ENGR-43_Lec-08-1_AC-Steady-State.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 43 Sinusoidal AC SteadySt

ENGR-43_Lec-08-1_AC-Steady-State.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Outline – AC Steady State  SINUSOIDS Review basic facts about sinusoidal signals  SINUSOIDAL AND COMPLEX FORCING FUNCTIONS Behavior of circuits with sinusoidal independent current & voltage sources Modeling of sinusoids in terms of complex exponentials

ENGR-43_Lec-08-1_AC-Steady-State.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Outline – AC SS cont.  PHASORS Representation of complex exponentials as vectors –Facilitates steady-state analysis of circuits Has NOTHING to do with StarTrek  IMPEDANCE AND ADMITANCE Generalization of the familiar concepts of RESISTANCE and CONDUCTANCE to describe AC steady-state circuit operation phasEr

ENGR-43_Lec-08-1_AC-Steady-State.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Outline – AC SS cont.2  PHASOR DIAGRAMS Representation of AC voltages and currents as COMPLEX VECTORS  BASIC AC ANALYSIS USING KIRCHHOFF’S LAWS  ANALYSIS TECHNIQUES Extension of node, loop, SuperPosition, Thevenin and other KVL/KCL Linear- Circuit Analysis techniques

ENGR-43_Lec-08-1_AC-Steady-State.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoids  Recall From Trig the Sine Function  Where X M  “Amplitude” or Peak or Maximum Value –Typical Units = A or V ω  Radian, or Angular, Frequency in rads/sec ωt  Sinusoid argument in radians (a pure no.)  The function Repeats every 2π; mathematically  For the RADIAN Plot Above, The Functional Relationship

ENGR-43_Lec-08-1_AC-Steady-State.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoids cont.  Now Define the “Period”, T Such That  How often does the Cycle Repeat? Define Next the CYCLIC FREQUENCY  From Above Observe  Now Can Construct a DIMENSIONAL (time) Plot for the sine

ENGR-43_Lec-08-1_AC-Steady-State.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoids cont.2  Now Define the Cyclic “Frequency”, f  Quick Example USA Residential Electrical Power Delivered as a 115V rms, 60Hz, AC sine wave  Describes the Signal Repetition-Rate in Units of Cycles-Per-Second, or HERTZ (Hz) Hz is a Derived SI Unit  Will Figure Out the  2 term Shortly RMS  “Root (of the) Mean Square”

ENGR-43_Lec-08-1_AC-Steady-State.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoids cont.3  Now Consider the GENERAL Expression for a Sinusoid  Where θ  “Phase Angle” in Radians  Graphically, for POSITIVE θ

ENGR-43_Lec-08-1_AC-Steady-State.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Leading, Lagging, In-Phase  Consider Two Sinusoids with the SAME Angular Frequency  Now if  >  x 1 LEADS x 2 by (  −  ) rads or Degrees x 2 LAGS x 1 by (  −  ) rads or Degrees  If  = , Then The Signals are IN-PHASE  If   , Then The Signals are OUT-of-PHASE  Phase Angle Typically Stated in DEGREES, but Radians are acceptable Both These Forms OK

ENGR-43_Lec-08-1_AC-Steady-State.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

ENGR-43_Lec-08-1_AC-Steady-State.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Useful Trig Identities  To Convert sin↔cos  To Make a Valid Phase-Angle Difference Measurement BOTH Sinusoids MUST have the SAME Frequency & Trig-Fcn (sin OR cos) Useful Phase-Difference ID’s   Additional Relations

ENGR-43_Lec-08-1_AC-Steady-State.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Phase Angles  Given Signals  Find Frequency in Standard Units of Hz Phase Difference  Frequency in radians per second is the PreFactor for the time variable Thus  To find phase angle must express BOTH sinusoids using The SAME trigonometric function; –either sine or cosine A POSITIVE amplitude

ENGR-43_Lec-08-1_AC-Steady-State.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – Phase Angles cont.  Convert –6V Amplitude to Positive Value Using  Then  It’s Poor Form to Express phase shifts in Angles >180° in Absolute Value  Next Convert cosine to sine using  So Finally  Thus v 1 LEADS v 2 by 120°

ENGR-43_Lec-08-1_AC-Steady-State.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

ENGR-43_Lec-08-1_AC-Steady-State.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Effective or rms Values  Consider Instantaneous Power For a Purely Resistive Load  Since a Resistive Load Dissipates this Power as HEAT, the Effective Value is also called the HEATING Value for the Time-Variable Source For example –A Car Coffee Maker Runs off 12 Vdc, and Heats the Water in 223s. –Connect a SawTooth Source to the coffee Maker and Adjust the Amplitude for the Same Time → Effective Voltage of 12V  Now Define The EFFECTIVE Value For a Time-Varying Signal as the EQUIVALENT DC value That Supplies The SAME AVERAGE POWER

