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To model this pattern, we assumed that if the graph is split up in pieces with the peaks and valleys as cutting points, each piece will be sinusoidal with a random period and amplitude. Furthermore, we made the assumption that the period is normally distributed with mean 90 minutes or 5400 seconds (mean duration of one sleep cycle) and standard deviation 30 minutes or 1800 seconds. At first, we assumed that the amplitude is uniformly distributed, but people sleep deeper just after going to sleep and lighter just before waking up, so we had to come up with our own probability distribution. We decided to use the following probability density function: | To model this pattern, we assumed that if the graph is split up in pieces with the peaks and valleys as cutting points, each piece will be sinusoidal with a random period and amplitude. Furthermore, we made the assumption that the period is normally distributed with mean 90 minutes or 5400 seconds (mean duration of one sleep cycle) and standard deviation 30 minutes or 1800 seconds. At first, we assumed that the amplitude is uniformly distributed, but people sleep deeper just after going to sleep and lighter just before waking up, so we had to come up with our own probability distribution. We decided to use the following probability density function: | ||
<math>y=r \left( x - \frac{1}{2} \right) + | <math>y=r \left( x - \frac{1}{2} \right) + 1 \qquad 0 \le x \le 1 \quad \and \quad -2 \le r \le 2</math> | ||
globals [y | globals [y |
Revision as of 18:00, 28 February 2016
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Brightness
Temperature
Sleep model
The Sleep Cycle App provides graphs to the user that show his measured sleep behavior during that night. Here's an example:
To model this pattern, we assumed that if the graph is split up in pieces with the peaks and valleys as cutting points, each piece will be sinusoidal with a random period and amplitude. Furthermore, we made the assumption that the period is normally distributed with mean 90 minutes or 5400 seconds (mean duration of one sleep cycle) and standard deviation 30 minutes or 1800 seconds. At first, we assumed that the amplitude is uniformly distributed, but people sleep deeper just after going to sleep and lighter just before waking up, so we had to come up with our own probability distribution. We decided to use the following probability density function:
[math]\displaystyle{ y=r \left( x - \frac{1}{2} \right) + 1 \qquad 0 \le x \le 1 \quad \and \quad -2 \le r \le 2 }[/math]
globals [y Tgem start T p rc x R time stage test] to Setup clear-all set y 100 set Tgem (90 * 60) reset-ticks end to Go foreach [1 2] [ set start y set T -1 while [T < 0] [set T (random-normal Tgem 1800)] set p (random-float 1.0) set rc (4 / (1 + exp((ticks - (2.5 * Tgem)) / 20000)) - 2) ifelse rc > 0 [set x (0.5 - (1 / rc) + sqrt((2.0 / rc) * p + (rc ^ -2) - (rc ^ -1) + 0.25))] [set x (0.5 - (1 / rc) - sqrt((2.0 / rc) * p + (rc ^ -2) - (rc ^ -1) + 0.25))] ifelse ? = 1 [set R (x * start / 2)] [set R ((x - 1) * (100 - start) / 2)] set time (n-values (T / 2) [?]) foreach time [ set y (R * (cos (360 * ? / T)) + start - R) set stage ((y - (y mod 25)) / 25 - 3) set test (1 / (1 + exp(ticks))) tick ] ] end