[Martin Taylor 2014.06.01.19.18]
I guess you aren't understanding me! I was setting up an extreme
case in which moving the mouse would immediately influence the
cursor, but that the loop transport lag would be very long, so that
the feedback effect would be delayed by hours.
Although in the long term, control would be established, at least
for very slowly moving disturbances, any more rapid mouse changes
would be closely correlated with cursor changes.
Yes. That has been the result whenever the simple “leaky-integral”
control loop has been used to model simple human tracking such as
you asked about. It usually seems to be about 18 msec (at least for
Bill and me). Physically, every connection in the loop MUST involve
some transport lag, but that’s probably not where most of the lag
occurs. It’s probably mostly processing time. The leaky integral
also imposes its own delay, but that’s not included in the transport
lag. The leaky integral just determines how fast the effect of a
change dies away. Transport lag determines how long it is before the
effect of the change begins to dies away.
That’s not right. There’s no such definition, and C must correlate
with both. Here’s a geometrical way of looking at why this is. If you are
visually inclined, as I am, this way of looking at it makes for much
clarity. If you aren’t, I guess it won’t help. But it’s worth a try.
If X = Y+Z, where X, Y, and Z are sequences of samples of
variables, X will be correlated with both Y and Z unless Y = -Z + k
(k is a constant), in which case X = k is constant. The degree of
correlation of X with each of the other two depends on their
relative range of variation and the correlation between them. For
example, if Y and Z are uncorrelated and have equal range of
variation, X is correlated 0.707 = 1/sqrt(2) with each.
-------- more detail ------
More generally, any sequence of N values identifies a point in
N-dimensional space, or a vector from zero to the point. Two such
sequences represent two vectors. Since any three points in space
define a plane, these two vectors lie in a plane. By doing the
algebra, you can prove that the cosine of the angle between them is
the correlation between the two sequences.
The difference between two vectors defines another vector that is
also in the same plane. So, if X = Y-Z, X is a vector that connects
the points defined by Y and Z. The points defined by Y, Z, and zero
are the corners of a triangle, whose sides are the vectors X, Y, and
Z. Hence, if one knows the lengths of Y and Z (square root of the
sum of squares of the sequence values y1, y2 … yN, and similarly
for Z) together with the correlation between them, one can find the
length of X and its correlation with Y and with Z using the formulae
(from Wikipedia "Solution of Triangles):
Wiki quote**
end Wiki quote*
In correlational terms using X, Y, Z as above, and taking “x”, “y”,
and “z” to be their lengths (or total energies if X, Y, and Z
represent signal values):
x = sqrt(y^2 + z^2 - 2yzcorr(y,z))
corr(x,z) = (x^2+z^2-y^2)/2(xz)
corr(x,y) = (x^2+y^2-z^2)/2(xy)
You can scale the lengths by dividing by the number of samples if
you want to use sequences extended indefinitely over time (using
power instead of total energy).
If Y and Z are uncorrelated and of the same length, the triangle is
a right-isosceles triangle. Scaling the lengths of Y and Z so that y
= z = 1, corr(x,z) = (2+1-1)/2(sqrt(2)) = 1/sqrt(2) = 0.707
-----------end extra detail--------
Now consider the control situation about which you asked. C = M+D corresponds to X = Y+Z above, but for the vectorial analysis
the triangle was described with X = Y-Z, so let’s put V = -D and say
C = M-V.
If control is very good, M and D have almost the same length and are
correlated nearly -1, so M and V are correlated nearly +1. Putting
these into the correlation formula in the “extra detail” section
above, we get
C = sqrt(M^2+V^2 - 2MV(1-eps)) where eps is a positive number much
less than 1.
Since M and V are almost the same length, C ~ Msqrt(2eps), which
is very small compared to M (or D), where ~ means “almost equal to”.
corr(C,M) = (C^2 + M^2 - V^2)/2(CM)
Since M and V are almost the same,
corr(C,M) ~ -corr(C,D) ~ C^2/2(CM) = C/2M = sqrt(eps/2)
For the trial in the diagram you showed in [From Rupert Young
(2014.05.31 17.00)], eps = 0.03, which gives corr
corr(C,M) = -corr(C,D) ~ 0.12
corr(C,D) is shown as -0.16, which, given the single-digit accuracy
of eps, is quite close.
But corr(C,M) is shown as 0.38, which is too big. Why is this?
The answer is implicit in the actual traces displayed. Sqrt(eps/2)
is a minimum value, which assumes that there is no autonomous
(noise) mouse motion. But in the traces shown, there is a lot of
variation in M that has no counterpart in D, and that variation,
which is reflected in C, is what caused you to ask your question in
the first place. This extra variation is additional to the variation
that opposes the disturbance, which means M^2 > D^2 (or V^2).
Looking back at the correlation formula, corr(C,M) was taken to be
nearly (C^2)/2(C*M), but that was under the assumption that M^2 =
D^2. Since M^2 > V^2, the difference is an added term that
increases corr(C,M).
The added variation in the displayed traces is clearly not only a
consequence of transport lag or of the delay induced by integration.
It’s too irregular for that, but they do have an effect that is
included in the correlation between cursor and mouse. As I said
earlier [Martin Taylor 2014.05.31.17.09]: “…rapid disturbance
changes are less well controlled than slow ones, and … more of the
cursor signal is due to the mouse at high frequencies than at low.
So, in the correlations, you see the effect of cursor changes more
than of cursor values.”
Martin
···
I'm not sure I am understanding you. You'd see the cursor moving,
but moving the mouse would have no effect.
Take the other extreme, with a loop transport lag
of zero. In this case, an arbitrary movement of the mouse would,
like a change in the disturbance, be immediately compensated,
and in fact you wouldn’t see it at all, except for the fact that
integration takes time (the output stage is usually taken to be
an integrator).
In the real case, there is a finite loop transport lag. In many
of Bill’s and my experiments, our results were best fitted by
loop transport lags of around 18 ms. This means that changes in
the mouse are indeed reflected momentarily (and it’s a short
moment) in the movement of the cursor, so that if you
differentiate the cursor trace and the mouse trace, they will
correlate more than will the simple values.
Do you mean there is a lag due to the processing time of the
system?
Or due to the non-zero resolution of the visual
system? Are you saying that there is a correlation?
Perhaps there is a simple answer to my question, in that C = f(M)
- f(D), so by definition C does not correlate with f(M)?
Two sides
and the included angle given (SAS)
Here the lengths of sides and the angle between these
sides are known. The third side can be determined from the law of
cosines:
Now we use law of cosines to find the second angle:
Finally,
[From Rupert Young (2014.06.01
17.00)]
([Martin Taylor 2014.05.31.17.09)
Rupert, imagine an extreme case. Suppose that the loop transport
lag were measured in hours rather than milliseconds. The cursor
position is the sum of the disturbance and the mouse. If, in
this extreme case, the mouse were to move 1 mm, no matter what
the disturbance did, the cursor would move 1 mm more in the
mouse direction than would be caused by the disturbance
movement, and that extra movement would not be compensated until
hours later. You would see a correlation between rapid (quicker
than hours-long) movements of the mouse and movements of the
cursor, regardless of what the disturbance does.
[From Rupert Young (2014.05.31 17.00)]
Hi Rick,
Looking at the tracking task I notice that although the
correlation C-M is poor it does look if the cursor trace goes
up and down when does the mouse. Has the correlation between
the change in the mouse value and the change in the cursor
value been computed? In other words, is there a good
correlation between the change in output and the change in
input? What would be the significance if this were the case?
