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No. Quoting Wikipedia's article Independence (probability theory): If $X$ and $Y$ are independent random variables, then the expectation operator $\operatorname{E}$ has the property $$\operatorname{E}[X Y] = \operatorname{E}[X]\operatorname{E}[Y].$$ Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $\operatorname{E}[X] \ne ... 4$x(t)$is a real-valued finite energy signal with Fourier transform$X(f). Its autocorrelation function is \begin{align} R_x(\tau) &= \int_{-\infty}^{\infty}x(t)x(t-\tau) \,\mathrm dt\\ &= \int_{-\infty}^{\infty}x(t)y(\tau-t) \,\mathrm dt & {\scriptstyle{\text{Define}~y(t)~\text{as the time-reversal}~x(-t)~\text{of}~x(t)}}\\ &= x\star y\big|... 4 $$\mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2 \pi f t} \ \mathrm{d}t$$ $$\mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty} X(f) \, e^{j2 \pi f t} \ \mathrm{d}f$$ Assumingx(t)$is real (which means that$R_x(\tau)is also real). \begin{align} S_x(f) &... 4 It's the empirical ACF computed using the sample. 4 The answer to the OP's question is more straightforward than rb-j's comments make it out to be. \{X(t)\colon -\infty < t < \infty\} is a continuous-time WSS random process with autocorrelation functionR_X(\tau) = E[X[t)X(t+\tau)], -\infty < \tau < \infty.$$On the other hand, \{Y(n)\colon n \in \mathbb Z\} where Y[n] = X(nT) is a ... 4 Very simple: your sampling rate is 512Hz and your sample time is the inverse of this: about 1.953 ms. Multiply with 506 and you get the time length of the fundamental: 0.988s. So you have one beat every 0.988s which is 1.0119Hz or 60.711 beats per minute. 3 \sigma_r^2=\sigma_y^2\sigma_v^2 iff 1) \mathbb{E}\{y\} = \mathbb{E}\{v\}=0. AND 2) y and v are independent, OR uncorrelated and their squares (y^2 and v^2) are also uncorrelated. Proof: assuming independence, \sigma_r^2=\mathbb{E}\{(yv-\mathbb{E}\{yv\})^2\} (by definition) ~~~~=\mathbb{E}\{(yv-\mathbb{E}\{y\}\mathbb{E}\{v\})^2\} (by ... 3 Well, by definition of the \delta distribution, you have: \int_{-\infty}^{\infty} f(t) \delta(t-T)\, \textrm{d}t = f(T) The autocorrelation of a function g(t) can be computed via: \int_{-\infty}^{\infty} g^{*}(t)g(t + \tau)\, \textrm{d}t, with g^* as the complex conjugate of g. Since \delta(t) is real-valued, this is conjugation can be ... 3 I believe that h(-t) means a "time-reversed" version of h(t). Your command: 'y = conv(r,-h);' computes the convolution of 'r' and negative 'h', and you don't want that. I think you want: y = conv(r,conj(fliplr(h))); 2 EDIT: sorry that I didn't fully get your question. Now I got your points of conufion and try to provide a shortest possible answer: First, you should better use xcorr function when estimating ACS r_x[k] from data sequence x[n]. The output of xcorr will return the lag-0 at the middle sample as you recognize. Look at the input argument scaleopt which ... 2 Are two signals are the same if their auto-convolution functions are the same? Almost. Look at the autoconvolution in the frequency domain where the autoconvolution of x (with itself) gives us (X(f))^2 in the frequency domain while the autoconvolution of -x (with itself) gives us (-X(f))^2 = (X(f))^2. So, given an autoconvolution function, there are ... 2 Because of the symmetry of the \operatorname{sinc}(t) we have$$R_{hh}(\tau)=\int_{-\infty}^{\infty}\operatorname{sinc}(t) \ \operatorname{sinc}(t-\tau) \ \mathrm{d}t=\int_{-\infty}^{\infty}\operatorname{sinc}(t) \ \operatorname{sinc}(\tau-t) \ \mathrm{d}t = \operatorname{sinc}(\tau) \ast \operatorname{sinc}(\tau) $$Analyzing in the frequency domain we ... 2 Whether that decorrelation happens depends on the signals you put in – it's not a general property of the DFT! Especially, when you model your signal as sum of narrowband signals, then you'll find that due to the DFT of a narrowband signal being concentrated on one or very few bins, and practically 0 elsewhere, the decorrelation simply stems from the fact ... 2 There are two mistakes in your code/method. The first is the term \sqrt{\Delta t} in your second formula; it should be replaced by \Delta t. The second is in the computation of the power spectrum from the estimated auto-correlation. What you do is square the result of the FFT Y to obtain mY, but that's not correct. First of all, Y is complex-valued, and ... 2 Here is a detour over the frequency domain. We'll use index k for frequency and index n for time. Capital letters for Frequency Domain variables The Fourier Transform (FT) of an impulse is R(k)=1 FT of the autocorrelation is the magnitude squared of the FT of the original signal, i.e R(k) = |S(k)|^2 Combining 1 and 2 we get |S(k)| = 1 ,i.e. S(k) ... 