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)23Lecture MaterialLast lectureFrequency-domain analysisBode plotsThis lectureMore Bode plotsSecond order functions4Get to know your logs!A “margin” of 6dB is a factor of 2 (power)!Knowing a few logs can help you calculate logs of different ratios by employing properties of log. For instance, knowing that the ratio of 2 is 6 dB, what’s the ratio of 4?dB ratio dB ratio-20 0.100 20 10.000-10 0.316 10 3.162-5 0.562 5 1.778-3 0.708 3 1.413-2 0.794 2 1.259-1 0.891 1 1.12235Bode Plot OverviewTechnique for estimating a complicated transfer function (several poles and zeros) quicklyBreak frequencies :)1()1)(1()1()1)(1()()(22210pmppznzzKjjjjjjjGHωτ+ωτ+ωτ+ωτ+ωτ+ωτ+ω=ωLLiiτ=ω16Summary of Individual Factors Simple Pole:Simple Zero:DC Zero:DC Pole:ωτ+ j11ωτ+ j1ωτjωτj1τω1=dB0dB0dB0dB090−90+90−90+τω1=47ExampleConsider the following transfer functionBreak frequencies: invert time constants)1)(1()1(10)(3125ωτ+ωτ+ωτ+ω=ω−jjjjjHps100ns10ns100321=τ=τ=τGrad/s10Mrad/s100Mrad/s10321=ω=ω=ω)1)(1()1(10)(3125ωω+ωω+ωω+ω=ωjjjjjH8Breaking Down the MagnitudeRecall log of products is sum of logsLet’s plot each factor separately and add them graphically)1)(1()1(10log20)(3125dBωω+ωω+ωω+ω=ωjjjjjH31251log201log201log2010log20ωω+−ωω+−ωω++ω=jjjj59Breaking Down the PhaseSinceLet’s plot each factor separately and add them graphically)1)(1()1(10)(3125ωτ+ωτ+ωτ+ω∠=ω∠−jjjjjHbaba∠+∠=⋅∠⎟⎟⎠⎞⎜⎜⎝⎛ωω+∠−⎟⎟⎠⎞⎜⎜⎝⎛ωω+∠−⎟⎟⎠⎞⎜⎜⎝⎛ωω+∠+⎟⎟⎠⎞⎜⎜⎝⎛ω∠=ω∠312511110)(jjjjjH10Magnitude Bode Plot: DC Zero80206040-20-60-80-4010410510610710810910101011ω510ωj0 dB611Phase Bode Plot: DC Zero1804513590-45-135-180-9010410510610710810910101011ω510ωj∠12Magnitude Bode Plot: Add First Pole80206040-20-60-80-4010410510610710810910101011ωdB510ωjdB71011ωj+Mrad/s101=ω713Phase Bode Plot: Add First Pole1804513590-45-135-180-9010410510610710810910101011ω510ωj∠71011ωj+∠14Magnitude Bode Plot: Add 2ndZero80206040-20-60-80-4010410510610710810910101011ωdB8101ωj+Mrad/s1002=ω815Phase Bode Plot: Add 2ndZero1804513590-45-135-180-9010410510610710810910101011ω8101ωj+∠16Magnitude Bode Plot: Add 2ndPole80206040-20-60-80-4010410510610710810910101011ωdB101011ωj+Grad/s103=ω917Phase Bode Plot: Add 2ndPole1804513590-45-135-180-9010410510610710810910101011ω10101ωj+∠−18Comparison to Actual Mag Plot1019Comparison to Actual Phase Plot20Power 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=∫∫∫1121Power 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 =⋅⋅=⋅=1223Second 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=ωZ1325ResonanceResonance 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=1427Second 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
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