# 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!

21. Jul 29, 2007

### T.Engineer

for n=1
∫e^(z^2)* fn(z) dz = -z^2
for n=2
∫e^(z^2)* fn(z) dz = $$\frac{3}{4}$$ z^3

So, should I continue for n=3,4,5,...

22. Jul 29, 2007

### EnumaElish

The AC function has a very specific definition; and it is defined for a single parent function.

I am going to look at your question that way. You can write the AC function of H1*H2 by following the general definition of AC for function F:

ACF(t) = LimT--->0 -TT F(t) F(t+s) ds / (2T)

and making the substitution F = H1*H2.

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

### EnumaElish

You can continue and hope that you will "see" the general solution eventually by induction and guesswork. But the more direct approach is to try to solve the general case algebraically.

24. Jul 30, 2007

### T.Engineer

The correlation is equal to the average of the product of two random variables and is defined as
cor(X,Y) = E[XY]
= ∫∞ −∞∫∞ −∞ xyf(x,y)dxdy

So, if the above equation seems to be correct, then I can define H1(t) as X and H2(t) as Y.
Can I work like this depending on the above?

And for a given set of variables, we use the correlation coefficient to give us the linear relationship between our variables. The correlation coefficient of two variables is defined in terms of their covariance and standard deviations, as seen below
ρ= cor(X,Y)/σχ σΥ

If there is no relationship between the variables then the correlation coefficient will be zero and if there is a perfect positive match it will be one.
If there is a perfect inverse relationship, where one set of variables increases while the other decreases, then the correlation coefficient will be negative one.

So, my second equastion is how to find the mean and variance for H1(t) and H2(t)
Hint: if Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

Thanks alot!!

Last edited: Jul 30, 2007
25. Jul 30, 2007

### EnumaElish

This is the definition of covariance, not correlation.

To obtain Cov(X,Y), yes.

The correct formula is ρ = Cov(X,Y)/(σχ σΥ) where ρ is the correlation coefficient between X and Y.

Correct.