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..

> but my mind has gone blank!

Maybe think about the problem a little longer. Then calculate the scalar
product again, and the answer should be obvious.

PS: How long is "a little bit longer"? Say, one sample duration :-) 

On 8/13/2012 1:03 AM, g...@gmail.com wrote:
> 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..
>

It's all in "k" and/or how you handle / express that.

Fred

On Tuesday, August 14, 2012 4:32:01 AM UTC+12, Fred Marshall wrote:
> On 8/13/2012 1:03 AM, g...@gmail.com wrote:
> 
> > 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..
> 
> >
> 
> 
> 
> It's all in "k" and/or how you handle / express that.
> 
> 
> 
> Fred

Yes, of course, it's for k=0,1,2 etc so the same thing

On 8/13/2012 12:03 PM, g...@gmail.com wrote:
> On Tuesday, August 14, 2012 4:32:01 AM UTC+12, Fred Marshall wrote:
>> On 8/13/2012 1:03 AM, g...@gmail.com wrote:
>>
>>> 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..
>>
>>>
>>
>>
>>
>> It's all in "k" and/or how you handle / express that.
>>
>>
>>
>> Fred
>
> Yes, of course, it's for k=0,1,2 etc so the same thing

I'm sorry, I don't understand what you mean by "so the same thing".
Does that mean you now have an answer to your question or you think the 
question is clear on this point and there has been no answer?

Fred

On Monday, August 13, 2012 8:03:57 PM UTC+12, gyans...@gmail.com wrote:
> 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..

I assumed you were saying that I need to take k=0,1,2,3 and hence have a vector solution and not
just a scalar.

g...@gmail.com writes:

> but my mind has gone blank!
>
> The output of an FIR filter is say s(k) = X'W(k)

You're really being quite sloppy here right at the start. Hint: rethink
where you are putting your time index k.

Also, strictly speaking, that is a dot product, or inner product, not an
FIR output.

> 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 

We do? Write the definition and think carefully about boundary
conditions.
-- 
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com

On Aug 13, 4:03 am, gyansor...@gmail.com wrote:
> 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..

k goes from 0 to n+m-2, or from 1 to n+m-1

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.

Cheers,
Dave

On 8/13/2012 6:27 PM, g...@gmail.com wrote:
> On Monday, August 13, 2012 8:03:57 PM UTC+12, gyans...@gmail.com wrote:
>> 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..
>
> I assumed you were saying that I need to take k=0,1,2,3 and hence have a vector solution
and not just a scalar.
>

If you're responding to me then that *is* what I was suggesting.  Yes.
There's really no other choice is there?

Fred

Dave <d...@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 Labs
http://www.digitalsignallabs.com