ENGR-43_Lec-08-1_AC-Steady-State.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis rms Values Cont.  For The Resistive Case, Define I eff for the Avg Power Condition  If the Current is DC, then i(t) = I dc, so  The P av Calc For a Periodic Signal by Integ  Now for the Time- Variable Current i(t) → I eff, and, by Definition

ENGR-43_Lec-08-1_AC-Steady-State.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis rms Values cont.2  In the Pwr Eqn  Examine the Eqn for I eff and notice it is Determined by Taking the Square ROOT of the time-averaged, or MEAN, SQUARE of the Current  In Engineering This Operation is given the Short-hand notation of “rms”  So  Equating the 1 st & 3 rd Expression for P av find  This Expression Holds for ANY Periodic Signal

ENGR-43_Lec-08-1_AC-Steady-State.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis RMS Value for Sinusoid  Find RMS Value for sinusoidal current Note the Period  Sub Above into RMS Eqn:  Use Trig ID:

ENGR-43_Lec-08-1_AC-Steady-State.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis RMS Value for Sinusoid  Sub cos 2 Trig ID into RMS integral  Integrating Term by Term

ENGR-43_Lec-08-1_AC-Steady-State.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis RMS Value for Sinusoid  Rearranging a bit  But the integral of a sinusoid over ONE PERIOD is ZERO, so the 2 nd Term goes to Zero leaving

ENGR-43_Lec-08-1_AC-Steady-State.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoidal rms Alternative  For a Sinusoidal Source Driving a Complex (Z = R + jX) Load  Similarly for the rms Current  If the Load is Purely Resistive  Now, By the “effective” Definition for a R Load

ENGR-43_Lec-08-1_AC-Steady-State.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoidal rms Values cont  In General for a Sinusoidal Quantity  Thus the Power to a Reactive Load Can be Calculated using These Quantities as Measured at the SOURCE Using a True-rms DMM –The rms Voltage –The rms Current Using an Oscilloscope and “Current Shunt” –The Phase Angle Difference  For the General, Complex-Load Case  By the rms Definitions

ENGR-43_Lec-08-1_AC-Steady-State.ppt 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  rms Voltage  Given Voltage Waveform Find the rms Voltage Value  During the 2s Rise Calc the slope m = [4V/2s] = 2 V/s  Thus The Math Model for the First Complete Period  Find The Period T = 4 s  Derive A Math Model for the Voltage WaveForm  Use the rms Integral T

ENGR-43_Lec-08-1_AC-Steady-State.ppt 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  rms Voltage cont.  Calc the rms Voltage  Numerically T 0

ENGR-43_Lec-08-1_AC-Steady-State.ppt 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Average Power  Given Current Waveform Thru a 10Ω Resistor, then Find the Average Power  The “squared” Version  Find The Period T = 8 s  Apply The rms Eqns  Then the Power

ENGR-43_Lec-08-1_AC-Steady-State.ppt 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Sinusoidal Forcing Functions  Consider the Arbitrary LINEAR Ckt at Right.  If the independent source is a sinusoid of constant frequency then for ANY variable in the LINEAR circuit the STEADY-STATE Response will be SINUSOIDAL and of the SAME FREQUENCY  Mathematically  Thus to Find i ss (t), Need ONLY to Determine Parameters A & 

ENGR-43_Lec-08-1_AC-Steady-State.ppt 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Single Loop  Given Simple Ckt Find i(t) in Steady State  Write KVL for Single Loop  In Steady State Expect  Sub Into ODE and Rearrange

ENGR-43_Lec-08-1_AC-Steady-State.ppt 28 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Single Loop cont  Recall the Expanded ODE  Equating the sin & cos PreFactors Yields  Solving for Constants A 1 and A 2  Recognize as an ALGEBRAIC Relation for 2 Unkwns in 2 Eqns  Found A 1 & A 2 using ONLY Algebra This is Good

ENGR-43_Lec-08-1_AC-Steady-State.ppt 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Loop cont.2  Using A 1 & A 2 State the i(t) Soln  Also the Source-V  If in the ID  =x, and ωt = y, then  Comparing Soln to Desired form → use sum-formula Trig ID  Would Like Soln in Form

ENGR-43_Lec-08-1_AC-Steady-State.ppt 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Loop cont.3  Dividing These Eqns Find  Now  Elegant Final Result, But VERY Tedious Calc for a SIMPLE Ckt  Not Good  Subbing for A &  in Solution Eqn  Find A to be

