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Q factor of an underdamped oscillator Damping time or "1/e" time is t = 1/b > 1/w0 (>> 1/w0 if b is very small) How many T0 periods elapse in the damping time? This number (times π) is the Quality factor or Q of the system. t w0 Q=p = T0 2 b large if b is small compared to w0 Max Amplitude Current Amplitude |I0| 1 of max amp 2 Dw w0 Q= Dw Driving Frequency------> Find frequencies where POWER drops to half maximum (current drops to 0.707 of max). These define w. Find resonant frequency, w0 Homework problem to show the two definitions are the same. QUETSION: FM radio stations have broadcast frequencies of approximately 100 MHz. Most radios use a series LRC circuit similar to the one you used in the lab as part of the receiver electronics. Estimate the spacing of the broadcast frequencies of FM stations if typical receivers have a Q of 500 or better. Explain your reasoning, and include a graph. station 1 w station 2 w w0 f0 Qº = Dw Df QUETSION: FM radio stations have broadcast frequencies of approximately 100 MHz. Most radios use a series LRC circuit similar to the one you used in the lab as part of the receiver electronics. Estimate the spacing of the broadcast frequencies of FM stations if typical receivers have a Q of 500 or better. Explain your reasoning, f0 and include a graph. Q= station 1 w Df = 500 station 2 f0 100 MHz Df = = = 0.2 MHz Q 500 Therefore, stations 99.3 and 99.5 FM are allowed, but 99.3 and 99.4 FM are not! They have cross-talk! You should be able to: • Calculate & plot the magnitude and phase of 1/Z • Convert between the mag/phase and Re/Im forms • Draw phasor diagrams of Vext, I, 1/Z (or Z) • Express 1/Z (or Z) in terms of R, L, C or w0, b You should be able to discuss: • The amplitude of the response and resonance • The phase of the response • The nature of the behavior at all frequencies • The transfer of the series LCR circuit analysis to analogous oscillatory systems Blackboard Example I = YV V ( t ) = V0 cos (w 0t ) Y (w ) V ( t ) = - V0 cos ( 2w 0t ) w f I (w ) V ( t ) = V0 cos ( 3w 0t ) w0 2w0 3w0 t q ( t ) = Re éë q0 e ùû iw t iw t Vext = Re éëV0 e ùû i (fq + p /2 ) iw t ù é I(t) = q ( t ) = Re w q0 e e ë û I0 = wV0 L ( é w2 -w 0 ë ) 2 2 1/2 + 4b w ù û 2 -2bw f I = + arctan 2 2 2 w0 - w 2 p 8 I0 = wV0 L ( é w -w ë 2 0 ) 2 2 1/2 + 4b w ù û 2 2 What is the best variable to plot for the LRC lab? ADMITTANCE Amplitude of I/Vapp Frequency, w -> p -2bw f I = + arctan 2 2 2 w0 - w 1 -wL tan f = w C R Phase of I (rel to Vapp) Frequency, w -> Response of a damped oscillator to a periodic driving force of arbitrary shape: We have learned that a damped oscillator produces a sinusoidal response to a sinusoidal driving force. In the language of your lab example: The LRC circuit responds to a sinusoidal driving voltage with a sinusoidal current (at the same frequency). That current is related to the voltage by the “admittance” Y by I = V Y = V(1/Z) Y is determined by the circuit parameters L, R, C, and is also dependent on the frequency of the driving force. We have learned that Y can be written as a complex number whose amplitude tells how large the current is for a given V and whose phase tells how much I(t) is shifted wrt V(t). (see previous notes and your class group exercise for derivation) The following page shows this graphically for a given circuit and for 3 different driving voltages. Response of a damped oscillator to a periodic driving force of arbitrary shape: If a system is a LINEAR system, it means that if several sinusoidal driving forces are added and applied at once, then the response is just the sum of the individual responses. It seems obvious that the circuit is linear, but there are many systems that are non-linear. The following page shows this graphically for the same circuit the driving voltage that is the sum of the three previous ones, and the resulting current which