# Tag Info

28

White Gaussian noise in the continuous-time case is not what is called a second-order process (meaning $E[X^2(t)]$ is finite) and so, yes, the variance is infinite. Fortunately, we can never observe a white noise process (whether Gaussian or not) in nature; it is only observable through some kind of device, e.g. a (BIBO-stable) linear filter with transfer ...

21

You are right, PSD has to do with calculating the Fourier Transform of the power of the signal and guess what.....it does. But first let's look at the mathematical relationship between the PSD and the autocorrelation function. Notations: Fourier Transform: $$\mathcal{F}[ x(t)] = X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$$ (Time) Auto-...

20

The fast Fourier transform ($\textrm{FFT}$) algorithms are fast algorithms for computing the discrete Fourier transform ($\textrm{DFT}$). This is achieved by successive decomposition of the $N$-point $\textrm{DFT}$ into smaller-block $\textrm{DFT}$, and taking advantage of periodicity and symmetry. Now, the $N$-point $\textrm{DFT}$ of a sequence $\{x[0], x[... 16 Power-Spectral Density is the distribution of power along the frequency axis. It is generally used for non-finite energy signals (mostly not limited in time signals), who aren't square-summable. The signal's PSD is the autocorrelation of the signal's Fourier Transform, as stated by the Wiener–Khinchin theorem. In Matlab: N = length(S); F = fft(S); F = F(1:N/... 9 Technically this is not a MATLAB-esque forum, but I can explain the steps in more detail for you: Suppose your input signal is$x[n]$, and its DFT is$X(f)$. For real signals you may use the one-sided DFT, since the other half would be redundant when you look at its Power Spectral Density. (PSD). Once you compute the DFT of your signal, the PSD is simply$|...

9

There is only one correct way of scaling DFT when calculating PSD with RMS values. Given input signal $x$ and its DFT $X$, the exact formula is: $$\mathrm{PSD}=\frac{2\cdot \hat{X}}{f_s\cdot S}$$ where: $\hat{X}=|X|^2=X\cdot X^*$ - squared spectrum magnitude $S=\sum_{i=1}^{N} w_{i}^{2}$ - scaling factor defined as sum of squared samples of window ...

8

Nice derivation but I think you can do this even easier Auto correlation $r(t) = x(t)*x(-t)$, it's the convolution of the signal with it's time flipped self. Convolution in the time domain is multiplication in the frequency domain. Time flip in the time domain is "complex conjugate" in the frequency domain. Hence we get $$R(\omega) = \mathcal{F}\{r(t)\} ... 8 The power spectrum is a general term that describes the distribution of power contained in a signal as a function of frequency. From this perspective, we can have a power spectrum that is defined over a discrete set of frequencies (applicable for infinite length periodic signals) or we can have a power spectrum that is defined as a continuous function of ... 7 I think this is simply an aspect of Parseval's Theorem (e.g. click me) It simply says: sum of squares in the time time domain equals sum of squares in the frequency domain. Substitute "sum" for "integral" if using the analog domain. In other words: total energy in the time domain equals total energy in the frequency domain. This can easily reproduce your ... 7$$\begin{align} R_x(\tau) &=\int_{-\infty}^\infty x(t)x^*(t-\tau)\,\mathrm dt\\ &= \int_{-\infty}^\infty \left[\int_{-\infty}^\infty X(f)e^{j2\pi ft}\,\mathrm df\right]x^*(t-\tau)\,\mathrm dt\\ &= \int_{-\infty}^\infty X(f) \left[\int_{-\infty}^\infty x^*(t-\tau) e^{j2\pi ft}\,\mathrm dt\right]\,\mathrm df\\ &= \int_{-\infty}^\infty X(f) \...

7

I don't really understand what do you mean by multiply them in the time domain and multiply them with window function. I think that you are trying to implement the Welch's PSD calculation. If so, steps should be: Split your data into possibly overlapping segments of length $N$ Preferably remove the mean of each segment. Window each of the segments by ...

7

As the documentation states, periodogram provides a power spectral density estimate pxx: [pxx, w] = periodogram(x); meaning that it shows how the total variance of the signal, var(x), is distributed over the frequency w. This implies that the integral of the estimated spectral density over the frequency axis trapz(w, pxx) should correspond to that total ...

7

The problem is not the spectrogram parameters, these are correct since they only depend on what resolution you want in time and frequency domain. Also, the spectrogram interpretation is correct, there are multiple frequency peaks. The problem may be: I expected to see one high power frequency after pressure rise, instead of multiple frequencies Why? If ...

6

To add to the above well stated explanation, in the case of wavelets, which are finite in time, it is more correct not to use the term 'power' but 'energy'. For Fourier who has as basis functions the sinusoid that extends infinite in time, power spectral density is the correct term. For wavelets, who has basis functions as finite in time deflections, we ...

6

Well, because in a lot of real world problems this If I have a deterministic signal with a fixed number of samples, shouldn't I be able to directly determine its spectral information? is just not the case. Very often, measured signals are more of a random process. A simple and common case would be to have the desired signal and some additive noise, very ...

