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Linear Circuit Analysis I

by: Cassidy Casper

Linear Circuit Analysis I ECE 20100

Cassidy Casper
GPA 3.59

Hong Tan

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Hong Tan
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This 33 page Class Notes was uploaded by Cassidy Casper on Saturday September 19, 2015. The Class Notes belongs to ECE 20100 at Purdue University taught by Hong Tan in Fall. Since its upload, it has received 112 views. For similar materials see /class/207890/ece-20100-purdue-university in Electrical Engineering & Computer Science at Purdue University.

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Date Created: 09/19/15
ECE201 Linear Circuit Analysis I Prof H Z Tan Topics Secondorder circuits RLC with constant inputs more examples Problem 1 Warmup In the circuit below R02 ohm C1 F L025 H iL08 A vc020 V Find vct for tgt0 Problem 2 Given that vc0 7 60 v and iL0 7 10 A Find 1L0 for tgt0 60V 9 continue for Problem 2 Problem 3 The switch in the circuit below has been open for a long time before it is closed at time t0 Find expressions for a the current ixt for tgt0 and b the voltage vxt for tgt0 0 19 x gt 2V 2H VA 19 1A 152 1H continue for Problem 3 ECE201 Linear Circuit Analysis I Prof H Z Tan Topic InductorCapacitor Combinations Duality 1 Inductor Combinations 1 Principles 1 Inductors in series add to form Leq Leq L1 L2 2 For n inductors in parallel l l l l Leq L1 L2 Ln Reciprocal of sum of reciprocals law Derivation of Principle 1 di t Vin t D Leq 1 iin Vin t V1 V2 V3 kVL dli L2 L3 dt dt d1 in L 1 dt LrLzLsdi i1 L Derivation of Principle 2 iin C di t di V Vin t D Leq lgt d L1 951 di t di V V t L 1 1 ln 1 dt L1 di2t diz Vin t L gt Vln 2 dt L2 di3t dis V39 V t L gti ln 3 dt L3 ii i1i2 i3 kCL diint di1t dim also dt dt dt dt V39 V39 V39 V39 In 1nJr 1nJr 1n Leq L1 L2 L3 1 1 1 1 Leq L1 L2 L3 Remark Inductor combinations satisfy the same formula as resistors 2 Voltage division formula wt d1 v1 L1 d V1 L t di Vin vi Leq f L2 v1 7 L1 L1 L3 ch LLZL3 General V k Lk n total ofL s V In 3 Current Division Formula General A n total of parallel inductors 1 2 Yb 39 l Derivation imo Leq 1L11L2 1L3391 1 t 11tL j mad 1 7w 1 t moi I w mad 6 q ilt 1 1 A w 2 Capacitor Combinations 1 Principles 3 For n capacitors in series 1 l l l Ceq C1 C2 C11 4 Capacitors in parallel add Ceq C1C2Cn Derivation of Principle 3 hum By de nition ii D Ceq 13 V1 Cl Vin vi v1 v2 v3 kVL I dvin 7 dvl dv2 dV3 dt dt dt dt d d 39 From 10C IE Deiivation of PIinciple 4 dvin Be de nition ii D Ceq dt ii i1i2i3 kCL dvin dvin dvin C C C 1 dt 2 dt 3 dt dvin dt C1C2C3 v ceq 2 Voltage Division Formula seiies circuit C 11 total of capamtors 1n senes 3 Examples 1 Find Leq and 10m 2 Find Ceq and vent ECE201 Linear Circuit Analysis I Prof H Z Tan Topic Phasors Ohm s Phasor Law for R L amp C KVL amp KCL 1quot Phasor 1 De nition a complex number representation of sinusoidal signals at a xed freq 2 Notation bold face capital letter underscored capital letter for handwriting e g V or X 3 rectangular vs polar coordinate polar is preferred eg X Vm 4 4 meaning a complex number used to represent a real sinusoidal function of time t at a fixed frequency of n I eg 1104 90quot y find Vt 1 First