ENGR-43_Lec-08-1_AC-Steady-State.ppt 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Exponential Form  Solving a Simple, One-Loop Circuit Can Be Very Tedious for Sinusoidal Excitations  To make the analysis simpler relate sinusoidal signals to COMPLEX NUMBERS. The Analysis Of the Steady State Will Be Converted To Solving Systems Of Algebraic Equations...  Start with Euler’s Identity (Appendix A) Where  Note: The Euler Relation can Be Proved Using Taylor’s Series (Power Series) Expansion of e j 

ENGR-43_Lec-08-1_AC-Steady-State.ppt 32 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Exponential cont  Now in the Euler Identity, Let  So  Notice That if  Next Multiply by a Constant Amplitude, V M  Separate Function into Real and Imaginary Parts  Now Recall that LINEAR Circuits Obey SUPERPOSITION

ENGR-43_Lec-08-1_AC-Steady-State.ppt 33 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Exponential cont.2  Consider at Right The Linear Ckt with Two Driving Sources  By KVL The Total V-src Applied to the Circuit  Now by SuperPosition The Current Response to the Applied Sources  This Suggests That the… General Linear Circuit

ENGR-43_Lec-08-1_AC-Steady-State.ppt 34 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Exponential cont.3  Application of the Complex SOURCE will Result in a Complex RESPONSE From Which The REAL (desired) Response Can be RECOVERED; That is  Thus the To find the Response a COMPLEX Source Can Be applied.  Then The Desired Response can Be RECOVERED By Taking the REAL Part of the COMPLEX Response at the end of the analysis General Linear Circuit

ENGR-43_Lec-08-1_AC-Steady-State.ppt 35 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Realizability  We can NOT Build Physical (REAL) Sources that Include IMAGINARY Outputs  We CAN, However, BUILD These  We can also NOT invalidate Superposition if we multiply a REAL Source by ANY CONSTANT Including “j”  Thus Superposition Holds, mathematically, for General Linear Circuit

ENGR-43_Lec-08-1_AC-Steady-State.ppt 36 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Single Loop  This Time, Start with a COMPLEX forcing Function, and Recover the REAL Response at The End of the Analysis Let  In a Linear Ckt, No Circuit Element Can Change The Driving Frequency, but They May induce a Phase Shift Relative to the Driving Sinusoid  Thus Assume Current Response of the Form  Then The KVL Eqn

ENGR-43_Lec-08-1_AC-Steady-State.ppt 37 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – RL Single Loop cont.  Taking the 1 st Time Derivative for the Assumed Solution  Then the Right-Hand- Side (RHS) of the KVL  Then the KVL Eqn  Canceling e j  t and Solving for I M e j 

ENGR-43_Lec-08-1_AC-Steady-State.ppt 38 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – RL Single Loop cont.2  Clear Denominator of The Imaginary Component By Multiplying by the Complex Conjugate  Then the Response in Rectangular Form: a+jb  Next A & θ

ENGR-43_Lec-08-1_AC-Steady-State.ppt 39 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – RL Single Loop cont.3  First A  And θ  A &  Correspondence with Assumed Soln A → I M θ →   Cast Solution into Assumed Form

ENGR-43_Lec-08-1_AC-Steady-State.ppt 40 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – RL Single Loop cont.4  The Complex Exponential Soln  Where  Recall Assumed Soln  Finally RECOVER the DESIRED Soln By Taking the REAL Part of the Response

ENGR-43_Lec-08-1_AC-Steady-State.ppt 41 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – RL Single Loop cont.5  By Superposition  Explicitly  SAME as Before 

ENGR-43_Lec-08-1_AC-Steady-State.ppt 42 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Phasor Notation  If ALL dependent Quantities In a Circuit (ALL i’s & v’s) Have The SAME FREQUENCY, Then They differ only by Magnitude and Phase That is, With Reference to the Complex-Plane Diagram at Right, The dependent Variable Takes the form  Borrowing Notation from Vector Mechanics The Frequency PreFactor Can Be Written in the Shorthand “Phasor” Form A b a Real Imaginary

ENGR-43_Lec-08-1_AC-Steady-State.ppt 43 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Phasor Characteristics  Since in the Euler Reln The REAL part of the expression is a COSINE, Need to express any SINE Function as an Equivalent CoSine Turn into Cos Phasor  Examples  Phasors Combine As the Complex Polar Exponentials that they are

ENGR-43_Lec-08-1_AC-Steady-State.ppt 44 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Single Loop  As Before Sub a Complex Source for the Real Source  The Form of the Responding Current  For the Complex Quantities  Then The KVL Eqn in the Phasor Domain  Recall The KVL Phasor Variables Denoted as BOLDFACE CAPITAL Letters

