How to forecast y(t+N) using removedelay function for NARX model

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Leon
Leon 2014년 9월 4일
답변: Greg Heath 2014년 9월 7일
In the Neural Network Toolbox there is a NARX model. After the model is created the model can be adjusted by “close” or “removedelay”.
When using “removedelay” there will be an one step ahead forecast: u(t) and y(t) resulting in y(t+1). So in principle y(t+1) is a forecast of one step ahead of the data. And this works well. However I would like to make a N step ahead forecast, y(t+N).
With the “removedelay” function there is a parameter to define N steps but when I define a number bigger than one I get an error message.
How can this be solve? What is the correct way to come to N-steps forecast for y(t+N)?

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Greg Heath
Greg Heath 2014년 9월 7일
There seems to be a misconception:
An openloop feedback design is not deployable because it depends on target feedback.
Therefore, it is only a step in one way to obtain a closeloop design (The other way of course, is to use much more training time and design a closeloop design from scratch).
Consequently, the option close OR remove delay makes no sense unless it means to either
a. use remove delay before closing the loop.
b. use remove delay after closing the loop
Either way the end result is the same.
Removedelay(net,M) subtracts M from every component in ID and FD. Therefore the smallest components that can exist in ID and FD are M and M+1 ,respectively.
Hope this helps.
Thank you for formally accepting my answer
Greg

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