treat 1 amp X as complex numbers 104 900 104 900 SEA 135 1 j J5445 Then treat X as a representation of Vt W Re gagW gt 515 cosat 135 y Notice Re and cat in X to Vt expansion 2 What is the difference between a phasor and a shorthand notation of a complex A phasor represents a sinusoidal function of time t The frequency of the sinusoidal function is not explicitly shown in the phasor notation 3 5 Phasors KCL amp KVL l Phasor currents satisfy KCL e g 11 12 1112l3 14 2 Phasor voltages satisfy KVL 3 Phasor currents and voltages have directions Use passive sign convention on phasors Ohm s Phasor Laws for R L amp C X Z003 1 assuming passive sign convention Z003 Impedance of a device with a unit of ohm R for R ij for L assuming Sinusoidal Steady State for C assuming SSS JmC If n 0 ZLGm 0 short circuit Zc m 00 open circuit 030 means dc or constant input signals 7 But we already know this If n oo ZLGm 00 open circuit Zc m 0 short circuit Equivalently l YGoa X passive sign convention YGm Admittance of device mho Y 3903 i R J R 1 YLD assumlng SSS JOJL Ycjn joaC assuming SSS Phase relationship between X and l for L amp C 1 For an inductor XL joaLlL mLz90 1L4 I oaLIL 44I 90 39VL leL V 190 Voltage phase leads the current phase by 90 Re 2 For a capacitor 1 1 o 10 o Kc RICEZ90 39Ic z EZG 90 I WES V 790 Voltage phase lags the current phase by 90 1m 6 Example Use phasors to solve the following circuit R 202 t 5 L L 373 H F 20cos20l 30 A c C 5 m b ZTHU D Yinow 1 At 03 20 radsec compute 2 Find Thevenin Equivalent Impedance ZTHGZO 3 Find the Thevenin Equivalent Open Circuit Phasor Xoc 4 Determine the Thevenin equivalent circuit in which ZTHQZO is a series combination of two circuit elements seen at aib ECE201 Linear Circuit Analysis I Prof H Z Tan Topic Second order circuits RLC constant inputs 1 Review of series or parallel source free RLC circuits Circuit C d2 x dx 2 c1rcu1t eguatlon 2 2039 mnx 0 dt dt characteristic eguation s2 203 a s SIXS 32 0 2 2 3172 039iquot039 aquot R 1 6 and n dam in coefficient 039 series or 039 arallel u p g M ZRC p 1 undamped oscillation frequency on E 3 forms ofresponses i 6 gt on overdamped xt K163quot K2e32 81 sz real lfc gt 0 then s1lt 0 s lt 0 ii 6 on critically damped xt K1 KJquot11 s1 sz 7 6 real Ifc gt 0 then s1 s lt 0 iii 6 lt on underdamped xt ed A cosadtB sinadt ma lm 0392 A amp B are real nal value of the response X1t For all 3 forms of Xt Xoo 0 ifs gt 0 In other words without any inputs any initial energy stored in the capacitor andor inductor will eventually be dissipated by the resistor 2 Now add a constant input to the RLC circuit VISA4m OR L Circuit V c 15 E ZrC dzx dx c1rcu1t equation 2 2039 mix F ch ch ie the only change is to add a constant F to the right side of the equation characteristic equation same as before ie the characteristic equation is determined by circuit elements R L C not inputs 6 and on same as before response xgtl Let Xnt be the response to the sourcefree circuit 2 d x3 20 61 wjxnU 0 ch ch Then xt x t X F where XF is a constant How to find XE i Xoo Xnoo X1 Assume 6 gt 0 then Xn00 0 regardless of 6Dn Therefore ie XF is the final value XF can be found by noting that for constant inputs capacitors look like open circuits amp inductors look like short circuits