Lecture 6: Resonance II. Announcements
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1 EES 5 Spring 4, Lecture 6 Lecture 6: Reonance II EES 5 Spring 4, Lecture 6 Announcement The lab tart thi week You mut how up for lab to tay enrolled in the coure. The firt lab i available on the web ite, it i an introduction to ppice Ppice i now available from the coure web ite.
2 EES 5 Spring 4, Lecture 6 ontext In the lat lecture, we dicued econd order tranfer function, circuit which have reonance Power into a load, and reactive power The concept of the of a reonator In thi lecture, we will review and complete the dicuion of circuit with reonance, in tandard form, and the Bode plot for the magnitude and phae of a circuit with an underdamped reonance EES 5 Spring 4, Lecture 6 Next Lecture In the next lecture, we will tart dicuing the phyical building block of integrated circuit Metal Inulator Semiconductor Reading aignment: hapter in the text
3 EES 5 Spring 4, Lecture 6 Second Order ircuit In the lat lecture we introduced the reonant circuit: EES 5 Spring 4, Lecture 6 And we derived the time domain DE for thi circuit: d dv L v t v t R v t dt dt Thi i the equation for a damped harmonic ocillator which i being driven. Thi we can write in a tandard form: d dt d dt v v R dv v v L L dt L dv v v dt L Where L R L 3
4 EES 5 Spring 4, Lecture 6 The behavior can alo be characterized by a damping factor intead of ζ In which cae the equation can be written: d dt v dv t v ζ dt v L EES 5 Spring 4, Lecture 6 Tranient olution To find the tranient olution, for t>, we tried a olution of the form: t v Ae V Where i a complex number Giving u a econd order equation for : ζ 4
5 EES 5 Spring 4, Lecture 6 We then ued the quadratic formula to find olution for, now in tandard form: ζ ζ ±, ζ EES 5 Spring 4, Lecture 6 General ae Solution are real or complex conj depending on if ζ > or ζ < ζ ± ζ τ t B t v V Aexp exp v V A B dv t i A exp t dt t A B A B V t B exp t 5
6 EES 5 Spring 4, Lecture 6 Solving for the coefficient: Solve for A and B: A A V V V B V A V A V V v V exp t exp t t t v V e e V EES 5 Spring 4, Lecture 6 Solution come in 3 kind: Depending on the damping factor, there are three type of olution: < ζ > Underdamped ritically Damped Overdamped 6
7 EES 5 Spring 4, Lecture 6 Overdamped ae ζ > Time contant are real and negative ζ ± τ ζ < τ V ζ EES 5 Spring 4, Lecture 6 ritically Damped ζ Time contant are real and equal ζ ± τ limv ζ V ζ τ t / τ t / τ e te τ V ζ 7
8 EES 5 Spring 4, Lecture 6 Underdamped ζ < Now the value are complex conjugate a jb a jb a jb t Bexp a jb t v V Aexp Aexp jbt B jbt at v V e exp A * V V V * * B * Aexp jbt A jbt at v V e exp EES 5 Spring 4, Lecture 6 Underdamped cont So we have: * Aexp jbt A jbt at v V e exp at v V e Re [ Aexp jbt ] φ at v V e A co t A V V φ 8
9 EES 5 Spring 4, Lecture 6 Underdamped Peaking For ζ <, the tep repone overhoot: τ V ζ.5 EES 5 Spring 4, Lecture 6 Extremely Underdamped τ V ζ. 9
10 EES 5 Spring 4, Lecture 6 Frequency domain analyi We then looked at a Phaor analyi: Z Z jl R j EES 5 Spring 4, Lecture 6 Power into an Impedance Z Letting the phae angle of the current be zero we found that the power into a circuit wa: P I Z Z cot Z int r r The firt term i the average power into the circuit, the econd i the unavoidable ripple in the power, ince thi i A, but the imaginary part of Z yield power which goe in at one part of the cycle, and then i returned at a later point. Thi i called reactive power. i
11 EES 5 Spring 4, Lecture 6 The Phaor reult At reonance the circuit impedance i purely real The reactive power for the ap i upplied from the inductor, and via vera. Imaginary component of impedance cancel out For a erie reonant circuit, the current i maximum at reonance V L V V V R EES 5 Spring 4, Lecture 6 Second Order Tranfer Function So we have: V o H V R j V jl R j To find the pole/zero, we put the H in canonical form: V j R H j V L j R
12 EES 5 Spring 4, Lecture 6 For a complex conjugate pair of pole in the denominator: Putting the denominator in tandard form, and factoring it: j j ± ± j 4 4 If >/ Pole are complex conjugate frequencie Im Re EES 5 Spring 4, Lecture 6 Magnitude Repone How peaked the repone i depend on R j j L H j R j j L H j H So at the peak Z: j H j j
13 3 EES 5 Spring 4, Lecture 6 How Peaky i it? Let find the point when the magnitude of the tranfer function quared ha dropped in half: j H / j H / EES 5 Spring 4, Lecture 6 Half Power Frequencie Bandwidth We have the following: / ± m a b b a > ± ± ± ± 4 / Four olution! < > < > b a b a b a b a Take poitive frequencie:
14 EES 5 Spring 4, Lecture 6 Second Order ircuit Bode Plot uadratic pole or zero have the following form: j j ζ The root can be parameterized in term of the damping ratio: ζ j j j damping ratio ζ > j ζ ± j ζ j ζ Two equal pole jτ jτ Two real pole EES 5 Spring 4, Lecture 6 Bode Plot: Damped ae The cae of ζ > and ζ i a imple generalization of imple pole zero. In the cae that ζ >, the pole zero are at ditinct frequencie. For ζ, the pole are at the ame real frequency: ζ j τ jτ jτ j τ jτ log j τ 4log jτ jτ jτ jτ j τ Aymptotic Phae Shift i 8 Aymptotic Slope i 4 db/dec 4
15 EES 5 Spring 4, Lecture 6 Underdamped ae For ζ <, the pole are complex conjugate: j j ζ j ζ ± For / <<, thi quadratic i negligible db For / >>, we can implify: ζ log j j ζ log j 4log In the tranition region / ~, thing are tricky! ζ ± j ζ EES 5 Spring 4, Lecture 6 Underdamped Mag Plot ζ. ζ ζ. ζ. ζ.4 ζ.6 ζ.8 5
16 EES 5 Spring 4, Lecture 6 Underdamped Phae The phae for the quadratic factor i given by: ζ j j ζ tan For / <, the phae hift i le than 9 For /, the phae hift i exactly 9 For / >, the argument i negative o the phae hift i above 9 and approache 8 Key point: argument hift ign around reonance EES 5 Spring 4, Lecture 6 Phae Bode Plot ζ ζ
17 7 EES 5 Spring 4, Lecture 6 Bode Plot Guideline In the tranition region, note that at the breakpoint: From thi you can etimate the peakine in the magnitude repone Example: for ζ., the Bode magnitude plot peak by log5 ~4 db The phae i much more difficult. Note for ζ, the phae repone i a tep function For ζ, the phae i two real pole at a fixed frequency For <ζ<, the plot hould go omewhere in between! j j j j ζ ζ ζ EES 5 Spring 4, Lecture 6 Energy Storage in Tank At reonance, the energy tored in the inductor and capacitor are t LI t i L w M L co t I t I d i t v w M M in in τ τ L I t t L I w w w M M L in co L I W W M S L,max
18 EES 5 Spring 4, Lecture 6 Energy Diipation in Tank Energy diipated per cycle: w D P T I R π M The ratio of the energy tored to the energy diipated per cycle i thu: w w D LI M L π I R R π π M S EES 5 Spring 4, Lecture 6 Phyical Interpretation of -Factor For the erie reonant circuit we have related the factor to very fundamental propertie of the tank: w π w The tank quality factor relate how much energy i tored in a tank to how much energy lo i occurring. If >>, then the tank pretty much run itelf even if you turn off the ource, the tank will continue to ocillate for everal cycle on the order of cycle Mechanical reonator can be fabricated with extremely high S D 8
19 EES 5 Spring 4, Lecture 6 thin-film Bulk Acoutic Reonator FBAR RF MEMS Agilent TechnologieIEEE ISS > Reonate at.9 GHz Pad Thin Piezoelectric Film an ue it to build low power ocillator R x R x L x EES 5 Spring 4, Lecture 6 rytal To etablih accurate clock, we need ocillator with narrow reonance, and therefore high Since mechanical reonator can have very high, becaue they can be built to looe very little energy on each cycle, mot clock are built around a mechanical reonance. 9
20 EES 5 Spring 4, Lecture 6 Feature of bilinear tranfer function For bilinear tranfer function, we can put them into the form: H j K j q j z j z j z3 L j j j L p Where the root in the numerator are called zero, and the root of the denominator are called pole. Since the root come from a real polynomial in j, they are either real, or come in complex conjugate pair. We can write thoe that come in pair: H j K j q p p3 j ζ j z L z z z L j L j ζ L p p p p EES 5 Spring 4, Lecture 6 Feature by Feature: Overall factor K D pole or Zero Overall decade for decade hift with frequency Each pole contribute: far below it corner frequency At the corner frequency, the repone i down by 3db /j far above it corner frequency -9 phae hift Each zero contribute: below it corner frequency At the corner frequency, the repone i up 3 db j above it corner frequency9 phae hift
21 EES 5 Spring 4, Lecture 6 Feature by Feature: Each complex conjugate pair of pole contribute: far below their corner frequency /j far above their corner frequency -8 phae hift At the reonance, the repone peak up by a factor Log r ζ ζ H j log Each complex conjugate pair of zero contribute: below their corner frequency j far above their corner frequency 8 phae hift At the reonance, the repone dip down by a factor Log r ζ ζ H j log ζ EES 5 Spring 4, Lecture 6 We will ue bilinear tranfer function for: Small ignal modeling of tranitor circuit Feedback and filtering in amplifier
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