Fund Electric Circuits
Fund Electric Circuits ECE 2260
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This 13 page Class Notes was uploaded by Shyanne Lubowitz on Monday October 26, 2015. The Class Notes belongs to ECE 2260 at University of Utah taught by N. Cotter in Fall. Since its upload, it has received 27 views. For similar materials see /class/230003/ece-2260-university-of-utah in ELECTRICAL AND COMPUTER ENGINEERING at University of Utah.
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Date Created: 10/26/15
COMPTUAL TOOLS By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS LC resonance EXAMPLE EX a Explain what single resonant circuit to use in the box marked Z to make the voltage across Z a minimum at a speci ed frequency 00g b Explain what single resonant circuit to use in the box marked Z to make the voltage across Z a maximum at a specified frequency 00g ANS a L in series with C b L in parallel with C SOL N a The voltage across Z is given by the voltage divider formula 70 H000 v where z is impedance in box We want Hjn 0 for specified 00g So we want 2 0 at 00g For L in series with C we have 2 jooL jnC z 0 at l 0 7 LC In other words the series LC looks like a wire at resonant frequency We set 000 mg or LC 2 10002 b The voltage across Z is given by the voltage divider formula 70 H000 v where z is impedance in box By Carl H Dmney and Neil E Cotter FILTERS RLC FILTERS LC resonance EXAMPLECONT We want Hjn as large as possible for specified mg The solution is to set 2 00 or 12 0 at 00g For L in parallel with C we have 12 1jtnL ij 12 0 at D 1 1LC 39 In other words the parallel LC looks like an open circuit at resonant frequency We set 000 mg or LC 2 10002 Note the same values of L and C will work for both cases The difference between the two cases is the configuration of the L and C COWGEPTUAL TOOLS FILTERS RLC FILTERS Qualitative response AMPLE 3 By Carl H Durney and Neil E Cotter EX SOL39N The circuit shown below is a wave trap used to prevent the signal from an amateur radio transmitter from entering the input of a TV receiver Choose proper values of L and C if the transmitter frequency is 50 MHz and R does not affect the resonant frequency appreciably Explain how the circuit works The bottom edge of Channel 2 is at 54 MHz What value of R would be required to make IZI at 54 MHz of one of the resonant circuits equal to 110 of its value at resonance and what value of impedance would that be at 54 MHz Note the frequencies here are in units of Hz rather than rads receiver antenna There is no unique answer For C 100 pF then L 0101 14H R 2 0492 Q IZI 205 Q at 54 MHz The idea is that L parallel C will act like an open circuit at resonance This prevents the interfering signal from reaching the receiver With the R included but relatively small the resonant frequency remains approximately the same By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLE3 CONT 1 xLC In units of Hz we have D f 2m 2m LC Any L and C satisfying the following equation will yield the correct resonant frequency C s2 101pps2 2 2 279 27550106 One solution is C 100 pF and L 0101 14H Now we find the value for R At resonance we have equal but opposite reactances for the L and C De ne the reactances at no to be jX and jX From the definition of no we have X xLC 318 9 At no the impedance zo of one side of the trap circuit ie one C in parallel with an L plus an R is RjX jX X2 jRX XX Z RjX jX R RJ39 For a frequency kooo our L and C reactances become ij and jXk and we have Z Rij jXk XZ jRXk X jRk k Rij jXk RjXk 1k RjXk 1k or L R k zk X X 1 397 k 1k JR We want IzkZOI 110 when k 54 MHzSOMHZ 108 If we define B E XR we have By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLE3 CONT 1 3 2 Jk 1 39Bk l E J k L zo B jl 1039 At this point it is prudent to attempt an approximation Because k