1EE105 - Fall 2005Microelectronic Devices and CircuitsLecture 17Frequency-Domain Analysis2AnnouncementsHomework 7 due todayHomework 8 due next weekLab 6 this weekReading: Chapter 10 (10.1)23Lecture MaterialLast lectureCommon drain amplifierReview phasorsThis lectureFrequency-domain analysisBode plots4RC Circuit with Sinusoidal Input R vc(t) C iR iC vs(t) + - vs(t)= Vs cos(ωt) : set phase of source to zero (use as the reference) vc(t)= Vc cos(ωt + φ) : solution is a sinusoidal signal with the same frequency, but with a different amplitude and phase-shifted with respect to the source35A Better TechniqueIt is much more efficient to work with imaginary exponentials as “representing” sinusoids, since these functions are direct solutions of linear differential equations:ddt----- ejωt()jω ejωt()=• Note that EEs use j = (-1)1/2rather than i, since thesymbol i is already taken for current6Using Imaginary ExponentialssccvvdtdvRC =+Substitute:tjssevtvω=)()()(φ+ω=tjccevtvResult:tjstjctjcevevevjωφ+ωφ+ω=+ωτ)()()(47Finding the Amplitude Ratiotjstjjcjceveevevjωωφφ=+ωτ ])([scjjvveej =+ωτφφ][)1(1ωτ+=+ωτ=φ−φφjeeejvvjjjscAmplitudeRatio:Answer is a real number, so take magnitude… use to find amplitude and phase()211ωτ+=scvv8Graphical Result for Amplitude Ratio ω 1/τ 10/τ 1/10τ 1.0 0.5 0.70759Amplitude: A New Representation• We are interested in very small ratios (e.g., Vc/Vs= 0.0001)• Therefore, we use a log plot … but we also define a new function called the deciBel (after Alex. Graham Bell)(Vc/Vs)dB= 20 log10(Vc/Vs)• Examples: Vc/Vs= 0.0001 Æ (Vc/Vs)dB= -80 dBVc/Vs= 0.707 Æ (Vc/Vs)dB= -3 dB10Finding the PhaseCollect real and imaginary parts; latter must be zero:Use Euler’s formula to convert to rectangular form:csjjvveej /][ =+ωτφφ(a real number)csvvjjj /)sin(cos)sin(cos=φ+φ+φ+φωτ0sincos)Im(=φ+φωτ=⋅ωτ−=φtan611Graphical Result for Phase φ ω 1/τ 10/τ 1/10τ 0 -45 -90 12Finding the “Real” WaveformHow to connect the imaginary exponential solution to the measured waveform v(t)? Conventionally, v(t) is the real part of the of the imaginary exponential)cos()Re()(φ+ω=φ+ωtvvetj713Pushing This Idea Further …There are two parameters needed to define a sinusoidal signal:* magnitude* phaseWhy not work with a complex number as the signal and eliminatethe imaginary exponential from the analysis (it cancels out)?Define the complex number consisting of the amplitude and phasea sinusoidal signal as a phasortjVetvtvtvω=⇔φ+ω= )()cos()(φ=jveV14Using Phasors: Capacitor CurrentCvc(t)+-ic(t)dtdvCticc=)(tjcceItiω=)(tjcceVtvω=)(Result:815Impedance of a CapacitorDefinition: the impedance Z of a two-terminal circuit element is the ratio of the phasor voltage to the phasorcurrent (positive reference convention)C+-IcVcZc=Admittance:Yc= 1 / Zc=16Using Phasors: Inductor Voltage+-LvL(t)iL(t)dtdiLtvLL=)(tjLLeItiω=)(tjLLeVtvω=)(Result:917Inductor ImpedanceZL=Admittance:YL= 1 / ZL=+-LVLIL18Circuit Analysis with PhasorsAssumption: sources are sinusoidal, steady-state!+-vs(t)RC+vc(t)-==scVVHRatio of output to input phasor is the transfer function of the circuit:1019Bode Plots1. Plot magnitude | H | in dB vs. ω (log scale)2. Plot phase H in degrees vs. ω (log scale)20Bode Plots for Low-Pass Filter1. Plot magnitude | H | in dB vs. ω (log scale)2. Plot phase H in degrees vs. ω (log scale)dBdBdBjjH⎥⎦⎤⎢⎣⎡+=+=|1||1||11|||ωτωτ1121Sketching the Magnitude Plot⎥⎥⎦⎤⎢⎢⎣⎡+=⎥⎦⎤⎢⎣⎡+=2)(11log20|1||1|||ωτωτdBdBjHLow-frequency (ωτ <<1) asymptoteHigh-frequency (ωτ >>1) asymptote22The Break Frequency ω-3dB= (1/τ) 1/τ10/τ 100/τ 1/(10τ) ω| H |dB -20 -40 -601223Finding the Phase Plot())arctan(011ωτ−=⎥⎦⎤⎢⎣⎡ωτ+∠=∠jHWhy?Low-frequency asymptoteHigh-frequency asymptoteApprox. linear with ω for 1/(10τ) < ω < 10/ τ24Rapidly Sketching the Phase Plot 1/τ10/τ 100/τ 1/(10τ) ωPhase H -45 -901325The High-Pass Filtervs(t)+-RC+vr(t)-+-R1/(jωC)+Vr-Vs==srVVH26Magnitude Bode PlotdBdBdBdBjjjjH |11||||1|||ωτωτωτωτ++=+=First term (numerator):1427Graphical Addition of Magnitudes | H |dB 1/τ10/τ 100/τ 1/(10τ) ω-20 -40 20 40 28Phase Bode Plot for HPF 1/τ10/τ 100/τ 1/(10τ) ωPhase H +90
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