Electronics Sci Instr
Electronics Sci Instr PHYS 3610
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Date Created: 10/26/15
Electronic 1 BBIIII B 15 TimeFrequency Domainscontinued Complex Impedance Transfer Function Complex Frequency Bode Plots S plane Laplace Transforms Read Bobrow chapters 5 and 7 Problems due Nov 4 5152545658514 TimeFrequency llomain Signals occur in time Circuits have well de ned frequency response Timedomain signal can be decomposed into frequency components Transform input signal from time domain into frequency domain Transfer function used to obtain frequency domain output signal Transform frequency domain output signal back into time domain uuk VatweAJ pram n n 25 a s a 735 T in mm 4 4111 mm gm um Shnvelhung sm rlgm 5 1st NH Hymn Huh 2st Figurc 5 Frequency Domain of AH AZHI s n Wm 32H complex lmnedance eactance mnlitulle and Phase Complex Impedance 2 R jX gt V ZlZei6 Z T X Reactance I Amplitude IZ I 1R2 X2 Phase 6 tan 1XR Imneuance aml Admittance J The equatiori39V39E Z1 is the phasor version of Ohm s law From this equation we 7 739 see that impedanceiis the ratio of voltage phasor to current phasor that is It follows that IVZ In this expression the reciprocal of impedance appears As this can occur frequently it is useful to give it a name We call the ratio of current phasor to voltage phasor admittance and denote it by the symbol Y that is Thus the admittances of an Reohm resistor L henry inductor and Cefarad capacitor are respectively Alternative forms of the phasor version of Ohm s law are IYV and VIY liain aml necillel Power Gain gain 10 log dB VOI39t39 39Gai mwquotquotquot gain 10 loglog g 20 loglo j dB Logarithmic scale dB Decade FI B IIIGIIBV BBSIIIIIISG Transfer Function Network Function four possibilities 1 voltage gain 2 current gain 3 transfer impedance 4 transfer admittance VoutDVin00 IoutDin00 VoutDin00 IoutDVin00 calculated for sinusoids has magnitude and phase has zeros and poles IGI 100 10 E 12 01 001 0001 lGldb 4o 20 Freuuencv BBSIIIIIISB KL szC JwC R Zquot 1 1ijc C ZRZC 2R Z gt T l R jwc 0 etan KcoRC 0 Fig 51 Parallel RC impedance Amplitude Response IZI WK V 1 0013C2 Frequency Response 9 tan39looRC R lZl V1 mRC2 39 n quotR mlw SI 1 RC I I L 0 RC RIM1 Z F AAA 0 52 A m39p39l39itudeLTespon39SeT Wi 7quot FI BIIIIBIIGV BBSIIIIIISE Illl I GiI GIIiI 6 tan391coRC v I 7 ll L i i i RC RC RC RC I Z gt ER C 450 I l olj Fig 51 Parallel RC impedance 900 Fig 53 Phase response Transfer Function quotillll Pass Filter Denoting the ratio of output voltage to input voltage by 1119 then 4 V1 V lel 39 nRC 1mm 2 Vi ivii 1 coRC Fig 55 RC circuit For 03 0 rads we have that Hjltol Hj0l 0 To determine the value of Hjw as h a 00 note that i Iqun weaivaaiw 03136 Va 1 H 39 ml 1 MC2 V 1 1mRC2 i 05 l l w A i RC RC RC RC Fig 56 Amplitude response of RC circuit Thus H 11 gt 1 as o gt oo Also note that for w 1RC H 12 1E The amplitude response of the function HUM called a system function or a transfer function is shown in Fig 56 Clearly at 1 RC is the half power frequency Complete Graphical description of steady state behavior of Circuit Amplitude response ain f 7 V1 ZO fogm l rLi quot7 V 7 Frequency response