6

Interesting question. I juts happen to live right outside a waterfall so I can relate some hands on experience. The sound depends A LOT and water flow and weather conditions. Water flow is the main engine but wind can generate significant modulations Depending on conditions you can get strong harmonics in the signal. Sometimes you can actually see standing ...

6

A simple, "non-technical" way of thinking of it is the fact that the Doppler frequency is proportional to $\cos\theta$. The amplitudes of cosine, however, are not uniformly distributed, but are heavily weighted towards $\pm 1$. Example plot to demonstrate, using Python/Pylab code: theta = linspace(0, 2*pi, 1001) x = cos(theta) hist(x) More rigor can be ...

6

Power-law behaviors in frequency can be found in several unrelated observations and systems. This is apparently the case for $1/f$ or flicker noise. Note that an exact $\alpha=1$ exponent might be too stringent, and people can be interested in a wider range, like $1/f^\alpha$ with $1/2\le\alpha\le3/2$. The plot above is from 1/f noise. Since only looking ...

5

I Just do here My source code: [x, Fs, nbits] = wavread('ederwander.wav'); winSize = 2048; n_samples = length(x); %50% overlap or 0 to not use overlap OverlapStep = 50; if OverlapStep > 0 Overlap = floor((OverlapStep*winSize) / 100); nFrames=floor(n_samples/Overlap)-1; else Overlap= winSize; ...

5

In principle you are doing the right thing. Step 4 should produce a real result unless there is a coding error. Sometimes you add up some residual imaginary part due to numerical noise but, if any, that should be very small. Here is an example %% random signal of length n n = 128; x = rand(n,1); % zero pad x = [x; zeros(n,1)]; % fft fx = fft(x); % mag ...

5

As Jason was saying in the comments, the power spectral density of white noise is flat. This is equivalent to saying that the autocorrelation of white noise is a delta dirac function (i.e. that there is no correlation, positive or negative, between one noise sample and another), not that the noise itself is a delta dirac function.

5

Suppose we have a discrete-time sequence $x[t]$ which is stationary, zero mean, white noise with variance $\sigma^2$. Then the autocorrelation of $x$ is: $$\begin{array} RR_{xx}[\tau] &=& E\left[ x[t] x[t+\tau] \right]\\ &=& \left \{ \begin{array} EE \left[ x[t]^2 \right], {\rm if\ }\tau=0 \\ 0, {\rm otherwise} \end{array} \right. \\ &=&... 5 If you sample a finite-power continuous-time random process x(t) you get a discrete-time random process y_k. If x(t) is wide-sense stationary (WSS) you get for the autocorrelation function of y_k$$R_y(k,l)=E\{X(kT)X^*(lT)\}=R_x((k-l)T)=R_x(mT)$$where T is the sampling period. Obviously, y_k is also WSS (it only depends on the difference m=k-... 5 The periodogram is simply the squared magnitude of the DFT. Since the periodogram is a rather poor estimate of the power density spectrum of a random process there are methods which use averaging of periodograms to obtain better estimates of the power spectrum. Two such methods are Welch's method and Bartlett's method. 5 You can compute the power of the process from its power spectrum as well as from its PDF. Equating the two gives you a relation between the constants A and B. More specifically you get$$\int_{-B}^BG_x(f)df=\frac{1}{2A}\int_{-A}^Ax^2dx If I'm not mistaken this should give $A=\sqrt{3B}$.

5

The difference is in the scaling of the power spectrum. I suppose you will find that the difference in scaling always equals $L/F_s$, which for the numbers in your question is indeed $8000/500=16$. Take a look at Power Spectral Density Estimates Using FFT for the correct scaling. If you normalize the FFT result by the FFT length, you need to scale the ...

5

As mentioned by @MBaz in comments, you could use the fact that filtering white noise with unitary power spectral density (for example from independent Gaussian samples zero mean and unitary standard deviation) with a filter having a frequency response $H(f)$ gives you colored noise with power spectral density $S_{xx}(f) = |H(f)|^2$. Then comes the question ...

5

Since the question has been raised as to whether the hint that I had given to the OP in a comment on the original question was appropriate for a newcomer to signal processing, here goes. Stripped of extraneous baggage and notation, the question is whether it is possible to determine the value of $E[X^2Y^2]$ straightforwardly where $X$ and $Y$ are zero-...

5

The cross-spectral density is in the frequency domain while the cross-correlation function is in the time domain. The two are Fourier Transform pairs, the FT of the cross correlation function is the cross-spectral density. The the two provide the same information, just that one is in the time domain and the other is in the frequency domain. This is just as ...

5

In addition to Carlos answer, I want to correct your general understanding: What I understand of Doppler spread is that the relative motion between Transmitter (TX) and Receiver (RX) change the exposing time of signal. In rapport to a constant-distance TX-RX, a moving toward each other TX-RX "compresses" signal in time (signal takes less time to ...

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