4/9/9 Page of 5 Reference:W:\Lib\MathCAD\Default\default.mcd. Objective a. Motivation. Finite circuit peed, e.g. amplifier - effect on ignal. E.g. how "fat" an amp do we need for audio? For video? For RF?. How do we characterize "fat"?. Check repone for "every poible" input i not practical. Solution: check repone for "repreentative ignal" that tand in for "every poible" ignal b. What i Frequency domain analyi?. Analyi technique. Not a new circuit element, it' applie to the circuit we already know. Widely ued in engineering (not jut circuit) - e.g. EE ++ - Underlying math: Laplace & Fourier tranform, (phaor). Analyi with inuoid. cacadable (hape retained). efficient: phaor analyi. univeral: Fourier analyi. Demo: http://www.faltad.com/fourier (live!) 3. Phaor / Laplace. Motivation: time domain diff eq, trigonometry: math problem a. Repreent inuoid with complex number b. Signal are not complex c. RC LPF example teady tate analyi d. Impedance, Admittance. RLC. generalized 4. Bode plot a. Ratio, db (why logarithmic) b. Bode plot c. Technique 5. Power (NR Ch ). intantaneou, average. rm. complex
4/9/9 Page of 5 Lecture 5: Sinuoidal Steady State (ee alo ppt) Motivation circuit 'deform' changing ignal e.g. digital ignal do not retain their "quare" hape amp alo cannot keep up with rapidly changing ignal (capacitor inide!) how determine e.g. if amplifier i fat enough? how do we even characterize how "fat" an amplifier i? what' a good metric? We don't want to deal at the component (chematic) level (not very hierarchical) Problem Example: RC LPF, (ideal) amp, RC HPF quare wave input (for implicity) "exponential" after LPF even more complicated waveform after HPF How to analyze thi??? Demo Concluion: inuoid retain hape --> modular!!! Sinuoidal Repone (AC analyi) Sinuoid: in or co amplitude frequency (f or omega) phae Initial tranient Example: RL HPF (ee NR text), calculate V R V m V input: f MHz ω π f v () t V m co ω t R 5kΩ L mh for t>, otherwie τ L R τ n T f KVL: olution: L di() t + R i() t v () t dt φ atan ω L R φ 5.488 v R_teady () t V m R R + ω L co ω t φ R R + ω L.63
4/9/9 Page 3 of 5 v R_tran () t V m R R + ω L co( φ) t τ e v R () t v R_tran () t + v R_teady () t.5 v () t v R_teady () t v R_tran () t v R () t.5 5. 7. 6.5. 6. 6.5. 6 t
4/9/9 Page 4 of 5 Lecture 6: Phaor Analyi technique:. efficiently analyze inuoidal teady tate repone. linear differential equation --> linear algebraic equation. ue complex number (ignal are real) Sinuoidal ignal: vt () V m co ω t + φ Phaor vt Re V m e j( ωt+ φ) Re V m e jφ e jωt Re V j ω V m e jφ Vjω ( e jωt ) abbreviation j ω - more compact equation without j - complex number how up in numeric reult - phaor analyi i pecial cae of Laplace tranform, --> conitency with other text on ubject Example: inductor diff eq analyi phaor analyi i L () t I Lm co ω t i L ( t) Re I L () e t d v L () t L t i d L() t L I Lm in ω t v L () t L d t Re I L() e t ReL I L () e t d v L ( t) Re V L () e t with V L () L I L () e.g. L mh I Lm ma f MHz ω π f j ω I L () I Lm V L () L I L () V L ( ) 6.83i V Impedance: Z L () V L () L I L () - univeral! Treat L, C like ''reitor" - complex number - unit: ohm! Admittance: Y L () I L () V L () L e.g. L L 6.83i kω 59.55i μs Let cla derive admittance of a capacitor.