ENGR-43_Lec-08-1_AC-Steady-State.ppt 45 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  RL Single Loop cont.  To recover the desired Time Domain Solution Substitute  Then by Superposition Take  This is A LOT Easier Than Previous Methods  The Solution Process in the Frequency Domain Entailed Only Simple Algebraic Operations on the Phasors

ENGR-43_Lec-08-1_AC-Steady-State.ppt 46 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Resistors in Frequency Domain  The v-i Reln for R  Thus the Frequency Domain Relationship for Resistors

ENGR-43_Lec-08-1_AC-Steady-State.ppt 47 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Resistors in  -Land cont.  Phasors are complex numbers. The Resistor Model Has A Geometric Interpretation  In the Complex-Plane The Current & Voltage Are CoLineal i.e., Resistors induce NO Phase Shift Between the Source and the Response Thus resistor voltage and current sinusoids are said to be “IN PHASE” R → IN Phase

ENGR-43_Lec-08-1_AC-Steady-State.ppt 48 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductors in Frequency Domain  The v-i Reln for L  Thus the Frequency Domain Relationship for Inductors

ENGR-43_Lec-08-1_AC-Steady-State.ppt 49 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductors in  -Land cont.  The relationship between phasors is algebraic. To Examine This Reln Note That  Thus  Therefore the Current and Voltage are OUT of PHASE by 90° Plotting the Current and Voltage Vectors in the Complex Plane

ENGR-43_Lec-08-1_AC-Steady-State.ppt 50 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductors in  -Land cont.2  In the Time Domain  Phase Relationship Descriptions The VOLTAGE LEADS the current by 90° The CURRENT LAGS the voltage by 90°  Short Example  In the Time Domain L → current LAGS

ENGR-43_Lec-08-1_AC-Steady-State.ppt 51 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitors in Frequency Domain  The v-i Reln for C  Thus the Frequency Domain Relationship for Capacitors

ENGR-43_Lec-08-1_AC-Steady-State.ppt 52 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitors in  -Land cont.  The relationship between phasors is algebraic. Recall  Thus  Therefore the Voltage and Current are OUT of PHASE by 90° Plotting the Current and Voltage Vectors in the Complex Plane

ENGR-43_Lec-08-1_AC-Steady-State.ppt 53 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitors in  -Land cont.2  In the Time Domain  Phase Relationship Descriptions The CURRENT LEADS the voltage by 90° The VOLTAGE LAGS the current by 90°  Short Example  In the Time Domain C → current LEADS

ENGR-43_Lec-08-1_AC-Steady-State.ppt 54 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis All Done for Today Root of the Mean Square rms voltage = peak voltagerms voltage = 1.11 average voltage peak voltage = rms voltagepeak voltage = 1.57 average voltage average voltage = peak voltageaverage voltage = 0.9 rms voltage

ENGR-43_Lec-08-1_AC-Steady-State.ppt 55 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work Text Problem 8.5

ENGR-43_Lec-08-1_AC-Steady-State.ppt 56 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 43 Appendix Complex No.s

ENGR-43_Lec-08-1_AC-Steady-State.ppt 57 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Numbers Reviewed  Consider a General Complex Number  This Can Be thought of as a VECTOR in the Complex Plane  This Vector Can be Expressed in Polar (exponential) Form Thru the Euler Identity  Where A b a Real Imaginary  Then from the Vector Plot

ENGR-43_Lec-08-1_AC-Steady-State.ppt 58 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Number Arithmetic  Consider Two Complex Numbers  The SUM, Σ, and DIFFERENCE, , for these numbers  The PRODUCT nm  Complex DIVISION is Painfully Tedious See Next Slide

ENGR-43_Lec-08-1_AC-Steady-State.ppt 59 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Number Division  For the Quotient n/m in Rectangular Form  The Generally accepted Form of a Complex Quotient Does NOT contain Complex or Imaginary DENOMINATORS  Use the Complex CONJUGATE to Clear the Complex Denominator  The Exponential Form is Cleaner See Next Slide

ENGR-43_Lec-08-1_AC-Steady-State.ppt 60 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Number Division cont.  For the Quotient n/m in Exponential Form  However Must Still Calculate the Magnitudes A & D...

ENGR-43_Lec-08-1_AC-Steady-State.ppt 61 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Phasor Notation cont.  Because of source superposition one can consider as a SINGLE source, a System That contains REAL and IMAGINARY Components  The Real Steady State Response Of Any Circuit Variable Will Be Of The Form  Or by SuperPosition  Since e j  t is COMMON To all Terms we can work with ONLY the PreFactor that contains Magnitude and Phase info; so