is also the sum of the three previous currents. No longer pure sinusoids! Notice that the shape of the current and the voltage are not the same anymore! It’s not true that there’s one simple scaling factor and one phase shift! V ( t ) = V0 cos (w 0t ) - V0 cos ( 2w 0t ) + V0 cos ( 3w 0t ) Y (w ) f I (w ) Vapp t V ( t ) = V0 cos (w 0t ) - V0 cos ( 2w 0t ) + V0 cos ( 3w 0t ) I ≠ Y Vapp Y (w ) f I (w ) t I ( t ) = V0 Y - V0 Y 2w 0 ( ) cos 2w 0t + f2w 0 + V0 Y 3w 0 w0 ( cos w 0t + fw 0 ( cos 3w 0t + f3w 0 ) ) Response of a damped oscillator to a periodic driving force of arbitrary shape: So now you can see how useful the admittance function is, and why it is important to know how its magnitude and phase shift vary with frequency: if there is a driving force that can be expressed as the sum of sinusoids, we simply use the admittance function to find the response at each of the driving frequencies, and add the responses to get the net response. It turns out that any periodic driving force can be decomposed into such a sum of sinusoidal components, and obviously it must be our job to learn how to break down such a periodic function into its component sinusoids. This technique is called Fourier analysis. What if V(t) is this? t V (t ) = 1 0 < t < 2p t p And periodic repetitions Use Fourier Analysis! Fourier Analysis • is easy • is a sensible thing to do • has a bad reputation (unjustly) • does not involve impossible integrals Before we tackle the problem we just posed, backtrack a bit to review some terminology involving Fourier series f t =1cost -cos2t +1cos3t 3 æ ç ç è ö ÷ ÷ ø æ ç ç è ö ÷ ÷ ø æ ç çç è ö ÷ ÷÷ ø æ ç çç è ö ÷ ÷÷ ø f(t) time f(t) time A Fourier Coefficients Height = amp position = freq “A” = code for cosine frequency g(t ) =1sin(t ) - sin(2t ) + 1 sin(3t ) 3 g(t) time g(t) B Fourier Coefficients time Height = amp position = freq “B” = code for sine frequency h(t ) = f (t) + g(t) h(t) time Fourier Spectrum A frequency B frequency Look at fundamental: Compute: 1cos (w t ) + 1sin (w t ) p æ -1ö 2 2 1 + 1 = 2;arctan ç ÷ = è 1ø 4 Aha! Fundamental also i (w t - p / 4 ) é ù Re ë 2e û Fourier Spectrum A frequency B frequency -ip /4 iw t i 2w t i 3w t é h(t ) = f (t) + g(t) = Re ë 2e e + ? e + ? e ùû h(t) time Fourier Spectrum A frequency B frequency What have we learned so far? • Odd periodic functions with period T=2π/w can be represented by a series of sin(nwt) functions • Even periodic functions with period T=2π/w can be represented by a series of cos(nwt) functions • If these functions represent physical motion, then we think of the motion as the superposition of the motions of SHOs with increasing frequencies (each a multiple of the fundamental one) • In these special odd/even cases the coefficient of the sin(nwt) (or cos) term represents the amplitude of that particular SHO • In these special odd/even cases, we can plot the size of a coefficient of the nwt term at the nwt frequency to give an alternative representation of the function What have we learned so far? • Periodic functions with period T=2π/w that are neither even nor odd can be represented by a series of sin(nwt) and cos(nwt) functions, all with different coefficients • If these functions represent physical motion, then we think of the motion as the superposition of the motions of SHOs with increasing frequencies (each a multiple of the fundamental one) • In these general cases the coefficients of the sin(nwt) cos(nwt) terms must be combined to represents the amplitude and phase of that particular SHO • In these general cases, we must make two plots to give an alternative representation of the function. We can chose to plot sin coeff and cos coeff, or magnitude and phase What have not learned yet? • How to find the coefficients if the function is not explicitly written in terms of sines and cosines (we will) • How to deal with exponential functions • Lots of practice needed, of course ….