6.002 Fall 2000 Lecture 1 17 6.002 CIRCUITS AND ELECTRONICS The Impedance Model 6.002 Fall 2000 Lecture 2 17  Sinusoidal Steady State (SSS) Reading 13.1, 13.2 + – O v tVv iI ω cos= + – R C  Focus on steady state, only care about v P as v H dies away.  Focus on sinusoids. Reading: Section 13.3 from course notes. SSS Review  Sinusoidal Steady State (SSS) Reading 13.1, 13.2 6.002 Fall 2000 Lecture 3 17 3 4 H v total Review V p contains all the information we need: p p V V ∠ Amplitude of output cosine phase sneak in V i e jωt drive complex algebra take real part The Sneaky Path p V tV i ω cos [] pp VtV ∠+ ω cos set up DE usual circuit model nightmare trig. 1 v P tj p eV ω RCj V i ω +1 2 6.002 Fall 2000 Lecture 4 17 i p V V transfer function () ω ω jH RCjV V i p = + = 1 1 ( ) ppO VtVv ∠+= ω cos 222 1 1 CR ω + break frequency Bode plot ω RC 1 = ω 1 RC 1 ω 2 1 remember demo ω RC 1 = ω 4 π − 2 π − 0 i p V V ∠ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − 1 RC tan 1 ω The Frequency View Review 6.002 Fall 2000 Lecture 5 17 Is there an even simpler way to get V p ? RCj V V i p ω + = 1 Divide numerator and denominator by jωC . R Cj Cj V V ip + = ω ω 1 1 Let’s explore further… Hmmm… looks like a voltage divider relationship. RZ Z VV C C ip + = 6.002 Fall 2000 Lecture 6 17 The Impedance Model Is there an even simpler way to get V p ? Consider: tj rR eIi ω = tj rR eVv ω = RR Riv = tj r tj r eRIeV ωω = rr RIV = R R i + – R v Resistor tj CC eIi ω = tj CC eVv ω = C C i + – C v Capacitor CC I Cj 1 V ω = dt dv Ci C C = tj C tj C ejCVeI ωω ω = C Z L L i + – L v tj lL eIi ω = tj lL eVv ω = dt di Lv L L = tj l tj l ejLIeV ωω ω = Inductor ll ILjV ω = L Z 6.002 Fall 2000 Lecture 7 17 In other words, For a drive of the form V c e jωt , complex amplitude V c is related to the complex amplitude I c algebraically, by a generalization of Ohm’s Law. inductor LjZ l ω = lll IZV = l I + – l V L Z resistor rrr IZV = RZ r = R Z r I + – r V capacitor Cj 1 Z C ω = cCc IZV = impedance c I + – c V C Z The Impedance Model 6.002 Fall 2000 Lecture 8 17 Impedance model: All our old friends apply! KVL, KCL, superposition… Back to RC example… i RC C ic V ZZ Z V R Cj 1 Cj 1 V + = + = ω ω ic V RCj1 1 V ω + = Done! + – C v I v + – R C + – c V i V + – RZ R = Cj Z C ω 1 = c I 6.002 Fall 2000 Lecture 9 17 Another example, recall series RLC: We will study this and other functions in more detail in the next lecture. R Cj Lj RV V i r ++ = ω ω 1 RCL Ri r ZZZ ZV V ++ = CRjLC CR jV V i r ωω ω ++− = 1 2 + – L r I C R + – r V i V tj r eV ω () rr VtV ∠+ ω cos tj i eV ω tV i ω cos Remember, we want only the steady-state response to sinusoid 6.002 Fall 2000 Lecture 10 17 The Big Picture… tV i ω cos [] pp VtV ∠+ ω cos set up DE usual circuit model nightmare trig. . 6.002 Fall 2000 Lecture 1 17 6.002 CIRCUITS AND ELECTRONICS The Impedance Model 6.002 Fall 2000 Lecture 2 17  Sinusoidal