z 1 we may approximate B jk as B j and cancel terms to obtain a simpler equation 1 1 4444444I6 Lk 1 39Bk l J k A Invertin g both sides and applying the de nition of magnitude we have 1 1 2 1 39Bk i 12 Bk i 10 J k k Solving for B we have 2 Bk l 99 or B 91 16458 k k 7 108 7 k 108 Using R XB 318 926458 we have R 0492 Q At 54 MHz we have 1 B k 7 l1 jBk 1kl 10 z X 3189 10 10 13 lzklX We obtain Izkl 205 Q at 54 MHz COWGEPTUAL TOOLS By Neil E Cotter LAPLACE TRANSFORM TABLE OF TRANSFORMS TOOL TABLE The following identity generates the basic Laplace transforms AsaB0 2 L 2 dsquot sa 00 X ix Inc atAcos 0t Bsin00tut sX 1quot dt where x 0 or 1 n a 00 2 0 ut is the unit step function and sgnn is the signum function as defined below 0 t 0 1 t lt 0 lt ut51 tgt0 sgnn 0 t0 1 t gt 0 v t VS condition 01 1 x1 n0 610 000 A1 30 1 l x0 110 610 000 A1 B0 S 1 t 7 x0 111 610 000 A1 B0 S n n t 7 x0 1120 610 000 A1 B0 Snl e 1 x0 n0 120 00 A1 B0 S a 16 2 x0 n1 120 00 A1 B0 s a I 6 x0 I120 120 00 A21 B0 s a cos00t 2 S 2 x0 n0 610 020 A21 B0 S D sin00t 2 0 2 x0 n0 a0 120A0 B1 S D Ae at cos00t A x0 n0 120 020 AA B0 s a 00 32 sin00t x0 n0 120 120A0 BB s a 00 REF 1 James A Nilsson Susan A Riedel Electric Circuits 8th Ed Upper Saddle River NJ Prentice Hall 2007 By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLE4 EX By L39 the 39 r J and J quot curves of series and parallel combinations of L39s and Cs and adding them appropriately we can find the approximate resonance characteIistics of multiple resonant circuits The figures show examples of these curves We sketch the approximate total reactance in Fig 1 by noting that XC dominates at low frequencies and XL dominates at high frequencies We sketch the approximate total susceptance in Fig 2 by noting that BL dominates at low frequencies and BC dominates at high frequencies Reactance Susce Dtance L Inductive Capacitive Region Region lt gt Fig 2 COWGEPTUALw By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLE4 CONT We sketch the approximate total reactance in Fig 3 by noting that XL dominates at low frequencies and XC dominates at high frequencies We sketch the approximate total reactance in Fig 4 by summing the curve from Fig 3 and the reactance curve for C2 We find that the reactance of C2 forms a resonance with the reactance of L1C1 at a frequency below the resonance of L1C1 This new resonace is such that the entire circuit acts like a wire There is also still a resonance at exactly the resonant frequency for L1 and C1 This resonance is such that the entire circuit acts like an open circuit as it did before adding C2 Reactance Total reactance Reactance Inductive Capacitive Fig 3 COWGEPTUAL TOOLS By Neil E Cotter LAPLACETRANSFORM IDENTITIES List TOOL The following list of identities allows one to generate complicated Laplace transforms from basic Laplace transforms Proofs follow from the definition of the Laplace transform vt a Vs a f 1 vte39 dt Other terms used are 6t which is the impulse or delta function and ut which is the unit step function 5a 0 t 0 39th f and 1 t 0 t lt 0 2 W1 u a 00 t 0 1 t 2 0 LIST Name tdomain sdomain nndi nn definition m Vs linearity av1t bv2t aV1s bV2s delay vt aut a eaSVs a 2 0 multiply by t tvt iVs ds at multiply by a e vt Vslsa replaces S a 2 0 vt so d1v1de by t T L V d d derivative E120 sVs vt 0 n n n l nth derivative d V S VS S VI t0 dt d squot27vt dt t0 l vt n l dt t0 t VS 1nte ral v t dt 7 g f0 S 1 time scaling vat gVs i replaces S a 2 0 ll REF 1 James A Nilsson Susan A Riedel Electric Circuits 8th Ed Upper Saddle River NJ Prentice Hall 2007 COMPTUAL TOOLS By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response AMPLE 1 EX Make calculations of IZI for enough values of n that you can plot a good graph of IZI versus 00 for each of the circuits shown in Figs 1 and 2 Then vary the values of the components