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

Download Impedance - Engineering Electrical Circuits - Lecture Slides and more Slides Electrical Circuit Analysis in PDF only on Docsity! Impedance KCL & KVL Docsity.com Review → V-I in Phasor Space • Resistors  Inductors  Capacitors R VI = °−∠= 90 Lω VI °∠= 90VI Cω No Phase Shift i(t) LAGS i(t) LEADS Docsity.com Impedance cont.2 • Thus  The Magnitude and Phase jXRZ z +=∠θ R X XRZ z 1 22 tan−= += θ  Where z z ZX ZR θ θ sin cos = = Cj Z LjZ Cj Lj C L RZRR ω ω ω ω 11 = = = = == IV IV IV ImpedanceEq. PhasorElement  Summary Of Passive- Element Impedance  Examine ZC C j Cjj j Cj ZC ωωω 1 1 − === C X C jZ CC ωω 11 − =⇒ − =∴ Docsity.com KVL & KCL Hold In Phasor Spc − + )(1 tv − + )(3 tv −+ )(2 tv )(0 ti )(1 ti )(2 ti )(3 ti 0)()()( 321 =++ tvtvtv :KVL 3,2,1,0,)( 0)()()()( )( 3210 == =+++− + keIti titititi ktj Mkk φω :KCL 3,2,1,)( )( == + ieVtv itjMii θω 0)( :KVL 321 321 =++ tjj M j M j M eeVeVeV ωθθθ 0332211 =∠+∠+∠ θθθ MMM VVV Phasors! 0321 =++ VVV − + 1V − + 3V −+ 2V 0I 1I 2I 3I 03210 =+++− IIII Similarly for the Sinusoidal Currents ... Docsity.com Series & Parallel Impedances • Impedances (which have units of Ω) Combine as do RESISTANCES – The SERIES Case • The Parallel Case I −+ 1V 1Z −+ 2V 2Z I 21 ZZZs += ∑= k ks ZZ 1Z 2Z − + V I I − + V 21 21 ZZ ZZZ p + = ∑= k kp ZZ 11 Docsity.com Complex Numbers in MATLAB • MATLAB recognizes complex numbers these in these forms – Rectangular – Exponential • Can Use “i” or “j” for √(- 1) • MATLAB Always returns “i” for √(-1) • Sometimes need “*” >> phiR = 23*pi/180 % 23deg in Rads phiR = 0.4014 >> Z1 = 7 + i*23 % if i or j BEFORE, then need * Z1 = 7.0000 +23.0000i >> Z2 = 11 - 13j Z2 = 11.0000 -13.0000i >> Z3 = 43*exp(j*phiR) % Need * Z3 = 39.5817 +16.8014i >> Z4 = 37*exp(0.61j) Z4 = 30.3270 +21.1961i Docsity.com Phasors in MATLAB • MATLAB Does NOT Recognize Phasor NOTATION – But it DOES handle Complex Exponentials • e.g.:   1753 4329 8 7 ∠−= −∠= Z Z >> phi7 = -43*pi/180 phi7 = -0.7505 >> phi8 = 17*pi/180 phi8 = 0.2967 >> Z7 = 29*exp(j*phi7) Z7 = 21.2093 -19.7780i >> Z8 = -53*exp(j*phi8) Z8 = -50.6842 -15.4957i >> Zsum = Z7 + Z8 Zsum = -29.4749 -35.2737i >> Zdif = Z7 - Z8 Zdif = 71.8934 - 4.2823i >> Zprod = Z7*Z8 Zprod = -1.3814e+003 +6.7378e+002i >> Zquo = Z7/Z8 Zquo = -0.2736 + 0.4739i Docsity.com Phasors in MATLAB • MATLAB Always returns Complex No.s in RECTANGULAR Form • Can Recover Magnitude & Phase Using Commands – abs(Z) – angle(Z) >> Zquo = Z7/Z8 Zquo = -0.2736 + 0.4739i >> Asum = abs(Zsum) Asum = 45.9674 >> phi_sum = angle(Zsum) phi_sum = -2.2669 >> phi_sumd = phi_sum*180/pi phi_sumd = -129.8824 >> Aquo = abs(Zquo) Aquo = 0.5472 >> phi_quo = angle(Zquo) phi_quo = 2.0944 >> phi_quod = phi_quo*180/pi phi_quod = 120.0000 >> Zquo_test = Aquo*exp(j*phi_quo) Zquo_test = -0.2736 + 0.4739i Docsity.com Phasor Diagrams • As Noted Earlier Phasors can be Considered as VECTORS in the Complex Plane – See Diagram at Right  See Next Slide for Review of Vector Addition • Text Diagrams follow the PARALLELOGRAM Method  Phasors Obey the Rules of Vector Arithmetic • Which were orignially Developed for Force Mechanics φ A b a Real Imaginary Docsity.com Vector Addition • Parallelogram Rule For Vector Addition • Examine Top & Bottom of The Parallelogram – Triangle Rule For Vector Addition – Vector Addition is Commutative – Vector Subtraction → Reverse Direction of The Subtrahend B B C C ( )QPQP −+=− PQQP +=+ Docsity.com Example  Phasor Diagram • For The Ckt at Right, Draw the Phasor Diagrams as a function of Frequency • First Write KCL  That is, we Can Select ONE Phasor to have a ZERO Phase Angle • In this Case Choose V  Next Examine Frequency Sensitivity of the Admittances  Now we can Select ANY Phasor Quantity, I or V, as the BaseLine ∑ ⇒=       ++= ++= k kS S S Y Cj LjR Cj LjR sAdmittance 11 VI VI VVVI ω ω ω ω Docsity.com KCL & KVL for AC Analysis • Simple-Circuit Analysis – AC Version of Ohm’s Law → V = IZ – Rules for Combining Z and/or Y – KCL & KVL – Current and/or Voltage Dividers • More Complex Circuits – Nodal Analysis – Loop or Mesh Analysis – SuperPosition Docsity.com Methods of AC Analysis cont. • More Complex Circuits – Thevenin’s Theorem – Norton’s Theorem – Numerical Techniques • MATLAB • SPICE Docsity.com Example • For The Ckt At Right, Find VS if  Then I2 by Ohm  Solution Plan: GND at Bot, then Find in Order • I3 → V1 → I2 → I1 → VS  I3 First by Ohm °∠= 458VoV ( ) ( )AVO °∠= Ω = 454 23 VI  Then V1 by Ohm ( ) °∠×°−∠=−= 454458)22( 31 IV j )(0314.111 V°∠=V )(90657.5 902 0314.11 2 1 2 A V j °−∠= °∠Ω °∠ = Ω = VI  Then I1 by KCL °∠+°−∠=+= 45490657.5321 III ))(828.2828.2(657.51 Ajj ++−=I )(829.2828.21 Aj−=I Docsity.com Nodal Analysis cont. • Solving for For V2  Recall  The Complex Arithmetic ( ) Ω = 1 2 V O VI  Or 11 62 11 11 11 1 2 jjj + +=      − ++ + V 11 6)11(2 )11)(11( )11()11)(11()11( 2 j j jj jjjj + ++ = −+ ++−++−V 28 1 4 2 jj += − V ( ) ( ) 2 3 2 5 4 1 1 28 2 j jj −= −+ =∴V )( 2 3 2 5 AjO       −=I °−∠= 96.3092.2 AOI Docsity.com Loop Analysis for AC Circuits • Same Ckt, But Different Approach to Find IO • Note: IO = –I3 • Constraint: I1 = –2A∠0  Simplify Loop2 & Loop3  The Loop Eqns  Two Eqns In Two Unknowns 0))(1(06))(1( :2 LOOP 3221 =+−+°∠−++ IIjIIj 01)()1( :3 LOOP 332 =Ω++Ω− IIIj )2)(1(6)1(2 )1(6)1(2 :L2 32 132 −+−=−+ +−=−+ jIjI IjIjI 0)2()1( :L3 32 =−+− IjIj  Solution is I3 = –IO  Recall I1 = –2A∠0 Docsity.com Loop Analysis cont • Isolating I3  The Next Step is to Solve the 3 Eqns for I2 and I3  So Then Note  Then The Solution ( ) )28)(1()2(2)1( 32 jjjj +−=−−− I 4 610 