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## Hot answers tagged stationary

21

There is no stationary signal. Stationary and non-stationary are characterisations of the process that generated the signal. A signal is an observation. A recording of something that has happened. A recording of a series of events as a result of some process. If the properties of the process that generates the events DOES NOT change in time, then the ...

15

@A_A's good answer misses one point: stationarity or nonstationarity are generally only applied to stochastic signals, not deterministic signals. In general, when statistical tests are applied for stationarity or nonstationarity, the deterministic component must be removed first. Hence, in my view, numbers 3, 4, and 5 are non-sensical questions because ...

10

A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where the $\{$ and $\}$ indicate that a set (or collection) of objects is being defined, and the interior says that a typical member of this set is denoted by $X(t)$ ...

6

$\sin(t)$ is no random process, because there's nothing random about it. You could add a random amplitude to get a random process: $$x(t)=A\sin(t)\tag{1}$$ This is a random process because $A$ is a random variable. However, $x(t)$ is not stationary, but it is cyclostationary, i.e., its statistical properties vary periodically. You can make the process $x(t)... 6 In the usual sense of the term, first-order stationarity means that the first-order distribution of all the random variables is the same: each$X_t$has the same CDF, and so the same pdf (or pmf) too if the random variables are continuous (or discrete). If the random variables have a mean, then they all have the same mean. But, a sequence of independent ... 5 What are reasons to choose for cross-correlation or cross-covariance when comparing signals with non-zero mean? Well, part of the issue is that cross-correlation as defined in your equation: $$(f \star g)[n]\ \stackrel{\mathrm{def}}{=} \sum_{m=-\infty}^{\infty} f^*[m]\ g[m+n].$$ will not exist (or be infinite) if$f$and$g$have non-zero mean. So, in ... 4 Note that in general the Fourier transform of a stationary process$x(t)$does not exist. The Wiener-Khinchin theorem only states that under certain conditions the power spectral density of$x(t)$exists, and it can be computed as the Fourier transform of the autocorrelation function of$x(t). Having said that, if for some reason one assumes that the ... 4 It's basically just about using the definitions and doing the math: \begin{align}R_{WW}[n]&=E\{W^*[k]W[k+n]\}\\ &=E\left\{\sum_mU^*[k-m]h^*[m]\sum_lU[k+n-l]h[l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]E\left\{U^*[k-m]U[k+n-l]\right\}\\ &=\sum_m\sum_lh^*[m]h[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\sum_lh[l]R_{UU}[n+m-l]\\ &=\sum_mh^*[m]\left(h\... 4 In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, instead of giving you e.g. a real value gives you a function every time you look at it. We call that realizations. If you now take a lot of these realizations, and ... 4 You're correct as the Cross Correlation function vanishes. This has the implicit assumption the process has zero mean (Actually, at least one of them). Namely, in order to have {R}_{XY} \left( \tau \right) = 0 having X \left( t \right) \perp Y \left( t \right) isn't enough but at least of them has zero mean (Namely, \mathbb{E} \left[ X \left( t \... 3 The mean of X(n) is always \mu_X =0, because the noise has zero mean. Thus we should check whether the autocorrelation corresponds to a WSS process.\begin{align} R_X(n_1,n_2) &=\mathbb{E}[X(n_1)X(n_2)]\\ &=\mathbb{E}[u(n_1)e^{-kn_1}F(n_1)u(n_2)e^{-kn_2}F(n_2)]\\ &=u(n_1)u(n_2)e^{-k(n_1+n_2)}\mathbb{E}[F(n_1)F(n_2)]\\ &=u(n_1)u(n_2)e^{-... 3 Your definitions are not correct. For a Strict Sense Stationary process (signal) the joint distribution of your process' value for all instants of time must be independent of time, in other words if x(t) were your process, P(x(t1),x(t2),x(t3),...) must be independent of time's origin. For a Wide Sense Stationary process the joint probability of process ... 