Phasors: Impedance and Circuit Anlysis. Phasors

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1 Phasors: Impedance and Circuit Anlysis Lecture 6, 0/07/05 OUTLINE Phasor ReCap Capacitor/Inductor Example Arithmetic with Complex Numbers Complex Impedance Circuit Analysis with Complex Impedance Phasor Equivalent Circuits Reading Hambley Lecture 6, Slide Phasors Assuming a source voltage is a sinusoid timevarying function v(t) cos (ωt θ) We can write: j( ωt θ) j( ωt θ) vt () cos( ωt θ) Re e Re e jθ Define Phasor as e θ Similarly, if the function is v(t) sin (ωt θ) π π j( ωt θ ) vt () sin( ωt θ) cos( ωt θ ) Re e Phasor π ( θ ) Lecture 6, Slide

2 Phasor: Rotating Complex ector jφ jwt t { e e } ( e j ω Re ) v( t) cos( ωt φ) Re Imaginary Axis Rotates at uniform angular velocity ωt cos(ωtφ) Real Axis The head start angle is φ. Lecture 6, Slide 3 Dervative Becomes Multiplication i( t) C dv( t) dt C d Re{ e dt } d Re{ M e C dt jω t jωt jθ } i( t) Re{ jωc M e d jω dt jωt jθ jωt jθ } Re{ Y e } Re{ Ye Y jωc Note: The differentiation and integration operations become algebraic operations M dt jω jωt } Lecture 6, Slide 4

3 Complex Exponentials We represent a realvalued sinusoid as the real part of a complex exponential after multiplying j t by e ω. 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 currentvoltage relationships for inductors and capacitors much like we express the currentvoltage relationship for a resistor. Lecture 6, Slide 5 Impedance AC steadystate analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm s law: IZ Z is called impedance. Lecture 6, Slide 6 3

4 oltage 7cos( ω t) 7 0 Phase inductor current 8 π π 7sin( ωt) 7cos( ωt ) capacitor current π π 7sin( ωt) 7cos( ωt ) 7 Lecture 6, Slide 7 Phasor Diagrams A phasor diagram helps to visualize the relationships between currents and voltages. Capacitor: I leads by 90 o Inductor: leads I by 90 o v( t) cos( ω t φ) ωt I CAP jwc ELI the ICE man. π/ π/ Lecture 6, Slide 8 I IND jωl Real Axis 4

5 Some 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 we use for DC steady state. Lecture 6, Slide 9 Example: Single Loop Circuit 0 0 0kΩ µf C f60 Hz, C? How do we find C? First compute impedances for resistor and capacitor: Z R R 0kΩ 0kΩ 0 Z C /j (πf x µf).65kω 90 Lecture 6, Slide 0 5

6 Impedance Example 0kΩ C.65kΩ 90 Now use the voltage divider to find C : C.65kΩ kΩ 90 0kΩ 0 C Lecture 6, Slide What happens when ω changes? 0 0 0kΩ µf C ω 0 Find C Lecture 6, Slide 6

7 Circuit Analysis Using Complex Impedances Suitable for AC steady state. KL v() t v() t v3() t 0 cos t cos t cos t 0 ( ω θ ) ( ω θ ) ( ω θ ) 3 3 Phasor Form KL j( ωt θ) j( ωt θ) j( ωt θ3 ) Re e e e 3 0 e e e 0 j( θ) j( θ) j( θ3 ) Phasor Form KCL I I I3 0 Use complex impedances for inductors and capacitors and follow same analysis as in chap. Lecture 6, Slide 3 SteadyState AC Analysis 0.µF kω Find v(t) for ωπ 3000 kω j530kω Lecture 6, Slide 4 7

8 Find the Equivalent Impedance ( j530) j Ω 6. 5mA Ω 6. I v( t).34 cos(π 3000t 6. ) Lecture 6, Slide 5 Change the Frequency 0.µF kω Find v(t) for ωπ kω j3.5ω Lecture 6, Slide 6 8

9 Find an Equivalent Impedance ( j3.5) j Ω mA 0 3.5Ω I 7.5m v( t) 7.5m cos(π t 89.8 ) Lecture 6, Slide 7 Series Impedance Z Z 3 Z For example: Z Z Z 3 L L jω(l L ) Z eq C C jωc jωc Lecture 6, Slide 8 9

10 Parallel Impedance Z Z Z 3 For example: / /Z /Z /Z 3 L L C C Z eq LL jω ( L L ) Z eq jω( C C ) Lecture 6, Slide 9 SteadyState AC Nodeoltage Analysis C I 0 sin(ωt) R L I cos(ωt) Nodal analysis or mesh? What are the nodes (or meshes)? What happens if the sources are at different frequencies? Lecture 6, Slide 0 0