Relationship for Circuit Elements Slide 2
3 Types of Circuit Excitation Linear Time- Invariant Circuit Steady-State Excitation (DC Steady-State) Linear Time- Invariant Circuitit Sinusoidal (Single- Frequency) Excitation AC Steady-State Step Excitation Digital Pulse Source Linear Time- Invariant Circuit OR Linear Time- Invariant Circuit Transient Excitations Slide 3
4 Sinusoids vt () V cos ωt θ Amplitude: V ( ) M M Angular frequency: ω 2π f Radians/sec Phase angle: θ Frequency: f 1/T Unit: 1/sec or Hz Period: T Time necessary to go through one cycle Slide 4
5 Sinusoids What is the amplitude, period, frequency, and radian frequency of this sinusoid? id? Slide 5
6 Sinusoidal Sources Create Too Much Algebra Guess a solution x P ( t) Asin( wt) B cos( wt) x dxp ( t) ( t) τ FA sin( wt) FB cos( wt) dt P d ( Asin( wt ) B cos( wt )) ( Asin( i( wt ) B cos( wt )) τ FA sin( i( wt ) FB cos( wt ) dt ( A τb FA)sin( wt) ( B τa FB )cos( wt) ( A τb FA ) 0 ( B τa F ) 0 B F τ A F A B τ 2 1 τ F A F B τ 2 1 Phasors s (vectors that rotate in the complex plane) are a clever alternative. B 0 Slide 6
7 y Complex Numbers (1) imaginary i x is the real part axis j ( 1) θ real x axis Rectangular Coordinates Z x jy Polar Coordinates: Z z θ Exponential Form: y is the imaginary part z is the magnitude θ is the phase x z cosθ 2 z x π j θ j θ j Z Z e ze 2 y Z z(cosθ jsin θ ) 2 j0 1 1e 1 0 j 1e 1 90 y z sinθ θ tan 1 y x Slide 7
8 Complex Numbers (2) Euler s Identities e cosθ e sinθ e jθ jθ jθ e 2 e 2 j jθ jθ cosθ jsinθ j 2 2 e θ cos θ sin θ 1 Exponential Form of a complex number Z j j θ θ Z e ze z θ Slide 8
9 Arithmetic With Complex Numbers To compute phasor voltages and currents, we need to be able to perform computation with complex numbers. Addition Subtraction ti Multiplication Division (And later use multiplication by jω to replace Differentiation Integration Slide 9
10 Addition Addition is most easily performed in rectangular coordinates: A x jy B z jw A B (x z) j(y w) A B Imaginary Axis B A Real Axis Slide 10
11 Subtraction Subtraction is most easily performed in rectangular coordinates: A x jy B z jw A - B (x - z) j(y - w) Imaginary Axis B A A - B Real Axis Slide 11
12 Multiplication Multiplication is most easily performed in polar coordinates: A A M θ B B M φ A B (A M B M ) (θ φ) A B Imaginary Axis B A Real Axis Slide 12
13 Division Division is most easily performed in polar coordinates: A A M θ B B M φ A / B (A M / B M ) (θ φ) Imaginary Axis B A A / B Real Axis Slide 13
14 Arithmetic Operations of Complex Numbers Add and Subtract: it is easiest to do this in rectangular format Add/subtract the real and imaginary parts separately Multiply and Divide: it is easiest to do this in exponential/polar format Multiply (divide) the magnitudes Add (subtract) the phases Z Z Z Z Z 1 jθ1 ze 1 z1 θ1 z1 cos θ1 jz1 sin θ1 jθ 2 ze z θ z cosθ jz sinθ Z 2 ( z 1 cos θ 1 z 2 cos θ 2 ) j ( z 1 sin θ 1 z 2 sin θ 2 ) Z ( z cosθ z cos θ ) j( z sinθ z sin θ ) j( θ1 θ2) ( ) ( ) ( θ θ ) Z z z e z z j( θ θ ) 2 z1 z2 e ( z1/ z2) ( θ1 θ2) 1 2 Z / Z ( / ) 1 Slide 14
15 Summary of Complex Numbers Slide 15
16 Phasor: Rotating Complex Vector { jφ jwt} ( j ωt Ve e Re Ve ) v( t) V cos( ωt φ) Re Imaginary Axis Phasor Rotates at uniform angular velocity ωt V cos(ωtφ) Real Axis The head start angle is φ. Slide 16
17 Phasors and Complex Exponentials A sinusoid can be described using a complex exponential: e jωt cos ωt j sin ωt So, v(t) V M cos (ωt θ) V M Re[e j (ωt θ) ] The phasor of v(t) is given by: V V M θ Slide 17
