6.002 CIRCUITS AND ELECTRONICS Sinusoidal Steady State 6.002 Fall 2000 Lecture 16 1 Review  We now understand the why of: 5V C R L v Today, look at response of networks to sinusoidal drive. Sinusoids important because signals can be represented as a sum of sinusoids. Response to sinusoids of various frequencies -- aka frequency response -- tells us a lot about the system 6.002 Fall 2000 Lecture 16 2 Motivation For motivation, consider our old friend, the amplifier: S V v O v i C v + – + – GS C R V BIAS Observe v o amplitude as the frequency of the input v i changes. Notice it decreases with frequency. Also observe v o shift as frequency changes (phase). Need to study behavior of networks for sinusoidal drive. Demo 6.002 Fall 2000 Lecture 16 3 Sinusoidal Response of RC Network Example: + – R i C + v I v C – v I ( t ) = V i cos ω t for t ≥ 0 ( V i real) = 0 for t < 0 v C (0) = 0 for t = 0 I v 0 t 6.002 Fall 2000 Lecture 16 4 1 1 1 1 e c t u r Example: + – R Our Approach i C + v I v C – Determine v C (t) I nd u l g e m e ! Effort l e c t u r e sneaky approach very sneaky Usual approach agony easy t e 0 : 0 : 0 0 2 1 0 : 2 l s hi T t x e N 6.002 Fall 2000 Lecture 16 5 Let’s use the usual approach… 1 Set up DE. 2 Find v p . 3 Find v H . 4 v C = v P + v H , solve for unknowns using initial conditions 6.002 Fall 2000 Lecture 16 6 Usual approach… 1 Set up DE RC dv C + v C = v I dt = V i cos ω t That was easy! 6.002 Fall 2000 Lecture 16 7 2 Find v p RC dv P + dt v P = V i cos ω t First try: v P = A Æ nope Second try: v P = A cos ω t Æ nope Third try: v P = A cos( amplitude φ ω + t frequency ) phase − RCA ω sin( ω t + φ ) + A cos( ω t + φ ) = V i cos ω t − RCA ω sin ω t cos φ − RCA ω cos ω t sin φ + A cos ω t cos φ − A sin ω t sin φ = V i cos ω t . . gasp ! . works, but trig nightmare! 6.002 Fall 2000 Lecture 16 8 6.002 ll 2000 Lecture 9 16 Let’s get sneaky! Try solution st pPS eVv = st i st p st p eVeV d t edV RC =+ st i st p st p eVeVesRCV =+ ip VV)1sRC( =+ sRC1 V V i p + = Nice property of exponentials ISPS PS vv d t dv RC =+ ( S : sneaky :-)) st i eV = Find particular solution to another input… p V complex amplitude Thus, st i PS e sR C1 V v ⋅ + = st i eV is particular solution to easy! where we replace s = jω ly tj i eV ω solution for tj i e RCj V ω ω ⋅ +1 Fa 2 Fourth try to find v P … using the sneaky approach Fact 1: Finding the response to V i e j ω t was easy. Fact 2: v I = V i cos ω t = real [ V i e j ω t ] = real [ v IS ] from Euler relation, j I v P v response IS v PS v response real part real part e j ω t = cos ω t + sin ω t an inverse superposition argument, assuming system is real, linear. 6.002 Fall 2000 Lecture 16 10 . 6.002 CIRCUITS AND ELECTRONICS Sinusoidal Steady State 6.002 Fall 2000 Lecture 16 1 Review. – R i C + v I v C – v I ( t ) = V i cos ω t for t ≥ 0 ( V i real) = 0 for t < 0 v C (0) = 0 for t = 0 I v 0 t 6.002 Fall 2000 Lecture 16 4 1 1 1 1 e