Phasor AnalysisImpedancePhasorsLoop Analysis ExampleNodal Analysis ExampleMultiple Source ExampleEquivalent Circuit ExampleEquivalent Circuit ExampleKevin D. Donohue, University of Kentucky 1Phasor Analysis Phasors, Impedance, SPICE, and Circuit AnalysisKevin D. Donohue, University of Kentucky 2ImpedanceThe conversion of resistive, inductive, and capacitive elements to impedance for a sinusoidal excitation at frequency ω is given by:e)(Resistanc )(Reactance 1)(Reactance RRCXLXCL=−==ωωIn general impedance is a complex quantity with a resistive component (real) and a reactive component (imaginary):∠+=+=−RXXRjXRZ122tanˆKevin D. Donohue, University of Kentucky 3PhasorsSources can be converted to phasor notation as follows:θθω∠⇔+ AtA )cos(o90)sin( −∠⇔+θθωAtAThis can be applied to all sources of the same frequency,where ω is used in the impedance conversion of the circuit.If sources of different frequencies exist, superposition mustapplied to solve for a given voltage or current:1. Select sources with a common ω and deactive all other sources.2. Convert circuit elements to impedances.3. Solve for desired voltage or current for selected ω.4. Repeat steps 1 through 3 for new ω until all sources have been applied.5. Add together all time-domain solutions solutions obtain in Step 3.Kevin D. Donohue, University of Kentucky 4Loop Analysis ExampleDetermine the steady-state response for vc(t) when vs(t) = 5cos(800πt) V 114.86 nF6 kΩ3 kΩvs(t)+ vc(t) -Show:V 6800cos8868.2)( 302.8868 j1.4434 - 2.5000 ˆ−=⇔−∠==ππttvVccoKevin D. Donohue, University of Kentucky 5Nodal Analysis ExampleFind the steady-state value of vo(t) in the circuit below, if vs(t) = 20cos(4t): vs10 Ω2 ix1 H0.5 H0.1 FixShow: v0(t) = 13.91cos(4t + 198.3º)+vo-Kevin D. Donohue, University of Kentucky 6Multiple Source ExampleFind io if is= 3cos(10t) and vs= 6cos(20t + 60º) vs5 Ω0.5 H0.01 F isioShow io= 0.54cos(20t+123.4º)+2.7cos(10t-153.4º)0 0.2 0.4 0.6 0.8 1 1.2-2-10123SecondsAmp sKevin D. Donohue, University of Kentucky 7Equivalent Circuit ExampleFind iosteady-state using Norton’s Theorem, if vs(t) = 2sin(10t): vs10 Ω.4 H0.01 Fio5 ΩShow is(t)= .2sin(10t); Zth= 3-j = 3.2∠-18.4º; io= 0.15cos(10t-153.4 º)Kevin D. Donohue, University of Kentucky 8Equivalent Circuit ExampleFind vosteady-state using Thévenin’s Theorem, if vs(t) = 20cos(4t): vs10 Ω2 ix1 H0.5 H0.1