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# Linear Circuit Analysis II ECE 20200

Purdue

GPA 3.59

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This 6 page Class Notes was uploaded by Cassidy Casper on Saturday September 19, 2015. The Class Notes belongs to ECE 20200 at Purdue University taught by Raymond Decarlo in Fall. Since its upload, it has received 89 views. For similar materials see /class/207903/ece-20200-purdue-university in Electrical Engineering & Computer Science at Purdue University.

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Date Created: 09/19/15

dot d39t u gtclgvmoLi l uwmnnmmw Impedance and Admittance 1 Zcs a YCS CS A mn Product Rule lmpedances in Parallel Z1SZ2S Zms Z18 Z2 3 Parallel Admittances Add Series lmpedances Add Voltage Division series elements n Lm total S Current Division parallel elements YK S 1ks 139 S Ytoml S m Initial Value Theorem Let Fs be a strictly proper rational function of s ie the numerator and denominator of Fs are both polynomials in s with the degree of the numerator less than that of the denominator Then limww SFS f0 FinalValue Theorem Suppose Fs has poles only in the open left half complex plane with the possible exception of a firstorder pole at sO Then limwo sFs 2 lime ft Feedback System X ref 3 135 1Hs Equivalent Circuits for Capacitors and Inductors Ca pacitor Transfer Functions Response designated as Ys Input Designated as Fs Ys HSFs Impulse Response Mt L 1HS Step Response Zerostate response ofthe circuit to a unit step function 1 d ImpR E StepR 5 Zl5 Zz 5 Zm 5 p1s p2 quot5 pn If ngtm there are nm infinite zeros Hs K If nltm there are mn infinite poles As a second illusilnilml consider he transfer I39umlinn a lw Ei iii1 Iln vll winch has me polcizem plot shown in gmc 152 Ix li i Poiseum mm m Hm givm 1 41am hen7MB y or pm oilhe polemzvro h wraxl reprcwms fumes he imaginary part or the pol or zero Poles and zeroes are of standard forms62 032 where o is the real part 0 is the imaginary part HSis stable only if every bounded input signal yields a bounded response signal Poles must lie on the open left half complex plane 1 Transfer Function Input 0 0 Output ytMmcosiiiiimtltpoo MUJKgtltHJ39UJ ltpoJLIIjn6 Transfer Function of Opamp Circuit Vent Zf5 Yinl5 Hs Vin Zin Magnitude Scaling Yo Znew SKm Zold 50rYnew S L6 Km Rnew KmRald I new KmLald C 1d Gold C 0 39 G new Km I new Km Current controlled voltage sources multiplied by Km Voltage controlled current sources divided by Km VCVS and CCCS are unchanged If Hs is a voltage or current ratio magnitude scaling has no effect on Hs If Hs has units of ohms then Hnew S KmHald S If Hs has 39 H unIts of seImens then Hnew s 0115 m Frequency Scaling S Inew S 2 Hold f c0 Km L0 Cnew KmldKf I new Kf 1d Convolution ya ha W f m rmr dr 2 fhtft TdT The convolution operation is commutative distributive and associative LhtftHSFS Sifting Property of Delta Function f ft6tT dtfT Time Shift Theorem ht T1ft T2 MIT T1 T2 Convolution Algebra ya mfmdr Bandpass Transfer Function l I 01 U U K K5 1 BF 52Bwswg 52mm 5w12n Qcir mm tUm Qm B 3w 26p Q m P H K m 20p 012 61 Parallel RLC QM wORC Series RLC Qm wOLR we 1 undamped natural frequency SeriesParallel Transformations wL Qwil R coul series resistance 5 Qmp wCRp parallelleakage resistance 39 E For Qwil gt 6 L39 m L R39 z Rs1 03ml For Qmp gt 6 C39 m C R R g P Heap Resonance Frequency at which steadystate voltage and current are in phase Zs Ys Hs must be completely real at sjuuR n I Cuzcqu m Ito milane quotmm muquot m mpm mda pm I my mm W gure I714 By Mr usual techniques 1 r 14 I v Yuwm rr myl 33mLy 5M 2 5 imgginmypqu 1 gen