Fall 2000EE201Phasors and Steady-State AC1 Phasors A concept of phasors, or rotating vectors, is used to find the AC steady-state response of linear circuits excited by sinusoidal voltages and/or currents: x(t) = X m cos(  t +  ) That is, x(t) is the real part of the complex quantity X m e j(  t +  ) Euler’s identity: Then,

Fall 2000EE201Phasors and Steady-State AC2 Vectors The quantity e j  is a unity vector (magnitude =1) pointing at an angle . Euler’s identity, in essence, allows one to relate polar form representation of the vector to the rectangular form representation. A shorthand form for representing e j  is 1 . j  (Imaginary axis) (Real axis) 1 sin  cos  The vector Xe j  =X  is stationary and constant. That is, the angle  never changes and the magnitude is always X. (polar form)(rectangular form)(Complex number)

Fall 2000EE201Phasors and Steady-State AC3 Phasors A phasor is a rotating vector. The magnitude never changes, but the angle is a function of time. The magnitude of the following phasor is X m, and the angle is  =  t + . The phasor is represented below at times t = 0 and t =  / . The value of X m cos(  t +  ) = Re[X m  (  t +  )] changes depending on the position of the phasor at time t. rotating at a velocity of  radians per second X m cos (  t +  )  XmXm (t = 0) XmXm X m cos (  t +  )  (  t =  )

Fall 2000EE201Phasors and Steady-State AC4 Solving circuits in the phasor domain In general, a circuit containing resistors, inductors, and capacitors and excited by time-varying voltage and current sources must be solved using differential equations. However, if the voltage and current sources are sinusoidal, then algebraic equations can be used to find the AC steady-state response of the circuit. These equations, however, do involve complex quantities and phasors. Observations: The steady-state response of a circuit is the response to an input after the input has been applied a “long time” and the circuit has been allowed to settle down. The steady-state value of all voltages and currents in a linear circuit excited by sinusoidal inputs will also be sinusoidal. The voltages and currents will be of the same frequency as the input; the magnitude and phase angle will, in general, be different. Linear Circuit v(t) = V m cos(  t +  v ) i(t) = I m cos(  t +  i )

Fall 2000EE201Phasors and Steady-State AC5 Solving circuits in the phasor domain Since all currents and voltages in the steady-state AC response have the same frequency, a short-hand notation, called phasor notation, is use to represent time- varying sinusoidal signals. By transforming voltages and signals from the time domain to the phasor domain, we can use algebraic equations and conventional circuit analysis techniques (KVL, KCL, nodal, mesh Thevenin, etc.) to solve for the steady-state voltages and currents in a circuit. In addition to transforming sources from sinusoidal functions of time to phasors, circuit components (R, L, and C) must be transformed, also. time domain representation phasor domain representation

Fall 2000EE201Phasors and Steady-State AC6 Steady-State AC Voltage and Current Phasors All steady-state voltages and currents are sinusoidal if the forcing functions (voltage and currents sources) are sinusoidal. These sinusoidal voltages and currents are represented as phasors (rotating vectors). vv VmVm 

Fall 2000EE201Phasors and Steady-State AC7 + v(t) - R i(t) Resistors Time Domain The voltage across a resistor is in phase with the resistor current.  v =  i Phasor Domain + - R Justification:

Fall 2000EE201Phasors and Steady-State AC8 + v(t) - L i(t) Inductors Time Domain The current through an inductor lags the voltage by 90 degrees. ii Justification: Phasor Domain + - jX L = j  L

Fall 2000EE201Phasors and Steady-State AC9 + v(t) - C i(t) Capacitors Time DomainPhasor Domain The current through a capacitor leads the voltage by 90 degrees jX C = 1/j  C vv Justification:

Fall 2000EE201Phasors and Steady-State AC10 Time Domain R + v(t) - i(t) C + v(t) - i(t) + v(t) - L Phasor Domain R j  L = jX L 1/j  C = -jX C Summary

Fall 2000EE201Phasors and Steady-State AC11 Impedance In dc circuits we have resistance, R: I +V-+V-  z > 0  R+jX, an inductive impedance.  z < 0  R-jX is a capacitive impedance In steady-state ac circuits we have impedance, Z: R = resistance X = reactance

Fall 2000EE201Phasors and Steady-State AC12 Impedance In dc circuits : B is called susceptance conductance is the reciprocal of resistance In ac circuits :admittance is the reciprocal of impedance Parallel and Series Combinations of Impedances

Fall 2000EE201Phasors and Steady-State AC13 Example Draw the frequency-domain network and calculate v o (t). Use a phasor diagram to determine v(t). v s (t) = 10cos(800  t) + v(t) - + v o (t) - -j25.14  j25.14  1k  1.11k  V=10  0  - 442mH

Fall 2000EE201Phasors and Steady-State AC14 Example (Continued) -j25.14  j25.14  1k  1.11k  V=10  0  - By Nodal Analysis:

Fall 2000EE201Phasors and Steady-State AC15 Example (Continued) -j25.14  j25.14  1k  1.11k  V=10  0  - By KVL: Substituting:

Fall 2000EE201Phasors and Steady-State AC16 Vo = 7.07  45   45  10  0  V= 10 - Vo = 7.07  -45  Finding V using vector addition

Fall 2000EE201Phasors and Steady-State AC17 E201 Circuits I Phasors Read chapter 8, Steady-State AC Analysis, pages HW Assignment #15 Chapter 8 – Sections 8.1 thru 8.4 Extension Problems: E8.1 thru E8.7 HW Assignment #16 Chapter 8 – Sections 8.5 and 8.6 Extension Problems: E8.8 thru E8.11 End-of- Chapter Problems: P8.27 HW Assignment #17 Chapter 8 – Sections 8.7 and 8.8 Extension Problems: E8.12, E.13, E8.14, E8.17 End-of- Chapter Problems:  P8.36(45   V)  P8.40 (8  -90  V)  P8.44 (8.5   V)  P8.46 (v o (t) = 6cos(10 4 t  )V v L (t) = 8cos(10 4 t  )V)  P8.49 (2.83  45  A)