18 Sinusoids, Complex Exponentials, Phasors Sinusoid: z cos (ωtθ) Complex exponential: Phasor: Ae jωt z e j(ωtθ) V z θ Slide 18
19 Sinusoids <--> Phasors Slide 19
20 Complex Exponentials Complex Exponentials provide the link between time functions and phasors. Allow dervatives 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. Slide 20
21 Complex Impedance AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm s law: V IZ Z is called impedance. Slide 21
22 I-V Relationship for a Capacitor i(t) C v(t) () - i ( t ) C dv( t) dt Suppose that v(t) is a sinusoid: v(t) Re{V M e j(ωtθ) } Find i(t). Slide 22
23 Capacitor i(t) i(t) C v(t ) dv( t) i ( t ) C dt - V j( ωt θ) j( ωt θ) vt () Vcos( ωt θ) 2 e e dv () t CV d j ( ω t θ ) j ( ω t θ ) CV j ( ω t θ ) j ( ω t θ ) it () C e e j e e dt 2 dt ω 2 ωcv e j( ωt θ) e j( ωt θ) π ω CV sin( ω t θ) ω CV cos( ω t θ ) 2 j 2 Z c V V θ V π 1 π 1 1 ( θ θ ) ( ) j I π ωcv 2 ωc 2 ωc jωc I θ 2 Slide 23
24 Capacitor Impedance i(t) C v(t - ) i( t) dv( t) C dt Phasor definition ω θ θ j( ωt θ) vt () Vcos( t ) Re Ve V V j ( ωt θ ) dv() t de j( ωt θ) it ( ) C Re CV Re jωcve I θ dt dt I V V θ V 1 Note: j ( e Zc ( θ θ) jπ/2 ) is lumped into the amplitude expression here I I θ jωcv jωc Slide 24
25 Inductor Impedance i(t) L v(t) () - v ( t ) L di( t) dt V jωl I Slide 25
26 lead Capacitor: I leads V by 90 o Inductor: V leads I by 90 o Voltage 7cos( ω t) 7 0 inductor current π π 7sin( ω t ) 7cos( ω t ) Behind 2 0 t capacitor current π π 7sin( ωt) 7cos( ωt ) Slide 26
27 RLC Complex Impedances Slide 27
28 Circuit Analysis Using Complex Impedances Real-time KVL v () t v () t v () t ( ω θ ) ( ω θ ) ( ω θ ) V cos t V cos t V cos t j 1 2 ( 3 ) ( t ) j ( t Re ) j t Ve ω θ 1 Ve ω θ 2 Ve ω θ 3 0 Phasor Form KVL Ve Ve Ve θ VVV 0 j( θ 1) j( 2 ) j( 3 ) 1 θ Phasor Form KCL I 1I 2 I3 0 Use complex impedances for inductors and capacitors and follow same analysis as in chap 2 of Hambley. Slide 28
29 Example : Voltage Divider R10 k Ω v ( ) ( ) t 10cos 2π60t - C1 µf _ v c ( t )? 1. Source Phasor jφ V V0 e 10 f 60 ω 2π 60 φ π 2. Complex impedance V 0 10 Z R 10 4 Ω, Z C 1 jωc 1 j120π 10 6 Slide 29
30 Example : Voltage Divider cont. () t 10cos( 2π t) v 60 R10 k Ω - C1 µf _ c ( t )? ω 2π 60 v c φ 0 φ0 V π V 3. Use KVL or voltage divider id formula c Z C 1 6 Z C j120 π V 10 tan (1.2 π ) Z 1 1 j R 4 π 10 1 ( 1.2π) 6 j120π Change phasor back to actual signal ( ) ( o t 2.56 cos 120 π t 75 ) v c 75 Slide 30
31 Example: Thevenin Equivalent Circuit Sinusoidal Source V C R L a Find the Thevenin equivalent ckt. Z TH a b 1. Thevenin impedance by shorting the source 1/jωC a R jωl b Z TH Z C R // Z R // Z jωlr jωl ω V TH L 2 LCR b 2. Thevenin voltage using voltage divider Z// VTH V 1 Z// jωc where Slide 31 Z // jωlr R jωl
32 Example : Voltage Divider using Polar Coordinates 20kΩ 0 10V 0 V C kΩ -90 Now use the voltage divider to find V C : V C V C 2.65kΩ V k Ω k Ω V Note: polar co-ordinate better only for multiply l and divide Slide 32
33 Example: Parallel Impedance 5mA µF 1kΩ V Find v(t) for ω2π mA 0 -j530kω 1kΩ V - Slide 33
34 Example: Parallel Impedance cont. 5mA 0 Z eq V Z eq Z eq ( j530) j Ω V IZ 5mA Ω eq V 2.34V v( t) 2.34V cos(2π 3000t 62.1 ) Slide 34
35 Example : Superposition v C ( t)? C - I ( t ) ( ) I sin ω t R L I ( t ) I cos ( ω t ) I1 1 ω1 1. Two sources superposition 2. For each source use phasors ( analyzed directly or using source transformation) 3. Sum up the results Note: The two sources have different frequencies. The impedance will have different values for the different sources! Slide 35
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