1EE105 - Fall 2005Microelectronic Devices and CircuitsLecture 18Frequency-Domain AnalysisSecond-Order Circuits2AnnouncementsHomework 8 due next TuesdayLab 6 this weekLab 7 next weekReading: Chapter 10 (10.1)3Lecture MaterialLast lectureFrequency-domain analysisBode plotsThis lectureMore Bode plotsSecond order functions20Power FlowThe instantaneous power flow into any element is the product of the voltage and current:For a periodic excitation, the average power is:In terms of sinusoids we have)()()(tvtitP =∫τττ=TavdviP )()()cos(2)sinsincos(cos2cossinsinsinsincoscoscos)sinsincos(cos)sinsincos(cos)cos()cos(22viviviTvivivvTiiTviavVIVIttctdVIdttttVIdtVtIPφ−φ⋅=φφ+φφ⋅=ωω+φφ+φφωτ⋅=τφω−φω⋅φω−φω⋅=τφ+ωφ+ω=∫∫∫21Power Flow with PhasorsNote that if , then From the previous slide:]Re[21]Re[21)cos(2**VIVIVIPvi⋅=⋅=φ−φ⋅=)cos(2viavVIP φ−φ⋅=Power Factor2)(π=φ−φvi0)2/cos(2=π⋅=VIPav22More PowerIn terms of the circuit impedance we have:Check the result for a real impedance (resistor)Also, in terms of current:]Re[2]Re[2]Re[2]Re[2]Re[21]Re[2122*222*212**ZZVZZVZZVZVVZVVIP====⋅=⋅=−]Re[2]Re[21]Re[212**ZIZIIVIP =⋅⋅=⋅=223Second Order CircuitsThe series resonant circuit is one of the most important elementary circuits:The physics describes not only electrical LCR circuits, but also approximates mechanical resonance (mass-spring, pendulum, molecular resonance, microwave cavities, transmission lines, buildings, bridges, …) 24Series LCR ImpedanceWith phasor analysis, this circuit is readily analyzed RCjLjZ +ω+ω=1⎟⎟⎠⎞⎜⎜⎝⎛ω−ω+=+ω+ω=LCLjRRCjLjZ2111011]Im[2=⎟⎟⎠⎞⎜⎜⎝⎛ω−ω=LCLZLC12=ωZ25ResonanceResonance occurs when the circuit impedance is purely real Imaginary components of impedance cancel outFor a series resonant circuit, the current is maximum at resonance+VR−+ VL –+ VC –VLVCVRVsVLVCVRVsVCVLVRVs+Vs−0ω<ω0ω=ω0ω>ω26Series Resonance Voltage GainNote that at resonance, the voltage across the inductor and capacitor can be larger than the input voltage:+VR−+ VL – + VC –sssLVjQLjRVLjZVLjIV×=ω=ωω=ω=0000)(sssCVjQLjRVjLZVCjIV×−=ω−=ωω=ω=0000)(1RZRCLRCLCRCRLQ0001111===ω=ω=27Second Order Transfer FunctionSo we have:To find the poles/zeros, let’s put the H in canonical form:One zero at DC frequency Æ no DC current through a capacitorRCjLjRVVjHs+ω+ω==ω1)(0+Vo−RCjLCCRjVVjHsω+ω−ω==ω201)(28Poles of 2ndOrder Transfer FunctionDenominator is a quadratic polynomial:LRjjLCLRjRCjLCCRjVVjHsω+ω+ω=ω+ω−ω==ω220)(11)(LRjjLRjjHω+ω+ωω=ω220)()(LC120≡ωQjjQjjH02200)()(ωω+ω+ωωω=ωRLQ0ω≡329Finding the poles…Let’s factor the denominator:Poles are complex conjugate frequenciesThe Q parameter is called the “quality-factor” or Q-factorThis is an important parameter:ReIm0)(2002=ω+ωω+ωQjj22−ω±ω−=ω−ω±ω−=ωQjQQQ4112420020200∞⎯⎯→⎯→0RQ30Resonance without LossThe transfer function can parameterized in terms of loss. First, take the lossless case, R=0:When the circuit is lossless, the poles are at realfrequencies, so the transfer function blows up!At this resonance frequency, the circuit has