EE 42 100 Lecture 16 Sinusoidal Steady State and Linear Systems ELECTRONICS Rev B 3 13 2012 2 23 PM Prof Ali M Niknejad University of California Berkeley Copyright A M Niknejad c 2012 by Ali M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 1 22 p The Magic of Sinusoids Phase Shift Any linear circuit With L C R M and dep sources Amp Scale When a linear time invariant LTI circuit is excited by a sinusoid it s output is a sinusoid at the same frequency Only the magnitude and phase of the output differ from the input Sinusoids are very special functions for LTI systems The Frequency Response is a characterization of the input output response for sinusoidal 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 LTI system to virtually any input is characterized by the frequency response of the system 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 Vs cos t We know that in SS the amplitude and phase will change vo t K Vs cos t z Vo The governing equations are vo t vs t i t R R dvo t i t C dt dvo t vo t vs t RC dt vo t vs t A M Niknejad vo vs C dvo t dt 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 satisfy KVL and KCL Vs cos t Vo cos t Vo sin t Use the following identities cos x y cos x cos y sin x sin y sin x y sin x cos y cos x sin y Vs cos t Vo cos t cos sin Vo sin t sin cos Since sine and cosine are linearly independent functions a1 sin t a2 cos t 0 implies 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 Vo sin Vo cos 0 this can be converted into tan The phase response is therefore tan 1 Likewise we have Vo cos Vo sin Vs 0 Vo cos sin Vs Vo cos 1 2 Vs Vo cos 1 tan Vs Vo 1 2 1 2 Vs The amplitude response is therefore given by Vo 1 p Vs 1 2 A M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 5 22 p LPF Magnitude Response 1 0 8 1 Vo p Vs 1 2 707 0 6 0 4 0 2 1 4 2 A M Niknejad 1 6 8 10 Filter Passband University of California Berkeley 10 EE 100 42 Lecture 16 p 6 22 p LPF Phase Response 0 20 6 Vo tan 1 Vs 40 45 60 90 80 2 A M Niknejad 4 6 1 8 10 University of California Berkeley 10 EE 100 42 Lecture 16 p 7 22 p dB Honor the inventor of the phone 0 10 dB 20 log Vo Vs 20 30 40 0 1 1 10 100 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 Vo Vs 2 1 Vo Vs dB 10 log Vo Vs At various frequencies we have 100 1000 20 log 3dB 40dB Vo Vs dB Vo Vs dB 60dB dB Observe slope of Signal attenuation is 20 dB decade in frequency Alternatively if you double the frequency the attenuation changes by 6 dB or 6 dB octave A M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 9 22 p Complex Exponential Eulor s Theorem says that ejx 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 e jx p cos2 x sin2 x 1 That means that ej is a point on the unit circle at an angle of from the x axis z z x jy y 6 z z 1 e j Any complex number z expressed as have a real and imaginary part z x jy can also be interpreted as having a magnitude and a phase p x2 y 2 and the phase The magnitude z z tan 1 y x can be combined using the complex exponential ej z z ej x 6 z mej z x jy z ej A M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 10 22 Eulor s Theorem and The Circle This implies that ej t is nothing but a point rotating on a circle on the complex plane The real part and imaginary parts are just projections of the circle which by trigonometry we know equal the cosine and sine functions We can also express cos and sin in terms of e as follows ejx e jx cos x 2 ejx e jx sin x 2j To see an animation of these equations click below ej t rotating around circlea e j t rotating around circleb ej t e j t oscillates on the real axisc a http rfic eecs berkeley edu ee100 pdf exp1 gif b http rfic eecs berkeley edu ee100 pdf exp2 gif c http rfic eecs berkeley edu ee100 pdf exp3 gif A M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 11 22 Why introduce complex numbers They actually make things easier Integration and differentiation are trivial with complex exponentials d j t e j ej t dt Z ej x dx 1 j t e j Any ODE is now trivial algebraic manipulations in fact we ll show that you don t even need to directly derive the ODE by using phasors phasor is essentially a shorthand notation for a complex number The key is to observe that the current voltage relation for any element can be derived for complex exponential excitation A M Niknejad University of California Berkeley EE 100 42 Lecture 16 p 12 22 Complex Exponential is Powerful To find steady state response we can excite the system with a complex exponential e j t LTI System H H e Magnitude Response j t 6 H Phase Response At any frequency the system response is characterized by a single complex number H The magnitude response is given by H The phase response is given by H We see that the complex exponential is an eigenfunction of the system It is used to probe the system Since a sinusoid is a sum of complex exponentials and because of linearity we can also probe a system by applying a real …
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