2 The autocorrelation function r_s[k] of a sequence \{s[n]\} is just the inner product (dot product) of \{s[n]\} with a shifted version of itself:$$r_s[k] = \sum_n s[n]\cdot s[n+k].$$Two important properties of finite-energy signals is that r_s[0] \geq 0 (indeed r_s[0] equals 0 only for the zero signal s[n]=0 for all n) and |r_s[n]| \leq r_s[... 2 GPS has a processing gain of 43 dB assuming you correlate over 20 consecutive symbols. (GPS sends 20 C/A code sequences for 1 data bit: Chip rate 1.023 MHz, data rate 50 bps). It looks like you are correlating over just one symbol which is at 1 KHz in which case the processing gain would only be 10Log_{10}(1023)= 30 \text{ dB}. However I don't think your ... 1 As suggested I am adding my comments as an answer. The sample autocorrelation function (ACF) for n observations is given by \hat{p}_x(m) = \frac{\sum_{n=0}^{N-m-1}(x_n - \bar{x})(x_{n+m} - \bar{x})}{\sum_{n=0}^{N-m-1}(x_n - \bar{x})^2} To understand why you have a difference between figure 1 and figure 2, lets assume we are looking at observations ... 1 to add to hot's answer, when using autocorrelation or an equivalent method like AMDF or ASDF, although the autocorrelation R_x[k] is evaluated only at integer values of lag k, which means the peak value and initial period length will be an integer number of samples, you can use quadratic interpolation using the point at the peak and the two neighboring ... 1 For higher resolution periodicity estimation, given properly band-limited data, the data can be upsampled (interpolated) before autocorrelation, and/or, the autocorrelation results can be interpolated (Sinc kernel or some order of polynomial). This, of course, assumes the signal has a high enough S/N, and that the periodicity itself is unmodulated in ... 1 Bcause the FFT is computed usingn N\log N arithmetic operations, whereas the DFT using N^2 operations, where N is the number of points in your discrete-time signal. So the number of operations is much lower if N is particularly high, which is often the case ... Watch this video if you want to know more about this https://www.youtube.com/watch?v=... 1 You are right, the derivation is full of typos. The first equation below Eq. (8.39) should read$$\int_{-\infty}^{\infty}x(t+\tau)e^{\color{red}{-}j\omega\tau}d\tau=X(\omega)e^{j\omega \color{red}{t}}\tag{1}$$Substituting into (8.39) gives$$\begin{align}\mathcal{F}\big\{R(\tau)\big\}&=\int_{-\infty}^{\infty}x(t)X(\omega)e^{j\omega t}dt\\&=X(\... 1 When the signal is assumed to be ergodic, then its ACF can be computed using time averages which can also be computed using the following convolution of lengthN$sequence$x[n]$by its conjugate symmetric version: $$\hat{r}_{xx}[m] = \frac{1}{N} ~~~x[m] ~~\star ~~ x^*[-m] ~~$$ Taking the$2N-1$point DFT of both sides yields the following: $$\hat{R}_{... 1 Since this does seem to answer the question, I'm reposting my comment as an answer: I guess you may have to distinguish continuous (analog) and discrete (digital) signals. For analog signals, the autocorrelation function is a Dirac if and only if the spectrum is flat for all frequencies. Typically though what we care about is whether or not our digital ... 1 It is not possible for a (WSS) random process \{X(t)\}to have an autocorrelation function of the form$$R_X(\tau) = E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & |\tau|\leq T,\\0, & |\tau| > T. \end{cases}$$Assuming a zero-mean process for simplicity, note that$R_X(0) = E[X^2(t)] = \sigma^2$where$\sigma^2$is the common variance of the ... 1 Walsh-Hadamard matrix is usually used for DS CDMA. In this case,$\mathbf{x}$contains symbols from different users. Since your matrix$\mathbf{H}$is$4\times 4$, then the vector$\mathbf{x}$must be of length of$4\times 1$. This means that you can support$4$users at any given time. So, the answer to your question is no, you cannot spread a signal of ... 1 Yes you can, because you do spreading for each bit of data. So, regardless the length of your data, you are going to spread every bit on your code. In case if you have$x = 10$you will have data after spreading with length 8, and if$x=101101$you will have data to send with length 4 x 6 = 24. Just try to understand the process of spreading (I think it's ... 1 All these notation should be considered within the context of the development. So$r_d[k]$may refer to the auto-correlations sequence of the desired signal whereas the$r_{dx}[k]$is the cross correlation between the input$x[n]$and desired signal$d[n]\$. So they are two different things. The latter is used in Wiener optimal filtering to solve the Wiener-...