Announcements Homework 8 due next Tuesday EE105 Fall 2005 Microelectronic Devices and Circuits Lab 6 this week Lecture 18 Reading Chapter 10 10 1 Lab 7 next week Frequency Domain Analysis Second Order Circuits 2 Lecture Material Power Flow The instantaneous power flow into any element is the product of the voltage and current P t i t v t For a periodic excitation the average power is Last lecture Frequency domain analysis Bode plots Pav i v d This lecture T More Bode plots Second order functions In terms of sinusoids we have Pav I cos t i V cos t v d T I V cos t cos i sin t sin i cos t cos v sin t sin v d T I V d cos 2 t cos i cos v sin2 sin i sin v c sin t cos t T 3 Power Flow with Phasors Pav I V 2 From the previous slide P 2 cos i cos v sin i sin v I V 2 cos i v 20 In terms of the circuit impedance we have 2 V 1 V 1 P Re I V Re V Re Z 1 2 2 Z 2 Power Factor Pav I V More Power cos i v Note that if i v then 2 I V 2 2 cos 2 0 2 2 V V V Z Re Re Z Re Z 2 2 2 2 Z 2Z 2Z Check the result for a real impedance resistor Also in terms of current I V 1 1 cos i v Re I V Re I V 2 2 2 2 I 1 1 P Re I V Re I I Z Re Z 2 2 2 21 22 1 Second Order Circuits Series LCR Impedance The series resonant circuit is one of the most important elementary circuits With phasor analysis this circuit is readily analyzed Z Z j L The physics describes not only electrical LCR circuits but also approximates mechanical resonance massspring pendulum molecular resonance microwave cavities transmission lines buildings bridges Z j L 23 Resonance 1 R j C 1 1 R R j L 1 j C 2LC 1 Im Z L 1 0 2LC 2 1 LC 24 Series Resonance Voltage Gain Resonance occurs when the circuit impedance is purely real Note that at resonance the voltage across the inductor and capacitor can be larger than the input voltage Imaginary components of impedance cancel out For a series resonant circuit the current is maximum at resonance VL VC VL Vs VR VC VC I VR VC Vs 0 VC 0 V Vs j 0L s j 0L R Z 0 jQ Vs VR Vs VR Vs VL I j 0L VC VL VL VR VL V 0L V 1 s s j 0L j 0C Z 0 j R jQ Vs 0 25 Second Order Transfer Function L LC 1 L 1 Z0 1 1 Q 0 R C R CR R 0C R 26 Poles of 2nd Order Transfer Function So we have Denominator is a quadratic polynomial Vo R j V0 j CR L H j R 1 Vs 1 2LC j RC j 2 j LC L R j L 1 H j 20 R 2 0 j 2 j LC L V R H j 0 Vs j L 1 R j C To find the poles zeros let s put the H in canonical form V j CR H j 0 Vs 1 2LC j RC One zero at DC frequency no DC current through a capacitor H j 27 j 0 Q 20 j 2 j 0 Q L Q 0 R 28 2 Resonance without Loss Finding the poles Let s factor the denominator The transfer function can parameterized in terms of Im loss First take the lossless case R 0 0 j j 20 0 Q 2 20 j 0 0 20 2 2 Q 4Q Q 20 1 0 20 0 j 0 1 2Q 2Q 4Q 2 4Q 2 Poles are complex conjugate frequencies The Q parameter is called the Im quality factor or Q factor Re This is an important parameter R 0 Q 29 Magnitude Response H j H j 0 1 20 R j 0 0 L 2 20 j 0 Q 2 2 H j H j 0 Q 100 31 0 We have the following 20 2 0 Q 2 m 0 20 0 Q 2 1 1 2 2 20 j 2 j a b 0 1 0 Q 32 Often a second order transfer function is characterized by the damping factor as opposed to the Quality factor Four solutions b a 2 1 2 2 2 0 1 0 Q 0 0 Q a b 0 1 j j 0 Q a b 0 a b 0 1 j 2 j 2 0 2 Take positive frequencies 0 Q 1 2 2 0 1 0 Q 2 More Notation Half Power Frequencies Bandwidth 0 0 20 a b 2Q 4Q 20 Q0 2 j 0 Q 1 20 20 j 0 0 Q 2 2 2 2 Q 10 2 2 0 1 0 Q 0 Q H j 2 30 Let s find the points when the transfer function squared has dropped in half j 0 Q Q 1 H 0 0 When the circuit is lossless the poles are at real frequencies so the transfer function blows up At this resonance frequency the circuit has zero imaginary impedance and thus zero total impedance Even if we set the source equal to zero the circuit can have a steady state response oscillates How Peaky is it The response peakiness depends on Q R j 0 0 L Re Q 33 1 0 1 2 34 3 Second Order Circuit Bode Plot Bode Plot Damped Case Quadratic poles or zeros have the following form The case of 1 and 1 is a simple generalization of simple poles zeros In the case that 1 the poles zeros are at distinct frequencies For 1 the poles are at the same real frequency j 2 j 2 1 0 damping ratio The roots can be parameterized in terms of the damping ratio 1 1 2 Asymptotic Slope is 40 dB dec 2 20 log 1 j 40 log 1 j j 2 j 2 1 1 j 1 1 j 2 1 j 2 1 j 1 j 2 1 j j 2 1 Two real poles j 2 j 2 1 1 j 2 1 j 2 1 j Two equal poles 1 j 2 j 2 1 1 j 2 Asymptotic Phase Shift is 180 35 Underdamped Case 36 Underdamped Mag Plot For 1 the poles are complex conjugates 0 01 j 2 j 2 1 0 0 1 j 2 1 j 1 2 0 2 0 4 0 6 For 1 this quadratic is negligible 0dB For 1 we can simplify 0 8 1 20 log j 2 j 2 1 20 log j 2 40 log In the transition region 1 things are tricky 37 Underdamped Phase 38 Phase Bode Plot The phase …
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