NN toolbox, Timedelay.net : removedelay?

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Edo
Edo on 12 Oct 2012
Commented: WT on 15 Dec 2014
Dear all,
I have an issue with the time-series prediction neural networks: I have a set of inputs, and I want to predict a related output at time (t+2). (I also have the data of my y up to time t, that's why I was using Narxnet, but apparently, probably for some errors in my network, it looks like it is taking into account only y(t) almost ignoring all the other inputs x(t) which I know for sure they would help improving the performance..so now I turned to timedelay so that it can't use the previous y).
I don't understand how the early prediction part works (with the command removedelay): do i get the same result fixing inputDelays = 1:d; ? I did some trials using y as my input, and apparently, I can guess that if: inputDelays = 0:d; then it is predicting y(t), having x(t)..x(t-d), in fact I got a pefect prediction with R=1, because i tried it y(t)as my input too(so it was actually doing y(t)=y(t)..y(t-d))
if: inputDelays = 1:d; so then I have my prediction one step ahead. I can notice it as the chart looks a bit shifted of one time step and R goes down to 0.8. So what's the point of the early prediction? Isn't it doing exactly the same thing of this?
To sum up, if I want my prediction 2 time steps ahead, can I simply use inputDelays = 2:d; ?
Thanks everybody for the support.
  1 Comment
WT
WT on 15 Dec 2014
Hi Edo,
May i know whether you have managed to obtain the predicted output for (t+2) cause i am still unable to predict the future outputs..I would appreciate if you could help me on this.
Thank you

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Accepted Answer

Greg Heath
Greg Heath on 15 Dec 2014
If dmin > = 0 then
net = timedelaynet( dmin:dmax)
assumes
y(t) = f(x(t-dmin),...,x(t-dmax))
or equivalently
y(t+dmin) = f(x(t),...,x(t-(dmax-dmin))
if dmin >= n >= 0
netr = removedelay(net,n);
causes
y(t) = f(x(t-(dmin-n)),...,x(t-(dmax-n)))
NOTE: Although
dmin >= 0 (NONNEGATIVE) for input delays
for feedback delays,
dmin > 0 % POSITIVE!
Hope this helps.
Thank you for formally accepting my answer
Greg

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