3 j− =I )( 2 3 2 5 0 Aj−=⇒ I 2I  Could also use a SuperMesh to Avoid the Current Source 0)1()(1 :3 MESH 0)(106)1( :SUPERMESH 02 :CONSTRAINT 323 321 12 =−+− =−+°∠−+ °∠=− III III II j j 32 III −=O Docsity.com AC Source SuperPosition cont. • Find I-Src Contribution to IO by I-Divider  The V-Src Contribution by V-Divider  Now Deactivate the I-Source (open it) Ω=1'Z )(01 11 102'0 A°∠=+ °∠=I )1(||1" jZ −= ( ) ( )j jjZ −+ −⋅ =−= 11 11)1(||1" )(06 1" " " 1 VjZ Z °∠ ++ =V )(06 1 1 " " " 1 " A jZ Z O °∠++ =Ω= VI Docsity.com AC Source SuperPosition cont.2 • Sub for Z”  The Total Response  Finally SuperPose the Response Components )(6 1 2 1 2 1 " 0 A j j j j j ++ − − − − =I 6 3)1( 1" 0 jj j ++− − =I )( 4 6 4 6" 0 Aj−=I )( 2 3 2 5 0 Aj−=∴ I ( )       −+=+= j 2 3 2 31"0 ' 00 III Docsity.com Multiple Frequencies • When Sources of Differing FREQUENCIES excite a ckt then we MUST use SuperPosition for every set of sources with NON-EQUAL FREQUENCIES • An Example 1V 2V  We Can Denote the Sources as Phasors  1050&0100 21 −∠=∠= VV VV  But canNOT COMBINE them due to DIFFERENT frequencies Docsity.com Source Transformation • Source transformation is a good tool to reduce complexity in a circuit – WHEN IT CAN BE APPLIED • “ideal sources” are not good models for real behavior of sources – A real battery does not produce infinite current when short-circuited • Resistance → Impedance Analogy + - Improved model for voltage source Improved model for current source SV VR SI IR a b a b SS IV RIV RRR = == WHEN SEQUIVALENT AREMODELS THE VZ IZ SS IV Z ZZZ IV = == Docsity.com Source Transformation • Same Ckt, But Use Source Transformation to Find IO • Start With I-Src  Then the Reduced Circuit jV 28' +=  Next Combine the Voltage Sources And Xform j j ZSeries S S + + == 1 28'VI Docsity.com Source Transformation cont • The Reduced Ckt  Now Combine the Series-Parallel Impedances  The Reduced Ckt j j S + + = 1 28I Ω= −++ − =−+= 1 11 1)1(||)1( 2 jj jjjZ p pZ 2 35 j O − =∴ I  IO by I-Divider 2 35 j− = ( )( ) ( )( )jj jj j j SO −+ −+ = + + =⋅= 11 14 1 4 2 1II Docsity.com Docsity.com abotcollege.edu SAR IZ_ Bruce Mayer, PE Engineering Instructor haba College KVL LooP-/ -244 20 + (Zi-T2) f2+\Va =o Kee Looh-2 —~Va + (De-L)gr-2) Tr + @I2 5 CB STRAIT . 22907 T2—Lh >. L > Zr-2l7e =T2-ZhAy Loor-(| £ Looh-> (242)D-2sT2+M = 24VL0 ey Z) ¢2Te +va = © 27, + (2-2)e = eevee @ 6 2.0(L- 24g) + 20(j-3) Tn = 2¢VA0 (#- 28) Da- ay Loo =2erho Doceity com (¢-24\) a-Z£2- ZEHt+ELID Te = @Etr4V ( “-2)r Tet 2ERBDKA YET Lea /-26.s Tet 6 ¥¢¥/ ALEC, 08 Diz (4¥ + 228)A THUS Vo = 2i1°L2 (@8 + C.¥5)V Vo = £0, 88V £36.08" —_ iin WhiteBoard Work 20Ω 6mH 10Ω is(t) 3.33 µF i1(t) i2(t)  See Next Slide for Phasor Diagrams  Let’s Work this Nice Problem ( ) ( ) °∠=⇒ °+= 13.8100 13.85000cos100 mA tmAtiS I Docsity.com