3 1 Proof You're misusing the word proof. Remember that a proof, even one led with stochastic methods, always leads to an absolute "If A, then B, no doubt" statement. Since yourx$is a realization of a stochastic process of which you don't know many properties, pretty much everything is possible – for example, a perfectly white process might have a ... 3 The paper deals in the lag operator$L$, rather than$z^{-1}$as DSP people tend to use. As a result, all the DSP results you know are inverted. So "inside the unit circle" becomes "outside the unit circle", black is white, and ... avoid zebra crossings. 3 I'm pretty sure that even algebraically,$Z$will not be Gaussian. You're multiplying a Gaussian by a random variable that has a distribution that is effectively the histogram of a (co)sine function. Let's do it empirically, though, following the example here. Looking first at the densities, you can see that the density estimate for$Z$(red) is much more ... 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 ... 2 Stationary processes have a spectrum that's time-independent. As the Wikipedia link shows, the colored noises each have a characteristic spectrum, are therefore not time-dependent, and thus stationary. 2 I’m not sure what you mean by the stationary Kalman filter, but it seems to be what I would call the steady-state Kalman filter. If that is the same thing, then you just solve for the Kalman gain at$t=\infty$and apply the normal Kalman filter equations. I’m not sure you mean “the Kalman gain goes to infinity” in a well-posed problem. 2 The decimator by integer$M$can be shown to be the following block: $$x[n] \longrightarrow \boxed{ \downarrow M } \longrightarrow y[n] = x[Mn]$$ Assuming that the input$x[n]is WSS it has the ACS as $$r_x[k] = E\{ x[n] x^*[n+k] \}$$ Then the output autocorrelation can be defined as: $$r_y[n,n+k] = E\{ y[n] y^*[n+k] \} = E\{ x[Mn] x^*[Mn+Mk] \} =... 2 It is not true that you can convert any non-stationary random process into a WSS random process by just adding a random phase. What is true is that you can make a (wide-sense) cyclostationary process WSS by adding a random phase. A common example is the PAM process mentioned in your question:$$x(t)=\sum_ka_kp(t-kT)\tag{1}$$The discrete values a_k in (... 1 You have described a discrete-time Gaussian white noise to a T. For a white stationary Gaussian random process n[k], the autocorrelation R_n[k] is:$$ R_n[k] = \mathbb E \left( n[m]n[m+k] \right) = \sigma^2 \delta[k], $$where \sigma^2 is the variance. The power spectral density P(\omega) is just the DTFT of the autocorrelation, which is:$$ P(\... 1 assuming ergodicity... ACF: \begin{align} R_x[k] &= \lim_{N \to \infty} \tfrac{1}{2N+1} \sum\limits_{n=-N}^{N} x[n] \, x[n+k] \\ &= \mathbb{E}\Big\{ x[n] \, x[n+k] \Big\} \\ &= \sigma_x^2 \ \delta[k] \\ &= 1\ \delta[k] \\ \end{align} PSD: $$S_x(e^{j\omega}) = \sum\limits_{n=-\infty}^{\infty} R_x[n] \, e^{-j \omega n}$$ 1 I believe that you are thinking that each value ofX_t$is determined by a different realisation of$Y$, which in this example is not true. Suppose that$Y$is the value that comes out from a dice throw. Thus,$Y$can achieve any integer value between$1$and$6$. Throw the dice once. Suppose that you get$2$. Then, according to the definition in the ... 1 You're on the right track. The cyclic autocorrelation can be estimated by inverse Fourier transformation of the cyclic periodogram. If$X_T(t, f)$is the Fourier transform of your data block (indexed by time$t$), and$\alpha\$ is your desired cycle frequency, you can form the cyclic periodogram by shifting the Fourier transform to the left and to the right ...

1

I'd recommend looking into the relation of correlation and covariance; the one is just the "bias-corrected" version of the other. Then, use the Wiener-Khinchin¹ theorem: If, and only if, the signal is weak-sense stationary, the Fourier transform of the autocorrelation of that signal is the same as Expectation value of the magnitude-squared Fourier transform ...

1

The FT of a finite length sine wave burst consists of more than just the one frequency of that sine wave. A finite width time domain signal has an infinite width FT. Same for an FFT, a rectangular window produces a full width periodic Sinc in the FFT result. So even if the phase of the 1 sine wave result bin stays constant, the phase of all the other ...

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