after a long time OR ii Plug Xt Xnt XF into the circuit equation we get 2 x x y20jwjxnmiXFF 0 This method requires thatF be computed first How to solve for Km 0 From the 6Dn relationship determine the proper form of Xnt 0 Determine XF using either method i or ii 0 Determine K1 K2 or AB from the initial conditions xt0 and x39t0 3 An Example V50 7 60 11H 60 ut 60 util Find vca for t 2 0 open at t25 sec V50V 120 60 0 1 t sec 60 Step 0 Relevant times Step 1 Analysis at t 0 O 69 Vc0 V vc0 vc0 60V 60 60V 0 i a iLO39 10A l39 039 39 I l iL039 Step 2 Analysis att 0 Equivalent circuit at t 0 Find ic0 because we need it later lt69 O Vc0 60v 9 lt39 0 iLO By replacing C amp L with and L at a particular time we emphasize the fact that the values of vc0 amp iL0 should be treated as given during circuit analysis Step 3 Find the characteristic equation and the form of vct Deactivate the independent source to nd equivalent RLC circuit 1 Step 4 Find the constants in the response for 0 S t lt 1 Treat the problem as though t 2 0 and VS E 60V Step 5 Analysis att 1 Step 6 Analysis att 1 Find ic1 l 7 Step 7 Find the response for 1 St lt 25 120V Step 8 Compute response fort 2 25 sec Capacitor Voltage in V Time in seconds ECE201 Linear Circuit Analysis I Prof H Z Tan Topic First order circuits step response Review 1 general Mt f0 dt 965 x0 2 wing 424 xt e xgL fre dr 2 zeroinputsourcefree W40 xt 6 x0 Hay 14 o v t ei vcag R It 8 L 1 RL circuits with constant input int Vs a VLO 1 Circuit Equation KVL amp Ohm s Law vLt VS R1 t 61139 t I d t t LL n uc 0r vL dt 61139 t LL V R39 t dt s 1L m 3mg dt L L Initial condition 139 L00 2 Response iLt for t 2 to 0720 V 7504 1Lte L 1Lt0I S e L allquot to L R R R 0720 V 72 L rt e L iLt0 se L eL L R In 75072 V 75072 e L iLt0ESl e L 75240 L l39it iLt0 e t to elapsed time 3 Interpretation of iLt r time constant R V L 139LcgtoEs assuming 17 gt0 tit 39iLtiLooiLtO iLooe I elapsed tim e nal initial nal e m constant value value value 4 Comments i ifl39iUD 500 use ME ii iLoo can be calculated directly by noting the L looks like a short circuit for a constant source at t 00 iii time constant can be computed using Rm and Leq 2 RC circuits with constant input l ice VS Von 1 Circuit Equation V t KCL amp Ohm s Law 151 4 d t capacitor iv 13950 CV45 d t V t Combine CLO 5 Vr dt R R w V t dt RC RC 2 V 3 V 4 Initial condition vt0 Response vct for t 2 t0 1 V Ji v te Rcvt I Le Rcdf 5 5 a to RC 2720 K mm Vs Interpretation of vct t to elapsed time 1 RC time constant vcoo VS tit Thus vct Vc 0 Vc toVc Oki RC I elapsed time nal e I value initial value final value Comments i if v50 vt use v50 ii vcoo can be directly computed by noting that C looks like an open circuit for a constant source at t 00 iii time constant can be computed using Rm amp Ceq iv iLt amp vct have the same form Example 1 Find vct for t Z 0 amp ict for t Z 0 Step 1 Find Thevenin Equivalent Circuit for the circuit to the le of AB 25V Step 2 Find vc0 Step 3 Find VCI for t 2 0 Step 4 Find ict for t 2 0 Example 2 15u tV Find iLt for tgt0 209 t01 sec QVAVA f 209 in 209 C 3H Relevant time intervals to consider Step 1 6 Step 2 t E 5ut0 2A Step 3 re Step 4 t e Example 3 Find iLt for t z 1 5ult C D A A 109 159 iLt


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