Sinusoidal - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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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 ωπω =+ tXtx M ωsin)( = 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 ωωπω =+=+ ) all(for ),() and 22 ttTtxx(t TT ∀+= =⇒= ω ππω  Now Can Construct a DIMENSIONAL (time) Plot for the sine ( )TtXtx M π2sin)( = 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)( πf T f πω π ω 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 or 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 • ω = 6.981 rad/s • θ = 1.257 rads = 72° • φ = 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 1105 PeriodmS PeriodmSφθ 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 πωω πωω ±−= ±−= tt tt       −=       += 2 cossin 2 sincos πωω πωω tt tt  Additional Relations βαβαβα βαβαβα sinsincoscos)cos( sincoscossin)sin( −=+ +=+ βαβαβα βαβαβα sinsincoscos)cos( sincoscossin)sin( +=− −=− (rads) rads 180(degrees) 360 radians 2 θ π θ π ° = °= 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 == = π ω ω 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 dttiT Rdttp T P 0 0 0 0 )(1)(1 2 22 2 0 0 0 0 )(11 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 RIRIdttiT RP ==        = ∫ + ∫ + = Tt t eff dttiT 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 22cos1 2 111 2 1       ++= ∫ ∫ dttTdtTII T T Mrms θω 0 212 1 12 111 2 1 21 21 0 21 0 M M T M T Mrms IT T It 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 RIR V P = ∆ = 2 2 2 1 2 1 M M av RIR VP == MMrms rmsM rmseffdcM av VVV VV R V R V R V R VP 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 == += θω )cos( 22 )cos( 2 1 iv MM ivMMav IV IVP θθ θθ −= −= )cos( ivrmsrmsav IVP θθ −= 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 dttuT 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 =Ω⋅== 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 & φ )sin()( )sin()( φω θω += ⇒+= 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 tdiLtRitv += )( [ ] [ ] tAtAt dt di tAtAti tAtAti tAtAti tAti ωωωω ωω ωφωφ ωφωφ φω cossin)( sincos)( sinsincoscos)( sinsincoscos)( sinysinx-cosycosx y)cos(x using )cos()( 21 21 +−= += −+= −= ⋅⋅=+ += tRAAL tRAAL tVM ωω ωω ω 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 & φ in Solution Eqn ( ) ( ) ( ) 2222 22 sincos sincos AA AA =+ =+ φφ φφ  Find A to be R L A A ω φ φφ −== cos sintan 22 )( LR VA M ω+ =       − + = − R Lt LR Vti M ωω ω 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 θθθ je j += • Where 1−=j  Note: • The Euler Relation can Be Proved Using Taylor’s Series (Power Series) Expansion of ejθ Docsity.com Complex Exponential cont • Now in the Euler Identity, Let  So  Notice That if tωθ = tjte tj ωωω sincos +=  Next Multiply by a Constant Amplitude, VM tjVtVeV MM tj M ωω ω sincos +=  Separate Function into Real and Imaginary Parts { } { } tVeV tVeV M tj M M tj M ω ω ω ω sinIm cosRe = = ( ) ( ) { } tVeVtv tVtv M tj M M ω ω ω 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 ωsin2 = ( ) tVtv M ωcos1 = General Linear Circuit tjV eV M tj M ω ω sin tV tV M M ω ω sin cos tjVtVeV MM tj M ωω ω 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 ω=)( tj M eVtv ω=)( ( ) ( ) tjjMtjM eeIeIti ωφφω == +)(  Then The KVL Eqn )()()( t dt diLtRitv += 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 ω=)(  Canceling ejωt and Solving for IMejφ ( ) ( )( ) ( )φωφω ω ++ == tjMtjM eIjeIdt d dt tdi ( )[ ] ( )[ ] ( )[ ] [ ]tjjM tj M tj M tj M eeIRLj eIRLj eIReIjL tRit dt diL ωφ φω φωφω ω ω ω )( )( )()( += += += + + ++ tj M tjj M eVeeIRLj ωωφω =+ )( RLj VeI MjM + = ω φ Docsity.com Example – RL Single Loop cont.4 • The Complex Exponential Soln  Where tj M eVtv ω=)( ( )    − − + = R Lj Mj M e LR VeI ω φ ω 1tan 22  Recall Assumed Soln ( ) R L LR VI MM ωφ ω 1 22 tan, −−= + =  Finally RECOVER the DESIRED Soln By Taking the REAL Part of the Response ( ) ( ) ( ) ( )[ ]φωφω φωωφ +++= == + tjtI eIeeIti M tj M tjj M sincos )( Docsity.com Example – RL Single Loop cont.5 • By Superposition  Explicitly tj M eVtv ω=)( ( )       − + = − R Lt LR Vti M ωω ω 1 22 tancos )(  SAME as Before ( ) )cos()( }Re{)( }Re{cos)( φω ω φω ω +=∴ =⇒ == + 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 ωφφω == +)( φ A b a Real Imaginary φφ ∠≡ 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 ω=→∠= 0V ( ) tjjMM eeIiI ωφφ =→∠=I ( ) ( ){ } ( ){ } ( )φω φω ωφ += = = + 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 ( ) ( ) ( ) θθ θω θωθω ∠=∠ = += = ++ MM tj M tj M M RIV eVeRI tIti tRitv Then cos if )()( )()( IV R= Docsity.com Resistors in ω-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 ω-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 ω I )(70 1020377 12 3 A°−∠×× =∴ −I  In the Time Domain ( ) )70377cos(593.1)( °−= tAti Ljω ω VI 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 dCeI θθ θω θωθω ω ω = = = + ++ or )( VI Cjω= Docsity.com Capacitors in ω-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 ω ω ω ω θ θ θθ  Therefore the Voltage and Current are OUT of PHASE by 90° • Plotting the Current and Voltage Vectors in the Complex Plane Docsity.com WhiteBoard Work 2Ω + _ v(t) i(t)  Let’s Work Text Problem 8.5 ()()     ° + =     ° + = 45 377 sin 12 180 377 cos 10 2 1 t s V t v t s V t v Docsity.com Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Appendix Complex No.s 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 φφ φ jA Aejba j += =+ jban += φ A b a Real Imaginary 11 −=⋅⇒−= jjj  Then from the Vector Plot a b baA 1 22 tan−= += φ Docsity.com Complex Number Division cont. • For the Quotient n/m in Exponential Form  However Must Still Calculate the Magnitudes A & D... ( )θφ θ φ −     == jj j e D A De Ae m n Docsity.com 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 ejωt is COMMON To all Terms we can work with ONLY the PreFactor that contains Magnitude and Phase info; so { }tjjM M eeU tUtu ωθ θω ⋅= += Re )cos()( )cos()( φω += tYty M }Re{}Re{ tjjM tjj M eeYeeU ωφωθ ⋅⇒⋅ )cos(}Re{)( )cos()( φω φθ θω +==→ ∠=⇒∠=→ += tYty YU tUtu M MM M Y YU Docsity.com