4/9/9 Page 5 of 5 Lecture 7: Network analyi with phaor - KVL and KLC till hold (with phaor) - repreent ignal with phaor (complex number) - Retance become Impedance, generalizable to L, C! - Ditto Admittance - Generalized ohm-meter exit alo: LCR meter, impedance / network analyzer Z R R Z L L Z C C Example: calculate magnitude and phae oc, vc Z tot () Z R () + Z C () Z tot () R + C + R C C I c () V () V () Z tot () C C + R with V m V R kω C pf R C n τ R C V () V m I c () V () C + R C I c I c j π MHz I c 8.34 + 45.48i μa I c.53 ma KVL, KCL, node voltage, divider - whatever, we get: 57.858 arg I c V c () V () H () + R C V c () V () H () + R C + R C H a H j π MHz H a arg H a H a H j π MHz H a.998 arg H a H a H j π MHz H a.847 arg H a H a H j π MHz H a.57 arg H a H a H j π GHz H a.6 arg H a 3.595 3.4 8.957 89.88
4/9/9 Page 6 of 5 Lecture 8: Frequency Repone N i.. N f linrange( Hz, MHz, N) H jπ.5. 7 4. 7 6. 7 8. 7. 8 ( ) arg H jπ 5. 7 4. 7 6. 7 8. 7. 8 N i.. N f logrange( khz, GHz, N) H jπ... 5. 6. 7. 8. 9
4/9/9 Page 7 of 5 ( ) arg H jπ 5. 5. 6. 7. 8. 9 Logarithmic axi: - more informative:. Hz reolution at low frequencie, but not at GHz!. many procee are logarithmic (cf. octave) - for frequency - beware of label log f; uually the freq and not the log i pecified - amplitude - phae cale i alway linear Interpretation H ab ω phi ω + ω τ V in V v in ω, t V out ω V in H ab ( ω) v out ω, t atan( ω τ) V in co( ω t) 5 ( + phi ( ω) ) ω τ V out ( ω) co ω t v in ω, t v out ω, t.5.5 5. 8. 7.5. 7 t
4/9/9 Page 8 of 5 deci-bel - for power ratio e.g. P antenna μw - for voltage ratio e.g. V antenna P antenna R antenna.5.55i H j π MHz db db( ) 6. db( r) log r P ref mw V ref P ref R antenna db( r) log r db( ) db(.5) 6. db P antenna P ref db P antenna P ref ( ) 6.7 db H j π MHz db db V antenna log V antenna V V ref ref db( ) 6. 3 db (dbm) db N i.. N f logrange( khz, GHz, N) ( ) db H jπ 4. 5. 6. 7. 8. 9 ( ) arg H jπ 5. 5. 6. 7. 8. 9 Contruct Bode-plot (ingle pole) from equation. Pole. 3dB Bandwidth. Filter LPF, HPF, BPF, BNF ideal, practical (approximate hape)
4/9/9 Page 9 of 5 Example: PWM DAC V dac D dac V ref wort cae ripple for.. 55 D dac 56 T t t D dac T (5% duty cycle) (time when Vout Vref) V ref V ω π MHz v out () t v () t V ref + V ref π V ref π in ω t n in ( n + ) ω t n + v 3 () t V ref π in 3 ω t 3 V ref t t.5 v out () t v () t v 3 () t.5. 7 4. 7 6. 7 8. 7. 6.. 6.4. 6 t Goal: amplitude of ripple < r*vref r % Ripple dominated by v V ref A π A.637 V ref < r Attenuation required a A V ref r a 6.366 db( a) 6.78 db a) graphical olution (from Bode diagram, determine wp /RC from w and a) b) algebraic Hj ( ω ) a + ω R C a RC ω C nf R RC. μ RC R. kω C
4/9/9 Page of 5 Pole of RC LPF: ω p RC uc output: RC LPF: v () t H f ( f, ) V ref ω p π Im V ref + π + ω f 59.54 khz f n e j( n + ) ω t n + ω p f.59 ω Filter output: v o ( tf, ) v n tn, V ref Im V ref π Im V ref + π n H f j ( n + ) ω, f H f j ( n + ) ω, f e j( n + ) ω t n + e j( n + ) ω t n + v () t v o ( t, ) v o ( t,.5) v o ( tf, ).5 v n ( t, ) v n ( t, ). 7 4. 7 6. 7 8. 7. 6.. 6.4. 6 t
4/9/9 Page of 5 Lecture 9: Bode Plot C R V x + + V in R C V out - - Given V x V in C. + V x R. V x V out + R V out V x + V out C R Find( V x, V out ) C. R. volt C. R. + C R C. R. + C. R. R C + + C R + R. C volt C. R. + C. R. R C + + C R + R. C e.g. f Hz f MHz ω p π f ω p π f R kω R kω C π f R C π f R C 5.95 nf C.6 nf Given Find( ω p, ω p ) H () + R C + R C + R C + R R C C + + R C ω p 99.899 +.583i 6. 7 + ω p + Hz π R C + ω p ω p ω p + ω p ω p ω p Pole plitting: wp << wp H () R C + R C + R C + R C + R R C C H () + R C ω p + ω p
