Phasor AnalysisImpedancePhasorsLoop Analysis ExampleNodal Analysis ExampleMultiple Source ExampleEquivalent Circuit ExampleSlide 8Kevin D. Donohue, University of Kentucky1Phasor Analysis Phasors, Impedance, SPICE, and Circuit AnalysisKevin D. Donohue, University of Kentucky2Impedance The conversion of resistive, inductive, and capacitive elements to impedance for a sinusoidal excitation at frequency  is given by:e)(Resistanc )(Reactance 1)(Reactance RRCXLXCL In general impedance is a complex quantity with a resistive component (real) and a reactive component (imaginary):RXXRjXRZ122tanˆKevin D. Donohue, University of Kentucky3PhasorsSources can be converted to phasor notation as follows: AtA )cos(90)sin( AtAThis can be applied to all sources of the same frequency,where  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  and deactive all other sources.2. Convert circuit elements to impedances.3. Solve for desired voltage or current for selected .4. Repeat steps 1 through 3 for new  until all sources have been applied.5. Add together all time-domain solutions solutions obtain in Step 3.Kevin D. Donohue, University of Kentucky4Loop Analysis Example Determine the steady-state response for vc(t) when vs(t) = 5cos(800t) VShow: V 6800cos8868.2)( 302.8868 j1.4434 - 2.5000 ˆttvVcc 114.86 nF6 k3 kvs(t)+ vc(t) -Kevin D. Donohue, University of Kentucky5Nodal Analysis ExampleFind the steady-state value of vo(t) in the circuit below, if vs(t) = 20cos(4t): vs10 2 ix1 H0.5 H0.1 FixShow: v0(t) = 13.91cos(4t + 198.3º)+vo-Kevin D. Donohue, University of Kentucky6Multiple Source ExampleFind io if is = 3cos(10t) and vs = 6cos(20t + 60º) vs5 0.5 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 1.2-2-10123SecondsAmpsKevin D. Donohue, University of Kentucky7 Equivalent Circuit ExampleFind io steady-state using Norton’s Theorem, if vs(t) = 2sin(10t):vs10 .4 H0.01 Fio5 Show is(t)= .2sin(10t); Zth = 3-j = 3.2-18.4º; io = 0.15cos(10t-153.4 º)Kevin D. Donohue, University of Kentucky8Equivalent Circuit ExampleFind vo steady-state using Thévenin’s Theorem, if vs(t) = 20cos(4t): vs10 2 ix1 H0.5 H0.1