derivative (chain rule) in DFT?

Hi,

I have a DTFT function Y(w) which is also a function of a: Y(w,a) (it's
actually a sum with a as coefficient)

Another function Yl(w) is the l-fold convolution of Y:

Yl(omega) = Y(w) * ... * Y(w)
             \             /
                 l-times

Now I want to take the derivative wrt. c, so I do:

dYl(w)/dc = dYl(w)/dY(w) * dY(w)/dc

Y and Yl are continuous functions, so the derivative operator is well
defined (given it exists, of course!)

However, now I have an N-periodic signal and therefore want to use the DFT.

The equation then becomes (w=pi2k/N):

dYr[k]/dc = dYr[k]/dY[k] * dY[k]/dc

While "dY[k]/dc" is trivial, I don't know how to deal with
"dYr[k]/dY[k]" - obviously this does not really make sense.

But there must be a way to generalize this also for the DFT!

Thanks
Peter

Hi,

I am still in big hope that someone has an idea about this.
Additional comments:

First: In my post below I used both "a" and "c" for the coefficient.
That's a mistake, there is just one coefficient (i.e., "a" and "c" are
the same).

Second: Meanwhile I tried two approaches.

1.)

dYr[k]/dY[k] = dYr[k]/dk * (dY[k]/dk)^-1
             = DFT{-j*n*yr[n])}/DFT{-j*n*y[n]}
             = DFT{n*yr[n])}/DFT{n*y[n]}

What's problematic here is that I just apply the differentiation as if
"k" would be a continous variable, i.e. I do as if I would use the DTFT.
[1] suggests this should work.

On the other hand, if I pretend that y[n] and yr[n] are infinite but
periodic signals and I write it using the DTFT and do the math I get

               Sum_{-infty}^infty -j n y[n] e^(-jwn)

which is not periodic any more, so I can't go back to DFT.

2.) Based on the first line I do the approximation

dY[k]/dk \approx (Y[k+1] - Y[k])/(k+1-k) = Y[k+1] - Y[k]

Then the result is (differently from above):

dYr[k]/dY[k] \approx (Yr[k+1] - Yr[k])/(Y[k+1] - Y[k])


Unfortunately both approaches do not get me to the expected result and
from these two (2) even less.

Thanks,
Peter


[1]
http://www.tricki.org/article/Use_Fourier_transforms_to_calculate_derivatives

On 2014-07-08 23:35, Peter Mairhofer wrote:
> I have a DTFT function Y(w) which is also a function of a: Y(w,a) (it's
> actually a sum with a as coefficient)
> 
> Another function Yl(w) is the l-fold convolution of Y:
> 
> Yl(omega) = Y(w) * ... * Y(w)
>              \             /
>                  l-times
> 
> Now I want to take the derivative wrt. c, so I do:
> 
> dYl(w)/dc = dYl(w)/dY(w) * dY(w)/dc
> 
> Y and Yl are continuous functions, so the derivative operator is well
> defined (given it exists, of course!)
> 
> However, now I have an N-periodic signal and therefore want to use the DFT.
> 
> The equation then becomes (w=pi2k/N):
> 
> dYr[k]/dc = dYr[k]/dY[k] * dY[k]/dc
> 
> While "dY[k]/dc" is trivial, I don't know how to deal with
> "dYr[k]/dY[k]" - obviously this does not really make sense.
> 
> But there must be a way to generalize this also for the DFT!
> 
> Thanks
> Peter
> 

On Tue, 08 Jul 2014 23:35:03 -0700, Peter Mairhofer wrote:

> Hi,
> 
> I have a DTFT function Y(w) which is also a function of a: Y(w,a) (it's
> actually a sum with a as coefficient)
> 
> Another function Yl(w) is the l-fold convolution of Y:
> 
> Yl(omega) = Y(w) * ... * Y(w)
>              \             /
>                  l-times
> 
> Now I want to take the derivative wrt. c, so I do:
> 
> dYl(w)/dc = dYl(w)/dY(w) * dY(w)/dc
> 
> Y and Yl are continuous functions, so the derivative operator is well
> defined (given it exists, of course!)

Actually, I don't think that what you're trying to do is "fair".  Yl(c, 
w) is not a function of Y(c, w) at w -- it is a function of Y(c, w) over 
all w.

I think you'd be much better served by writing out the convolution 
integral for Yl, then investigating the effects of taking the derivative 
w.r.t c.  It may be a "you can't get there from here" thing.

> However, now I have an N-periodic signal and therefore want to use the
> DFT.
> 
> The equation then becomes (w=pi2k/N):
> 
> dYr[k]/dc = dYr[k]/dY[k] * dY[k]/dc
> 
> While "dY[k]/dc" is trivial, I don't know how to deal with
> "dYr[k]/dY[k]" - obviously this does not really make sense.
> 
> But there must be a way to generalize this also for the DFT!

Again, your Yr(c, k) isn't a function of just Y(c, k) at k -- it's for 
all k.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On 2014-07-09 16:34, Tim Wescott wrote:
> On Tue, 08 Jul 2014 23:35:03 -0700, Peter Mairhofer wrote:
> [...]
> Again, your Yr(c, k) isn't a function of just Y(c, k) at k -- it's for 
> all k.

Hi Tim,

Really great, thank you! You fixed it. I worked it out using the
convolution integral and everything makes sense.

Regards,
Peter