4/9/9 Page of 5 Bode Plot N i.. N f logrange 3dB bandwidth f, f, N ( ) db H iπ 4. 3. 4. 5. 6. 7. 8. 9 9 45 ( ) arg H iπ 45 9. 3. 4. 5. 6. 7. 8. 9 Bandpa characteritic: reject low and high frequency, pae mid-band Bode plot capture ignificant information about behavior of circuit. Need efficient way to contruct and draw them. Application: Phone Bapand from 3Hz... 3kHz
4/9/9 Page 3 of 5 Lecture 3: Contructing Bode Plot Tranfer function: H () N () D () Need root of N (zero of H) and root of D (pole of H) for Bode Plot. Zero and pole capture many important characteritic of ytem (ee other coure). Then rewrite H() with pole and zero explicitly:. Standard form H () K r m n z m p n z m zero of H() p n pole of H(). Magnitude repone db( H( j ω) ) H m ω log( K ) H m ω log K r m n z m p n j ω + r log( ω) + m 3. Phae repone n log + j log + j ω z m ω z n um individual term (thank to log!) + r 9 φω ( ) + m φ j ω z m n φ j ω p n um individual term
4/9/9 Page 4 of 5 Example : R kω C pf H () + R C K p K p R C p f π f.59 4 khz Magnitude repone: H m ω log j ω p H m ω log + ω p for ω<<p : H m_low ( ω) for ω>>p : at ωp : H m_high ω 3. H m p log ω p N i.. N f logrange f, f, N 3-dB frequency f 3dB H m_low π H m_high π H m π 4 6 db per decade rolloff 8. 4. 5. 6. 7. 8. 9.. Straight line approximation: line interect at p /π
4/9/9 Page 5 of 5 Phae repone: arg φω j ω φω p atan ω p 5 3 φ π 45 6 75 9. 4. 5. 6. 7. 8. 9.. Straight line approximation: 9 over decade, -45 @ f Example : H () H m ω K K π Hz K ω N i.. N f logrange( Hz, khz, N) 4 ( ) db H m π ( ) arg H iπ 4 6 8. 3. 4. 5
4/9/9 Page 6 of 5 Example 3: H () R C + R C + R C + R C + R R C C Find root of denominator: + R C + R C + R C + R R C C a R R C C a.533 C R V x + + V in R C V out - - b R C + R C + R C b.593 3 b + b 4 a p p f f 99.9 Hz a π b b 4 a p p f f. MHz a π check root: D () + R C + R C + R C + R R C C.694 Dp hould be (nearly) zero.9 Dp tandard form: H () R C p p r K R C H () R C π R C Hz H () p H 3 () p N 3 i.. N f logrange f, 9 f, N
4/9/9 Page 7 of 5 4 ( ) db H iπ ( ) db H iπ ( ) db H 3 iπ ( ) db H iπ 4 6 8. 3. 4. 5. 6. 7. 8. 9. 8 ( ) arg H iπ ( ) arg H iπ ( ) arg H 3 iπ ( ) arg H iπ 6 6 8. 3. 4. 5. 6. 7. 8. 9.
4/9/9 Page 8 of 5 Lecture 3: Power Reitive load V V f 6Hz v () t V co ω t p () t ω π f v () t R R Ω T f v () t p () t...3 t T P rm p () t dt T T T ( v () t ) R dt V rm R T V rm T ( V co( ω t) ) dt V V V rm V rm P rm R Capacitive Load farad d C i () t C t v () t ω d p () t v () t i () t P rm.5 W v () t i () t p () t.5..5..5.3 t
4/9/9 Page 9 of 5 Power Analyi V V I.7A φ 7 Phae: φ Δφ φ v φ i v () t V co ω t i () t I co ω t φ p a () t V I co ω t φ co ω t after ome work... p b () t V I ( + co( ω t) ) co φ + in φ in( ω t) v () t i () t p b () t.5.5.5..5..5.3 t Nomenclature V I Average (real) power: P co( φ ) power factor pf co( φ ) pf.34 V I Reactive power: Q in( φ ) rf in( φ ) rf.94 pf + rf p () t P ( + co( ω t) ) + Q in( ω t) a) Purely reitive circuit V I φ P Q [W] P 35 mw Actually diipated in load (ha a w ripple on top of average). b) Purely reactive circuit V I φ 9 P Q [VAR] Q 35 mw Average i zero. Only heat wire (which add a reitive component).
