RE: PCT Modeling - changing signals
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----- Original Message -----
From: isaac
To: Control Systems Group Network (CSGnet)
Sent: Friday, January 05, 2001 11:20 AM
Subject: Re: Re: PCT Modeling - changing signals
i.kurtzer (2000.01.05.1000)
I'm writing this for Rupert as well as other other people that might need
some brushing up in Calculus or have not yet been exposed to it. Rupert has
a series of timemarked snapshots of some value. You could imagine that
these are the series of a reference value or a ball's position after being
thrown. There are two things. The value and the timemark. We can see that
the values are not constant for all the timemarks. If they were then
plotting this would yield a flat line, i.e. parallel with the X-axis.
Instead, at later and later timemarks the values are larger and larger.
They change and in a particular direction, increasing. However, we can also
see that during some intervals the values have changed by a greater amount
than in other intervals. For example, in the first time interval
(10.1-10.3) the value changed by .2. Therefore, the rate is the change in
the value divided by the change in the timemark, or .2/.2 = 1. The slope at
this point would be 1 since the value is increasing in equal amounts to the
time. Instead, in the interval (10.9-11) the values changes by .5. This is
a change of .5/.1, or 5 so the slope at this point would be 5 since the
value increases by five for each unit increase in time.
So we can see there is a change in the slope over time as well a change in
value. We can therefore ask what is the slope of the change in slope. Each
time we ask what is the slope of some function when the time intervals
approaches the limit of an instantaneous change then we perform a
derivative. Here the velocity would be the first derivative and the
acceleration, or change in velocity, would be the second derivative.
So lets look the same time intervals again and see what is their second
derivative, or rate of change of rate of change. In the interval in the
(10.1-10.3) the slope was 1, in the interval just ahead of that (10.3-10.5)
the slope was also 1. So the change in the slope between those intervals is
zero. For the interval (10.9-11) the slope was 5 while the following
interval (11-11.2) has a slope of 7.5. So the change in the slope is
(7.5-5)/(11-10.9) = 25. Notice how the slopes were across the intervals
(10.9-11.2) but the time inteval in question is (10.9-11). We can see the
result of this moving down the chart. We can see a sucession of slopes
(D')of 1 and then the next is 5. Right of that (D'') is a sucession of 0's
then a 10 DURING a slope of 1. That is because derivatives are phased
advanced from their input. That probably enough for now.
To your original question to get the average velocity over some time
interval you ONLY use the change in values over the change in the time
interval. This is because here the values are supposed to be continuous.
In this case the answer is (27-21.6)/(11.5-10.5) or 5.4. Also, the average
velocity approaches the instantaneous velocity as the interval becomes
shorter and shorter.
For anyone interested I recommend Finney and Thomas's "Calculus" or anything
you can get your hands on. When receiving my undergrad degree in
psychology, basic math was not emphasized. Rather I enjoyed a steady diet
in courses that self-justified themselves as science. Notice that ONLY
psychology and its allied garbage fields have Experiemental Methods courses.
None in Chemistry, Biology, Geology, or Physics. Mmm. I am still
suffering from this neglect but the return I've seen so far is worth it.
And anyone that can show that they have improved their math-lot in the past
year can get a kudos and a beer from me at the next conference.
i.
----- Original Message -----
From: Rupert Young
To: CSGNET@POSTOFFICE.CSO.UIUC.EDU
Sent: Friday, January 05, 2001 7:45 AM
Subject: Re: PCT Modeling - changing signals
[From Rupert Young (2001.01.05.1300 UT)]
i.kurtzer (2001.01.03.1530)
A derivative is the instaneous rate of change of a function. For a short
time interval this is the rise over run, or change in y over change in x.
With your data you would apply the formula (Yi+1-Yi)/(Xi+1-Xi). If the time
sample is larger then the slope is the average rate of change over that
interval, although the instanteous rate of change may have only occured at
that value once, but it HAS to at least once. I typed in the answers in
your post.
Thanks.
In the table below I can see you got the 7.5 in V' from,
24 - 22.5
--------- = 7.5
11.2 - 11
but how did you get the 12.5 in V'' ?
Also to get the velocity over the period 10.5 to 11.5 do I just take the
average of the V' values, ie. (1 + 5 + 7.5 + 10.5)/4 = 6 ?
Time (mins) Value V' V''
10.1 21.2 1 0
10.3 21.4 1 0
10.5 21.6 1 10
10.9 22.0 5 25
11.0 22.5 7.5 12.5
11.2 24.0 10
11.5 27.0
Cheers,
Rup