Convolution of a signal with the time reversed version of itself

Hi,

I am somewhat new to the world of DSP and my math skill need a lot of
push-forward. So the question may be fairly basic and the answer is
probably quite simple but I am having a hard time to figure it out.

So say I have a signal x(n) and I want to convolve it with x(-n).
Let's call the operator sum from k = -INF to INF as S.
So if I take the first variable as x(-n) then from the conv. equation the
answer is:

  S(x(-k) * x(n-k))

Now if I take the first variable as x(n) then I get into trouble:

conv(x(n),x(-n))= S(x(k)x(k-n)). 
The way I got the above is as follows: I take p(n) = x(-n).
Then from the conv equation I get:

conv(x(n),x(-n)) = S(x(k)* p(n-k)). Now p(n-k)=x(k-n) (that's what I
think).

Now the two answers can't be different because convolution is commutative
so I am having a hard time to convince myself. anyone put a light on why
they are different and what I am missing here?

Thanks

Also is there any way in this forum to insert mathematical symbols like sum
from -INF to INF etc. I am new to this forum so not sure if copy pasting
from equation editor would work.

On Mar 2, 7:29&#4294967295;am, "aru_3D" <arunavdev@n_o_s_p_a_m.yahoo.com> wrote:
> Hi,
>
> I am somewhat new to the world of DSP and my math skill need a lot of
> push-forward. So the question may be fairly basic and the answer is
> probably quite simple but I am having a hard time to figure it out.
>
> So say I have a signal x(n) and I want to convolve it with x(-n).
> Let's call the operator sum from k = -INF to INF as S.
> So if I take the first variable as x(-n) then from the conv. equation the
> answer is:
>
> &#4294967295; S(x(-k) * x(n-k))
>
> Now if I take the first variable as x(n) then I get into trouble:
>
> conv(x(n),x(-n))= S(x(k)x(k-n)).
> The way I got the above is as follows: I take p(n) = x(-n).
> Then from the conv equation I get:
>
> conv(x(n),x(-n)) = S(x(k)* p(n-k)). Now p(n-k)=x(k-n) (that's what I
> think).
>
> Now the two answers can't be different because convolution is commutative
> so I am having a hard time to convince myself. anyone put a light on why
> they are different and what I am missing here?
>
> Thanks
>
> Also is there any way in this forum to insert mathematical symbols like sum
> from -INF to INF etc. I am new to this forum so not sure if copy pasting
> from equation editor would work.

I don't understand what your mathematical notation
means, and cannot tell you where you are going astray.

Convolution of a signal with its time-reversed
version is called autocorrelation of the signal.
Autocorrelation functions are _even_ functions:

R_x[n] = R_x[-n] for all n

with a positive peak at the origin: R_x[0] > |R_x[n]|
for n not equal to zero except that if x is periodic with
period N, then R_x is also periodic with period N.

R_x[n] = sum from k = -infinity to +infinity of x[k]x[k-n]

= sum from k = -infinity to +infinity of x[k]y[n-k]

= x*y

where y is defined via y[t] = x[-t], that is, y is the
time reverse of x (and so we replaced x[k-n] by y[n-k]
in the above.

Dilip Sarwate

On 3/2/12 9:23 AM, dvsarwate wrote:
> On Mar 2, 7:29 am, "aru_3D"<arunavdev@n_o_s_p_a_m.yahoo.com>  wrote:
>>
>> Also is there any way in this forum to insert mathematical symbols like sum
>> from -INF to INF etc. I am new to this forum so not sure if copy pasting
>> from equation editor would work.

have you used LaTeX or the equation syntax at sites like Wikipedia? 
that's one method (one that i am not fond of for here).  i like to use 
what i call "ASCII math" (in the sense of "ASCII art").  i'll show you...

>
> I don't understand what your mathematical notation
> means, and cannot tell you where you are going astray.
>
> Convolution of a signal with its time-reversed
> version is called autocorrelation of the signal.
> Autocorrelation functions are _even_ functions:
>
> R_x[n] = R_x[-n] for all n
>

R_x means "R sub x".  i (and Dilip) like to stick with a convention 
that's a couple of decades old where discrete functions of time (or lag 
or frequency) have the independent variable n surrounded by square 
brackets x[n] or X[k] instead of round parenths as would be the case for 
continuous-time or continuous-frequency, x(t) or X(f).

> with a positive peak at the origin: R_x[0]>  |R_x[n]|
> for n not equal to zero except that if x is periodic with
> period N, then R_x is also periodic with period N.
>
> R_x[n] = sum from k = -infinity to +infinity of x[k]x[k-n]
>

my ASCII notation here would be this (assuming the reader is using a 
mono-spaced font):

                 +inf
      R_x[n]  =  SUM{ x[k] * x[k-n] }
                k=-inf

> = sum from k = -infinity to +infinity of x[k]y[n-k]
>
> = x*y

and, because we use it so for programming languages, i would leave the 
naked asterisk * for multiplication, not convolution.  for convolution, 
i would jazz it up a little:  x (*) y


> where y is defined via y[t] = x[-t], that is, y is the
> time reverse of x (and so we replaced x[k-n] by y[n-k]
> in the above.

one thing, Dilip, that i would add is that the above definition of 
auto-correlation doesn't work for periodic x[n] or any other 
finite-power x[n] (it blows up).  to compute the auto-correlation for a 
finite-power signal, including periodic, would be like:


                                  +K
      R_x[n]  =  lim   1/(2K+1) * SUM{ x[k] * x[k-n] }
                K->inf           k=-K


so, aru_3D, can you read that math sufficiently?  using a mono-spaced font?

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

On 02.03.2012 13:29, aru_3D wrote:
> Hi,
>
> I am somewhat new to the world of DSP and my math skill need a lot of
> push-forward. So the question may be fairly basic and the answer is
> probably quite simple but I am having a hard time to figure it out.
>
> So say I have a signal x(n) and I want to convolve it with x(-n).
> Let's call the operator sum from k = -INF to INF as S.
> So if I take the first variable as x(-n) then from the conv. equation the
> answer is:
>
>    S(x(-k) * x(n-k))
>
> Now if I take the first variable as x(n) then I get into trouble:
>
> conv(x(n),x(-n))= S(x(k)x(k-n)).
> The way I got the above is as follows: I take p(n) = x(-n).
> Then from the conv equation I get:
>
> conv(x(n),x(-n)) = S(x(k)* p(n-k)). Now p(n-k)=x(k-n) (that's what I
> think).
>
> Now the two answers can't be different because convolution is commutative
> so I am having a hard time to convince myself. anyone put a light on why
> they are different and what I am missing here?

All all you need to do is to relabel the indices. The first sum equals to

\sum_k x(-k) x(n-k)

and the second to

\sum_k x(k) x(k-n) = \sum_j x(-j) x(-j-n) = \sum_l x(-l+n) x(-l) = 
\sum_k x(-k) x(n-k)

the first is the substitution k -> -j, the second j -> l-n (thus -j = 
n-l => j+n = l => -l = -n -j ) and the last just renamed l to k.

HTHH,
	Thomas