and make enough further calculations that you can get a qualitative feel for how the size and shape of the curve depends on the parameters without making numerical calculations For example if IZI versus 00 is given for the circuit shown in Fig 3 explain how the curve would change if R were doubled with nothing else changed Explain how the curve would change if L were doubled with nothing else changed wv 40 Q 10 mH 1000 pF Z Fig 1 Fig 2 L C R Fig 3 ANS Fig 1 plot of Izl will be bowl shaped with a minimum value of R at 000 Fig 2 zR40 2atrn0 Izl increases to a resonant peak of approximate height kR 625 k 40 Q 250 kg The resonant frequency is lxLC 1 MxIO rads Izl then decreases toward 0 as n gt 00 COWGEPTUAL TOOLS By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLEl CONT Fig 3 SOL N Fig 1 Fig 2 Double R Izl starts at twice the height but the resonant peak height is only half as high The frequency of the resonant peak is unaffected Double L then the curve for Izl still starts at height R but the resonant peak height is about twice as high The frequency of the resonant peak is lowered by a factor of xZ z R ij jooC R jooL 1ooC IzI2 R2 00L 100C2 We may visualize Izl as the length of the hypotenuse of a right triangle in the complex plane with sides of length R and 00L 100C At the resonant frequency of the L and C 0002 1LC the L and C act like a wire 00L 100C 0 Thus 1 R at 000 For 1 above or below 000 the magnitude of 00L 100C increases Thus our plot of Izl will be bowl shaped with a minimum value of R at 000 z R ij 1ij Consider the qualitative behavior at key frequencies Note that the lower magnitude impedance dominates in a parallel conf1 guration At 00 0 R ij R and 1jtnC 00 Thus 1 R At 00 RL R jooL R1j and 1jooC jRLRRC Thus 1 R1 j jLRRC and 1 will behave more like the smaller of the parallel terms If LR 2 RC however we will have Z R1 j H j Rl j For the component values given LR 14 ms RC 40 ns and LRRC 625 k It follows that Z R1j at n RL for this circuit COWGEPTUAL TOOLS FILTERS RLC FILTERS Qualitative response EXAMPLEl CONT By Carl H Durney and Neil E Cotter Fig 3 Asm gt00RjnL gtRj00and1jrnC gt0 Thus 1 gt 0 Now we examine the behavior versus frequency more carefully Our calculations are driven by the thought that the resonance of the L and C may cause the L and C to become almost an open circuit The R however will affect the resonance To simplify the algebra we use normalized and unitless frequency con 2 nRL We have 1 R1 joon jLRRCnn Define k E LRRC Then we have 1 R1 jtnn jknn or z R jkoon k1 joon koon Define Dno E xk At run 2 one we have n no 2 lxLC one knnoy and z R juno k R jxk k If kltlt 1 then 1 Rkltlt R and the L and C do not act like an open circuit at resonance Instead the circuit acts like the C is nearly shorted If kgtgt 1 then 1 Rkgtgt R and the L and C almost act like an open circuit at resonance This case applies to Fig 2 Now we have enough information to describe the shape of Izl versus 00 as 00 goes from 0 to 00 zR40 2at000 Izl increases to a resonant kR 625 k 40 Q 250 kg lxLC 1 MxIO rads peak of approximate height The resonant frequency is Izl then decreases toward 0 as n gt 00 Based on calculations for Fig 2 if we double R then the curve for z starts at twice the height but the resonant peak height of Rk is only half as By Carl H Durney and Neil E Cotter FILTERS RLC FILTERS Qualitative response EXAMPLEI CONT high since Rk behaves like lR The frequency of the resonant peak is unaffected If we double L then the curve for z still starts at height R but the resonant peak height of Rkis about twice as high since k is proportional to L The frequency of the resonant peak is lowered by a factor of xZ Note These results only apply to the case k gtgt 1 If R becomes too large then our approximations break down and the resonant peak may disappear entirely