Resonant nmux when Y L rcul LE when quot11 mm unC R u L r1 Solving far In and than cxprcssing n 14 a Function tum chldgt CR3 i 7 7 7 ml 1 1 VLF L W L The rightmost term Show how the remnant cquuwy l I li39fl Way from Um parallel or Series ideal cam whm m mu 5M 1 Tu ublain the values of ihe admilmncc md impcdauw ll insununce substitute his value mm mm equation 17 3 to obtain RC 1 YuT And Zlw Brickwall Specs i q a gt c i1 rmiliml u ilu ifzinhnml m u in mm li siuplmml i A W me minimum lllnnm 39 m i an lc um i39w S hk uncmmlmn m u r upllhllll m 5 lion hziml mu Laxm tin quotT a 4 mm 5W b u lulml Butterworth Loss Functions TAELE 211 Normalized Butterworth Lass Functions n1 g 7 n Hm I Y l l 2 x1 V l 3 x39r39l 4 31 076537quot l39318477m7 1 5 x1v1 061803x l 1w I1mgt3 l 6 3391 051 ml Jil 1m t39x39l 1 Procedure for Computing Butterworth Loss Function 1 Identify filter specs up Amax ms Am 2 Set 15 and compute filter order via P 100391Amin 1 10 Ama 10 10 as 3 Look up 3dB normalized Butterworth 0112 polynomial in table 211 4 Choose wc ors so that loss magnitude response falls within the range permitted by brick wall specs 0114 39 w lt 8 lt 1001Amax 1 nquot 5 I I o p S 06 S 7n 5 2 quot1001Amax 1 5 Frequency scale the loss function by wc 1001Amm 1 P g1n 2quotd Order Chebeyshev Filter HNLP S kw 52sw5 Adjust gain of circuit by using input attenuation Frequency scale by mo Magnitude scale from given component valuesCommonly realized by Sallen and Key circuit Procedure for HighPass Filter Realization umqm an I um In ma 1 mm Wuhan spmymmm ll l u mm Flnmuul 1anrmMilhmmw Itlnlnmlpaullllul Wu my 1 mm a Hum mm 1 Compute equivalent low pass specs using the on frequency transformation equation 1 pJD 2 Determine filter order same equation as before and refer to polynomial table 211 3 Determine 3dBfrequency C DCmp 1 1 g E 4 Frequency scale circuit elements so that 3dB frequency becomes QC 5 Execute LP gtHP transformation on individual circuit elements Capacitors become inductors of value 1CJD Inductors become capacitors of p value 1 Resistors are unchanged Loop 6 Magnitude scale circuit to obtain desired load resistances Magnetically Coupled Circuits vie 9 L l v Dot Convention A current entering the dotted terminal of one coil induces a positive voltage with respect to the dot ofthe other coil V1 SL1 511SiM5125 V2 SL2512SiMSI1 5 M k 1 E Wt Li2 tfor inductors WtMiltiZ tfrom coupling Series Equivalent Inductance 1 hi mun vmi whirlan mum m m i l i am ii m Magnetically Coupled Equivalent Circuits 74 a m g L a Q m 5quot L La Two useful equivalent circuits to the above circuit two more in DeCarlo s handout TEquivale nt A HMS lm V2 Es E 12 TTEquivalent 392 LJ a LB 9 5 quot M N 1 1 1 11 7 LAS L35 L35 V1 1 7 1 1 1 L35 L35 L55 Ideal Transformers Finding Thevenin Equivalent Example General Form V5 Zthls Vac 1 0 V5 315 1V1 0 V1 RIM 71211 mm aRIs 0 V5 ls aR1m aR15 0 V5 Ra215 aR1m Two Part Networks A Vita 12 v1 1 v2 yQarameters 11 Y11V1 Y12V2 I2 Y21V1 y22V2 y 1 1l 11 12 V1V20 V2 VFO 1 1239 y 2I 21 22 V1 Vz0 V2 VFO y 37123721 m 3 11 y zz YL y 37123721 W Y Y11 Y5 V1 Zin Ys Gm Vs Zin Zs Ys Ym G E 3 21 V2 V1 3 22 YL zQarameters V1 21111 21212 V2 Z2111 Z2212 z Vll z V1 11 12 1 120 12 10 2 V2 z 2l 21 2 1 120 12 10 Z12Z21 Zm Z11 Z22 ZL Z12Z21 Z I 222 W Z11 Zs G V1 Zin ys 11 Vs Zin Zs Ys Yin V2 L Z21 sz V1 ZL Z22 Zin hQarameters V1 hull h12V2 I2 112111 11sz2 V1 V1 I 7 1 Vz0 2 110 12 12 7 1 VZ0 2110 h12h21 Zin h11 m h12h21 Y h Wt 22 h11 Zs a i L Vs 2m 25 mym V 4121 0V2 h Vl Zin 22 YL mm V1 t11V2 1212 1 21V2 2212 V 1 1 1 V 12 j 1120 2 VFO I1 1 t V 22 j 2120 2 Vz0

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