The diagram shows a relaxation oscillator. The charge q on the capacitor builds up until the...

 

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The diagram shows a "relaxation" oscillator. The charge q on the capacitor builds up until the neon tube fires and discharges the capacitor (we assume instantaneously). Then the cycle repeats itself over and over. (a) The charge q on the capacitor satisfies the differential equation R + 9 = V₁ dq dt where R is the resistance, C is the ca- pacitance, and V is the constant d-c R 9 Neon tube voltage, as shown in the diagram. Show that if q = 0 when t = 0, then at any later time t (during one cycle, that is, before the neon tube fires) q=CV (1-e-t/RC). (b) Suppose the neon tube fires at t = RC. Sketch q as a function of t for several cycles. (c) Expand the periodic q in part (b) in an appropriate Fourier series. The diagram shows a "relaxation" oscillator. The charge q on the capacitor builds up until the neon tube fires and discharges the capacitor (we assume instantaneously). Then the cycle repeats itself over and over. (a) The charge q on the capacitor satisfies the differential equation R + 9 = V₁ dq dt where R is the resistance, C is the ca- pacitance, and V is the constant d-c R 9 Neon tube voltage, as shown in the diagram. Show that if q = 0 when t = 0, then at any later time t (during one cycle, that is, before the neon tube fires) q=CV (1-e-t/RC). (b) Suppose the neon tube fires at t = RC. Sketch q as a function of t for several cycles. (c) Expand the periodic q in part (b) in an appropriate Fourier series.

Answer rating: 100% (QA)

What you have here is a problem involving a relaxation oscillator circuit which makes use of a resistor a capacitor and a neon tube that fires when a certain voltage across it is reached instantaneously discharging the capacitor Lets address each part of the question in steps a Deriving the charge equation on the capacitor To derive the given equation we need to solve the provided differential equation The differential equation relating charge qt on the capacitor with time is Rdqdt qC V Notice that it is a firstorder linear differential equation which can be solved using the integrating factor method or by separation of variables Lets go with separation of variables 1 Separate variables dqCV q dtRC 2 Integrate both sides 1CV q dq 1RC dt Lets make a substitution to simplify the integral Let u CV q Hence du dq Now the integral becomes 1u du 1RC dt The integral of 1u is lnu so we get lnCV q tRC C1 where C1 is the constant of integration Multiply by 1 to get lnCV q tRC C1 3 Remove the absolute value and solve for q elnCV q etRC C1 CV q eC1etRC We can write eC1 as another constant say C2 eC1 q CV C2etRC 4 Apply the initial condition When t 0 q 0 so 0 CV C2e0 C2 CV 5 Insert C2 into the equation q CV CVetRC q CV1 etRC So we have derived the given equation b Sketching q as a function of t for several cycles Now for the neon tube firing at t 12 RC we would sketch the graph of qt knowing that it increases exponentially from zero to CV as given by the equation with a time constant RC until reaching the firing voltage at t 12 RC At this point the charge on the capacitor would drop instantly to zero as the neon tube fires and discharges the capacitor This cycle would then repeat itself Heres a qualitative description of the graph Starting at q 0 when t 0 Increases exponentially towards CV as given by the equation q CV1 etRC reaching close to CV around t RC since e1 is roughly 037 meaning that the charge has reached about 63 of CV at t RC At t 12 RC q rapidly drops to zero as the neon bulb fires and the cycle starts again c Expanding the periodic q in part b into a Fourier series Fourier series are used to represent a periodic function as a sum of simple sine and cosine functions Since the charging part of the qt function is exponential and not a standard sinusoidal function you would first need to determine the period of the function which would be related to the firing time of the neon bulb and how it repeats over time Then you have to integrate over one period of the function to determine the Fourier coefficients Without actually doing the math because there is no explicit function to integrate over multiple periods heres a qualitative explanation The Fourier series would consist of a fundamental frequency corresponding to the period of the oscillation and harmonics of that frequency The coefficients of the series would be determined by how closely the waveforms of sine and cosine can approximate the piecewise linear instantaneous ... View the full answer