zero imaginary impedance and thus zero total impedanceEven if we set the source equal to zero, the circuit can have a steady-state response (oscillates)ReIm02020042ω±=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛ω−ω±ω−=ω∞→2jQQQ31Magnitude ResponseThe response peakiness depends on QQjQjLRjLRjjH022000022000)(ωω+ω−ωωω=ωωω+ω−ωωωω=ω1=Q10=Q100=Q0ω1)(002020200=ωω+ω−ωω=ωQjQjjH1)(0=ωjH0)0( =HωΔ32How Peaky is it?Let’s find the points when the transfer function squared has dropped in half:()21)(202220202=⎟⎠⎞⎜⎝⎛ωω+ω−ω⎟⎠⎞⎜⎝⎛ωω=ωQQjH211/1)(202202=+⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ω=ωQjH1/20220=⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ωQ33Half Power Frequencies (Bandwidth)We have the following:1/0220±=ωωω−ωQ02002=ω−ωωωQmabbaQQ>±±=ω+⎟⎠⎞⎜⎝⎛ω±ω±=ω2020042Q0ω=ω−ω=ωΔ−+1/20220=⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ωQQ10=ωωΔFour solutions!0000<−−>+−<−+>++babababaTake positive frequencies:34More “Notation”Often a second-order transfer function is characterized by the “damping” factor as opposed to the “Quality”factor0)(0220=ωω+ω+ωQjj0)(12=ωτ+ωτ+Qjj01ω=τ02)()(12=ζωτ+ωτ+ jjζ=21Q435Second Order Circuit Bode PlotQuadratic poles or zeros have the following form:The roots can be parameterized in terms of the damping ratio:012)()(2=+ζωτ+ωτ jj22)1(12)()(1 ωτ+=+ωτ+ωτ⇒=ζ jjjdamping ratioTwo equal poles1)1)(1(12)()(12212−ζ±ζ−=ωτωτ+ωτ+=+ζωτ+ωτ⇒>ζjjjjjTwo real poles36Bode Plot: Damped CaseThe case of ζ >1 and ζ =1 is a simple generalization of simple poles (zeros). In the case that ζ >1, the poles (zeros) are at distinct frequencies. For ζ =1, the poles are at the same real frequency:22)1(12)()(1 ωτ+=+ωτ+ωτ⇒=ζ jjj221)1( ωτ+=ωτ+ jjωτ+=ωτ+ jj 1log401log202()()()ωτ+∠=ωτ+∠+ωτ+∠=ωτ+∠ jjjj 1211)1(2AsymptoticSlope is 40 dB/decAsymptotic Phase Shift is 180°37Underdamped CaseFor ζ <1, the poles are complex conjugates:For ωτ << 1, this quadratic is negligible (0dB)For ωτ >> 1, we can simplify:In the transition region ωτ ~ 1, things are tricky!22211012)()(ζ−±ζ=−ζ±ζ−=ωτ=+ζωτ+ωτjjjjωτ=ωτ≈+ζωτ+ωτ log40)(log2012)()(log2022jjj38Underdamped Mag Plot ζ=1ζ=0.01ζ=0.1ζ=0.2ζ=0.4ζ=0.6ζ=0.839Underdamped PhaseThe phase for the quadratic factor is given by:For ωτ < 1, the phase shift is less than 90°For ωτ = 1, the phase shift is exactly 90°For ωτ > 1, the argument is negative so the phase shift is above 90° and approaches 180°Key point: argument shifts sign around resonance()⎟⎟⎠⎞⎜⎜⎝⎛ωτ−ωτζ=+ζωτ+ωτ∠−212)(12tan12)()( jj40Phase Bode Plotζ=0.010.10.20.40.60.8ζ=1541Bode Plot GuidelinesIn the transition region, note that at the breakpoint:From this you can estimate the peakiness in the magnitude response. Example: For ζ = 0.1, the Bode magnitude plot peaks by 20 log(5) ~ 14 dBThe phase is much more difficult. Note for ζ = 0, the phase response is a step functionFor ζ = 1, the phase is two real poles at a fixed frequencyFor 0 < ζ < 1, the plot should go somewhere in
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