# Study the autocorrelation function

1. Jul 20, 2007

### T.Engineer

I would like to know how to find the autocorrelation function.
Thanks!

2. Jul 20, 2007

### EnumaElish

3. Jul 20, 2007

### T.Engineer

4. Jul 21, 2007

### EnumaElish

5. Jul 21, 2007

### T.Engineer

I am going to find the autocorrelation function for the following equation
Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2)

So, I didnt find the previous links can help me to solve this problem.

Last edited: Jul 21, 2007
6. Jul 22, 2007

### EnumaElish

You need to solve

Lims-->0-ss Hn(t)Hn(t+s)ds / (2s).

Note,
-ss Hn(t)Hn(t+s)ds = ∫t-zz-t Hn(t)Hn(z)dz

Last edited: Jul 22, 2007
7. Jul 23, 2007

### T.Engineer

You need to solve

Lims-->0∫-ss Hn(t)Hn(t+s)ds / (2s).

Note,
∫-ss Hn(t)Hn(t+s)ds = ∫t-zz-t Hn(t)Hn(z)dz

but I want to understand why you suggest this. I want to understad exactly by steps how can I find a solution.
Thanks alot.

8. Jul 23, 2007

### EnumaElish

9. Jul 23, 2007

### T.Engineer

Thanks alot EnumaElish!
Thanks alot.

10. Jul 25, 2007

### T.Engineer

I couldnt solve the equation
t-zz-t Hn(t)Hn(z)dz

may be I should to follow such courses that explain Autocorrelation function or may be I must make reveiw for the integrals.

11. Jul 25, 2007

### EnumaElish

For starters,

∫Hn(t)Hn(z)dz = Hn(t)∫Hn(z)dz.

Then, insert Hn(z) = (-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2). Now you have:

Hn(t)∫(-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2)dz = Hn(t)(-1)^n ∫e^(z^2)* fn(z) dz

where fn(z) = d^n/dz^n * e^(-z^2).

You need to work on solving ∫e^(z^2)* fn(z) dz.
___________________
CORRECTION: Lims-->0-ss Hn(t)Hn(t+s)ds should have been LimT-->0-TT Hn(t)Hn(t+s)ds.

Then, defining z = t+s, we have-TT Hn(t)Hn(t+s)ds = ∫-TT Hn(t)Hn(z)dz.

This does not change the substance of the advice I have given until now.

Last edited: Jul 25, 2007
12. Jul 27, 2007

### T.Engineer

let g(z)=∫e^(z^2)* fn(z) dz
for n = 1

fn(z)= d/dz e^-(z^2)= -2 z e-(z^2)
So, g(z) = ∫-2 z * e^(z^2)* e-(z^2) dz
= ∫-2 z *e^[(z^2)-(z^2)] dz
= ∫-2 z dz
=-z^2

is not that true?

13. Jul 27, 2007

### EnumaElish

Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.

14. Jul 28, 2007

### T.Engineer

Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.[/QUOTE]

we have fn(z) = -Z^2 for n =1
by multiplying it by (-1)^n
we will have Z^2
by multiplying it by Hn(t),
where Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)]
Then, we will have:
Z^2 * (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)], where z = t+s
So, how I will find the integral and I have two variables (t and Z)??

Last edited: Jul 28, 2007
15. Jul 28, 2007

### EnumaElish

Remember, the integral is a definite integral. It has limits "from -T to T." You need to evaluate the integral according to the limits.

Before you multiply the integral with Hn(t), you need to apply the integ. limits to -z^2. This will make z "disappear." The z^2 will be replaced with a quadratic function of T, say q(T).

Do you understand what q(T) is? Can you derive it explicitly?

16. Jul 29, 2007

### T.Engineer

OK, I will have:
-z^2$$\left|^{T}_{-T}$$
then I will get
-(T)^2 - [- (-T)^2]= 0
So, the autocorelation for
Hn(t)Hn(t+s)ds = 0

is that true!!!

Last edited: Jul 29, 2007
17. Jul 29, 2007

### EnumaElish

It seems to be true for n=1.

Can you solve the general case for arbitrary n?

18. Jul 29, 2007

### T.Engineer

Actually, this is exactly what I am going to ask you about.
So, can you just help me a little bit?
And I have another question:
Hn(t)Hm(t)dt , where m not equal to n

19. Jul 29, 2007

### EnumaElish

I cannot see the LaTex equation for m. There seems to be a typesetting error.

EDIT: I can see it now.

For the Hn(t)Hn(t+s) case, you need to work on solving ∫e^(z^2)* fn(z) dz for arbitrary n.

The Hn(t)Hm(t)dt case is not a regular AC function. Go back to the MathWorld link (http://mathworld.wolfram.com/Autocorrelation.html) and review the definition. The AC function has a single argument (a single function), for example, ac(t) = AC(Hn(t)) or equivalently ac = AC(Hn). What you are proposing has two arguments, Hn and Hm. So it doesn't fit the definition of AC.

Last edited: Jul 29, 2007
20. Jul 29, 2007

### T.Engineer

when I written:the autocorrelation of Hn(t)Hm(t)dt
I meant that what about the autocorrelation function between the first derivative of H(t) and the second derivative of H(t)
Such as :
the autocorrelation function of H1(t)H2(t).
Thanks alot!