Sinusoidal AC Steady State - Engineering Circuit Analysis - Lecture Slides, Slides of Electrical Circuit Analysis

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Engineering 43 Sinusoidal AC SteadySt Docsity.com 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 Docsity.com Sinusoids • Recall From Trig the Sine Function  Where • XM  “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 )()2( txtx   tXtx M sin)(  Docsity.com Sinusoids cont. • Now Define the “Period”, T Such That  How often does the Cycle Repeat? • Define Next the CYCLIC FREQUENCY  From Above Observe   )()()2( txTtxtx   ) all(for ),() and 2 2 ttTtxx(t TT       Now Can Construct a DIMENSIONAL (time) Plot for the sine  TtXtx M 2sin)(  Docsity.com Sinusoids cont.2 • Now Define the Cyclic “Frequency”, f  Quick Example • USA Residential Electrical Power Delivered as a 115Vrms, 60Hz, AC sine wave  Describes the Signal Repetition-Rate in Units of Cycles-Per-Second, or HERTZ (Hz) • Hz is a Derived SI Unit    tVtv tVtv residence residence   99.376sin6.162)( 602sin1152)( f T f    2 2 1    Will Figure Out the 2 term Shortly • RMS  “Root (of the) Mean Square” Docsity.com Sinusoid Phase Difference -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Time (S) x i ( V o r A ) x1 (V or A) x2 (V or A) file =Sinusoid_Lead-Lag_Plot_0311.xls Out of Phase * 0.733 rads * 42° * 105 mS PARAMETERS • T = 900 mS • f = 1.1111 Hz •  = 6.981 rad/s •  = 1.257 rads = 72° •  = 0.5236 rads = 30° x1 LEADs x2 x2 LAGs x1 For Different Amplitudes, Measure Phase Difference • peak-to-peak • valley-to-valley • ZeroCross-to-ZeroCross    42 1 360 900 1 105 PeriodmS Period mS Docsity.com 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  )sin(sin )cos(cos     tt tt               2 cossin 2 sincos     tt tt  Additional Relations   sinsincoscos)cos( sincoscossin)sin(     sinsincoscos)cos( sincoscossin)sin(   (rads) rads 180 (degrees) 360 radians 2        Docsity.com 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 )301000cos(6)( )601000sin(12)( 2 1   ttv ttv Hz 2.159 2 )Hz( S 1000 -1      f Docsity.com 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 )(ti R Rtitp )()( 2 Docsity.com rms Values Cont. • For The Resistive Case, Define Ieff for the Avg Power Condition  If the Current is DC, then i(t) = Idc, so  The Pav Calc For a Periodic Signal by Integ  Now for the Time- Variable Current i(t) → Ieff, and, by Definition )(ti R RIP Rtitp effav 2 2 )()(              Tt t Tt t av dtti T Rdttp T P 0 0 0 0 )( 1 )( 1 2 22 2 0 0 0 0 )(1 1 1 dc Tt t dc Tt t dcav RIdtt T RI dtI T RP                        avdceffav PRIRIP  22 Docsity.com rms Values cont.2 • In the Pwr Eqn  Examine the Eqn for Ieff 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 1st & 3rd Expression for Pav find 222 0 0 )( 1 effdc Tt t av RIRIdtti T RP                Tt t eff dtti T I 0 0 )( 1 2  This Expression Holds for ANY Periodic Signal rmseff II  Docsity.com RMS Value for Sinusoid • Rearranging a bit • But the integral of a sinusoid over ONE PERIOD is ZERO, so the 2nd Term goes to Zero leaving   21 0 0 22cos 1 2 1 1 1 2 1          dttT dt T II T T Mrms  0 212 1 12 1 1 1 2 1 21 21 0 21 0 M M T M T Mrms IT T I t T Idt T II                               Docsity.com 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 2 , 2 , 2 1 2 1 resM resM av RI R V P    2 2 2 1 2 1 M M av RI R V P  MMrms rmsM rmseffdcM av VVV VV R V R V R V R V P 707.02 2 2 1 22 2222    MMrms rmsM rmseffdcMav III II RIRIRIRIP 707.02 2 2 1 22 2222    Docsity.com 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 2 is valueeffective theand )cos()( Mrmseff M UUU tUtu    )cos( 22 )cos( 2 1 iv MM ivMMav IV IVP     )cos( ivrmsrmsav IVP   Docsity.com 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    Tt t rms dttu T U 0 0 )( 1 2 RIP rmsav 2   2 6 4 2 2 0 22 844 8 1 Adtdt s I s s s rms         WARIP rmsav 80108 22  Docsity.com 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 iss(t), Need ONLY to Determine Parameters A &  )sin()( )sin()(     tAti tVtv SS M Docsity.com 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     dt tdi LtRitv  )(     tAtAt dt di tAtAti tAtAti tAtAti tAti      