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Write two vectors as convolution

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0

How to write two column vectors as an analytic convolution so that the discrete FFT may be used. MATLAB syntax is used.

Consider:

1. 
   

   a set of vectors which, when sorted into a step function appears as any of the following:

   
   
   

   [1,1,1,1,0,0,0,0], or [1,1,1,1,1,0,0,0], or [1,1,1,1,1,1,0,0]

   
   
   

   (...the location at which the function "steps up" varies over members of this set)

   
   



2. The other is random vec=[1,0,1,0,1,1,1,0], and obviously both contain only 0s and 1s.

Is it possible to write these vectors as an analytic convolution? I would like the 1st, 2nd, 3rd, 4th... entries of the convolution to have values of:

sum(vec.*[1,0,0,0,0,0,0,0])
sum(vec.*[1,1,0,0,0,0,0,0])
sum(vec.*[1,1,1,0,0,0,0,0])
sum(vec.*[1,1,1,1,0,0,0,0])
... 
sum(vec.*[1,1,1,1,1,1,1,1])

For speed, I am trying to avoid use of a for-loop. I cannot vectorize because this requires terabytes of RAM. (I work with vectors that are not of length 8, but rather length nearly a million).

The convolution theorem gives the function R from the convolution of functions L and 1/w from the Fourier transform F and its inverse F^-1 as,

Clearly, the function 1/(w-w') in the convolution is from 1/w under F; it's as if you just set w'=0. But if I use analogous reasoning in my [1,1,1,1,0,0,0,0], I get either [1,1,1,1,1,1,1,1], the identity under .* in MATLAB or [0,0,0,0,0,0,0,0](a very boring result).

What is the mistake in reasoning I've made?

fft convolution

share|improve this question

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think maybe you need to add some code to this, because I can't figure out what it is that you are trying to do. Yes, you can write any vector as a convolution of other vectors. What you describe with sum(vector.*[...]) looks to be just cumsum(vec), I'm not sure I understand what you're looking for. I also don't understand how the vectors you show relate to 1/w.
  
  – Cris Luengo
  Mar 23 at 4:06
  
  

  
  
  
  

add a comment | 

0

How to write two column vectors as an analytic convolution so that the discrete FFT may be used. MATLAB syntax is used.

Consider:

1. 
   

   a set of vectors which, when sorted into a step function appears as any of the following:

   
   
   

   [1,1,1,1,0,0,0,0], or [1,1,1,1,1,0,0,0], or [1,1,1,1,1,1,0,0]

   
   
   

   (...the location at which the function "steps up" varies over members of this set)

   
   



2. The other is random vec=[1,0,1,0,1,1,1,0], and obviously both contain only 0s and 1s.

Is it possible to write these vectors as an analytic convolution? I would like the 1st, 2nd, 3rd, 4th... entries of the convolution to have values of:

sum(vec.*[1,0,0,0,0,0,0,0])
sum(vec.*[1,1,0,0,0,0,0,0])
sum(vec.*[1,1,1,0,0,0,0,0])
sum(vec.*[1,1,1,1,0,0,0,0])
... 
sum(vec.*[1,1,1,1,1,1,1,1])

For speed, I am trying to avoid use of a for-loop. I cannot vectorize because this requires terabytes of RAM. (I work with vectors that are not of length 8, but rather length nearly a million).

The convolution theorem gives the function R from the convolution of functions L and 1/w from the Fourier transform F and its inverse F^-1 as,

Clearly, the function 1/(w-w') in the convolution is from 1/w under F; it's as if you just set w'=0. But if I use analogous reasoning in my [1,1,1,1,0,0,0,0], I get either [1,1,1,1,1,1,1,1], the identity under .* in MATLAB or [0,0,0,0,0,0,0,0](a very boring result).

What is the mistake in reasoning I've made?

fft convolution

share|improve this question

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think maybe you need to add some code to this, because I can't figure out what it is that you are trying to do. Yes, you can write any vector as a convolution of other vectors. What you describe with sum(vector.*[...]) looks to be just cumsum(vec), I'm not sure I understand what you're looking for. I also don't understand how the vectors you show relate to 1/w.
  
  – Cris Luengo
  Mar 23 at 4:06
  
  

  
  
  
  

add a comment | 

How to write two column vectors as an analytic convolution so that the discrete FFT may be used. MATLAB syntax is used.

Consider:

1. 
   

   a set of vectors which, when sorted into a step function appears as any of the following:

   
   
   

   [1,1,1,1,0,0,0,0], or [1,1,1,1,1,0,0,0], or [1,1,1,1,1,1,0,0]

   
   
   

   (...the location at which the function "steps up" varies over members of this set)

   
   



2. The other is random vec=[1,0,1,0,1,1,1,0], and obviously both contain only 0s and 1s.

Is it possible to write these vectors as an analytic convolution? I would like the 1st, 2nd, 3rd, 4th... entries of the convolution to have values of:

sum(vec.*[1,0,0,0,0,0,0,0])
sum(vec.*[1,1,0,0,0,0,0,0])
sum(vec.*[1,1,1,0,0,0,0,0])
sum(vec.*[1,1,1,1,0,0,0,0])
... 
sum(vec.*[1,1,1,1,1,1,1,1])

For speed, I am trying to avoid use of a for-loop. I cannot vectorize because this requires terabytes of RAM. (I work with vectors that are not of length 8, but rather length nearly a million).

The convolution theorem gives the function R from the convolution of functions L and 1/w from the Fourier transform F and its inverse F^-1 as,

Clearly, the function 1/(w-w') in the convolution is from 1/w under F; it's as if you just set w'=0. But if I use analogous reasoning in my [1,1,1,1,0,0,0,0], I get either [1,1,1,1,1,1,1,1], the identity under .* in MATLAB or [0,0,0,0,0,0,0,0](a very boring result).

What is the mistake in reasoning I've made?

fft convolution

share|improve this question

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

sum(vec.*[1,0,0,0,0,0,0,0])
sum(vec.*[1,1,0,0,0,0,0,0])
sum(vec.*[1,1,1,0,0,0,0,0])
sum(vec.*[1,1,1,1,0,0,0,0])
... 
sum(vec.*[1,1,1,1,1,1,1,1])

share|improve this question

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

share|improve this question

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

edited Mar 23 at 4:02

Cris Luengo

24k62254

edited Mar 23 at 4:02

Cris Luengo

24k62254

asked Mar 23 at 2:19

ignoramusignoramus

81

asked Mar 23 at 2:19

ignoramusignoramus

81