The Magic of SinusoidsExample: Low Pass Filter (LPF)LPF the ``hard way"LPF: Solving for responseLPF Magnitude ResponseLPF Phase ResponsedB: Honor the inventor of the phone...Why 20? Power!Complex ExponentialEulor's Theorem and The CircleWhy introduce complex numbers?Complex Exponential is PowerfulLPF Example: The ``soft way''Magnitude and Phase ResponseWhy did it work?And yet another perspectiveAnother perspective (cont.)``Proof'' for Linear Systems``Proof'' (cont.)PhasorsCapacitor $I$-$V$ Phasor RelationEE 42/100Lecture 16: Sinusoidal Steady-State and Linear SystemsELECTRONICSRev B 3/13/2012 (2:23 PM)Prof. Ali M. NiknejadUniversity of California, BerkeleyCopyrightc 2012 by Ali M. NiknejadA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 1/22 – p.The Magic of SinusoidsAny linear circuit With L,C,R,M and dep. sources Amp Scale Phase Shift •When a linear, time invariant (LTI) circuit is excited by a sinusoid, it’s output is asinusoid at the same frequency. Only the magnitude and phase of the output differfrom the input. Sinusoids are very special functions for LTI systems.•The “Frequency Response” is a characterization of the input-output response forsinusoidal inputs at all frequencies.•Since most periodic (non-periodic) signals can be decomposed into a summation(integration) of sinusoids via Fourier Series (Transform), the response of a LTIsystem to virtually any input is characterized by the frequency response of thesystem.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 2/22 – p.Example: Low Pass Filter (LPF)•Input signal: vs(t) = Vscos(ωt)•We know that in SS the amplitude and phase will change:vo(t) = K · Vs|{z }Vocos(ωt + φ)•The governing equations are:vo(t) = vs(t) − i(t)Ri(t) = Cdvo(t)dtvo(t) = vs(t) − RCdvo(t)dtvo(t) = vs(t) − τdvo(t)dtvsvoRCA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 3/22 – p.LPF the “hard way"•Plug the known form of the output into the equation and verify that it can satisfyKVL and KCLVscos(ωt) = Vocos(ωt + φ) − τωVosin(ωt + φ)Use the following identities:cos(x + y) = cos x cos y − sin x sin ysin(x + y) = sin x cos y + cos x sin yVscos(ωt) = Vocos(ωt)(cos φ − τ ω sin φ) − Vosin(ωt)(sin φ + τω cos φ)•Since sine and cosine are linearly independent functions:a1sin(ωt) + a2cos(ωt) = 0implies that a1= 0 and a2= 0.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 4/22 – p.LPF: Solving for response•Applying the linear independence gives us−Vosin φ − Voτω cos φ = 0this can be converted intotan φ = −τ ω•The phase response is thereforeφ = − tan−1τωLikewise we haveVocos φ − Voτω sin φ − Vs= 0Vo(cos φ − τω sin φ) = VsVocos φ(1 − τω tan φ) = VsVocos φ(1 + (τω)2) = VsVo(1 + (τω)2)1/2= Vs•The amplitude response is therefore given byVoVs=1p1 + (τω)2A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 5/22 – p.LPF Magnitude Response246 8 100.20.40.60.81Filter Passbandω =1τω =10τ≈ .707≈ .1VoVs=1p1 + (τω)2A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 6/22 – p.LPF Phase Response246 8 10-80-60-40-200ω =1τω =10τ6VoVs= − tan−1ωτ−45◦−90◦≈A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 7/22 – p.dB: Honor the inventor of the phone...0.1 1 10 100-40-30-20-100dB = 20 logVoVs•The LPF response quickly decays to zero•We can expand range by taking the log of the magnitude response•dB = deciBel (deci = 10)A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 8/22 – p.Why 20? Power!•Why multiply log by “20” rather than “10”?•Power is proportional to voltage squared:dB = 10 log„VoVs«2= 20 log„VoVs«•At various frequencies we have:ω = 1/τ →„VoVs«dB= −3dBω = 100/τ →„VoVs«dB= −40dBω = 1000/τ →„VoVs«dB= −60dB•Observe: slope of Signal attenuation is 20 dB/decade in frequency.•Alternatively, if you double the frequency, the attenuation changes by 6 dB, or 6dB/octave.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 9/22 – p.Complex Exponential•Eulor’s Theorem says thatejx= cos x + j sin x•This can be derived by expanding each term in a power series.•If take the magnitude of this quantity, it’s unity|ejx| =pcos2x + sin2x = 1•That means that ejφis a point on the unit circle at an angle of φ from the x-axis.Any complex number z, expressed as have a realand imaginary part z = x + jy, can also be in-terpreted as having a magnitude and a phase.The magnitude |z| =px2+ y2and the phaseφ = ∠z = tan−1y/x can be combined using thecomplex exponentialx + jy = |z|ejφℜ(z)ℑ(z)z = x + jy|z|φejφ|ejθ| = 1z = |z|ej6z= mejφxyφ =6zA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 10/22 – pEulor’s Theorem and The Circle•This implies that ejωtis nothing but a point rotating on a circle on the complexplane. The real part and imaginary parts are just projections of the circle, which bytrigonometry we know equal the cosine and sine functions.•We can also express cos and sin in terms of e as followscos x =ejx+ e−jx2sin x =ejx− e−jx2j•To see an animation of these equations, click below:ejωtrotating around circleae−jωtrotating around circlebejωt+ e−jωtoscillates on the real axiscahttp://rfic.eecs.berkeley.edu/ee100/pdf/exp1.gifbhttp://rfic.eecs.berkeley.edu/ee100/pdf/exp2.gifchttp://rfic.eecs.berkeley.edu/ee100/pdf/exp3.gifA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 11/22 – pWhy introduce complex numbers?•They actually make things easier!•Integration and differentiation are trivial with complex exponentials:ddtejωt= jωejωtZejωxdx =1jωejωt•Any ODE is now trivial algebraic manipulations ... in fact, we’ll show that you don’teven need to directly derive the ODE by using phasors (phasor is essentially ashorthand notation for a complex number)•The key is to observe that the current/voltage relation for any element can bederived for complex exponential excitationA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 16 p. 12/22 – pComplex Exponential is Powerful•To find steady state response we can excite the system with a complex exponentialLTI System Hejωt|H(ω)|ej(ωt+6H)Magnitude ResponsePhaseResponse•At any
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