A daft question

>but my mind has gone blank!
>
>The output of an FIR filter is say s(k) = X'W(k)
>
>where X is the input vector and W(k) the weight vector. This is
convolving the two vectors X and W to give a scalar s(k) ie the dot product
of two vectors.
>
>However, we know that convolving a vector of dimension n with another one
of dimension m gives an output like multiplying two polynomials and is of
dimension n+m-1 so where has the other terms gone..
>

If I understood you, it could be you are comparing convolution of two
finite streams with that of filtering(coefficients stream and data
stream). It is true that convolving two finite streams of length n & m 
results in vector of length n+m-1.
However in the case of filtering, the convolution is between the finite
coefficient set and running data. The output rises up from zero state and
indeed if you stop your data then you do get the m+n-1 length as it tails
back towards zero. Though some filter functions may ignore the tail
samples that head towards zero.

Kadhiem 

On Aug 14, 2:11=A0pm, Randy Yates <ya...@digitalsignallabs.com> wrote:
> Dave <dspg...@netscape.net> writes:
> > So to get the convolution you need to think about how the content of
> > the vector W(k) changes at each value of k i.e. you actually have n
> > +m-1 different W vectors.
>
> Why would the weight vector change at all?
> --
> Randy Yates
> Digital Signal Labshttp://www.digitalsignallabs.com

From the notation in the original post - the X vector is the input
data vector - since it isn't given an index I'm assuming it is
constant.

As for W(k), if it doesn't vary with the index (k) then it would be
constant too - meaning s(k) doesn't depend on k at all and the output
would be constant for all k.

Trying working the equation with a simple input data vector, and for a
filter just use a simple impulse. Then to get the output you need W(0)
=3D [1 0 0 0 0 ...]' then for W(1)=3D[ 0 1 0 0 0 ...]' and so on.

Cheers,
Dave

Dave <dspguy2@netscape.net> writes:

> On Aug 14, 2:11&nbsp;pm, Randy Yates <ya...@digitalsignallabs.com> wrote:
>> Dave <dspg...@netscape.net> writes:
>> > So to get the convolution you need to think about how the content of
>> > the vector W(k) changes at each value of k i.e. you actually have n
>> > +m-1 different W vectors.
>>
>> Why would the weight vector change at all?
>> --
>> Randy Yates
>> Digital Signal Labshttp://www.digitalsignallabs.com
>
> From the notation in the original post - the X vector is the input
> data vector - since it isn't given an index I'm assuming it is
> constant.
>
> As for W(k), if it doesn't vary with the index (k) then it would be
> constant too - meaning s(k) doesn't depend on k at all and the output
> would be constant for all k.
>
> Trying working the equation with a simple input data vector, and for a
> filter just use a simple impulse. Then to get the output you need W(0)
> = [1 0 0 0 0 ...]' then for W(1)=[ 0 1 0 0 0 ...]' and so on.

Typically the weight vector is fixed and the data "flows through" the
input vector:

  w = [w_0 w_1 w2 ... w_(N - 1)]

  x(k) = [x[k] x[k - 1] x[k - 2] ... x[k - N + 1]]

so that 

  y(k) = x(k).w
       = x'(k) * w, x real,

where "." is the dot product operator and * is standard matrix multiplication.
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

Randy Yates <yates@digitalsignallabs.com> writes:

> Dave <dspguy2@netscape.net> writes:
>
>> On Aug 14, 2:11&nbsp;pm, Randy Yates <ya...@digitalsignallabs.com> wrote:
>>> Dave <dspg...@netscape.net> writes:
>>> > So to get the convolution you need to think about how the content of
>>> > the vector W(k) changes at each value of k i.e. you actually have n
>>> > +m-1 different W vectors.
>>>
>>> Why would the weight vector change at all?
>>> --
>>> Randy Yates
>>> Digital Signal Labshttp://www.digitalsignallabs.com
>>
>> From the notation in the original post - the X vector is the input
>> data vector - since it isn't given an index I'm assuming it is
>> constant.
>>
>> As for W(k), if it doesn't vary with the index (k) then it would be
>> constant too - meaning s(k) doesn't depend on k at all and the output
>> would be constant for all k.
>>
>> Trying working the equation with a simple input data vector, and for a
>> filter just use a simple impulse. Then to get the output you need W(0)
>> = [1 0 0 0 0 ...]' then for W(1)=[ 0 1 0 0 0 ...]' and so on.
>
> Typically the weight vector is fixed and the data "flows through" the
> input vector:
>
>   w = [w_0 w_1 w2 ... w_(N - 1)]

Correction: 

  w = [w_0 w_1 w2 ... w_(N - 1)]'

>
>   x(k) = [x[k] x[k - 1] x[k - 2] ... x[k - N + 1]]

Correction:

  x(k) = [x[k] x[k - 1] x[k - 2] ... x[k - N + 1]]'

> so that 
>
>   y(k) = x(k).w
>        = x'(k) * w, x real,
>
> where "." is the dot product operator and * is standard matrix
> multiplication.

And where "'" denotes matrix transpose. 
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On Aug 15, 3:35=A0pm, Randy Yates <ya...@digitalsignallabs.com> wrote:
> Dave <dspg...@netscape.net> writes:
> > On Aug 14, 2:11=A0pm, Randy Yates <ya...@digitalsignallabs.com> wrote:
> >> Dave <dspg...@netscape.net> writes:
> >> > So to get the convolution you need to think about how the content of
> >> > the vector W(k) changes at each value of k i.e. you actually have n
> >> > +m-1 different W vectors.
>
> >> Why would the weight vector change at all?
> >> --
> >> Randy Yates
> >> Digital Signal Labshttp://www.digitalsignallabs.com
>
> > From the notation in the original post - the X vector is the input
> > data vector - since it isn't given an index I'm assuming it is
> > constant.
>
> > As for W(k), if it doesn't vary with the index (k) then it would be
> > constant too - meaning s(k) doesn't depend on k at all and the output
> > would be constant for all k.
>
> > Trying working the equation with a simple input data vector, and for a
> > filter just use a simple impulse. Then to get the output you need W(0)
> > =3D [1 0 0 0 0 ...]' then for W(1)=3D[ 0 1 0 0 0 ...]' and so on.
>
> Typically the weight vector is fixed and the data "flows through" the
> input vector:
>
> =A0 w =3D [w_0 w_1 w2 ... w_(N - 1)]
>
> =A0 x(k) =3D [x[k] x[k - 1] x[k - 2] ... x[k - N + 1]]
>
> so that
>
> =A0 y(k) =3D x(k).w
> =A0 =A0 =A0 =A0=3D x'(k) * w, x real,
>
> where "." is the dot product operator and * is standard matrix multiplica=
tion.

Randy, Yes I've normally seen the data vector indexed, but that's not
the way notation used in the OP. So, my explanation with given based
on what the OP gave me.

Dave