Whammy 19me 0 1 w o 5 m an 2 u 10 w u 0 dB l 1 dB l 3 dB I dB l 7 dB Highfrequency asymptore Actual response 20 10gm mRC Slope 720 dBdecade 76 dBociave 20113 Fig 59 Amplitude response of lowpass filter OdB 1dB 3dB 76dB 7 dB 20 dB Va 20 Iogm V Evilfrequency asymptote 05 a cut q Highfrequency asymptote Actual response 20 mm DEC Slope 70 dBdecade 6 EBoctave Fig 59 Amplitude response of lowpass filter GjaKgtltG1jagtltG2jaxgtltGnja with Gkjw IGkjwIej quotw Amplitude response Gja K XG1ja x G2jw x x Phase response ej w ej 1w 2w nw PI IIIIIIBI 9 logarithm Amplitude Gja K xG1ja x G2ja xgtlt G jw 2010g10Gja 2010g10K 2010g10G1ja 2010g10G2ja 2010g10Gn JUDI Phase a 1a 2a na Sum 9 Superposition Bode Plots Ill Ilaml Decompose transfer function such that Gs KA1SA2S Ans K is a constant and the Aks are of one of the three following forms 2 1 or SOt1 S22wngsa wn 5000ss 2 CS s 10 s 8s 25 can be rewritten as GS4XSXS2X i xix 10 j 2 2852 s10 s10 Note the ordering 1 constant term 2 squot391 terms 3 remaining according to value 9 2 5 10 10 in this example IIIISIEIIII TBI III K K KAO 20dB Gain 2010g10K dB 0dB Phase 00 GAIN CHARACTERISTIC 20dB frequency rads Term 3 Alja ja 60490quot Gain 2010g10w dB Phase 90 Gain change 20dBdecade Gain a lrads 0dB 20dB gain0dB 1 rads 0dB GAIN CHARACTERISTIC 20dB 01 1 10 rads 100 90 0 CHARACTERI STIC 90 Term 3quot A1001 i14 90 JO 3 Gain 2010g101w dB Phase 90 Gain change 20dBdecade Gain a lrads 0dB 20dB 1 rads 20dBdecade 20dB GAIN CHARACTERI STIC 0dB 39 gain0dB 01 1 10 rads 100 90 0 90 01 1 10 rads 100 TBI III 33 a 2 O a a Ai1w aj 41 Araco 2 Gain 2010g101 1 w 2 dB or low frequency 9 OdB high frequency 9 20log10uordB a 20 4a 100 rads 100a corner frequency 00 0L rads 2 corner 2010g1011 3 dB 3dB TBI III 83 a lillllell 2 O a 1 Am a 11 a2 Aram Phase tan391 or first approximation 90 low frequency 9 q 0 deg 45 high frequency 9 q 90 deg phase corner 45 deg 01a 0 2a 4a 10a 100a rads 1st order for slope 45 degdecade TBI III allsal 1 OC 2 4940 OCJOU w OC 2 I w Gain 2010g10 1 dB Aijw low frequency 9 OdB lf asymptote 3dB OdB hf asymptote 20dBdecade Exact gain 20dB characteristic 40dB 010 0 2a 40 10a rads 1000 high frequency 9 ZOlog1000OLdB TBI III allsal lillllell 0c 1 AiJw 39724 kw OCJOU w 1 2 I 06 00 g I First approximation I l 9 Ph t 1 9 45 age an a a PHASE CHARACTERISTICl e l Jquot S i 90 3 Line of slope 45 decade 0391 0 20 40 10 rads 1000 Step 1 Analyze the transfer function 10s2 1 10 H2 Gs SCS10 ss410 l 10 522 SS10 2 21s2 10 s 2 s10 Q Where are the corner points A 2 rads amp 10 rads Examnle continued make double log sketch for gain 1 s 2 10 Gs 2 s 2 5 10 20 dB v low frequency limit 1 o GS 2 0 dB 4 S GainK 2010g102 dB o x PhaseK 0 20 dB Gains1 2010g101a dB 01 10 10 100 Ph 1 90 ages gam 1 rads a 6dB If Ilmlt corners 2 rads amp 10 rads 9 3 decades 01 100 Gs2ls2i S 2 S1O wz 0t 20 4 radS 100a Gain 2010 1 dB a g j a2 3dB 2 Gain 2010g1011 7 dB hf asymptote O Exact gain Examnle Putting It Together Gain 1quot 20dB APPROXIMATION OdB 20dB rads 02 04 4 10 20 40 100 1s2 10 G 2E 2 5 10 Examnle Putting It Together Phase 0 45 decade 90 02 04 2 4 10 20 40 100 GS21s2 10 s 2 s10 Term with 32 602 1 1 a a 14100 2 quot2 7 quot s 2 wnswn 121 