Then both of the above problems can be solved in one

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, then both of the above problems can be solved in one single command: >> x =(0: s i z e ( y ) - 1)/ S ; 2 Underdamped The underdamped plot is a bit more complex than previous weeks. We’re trying to model the data with the following equation: V ( t ) = V 0 e - t / 2 τ sin ωt τ = L R ω = r 1 LC - R 2 4 L 2 >> modelUnder = @ ( p , x ) p (1) . exp ( - x ./(2 p (2) ) ) . s i n ( p (3) . x + p (4) ) + p (5) ; 1

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p(1): V 0 ; notice this is not simply V ( t = 0) = 0. You may guess the value of your first peak, but this isn’t quite right either (although it should be close enough). p(2): τ , it can be estimated as in labs 2 and 4 by looking at the peaks in your data as an exponential decay (note that the actual decay time is 2 τ , but this factor should be irrelevant for your guessing). p(3): ω (= 2 π T ) where T is the period. The period is easily estimated as the time difference between consecutive peaks in your data. p(4): This is the phase φ , it is approximately zero if you selected your data to have (0,0) as the first data point. p(5): A vertical shift in the data that is approximately zero. 3 Overdamped This is almost exactly the same as the previous model, except the hyperbolic sine is used. There is not simply one set of criteria to use for guessing the following, it depends on where you ran the selection over the data. Let’s narrow it to two cases:
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