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# 30 Note for ECE 20200 with Professor Decarlo at Purdue

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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 61 views.

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Date Created: 02/06/15

dot d39t umc 8nmL uwmnnmmw Impedance and Admittance 1 Zcs a YCS CS 1 A mn Product Rule lmpedances in Parallel Z15Z25 Zms Z18 Z2 3 Parallel Admittances Add Series lmpedances Add Voltage Division series elements Z n z m Current Division parallel elements Y uwf u 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 Es quotquot Xrefs Ys X ref 3 135 1Hs Equivalent Circuits for Capacitors and Inductors Ca pacitor mow 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 r251 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 illusumion consider Lhe transfer l39umlim a I m Fl in 1V lllAle which has me polerzem plal shown in gulc 152 IIIH 153 m 3 Poleum plow Hm given a nun i51rgainiheirraxis HP a h 4 3 3 391 IpruunsllwrrmlmnallhepolemE ro r 7 md he Maxis reprcwnls fumes lhu imaginary part in the pull 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 Transfer Function Input 0 0 Output ytMoocosl23llmtltpm MUJKgtltHJ39UJ ltpoJLIIjm6 Transfer Function of Opamp Circuit Vent Zf 5 Yin vi s 2mm ms Hs Magnitude Scaling Y0 Znew SKm Zold 50rYnew S 1155 Rnew KmRald I new KmLald c zcold G 2 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 Hold 5 units of seimens then Hnew S K 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 ll Ijml Hquot 39 39 39 39 39 V quot V V 7 V V V 7 39 39 39 V 7 39 39 39 39 39 quot Em 13 Is l 39 i i Y i r l r i i l l m 111 U U H K5 K5 BF 52Bwswg 52mmsw12n Qcir U U QmB Bw2crpQ 39quot w P 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 For Qwil gt 6 L39 m L R39 Rs1 03ml For Qmp gt 6 C39 m C R R P Heap Resonance Frequency at which steadystate voltage and current are in phase Zs Ys Hs must be completely real at sjuuR II I39H39N I minim 11M HJr l39llwlL quotmin g1r l7 lBylhr usual techniques g Hlln39 we AIIPHI i HIl39 pail Hi we mm M L 39 r39 J 7 In F M 1 2010 lmr 39 E imi quot R mL3 5w 2 5 imaginary ya in mi Resonant ni39cun when Y 1239 mm it when mL 7 0 Ilel unC R2 my quot1 Solving For m and then expressing it as u runcu39on um yiclih i I Ciel 7 7 7 mi 7 lLr I W L The rightmost term shows how he mung cquenLy l mini away from ll parallel or Series ideal mm whnrr m um 511739 Tu obtain the values ofihe admillimcc iind impedance ill momma subsrimia us value a in into equation 7 3 lo obtain RC 1 ViT and zimnz Brickwall Specs l equiilill 0A l1vi lt l 1 twisl k hln 1v umln iii utluk si iiinu mums u lllr minimnii I39rcqucncy slnphilml 39 in minimum nllimuhlc n luilllun m in whim lion bitml 4quot 5 HiVllAllC ll l11 IIlli mm m an A l l7 gig Butterworth Loss Functions TA LE 211 Normalized Butterworth Loss Functions Hm I x l l 2 x3 V 1 3 Ss39l 4 sl 076537s l39118477m7 5 v 1m 061803x me 1131303 I 6 x1 0576x 1m x 1m Hum 1 Procedure for Computing Butterworth Loss Function 1 Identify filter specs up Amax ms Am 2 Set 15 and compute filter order via P 10 10 as 3 Look up 3dB normalized Butterworth polynomial in table 211 4 Choose wc ors so that loss magnitude response falls within the range permitted by brick wall specs o 100Zquotm 1 S 5 S 17 O 7 quot1001Amax 1 5 Frequency scale the loss function by wc 1quot looMm 1 0 s S 7n 1001Amin 1 Dc 2quotd Order Chebeyshev Filter HNLP S s Frequency scale by mo Magnitude scale from given component valuesCommonly realized by Sallen and Key circuit Procedure for HighPass Filter Realization AlllHlLu an Am mm W n l 39n l s l quot1 U g n m w v m mum mm m mu hm mm m n my p14 llllnl mum Hum mm w mmms 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 111 Resistors are unchanged p 6 Magnitude scale circuit to obtain desired load resistances Magnetically Coupled Circuits i l quotor m JD c Ll l w 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 L1112 Wt Li2 tfor inductors WtMiltiZ tfrom coupling 1 Hum n u lemllluu vuwpm mumlmu symmwlmis Series Equivalent Inductance Two useful equivalent circuits to the above circuit two more in DeCarlo s handout TEquivale nt 12 HMS lm v2 as am TTEquivalent 392 LJ Q L B a 7 1 I ll N a 1 1 1 11 7 LAS L35 L35 V1 1 7 1 1 1 V2 L35 L35 LE s Ideal Transformers Finding Thevenin Equivalent Example General Form V5 Zthls Vac ll 397 iquot 0 V5 15 VZ and V2 1V1 idealtransformer 1 0 V5 315 aV1 0 V1 RIM 71211 mm aRIs 0 V5 ls 1R1m aRIs 0 V5 Ra215 aR1m Two Part Networks yQarameters I1 Y11V1 Y12V2 I2 Y21V1 Y22V2 I I 3 11 3 12 V1 V270 V2 V10 2 2 3 21 3 22 V1 V2 0 V2 V10 37123721 Y m 3 11 y22 YL Y 37123721 Amt Y22 y11ys V1 Zin Y5 G 7Z Z Y Y s m s s m G z zi V2 V1 3 22 YL zQarameters V1 Z11I1 Z12I2 V2 Z2111 Z2212 I V1 211 12 1 1270 12 1170 V l Vzl Z21 22 I1 12 0 2 110 Z12Z21 Zin Z11 222 ZL 212221 2 Amt Z22 Z ZS V1 Zin Y5 G 7Z Z Y 11 5 m s s m V2 ZL Z21 sz V1 ZL Z22 Zin hQarameters V1 h1111 h12V2 I2 h2111 h22V2 V1 V1 1111 7 h12 7 1 1er 2 1170 12 12 1121 TI 1122 7 1 VZ 0 211 h12h21 Zin 1111 m h12h21 Y h mu 22 h11ZS G E Zin Y5 Vs 2m 25 YsYin V2 h21 sz h V1 Zin 22 YL tQarameters V1 t11V2 1212 I1 t21V2 2212 V1 V1 7 1120 2 Vz0 11 11 7 2120 2 Vz0

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