0
$\begingroup$

I am following the arguments presented in the paper AN-255 Power Spectra Estimation, from Texas Instruments, to learn how to derive the power spectral density for a stationary stochastic process, and do not understand one of their steps. I am looking at Page 4 only of the the document for the purposes of this question.

The derivation is done by considering only a single realisation (sample function), denoted by $x(t)$, of the stochastic process. Then they do the usual thing of only considering a truncated version of this signal $x_T(t)$, which allows the Fourier transform of it to exist, denoted by $X_T(f)$ (because one of the requirements for FT to exist is absolute integrability, which isn't satisfied for random signals).

I am confused with how they justify taking the limit $T\rightarrow\infty$ in the right hand column of page 4, shown here:

$\hskip2in$enter image description here

$\hskip2in$enter image description here

I understand that you cannot take the limit $T\rightarrow\infty$ directly on the truncated sample function's Fourier transform $X_T(f)$, because it is not defined in that limit (indeed, that was the reason for doing the truncation in the first place).

I also understand that $X_T(f)$ results from a single realisation only of the stochastic process, and if you were to repeat the experiment again you would obtain a different $X_T(f)$. As such, this function is actually a random variable itself (at each frequency $f$), and therefore it makes sense to do an ensemble average over many realisations, and consequently motivates taking an expectation.

What I don't understand is why the expectation solves the problem, and allows you to take the limit $T\rightarrow\infty$ after you have taken the expectation:

$\hskip2in$enter image description here

What is it about the expectation of $X_T(f)$ that means the limit $T\rightarrow\infty$ can then be taken? I'm having a hard time seeing it... Could someone spell this out plainly for me?

$\endgroup$
2
  • $\begingroup$ I think I see your point; and would there need to be an additional requirement that $|X_T(f)|^2$ would need to be ultimately decreasing at the rate of $1/f$ or more as $f -> \infty$ otherwise the Expectation integral blows up, right? $\endgroup$ Apr 8 '20 at 3:47
  • $\begingroup$ Possibly, something like that could make sense to me, but I haven't seen anything like that in any other derivation. I am learning this for the first time, so feel like I'm just missing something. What you said about it blowing up is what I was thinking - if the individual realisations of $X_T$ are not defined in the limit, I don't see what would be so different about the expectation of many. What happens for example if you coincidently obtain all ideantical spectra for every realisation - the expectation should do nothing in that case, i suppose. $\endgroup$
    – teeeeee
    Apr 8 '20 at 7:17
2
$\begingroup$

It is not the expectation operator that makes sure that the limit exists. The expectation just results in an ensemble average, which we need to obtain a deterministic function $S(f)$ for the power spectrum.

Assume we're given a deterministic power signal $x(t)$, i.e., a signal with finite non-zero power, and, consequently, infinite energy. Its Fourier transform generally doesn't exist. We can define a truncated Fourier transform

$$X_T(\omega)=\int_{-T}^Tx(t)e^{-j\omega t}dt\tag{1}$$

By assumption, $\lim_{T\to\infty}X_T(\omega)$ doesn't exist. However, the limit

$$\lim_{T\to\infty}\frac{\left|X_T(\omega)\right|^2}{2T}\tag{2}$$

exists because of the finite power assumption, and this is also the way the power spectrum is defined for such deterministic power signals.

If $x(t)$ is modeled as a random signal, we only need to modify $(2)$ by taking the expectation of the numerator to obtain the power spectrum of $x(t)$.

$\endgroup$
7
  • $\begingroup$ Okay, I follow that - you are saying it is nothing to do with the fact that the signal is stochastic. So then my question becomes: Can you help me to see/recognise easily why $$\lim_{T\to\infty}\frac{\left|X_T(\omega)\right|^2}{2T}$$ exists, when $\lim_{T\to\infty} X_T(\omega)$ does not? $\endgroup$
    – teeeeee
    Apr 8 '20 at 8:04
  • $\begingroup$ @teeeeee: Because we divide by $2T$ and $T$ goes to infinity, AND because we assume that the process has finite power. In general we cannot know if that limit exists, but if we assume that a power spectrum exists, then the limit must also exist. $\endgroup$
    – Matt L.
    Apr 8 '20 at 10:15
  • $\begingroup$ Ok, I buy that. But it still doesn't explain the statement in the paper (see the screenshot I included in the OP) that "at this particular point the limit $T\rightarrow\infty$ cannot be taken", with reference to Eq. (15). However, as you can see, Eq. (15) already includes the $\frac{1}{T}$ factor, as you said was required. $\endgroup$
    – teeeeee
    Apr 8 '20 at 10:49
  • $\begingroup$ So, just be clear for me, are you disagreeing and claiming that the limit $T\rightarrow\infty$ can in fact be taken already directly on Eq. (15) ? $\endgroup$
    – teeeeee
    Apr 8 '20 at 10:50
  • $\begingroup$ @teeeeee: For a single realization of $x(t)$ or for deterministic $x(t)$ that expression is just the power of $x(t)$, and if we assume that the power is finite, then the limit exists. For random $x(t)$ the question is how to interpret that limit because it is in itself a random variable, so you need the expectation operator to obtain a deterministic result. But it's not the case that the limit doesn't exist initially, and the expectation operator in some magical way changes that. $\endgroup$
    – Matt L.
    Apr 8 '20 at 11:10
0
$\begingroup$