4/9/9 Page of 5 Complex Power Efficient way to calculate P & Q. Complex power: S P + j Q [VA] Apparent power: S P + Q [VA] Power handling capacity of ditribution network (e.g. power cord). S ()... ome math... V () I () (complex conjugate of current) Example : computer upply R // C V V V 55.563 V f 6Hz ω π f R R kω C μf Z () drop at high freq!!! + R C V () V () V I () increae at high frequency! Z () S () V () I () S ( Hz). W only reitive part matter VA W mva mw. 9.3i S j ω VA S ( j ω ) 9.3 VA (!!!) ( ). W Re S j ω V ( j ω ) 55.563 V Im S j ω I j ω ( ) 9.3.83 A VAR (!!!) kva kw VAR W V ( Hz) R 55.563 ma Current dominated by reactive part
4/9/9 Page of 5 Example : loudpeaker with(out) erie capacitor V dc 5V 8Ω L ω Z () R + L V V L.894 mh I () f 44Hz V Z () R Ω ω π f S () V I () S ( Hz) VA jut heat peaker! (nobody can hear DC) 3.769 46.54i S j ω + mva audio power i much maller (than heat) with erie capacitor: C 5mF Z c () R + L + C I c () V Z c () S c () V I c () VA S c j Hz no heat (DC term) 3.35 48.33i S c j ω + mva ame a w/o C! N 3 i.. N f logrange( Hz, khz, N) Z jπ Z c jπ. 3 S jπ S c jπ.. 3
4/9/9 Page of 5 Example 3: correcting the power factor (zero reactive power) V V Inductive load ω π 6Hz Z L 4Ω + j Ω without correction: V S a Z L S a 484 + 4i VA V I a Z L I a 3..556i A I a 3.479 A with correction: Y c Im Z L Z L Z c Z tot Z L + Z c V S b Z tot Z c Y c j Z tot 5 Ω S b 484 VA Z c i Ω capacitor! Example 4: voltage drop in inductive power upply wiring (model current a inuoid at f)
4/9/9 Page 3 of 5 Lecture 3: Finite Opamp Bandwidth Model: a () ω b a ajω b like ideal opamp Note: can model finite gain and bw imultaneouly - more complicated math - rather ue uperpoition for mall error (ytem linear) Given V x V. V x V + R. R. ω b V V x A () V () V () ω b R ( R + R ) + R ω b Find( V, V x ) ω b R. R. V. V. R. + R. + ω b R. R. + R. + ω b R. A () R A o R R + R + ω b R + + A o ω b + A o ω p ω p ω b + A o AA, o, f p + A o A o + π f p ( ) 6.556 db A j π khz,, MHz db N 3 i.. N f logrange( Hz, MHz, N) 4 ( ) ( (,, MHz) ) ( (,, MHz) ) db A jπ,, MHz db A j π db A j π. 3. 4. 5. 6. 7
4/9/9 Page 4 of 5 Lecture 33: Electronic Noie Digital: finite # bit limit reolution Analog: many error, interference etc --> hield or ue other method to reduce ultimate: finite charge of electron Lecture 34: Filter Example: Sampling.8.6.4 Amplitude [V]. -. -.4 -.6 -.8-3 4 5 6 7 8 Time [] x -4 Show pectrum onput ignal and alia. Sampling theorem: frequencie > f/ alia --> filter out! f 44.kHz B khz ω p π B 3dB bandwidth H () Hj π + ω p f.67 db H j π f 3.455
4/9/9 Page 5 of 5 http://www.maxim-ic.com/appnote.cfm/an_pk/76 ee alo MAX 74/5 Sallen Key: H () + τ + τ Given Find() + τ + τ + i 3 τ i 3 τ