cossin)( sincos)( sinsincoscos)( sinsincoscos)( sinysinx-cosycosx y)cos(x using )cos()( 21 21       tRAAL tRAAL tVM    cos)( sin)( cos 12 21    Docsity.com 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       2222 22 sincos sincos AA AA      Find A to be R L A A      cos sin tan 22 )( LR V A M             R L t LR V ti M    1 22 tancos )( )( Docsity.com 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) sincos  je j  • Where 1j  Note: • The Euler Relation can Be Proved Using Taylor’s Series (Power Series) Expansion of ej Docsity.com Complex Exponential cont • Now in the Euler Identity, Let  So  Notice That if t  tjte tj  sincos   Next Multiply by a Constant Amplitude, VM tjVtVeV MM tj M   sincos   Separate Function into Real and Imaginary Parts     tVeV tVeV M tj M M tj M     sinIm cosRe         tVeVtv tVtv M tj M M    cosRe Then cos    Now Recall that LINEAR Circuits Obey SUPERPOSITION Docsity.com 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   tjVtv M sin2    tVtv M cos1  General Linear Circuit tjV eV M tj M   sin tV tV M M   sin cos tjVtVeV MM tj M   sincos  Docsity.com 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 tj M eVtv )( tj M eVtv )(     tjjMtjM eeIeIti   )(  Then The KVL Eqn )()()( t dt di LtRitv  Docsity.com Example – RL Single Loop cont. • Taking the 1st Time Derivative for the Assumed Solution  Then the Right-Hand- Side (RHS) of the KVL  Then the KVL Eqn tj M eVtv )(  Canceling ejt and Solving for IMe j           tjMtjM eIjeI dt d dt tdi           tjjM tj M tj M tj M eeIRLj eIRLj eIReIjL tRit dt di L       )( )( )()(       tj M tjj M eVeeIRLj   )( RLj V eI MjM     Docsity.com Example – RL Single Loop cont.4 • The Complex Exponential Soln  Where tj M eVtv )(             R L j Mj M e LR V eI    1tan 22  Recall Assumed Soln   R L LR V I MM    1 22 tan,     Finally RECOVER the DESIRED Soln By Taking the REAL Part of the Response              tjtI eIeeIti M tj M tjj M sincos )( Docsity.com Example – RL Single Loop cont.5 • By Superposition  Explicitly tj M eVtv )(             R L t LR V ti M    1 22 tancos )(  SAME as Before   )cos()( }Re{)( }Re{cos)(         tIti eIti eVtVtv M tj M tj MM  Docsity.com 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     tjjtj eAeAetx   )(  A b a Real Imaginary   AAe j Docsity.com 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 tj M VevV  0V   tjjMM eeIiI  I                 tI eI eeIti M tj M tjj M cos Re Re Docsity.com Resistors in Frequency Domain • The v-i Reln for R  Thus the Frequency Domain Relationship for Resistors               MM tj M tj M M RIV eVeRI tIti tRitv Then cos if )()( )()( IV R Docsity.com 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 Docsity.com 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 )( Find )20377cos(12)(,20 Given ti ttvmHL          90 2012 901 with so AVL V j  I )(70 1020377 12 3 A    I  In the Time Domain   )70377cos(593.1)(  tAti Lj  V I V    2012 377 L → current LAGS Docsity.com Capacitors in Frequency Domain • The v-i Reln for C  Thus the Frequency Domain Relationship for Capacitors      vi v vi j M j M tj M tj M tj M eCVjeI eCVj eV dt d CeI           or )( VI Cj Docsity.com Capacitors in -Land cont. • The relationship between phasors is algebraic. – Recall  Thus  909011 jej         90 90 90 VI C eCV eCVe eCVjeI v v vi j M j M j j M j M         Therefore the Voltage and Current are OUT of PHASE by 90° • Plotting the Current and Voltage Vectors in the Complex Plane Docsity.com 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 )sin(cos   jA Aejba j   jban   A b a Real Imaginary 11  jjj  Then from the Vector Plot a b baA 1 22 tan   Docsity.com Complex Number Arithmetic • Consider Two Complex Numbers  The SUM, Σ, and DIFFERENCE, , for these numbers  The PRODUCT n•m   j j Dejdcm Aejban    Complex DIVISION is Painfully Tedious • See Next Slide        dbjcamn dbjcamn                  jjj ADeDeAemn adbcjbdac bdjadbcjac jdcjbamn 2 Docsity.com 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 jdc jba m n     The Exponential Form is Cleaner • See Next Slide         22 22 2 dc adbcjbdac m n dcddcjc bdjadbcjac m n jdc jdc jdc jba m n             Docsity.com