For stationary stochastic signal, taking limit after expectation works as explained below.

You have a windowed random process $x_T(t)$ from which you want to derive the PSD $S_{xx}(f)$ by letting $T \rightarrow \infty$. There are two approaches to this.

(1) You take the $lim_{T \rightarrow \infty} $ on $x_T(t)$ , then compute $|X_T(f)|^2$ and then take Expectation of the it $E(|X_T(f)|^2)$. This fails because $X_T(f)$ does not even exist when $lim_{T \rightarrow \infty}$.

OR

(2) You compute $\frac{1}{2T}|X_T(f)|^2$ from $x_T(t)$ , take expectation over it, i.e., $\frac{1}{2T}E(|X_T(f)|^2)$, then let $lim_{T \rightarrow \infty} \frac{1}{2T}E(|X_T(f)|^2)$.

The second method of taking limit after expectation works because of following reason.

The quantity which you can compute, $ E(|X_T(f)|^2)$, corresponds to $R_{xx}(\tau)W(\tau)$ , which is the expectation of convolution of ACF of $x_T(t)$ and triangular function $w_T(\tau)$. (Remember we did rectangular windowing, so when you compute $|X_T(f)|^2$ you are doing convolution in time domain between $x(t)w_T(t)$ and $x(-t)w_T(-t)$).

When you let $lim_{T \rightarrow \infty}$ the triangular window $W$ becomes square window which results in ACF being $R_{xx}(\tau)$ itself. On the other side of equation, you have $lim_{T \rightarrow \infty}\frac{1}{2T}E(|X_T(f)|^2)$. So, $$ R_{xx}(\tau) \leftrightarrow lim_{T \rightarrow \infty}\frac{1}{2T}E(|X_T(f)|^2) $$

Appendix: $$ |X(f)|^2 \leftrightarrow [x_T(t)w_T(t)]*[x_T(-t)w_T(-t)]\\ E(|X(f)|)^2 \leftrightarrow R_{xx}(\tau)W(\tau)\\ $$ Reference : https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-011-introduction-to-communication-control-and-signal-processing-spring-2010/readings/MIT6_011S10_chap10.pdf

For deterministic signals, again approach (1) would fail because $x$ being a finite power signal has infinite energy. Here no assumption is made about $x$ so it is assumed to exist $-\infty \lt t \lt +\infty$. So $lim_{T \rightarrow \infty}$ does not exist.

$\endgroup$
3
  • $\begingroup$ This is not the same reason as the one given by Matt in his answer. He says it is nothing to do with the fact that you have taken an expectation (indeed that you would run into the same problem even if you were working with a deterministic signal). Whilst you are saying "taking the limit after expectation works"... I am confused... $\endgroup$
    – teeeeee
    Apr 8 '20 at 9:27
  • $\begingroup$ Your explanation about the triagngular convolution is not very clear to me. I would like to understand this without making reference ot the autocorrelation function. This is something that seems be because there is a general difference between taking limit $T\rightarrow\infty$ on the quantity $X_T(\omega)$ versus taking the limit on the quantity $\frac{1}{2T}|X_T(\omega)|^2$. $\endgroup$
    – teeeeee
    Apr 8 '20 at 9:37
  • $\begingroup$ I have added reference to the triangular operation see sec 10.2 in that pdf. In the beginning of your question, you mentioned stochastic signals, that is why my answer is based on expectations. For deterministic signal, you can skip the expectation operator. Let me modify the answer for that. $\endgroup$
    – jithin
    Apr 8 '20 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.