6.002 CIRCUITS AND ELECTRONICS Sinusoidal Steady State Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Review We now understand the why of: v 5V R L C 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 Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Motivation For motivation, consider our old friend, the amplifier: V S + – vi VBIAS + – vO vC R Demo CGS Observe vo amplitude as the frequency of the input vi changes. Notice it decreases with frequency. Also observe vo shift as frequency changes (phase). Need to study behavior of networks for sinusoidal drive. Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Sinusoidal Response of RC Network Example: vI + – iC R vI (t ) = Vi cos ω t =0 vC (0) = 0 + vC – for t ≥ 0 (Vi real) for t < 0 for t = 0 vI t 0 Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Our Approach Example: vI + – iC R + vC – Effort Determine vC(t) agony Usual approach sneaky approach very sneaky t Th is le ct ur 11 e :0 0 11 :2 0 12 N ex :0 0 t le ct ur e easy ! e m e g l u d In Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Let’s use the usual approach… 1 Set up DE. 2 Find vp. 3 Find vH. 4 vC = vP + vH, solve for unknowns using initial conditions Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Usual approach… 1 Set up DE RC dvC + vC = vI dt = Vi cos ω t That was easy! Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 2 Find vp dvP RC + vP = Vi cos ωt dt First try: Æ nope vP = A Second try: vP = A cos ωt Third try: vP = A cos(ωt + φ ) frequency amplitude phase Æ nope − RCAω sin(ωt + φ ) + A cos(ωt + φ ) = Vi cos ωt − RCAω sin ωt cos φ − RCAω cos ωt sin φ + A cos ωt cos φ − A sin ωt sin φ = Vi cos ωt .. . gasp ! works, but trig nightmare! Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Let’s get sneaky! Find particular solution to another input… dvPS + vPS = vIS RC dt = Vi e st st Try solution vPS = V p e RC dV p e st (S: sneaky :-)) + V p e st = Vi e st dt sRCV p e st + V p e st = Vi e st Nice property of exponentials ( sRC + 1 )V p = Vi Vp = Vi 1 + sRC Vi ⋅ e st Thus, vPS = 1 + sRC is particular solution to Vi e st ly Vi ⋅ e jω t 1 + jωRC easy! jω t solution for Vi e where we replace s = jω complex amplitude Vp Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 2 Fourth try to find vP… using the sneaky approach Fact 1: Finding the response to Vi e jω t was easy. Fact 2: vI = Vi cos ωt = real[Vi e jω t ] = real[vIS ] from Euler relation, e jω t = cos ωt + j sin ωt real part vI response vP vIS response vPS real part an inverse superposition argument, assuming system is real, linear. Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 2 Fourth try to find vP… so, complex vP = Re[vPS ] = Re[V p e jωt ] ⎡ Vi ⎤ = Re ⎢ ⋅ e jω t ⎥ ⎣1 + jωRC ⎦ Vi (1 − jωRC ) jω t ⎤ ⎡ = Re ⋅e ⎢⎣ 1 + ω 2 R 2C 2 ⎥⎦ Vi ⎡ j φ jω t ⎤ = Re ⎢ ⋅ e e ⎥ , tan φ = −ωRC 2 2 2 ⎣ 1+ω R C ⎦ Vi ⎡ j ( ωt +φ ) ⎤ = Re ⎢ ⋅ e ⎥⎦ ⎣ 1 + ω 2 R 2C 2 vP = Vi 1+ω R C 2 2 2 ⋅ cos( ωt + φ ) Recall, vP is particular response to Vi cos ωt . Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 3 Find vH Recall, vH = Ae −t RC Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 4 Find total solution vC = vP + vH vC = Vi 1+ω R C 2 2 2 cos( ωt + φ ) + Ae − t RC where φ = tan −1 ( −ωRC ) Given vC(0) = 0 for t = 0 so, Vi A=− cos(φ ) 2 2 2 1+ ω R C Done! Phew ! Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Sinusoidal Steady State We are usually interested only in the particular solution for sinusoids, i.e. after transients have died. Notice when t → ∞, vC → vP as e vC = Vi 1+ω R C 2 2 2 − t RC cos( ωt + φ ) + Ae − →0 t RC 0 where φ = tan −1 ( −ωRC ) Vi A=− cos(φ ) 2 2 2 1+ ω R C Vp ∠Vp Described as SSS: Sinusoidal Steady State Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Sinusoidal Steady State All information about SSS is contained in Vp , the complex amplitude! Steps 3 , 4 were a waste of time! Vi Vp = 1 + jωRC Recall Vp 1 = Vi 1 + jωRC Vp Vi magnitude = Vp phase φ : ∠ Vi Vp Vi 1 1 + ω 2 R 2C 2 = e jφ where φ = tan −1 − ωRC 1 1 + ω 2 R 2C 2 = − tan −1 ωRC Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Sinusoidal Steady State Visualizing the process of finding the particular solution vP Vi cos ωt drive D.E. + nightmare trig. V p cos[ωt + ∠V p ] particular solution algebraic take equation real + part complex algebra V p e jω t sneak in Vi e jωt drive the sneaky path! Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Magnitude Plot transfer function Vp H ( jω ) = Vi Vp Vi Vp Vi = 1 1 + ω 2 R 2C 2 1 log scale log scale 1 ω= RC ω From demo: explains vo fall off for high frequencies! Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16 Phase Plot φ = tan −1 − ωRC φ =∠ Vp Vi ω= 0 − − 1 RC ω log scale π 4 π 2 Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.002 Fall 2000 Lecture 16
