Autocorrelation and Spectral Density

[CivilSigma]

Homework Statement

For a constant power signal x(t) = c, determine the auto correlation function and the spectral density function.

Homework Equations

The auto correlation function is:

$$R_x (\tau) = \int_{-\infty}^{\infty} E(x(t) \cdot x(t+\tau)) d\tau$$

To my understanding, here to find the expected value of the signal we must also multiply the function x(t)x(t+tau) by the probability function which is taken to be 1/period. Then, we change the integral limits to one period and get:

$$R_x (\tau) = \int_{-T/2}^{T/2} E(x(t) \cdot x(t+\tau)) \cdot \frac{1}{T} d\tau$$

Is my understanding of the auto-correlational function correct?

For the spectral density function :

$$S_x (\omega) = \int_{-\infty}^{\infty} R_x(\tau) \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$

Here, we don't change the integral limits and consider all the possible values of tau.

The Attempt at a Solution

The auto-correlation is:

$$R_x (\tau) = \int_{-T/2}^{T/2} E(x(t) \cdot x(t+\tau)) \cdot \frac{1}{T} d\tau$$
$$R_x(\tau) = \int_{-T/2}^{T/2} c^2 \cdot \frac{1}{T} d\tau$$
$$R_x(\tau)=c^2$$

The spectral density is then:
$$S_x (\omega) = \int_{-\infty}^{\infty} R_x(\tau) \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$
$$S_x (\omega) = \int_{-\infty}^{\infty} c^2 \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$
$$S_x (\omega) = \frac{-c^2}{2\pi \cdot i \omega} \times (e^{-i \omega \tau})|_{-\infty}^{\infty}$$

$$S_x (\omega) = \frac{-c^2}{2\pi \cdot i \omega} \times (0 - \infty)=?$$

However, my lecture notes suggest that the answer is :
$$S_x(\omega) = c^2 \cdot \delta(t)$$How did they get to here? How is the Dirac function obtained from evaluating the integral? Moreover, what happened to the other constants such as 2pi, i and omega from my original solution? If someone can shed some light on this mystery I would really appreciate it.

 

[Charles Link]

$\delta(x)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} e^{i kx} dk$. This is a very well-known result. At the moment, I don't have a proof on my fingertips, but this is a very well-known result.

 

[CivilSigma]

$\delta(x)=\frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty} e^{i kx} dk$. This is a very well-known result. At the moment, I don't have a proof on my fingertips, but this is a very well-known result.

I found it on WikiPedia https://en.wikipedia.org/wiki/Dirac_delta_function but with no derivation. I had originally read the Wiki, but I guess I did not read it well the first time.

 

[CivilSigma]

I noticed that in the definition, the exponential is positive. However, in my problem the exponential is negative.

Can I then say that:

Dirac (-x) is the solution to my original problem?

But really, the negative is not necessary because the function is defined at x=0 and will approach infinity.

So, Dirac (-x) = Dirac (x) ?

 

[rude man]

Homework Statement

For a constant power signal x(t) = c, determine the auto correlation function and the spectral density function.

Homework Equations

Something funny here.
R has the dimensions of c²
So S has the dimensions of c²T
But the dimension of δ(t) is T^-1.

 

[rude man]

First off, the answer for S is not as stated by the author. It should probably have read S = c²δ(ω). But that's still not exactly what I got.

You can derive this by using the fact (Wiener-Khintchine relation) that S is the Fourier integral of R, which you have already attempted.

The Fourier transform of c² would be 2πc²δ(ω). So that's not the same as the answer either.

 

[Charles Link]

@rude man Thanks for pointing that out, that it should be $\delta(\omega)$ rather than $\delta(t)$. I looked at it very quickly.

 

[rude man]

@rude man Thanks for pointing that out, that it should be $\delta(\omega)$ rather than $\delta(t)$. I looked at it very quickly.

OK but I still didn't get their answer. Did you try it? BTW if anyone asked me to derive the Fourier transform of e^{-jω₀(t-τ)} I would just tell them what it is & they can convince themselves that the inverse transform of that is indeed e^{-jω₀(t-τ)}!

 

[Charles Link]

OK but I still didn't get their answer. Did you try it? BTW if anyone asked me to derive the Fourier transform of e^{-jω₀(t-τ)} I would just tell them what it is & they can convince themselves that the inverse transform of that is indeed e^{-jω₀(t-τ)}!

There are different conventions used in Fourier transforms for when the $\frac{1}{2 \pi}$ is inserted. I normally insert this factor when doing the inverse transform, but here they inserted it when doing the primary transform. Some authors choose to make this symmetric and use $\frac{1}{\sqrt{2 \pi}}$ in both cases.