1 a a 12 a a 602 Gain 2010 g10w w22j wnw asymptotic gain low frequency 9 0 dB high frequency 9 40 Iog10unudB Phase 9O tan 1 i a 3 2C a a Term with 32 continued Gain wnZs22 wns wnz 1 20L0 g10 2C OdB asymptote 20dB 40dB 0160 a 10m Term with 32 continued Gain s22 wnswn2 can2 40dB 20dB 1f asymptote OdB ZOLOg 10 2C la a 10m radS Term with 32 continued Phase wnZs22 wns wnz Initial asymptotic 0 approximation 1f decreasing C 90 hf asymptote 180 0102 a 10a radS comnlex HGIIIIGIIGV Consider damped sinusoidal functions A601 cosat 9 Complex Frequency s O ja Impedances comnlex H BIIIIBIIGV circuit BBSIIIIIISB v 4 quot39s39in 2t Fig 521 Circuit with dampedsinusoidal Fig 522 Frequencydomain representation excitation 44 4 ZRCV 44s s1 ZRC V 4s1 39 2V ZRcZL 4Sl25 s2s2 Vc From the fact that V 4 90 and s 1 j2 we get 24g 90quot 7 2quot O V 1 j2 12 2 2V 45 Vc Thus we deduce that for the circuit shown in Fig 521 vct 2V5 equot cos2t 45 V Poles and lanes 1 1 Zero value of s where Hgt0 Pole value of s where HgtOO Fig 522 Frequencydomain representation For the circuit given in Fig 522 if we de ne the voltage transfer function HCs to be the ratio of VC to V then from Example 56 we have that 2 2 r s2 s 25 S 12 jF72s 12 jW2 Vc H CS V Thus we see for the special case that 2 or S is H Q s l39 2 J the transfer function becomes in nite which implies that the forced response is in nite We say that the function Hcs has a pole at each of these two complex frequencies eSl z Imaginary J U I K damped K pure exponentially location of S X Sinusoul smusmd gi owmg smusmd pole OI zero Real damped growing exponential exponential sX MX jm or Mx jm sX Kjm slltjco s2ltjm sm39 FOOD JmSnJm Sb 39lw 3n 1 quot iv Wm I MIGIIV12e I Mme M1M2M3 Mm ej l39 i z 391 l a l IW39J MaeJ l aneJ l39h MM er 1ah1cv1n Can express transfer function as a function of frequency as a product of distances from zeroes to point on imaginary axis that represents the chosen frequency divided by the distances from poles to that same point Multiply this by a phase term that is the sum of the phases associated With each zero and the negative values of the phases associated With each pole SIllalle BIIIISII IIBIiIIII When we know the locations of the poles and zeros of F we have graphical information about the M39s and y39s Let us apply this method to our specific case zero at s 0 Fs R2 S R pom m s l s l RIC RIC Poles are usually marked with X39S and zeros with 039s 801T C 0 real Consider how the values B4amp wahdh q vary as a varies ie moves along the j axis 1 Amplitude Phase 5 Smallest Total M ltenn wa w ltenn y1 1 m K w Ma 3 332 O 0 O R I i 7 0 7c2 7t2 RIC small w RoCm i n 0 HQ n0 m I amp R C 4 7 3 4 RIC R J5 L 7t 7t 7r 7t RIC R R o a no 7 2 i 1 rtZ 7r2 7r R m RI A circuit should not have a pole in the right half plane if it is to be stable Such a pole implies an infinite response to an exponentially growing sinusoid Such a response can start from noise and build up to saturation levels sinusoidal A pole on the pure imaginary axis implies a stead have this Y output in the absence of an input An oscillator should behavior lanlace Transforms Read Bobrow chapters 5 and 7 Problemsdue Nov 4 5152545658514