# Tag Info

12

Sorry for the long winded answer, but making a good acoustic measurement is hard. Here are some of the steps involved (and at some point, we'll actually get to the direct answer of your question). First, you have to make sure that the system you are measuring is actual linear and time invariant (otherwise you can't deconvolve). This can be tricky on a PC if ...

10

The elements building a lumped system are thought of being concentrated at singular points in space. The classical example is an electrical circuit with passive elements like resistor, inductance and capacitor. The physical quantities current and voltage are functions of time (only). E. g. the current at a capacitor with capacity $C$ is given by $$i(t) = C\... 7 A causal system does not need to know the future in order to compute its output. A memoryless system computes the output only from the current input. A memoryless system is always causal (as it doesn't depend on future input values), but a causal system doesn't need to be memoryless (because it may depend on past input or output values). The system$$y[n]=x[...

7

Short answer: You can't. If an attacker can insert a signal that covers the whole bandwidth (e.g. a white signal, or at least one that has no spectral zeros) into the system (and he can do that over an arbitrarily long time, or add up observations), they will get an output, and can through the magic of correlation get the impulse response.

6

It is important to realize that the terms lumped or distributed are not properties of the system itself. These properties are related to the size of the system compared to the wavelength of the voltages and currents passing through it. So a resistor is or isn't a lumped element (even though it is usually meant to be one), depending on the frequency of the ...

6

The theory behind sweep-sine measurements of LTI systems requires a signal with constantly changing the frequency. You cannot simply playback few tones - the whole frequency range is necessary. So that if you want to identify your system with the impulse response $h[n]$, you feed the sweep sine signal $s[n]$ into it and record the output. Obviously output ...

5

You can try to remove noise (especially occasional spikes) by non-linear filter. I suggest to use median filter (http://en.wikipedia.org/wiki/Median_filter), as your spikes has length only few points (as I see). There is matlab function - medfilt1. It is probably that such filtration will be enough.

5

I believe there's either a mistake in the presentation or the presentation is using a different definition of linear. For example, the system is linear in $x$ from a system perspective, but it's affine in $x[n]$ (and, therefore not linear) because of the $bx[n-3]$ offset. On this site, we tend to go with the system definition rather than split hairs about ...

4

No, it can be rewritten as $y[n] = x[n]/2$. Basically, the output is the input divided by 2...

4

In the following, I suggest that, before using the generic $T(\alpha_1 x_1+\alpha_2 x_2)$ versus $\alpha_1 T( x_1)+\alpha_2T( x_2)$, it can be more informative to try with simpler partial tests, or try counterexamples, based on your intuition. I don't really understand the motivation behind the second group of equations, or why $b$ gets multiplied by $\... 4 [Note: it may happen that a teacher makes a oral mistake, that puzzles the audience. So here is an alternative explanation on this system being non-something] This system is, as far as Peter K., Matt L. and I know, nicely linear. You already did the computations. With a little more work, among classical properties, it is also time-invariant, causal, stable. ... 4 The system $$y[n] = x[n] \star (u[n]-u[n-2])$$ where$u[n]$is the unit step function, has memory. Indeed the system is equivalent to $$y[n] = x[n] \star ( \delta[n] + \delta[n-1] ) \implies y[n] = x[n] + x[n-1]$$ and as it's clear from the given I/O relationship, the current value of the output$y[n]$, depends on the values input$x[n]$at other times ... 3 I think your E-step is correct(only one term missing in the last expression:$-N\ln\sigma_w$). To obtain the M-step you have to differentiate with respect to all your parameters. You don't have to include$u(n)$in$\theta$, since it's defined by$p$. So you compute$\frac{\partial Q}{\partial\theta}=0$. Solving for$\theta$, and this is your new$\theta$... 3 conceptually, any driving signal with bandwidth that covers the maximum expected bandwidth of the system being identified is sufficient. you drive input and synchronously measure input and output, use the FFT to compute the spectra of both input and output, divide the output spectrum by the input spectrum, inverse FFT the resulting spectrum and you have the ... 3 The Periodogram implicitly uses a window to get an idea of the spectrum. Implicitly this is either a rectangular window, with frequency response which is like$\dfrac{\sin(\omega n)}{N\sin(\omega)}$, or a triangular window$(N-k)/N$. Roughly, the variance of the periodogram is the square of the Fourier transform multiplied by the square of the window. Note ... 3 Taking a quick look at your acquisition: this doesn't look too terrible in terms of noise, however, the acquisition window looks way too short. With a 60ms window your frequency resolution is only 160 Hz. From the looks of it, you will also get a massive truncation error. As a rough rule of thumb, you want the acquisition window long enough so that the ... 3 Batman has given a great answer. You need to go through the recommended book in order to understand the concepts mentioned. Let me try to simplify it. BIG PICTURE: De-convolution or inverse filtering is required to retrieve an estimate of the original signal that went through an unknown linear system. Basically, we have a signal which went through an ... 3 If by system identification you mean determining the impulse response of a linearized model of your actual system, then pseudorandom binary sequence (PRBS) signals are a good way to go. With chipping rate$T^{-1}$and$N$chips in each period of the PRBS, the PRBS signal has period$NT$seconds, and it is important to choose$N$and$T$so that period of ... 3 For linear systems, you can completely characterize the transfer function using its frequency response, so a frequency sweep would be one possible choice. However, you would need to ensure that at each test frequency, you allow time for the system's transient response to die out before measuring its steady-state amplitude/phase response. 3 In any signal processing problem, there are usually two components: the signal model and the channel model. The signal model is the mathematical description of how your ideal signal, call it$s(t)$is generated. The channel model is the mathematical description of how your channel corrupts or alters the signal. One aim in telecommunications is to look at ... 3 Knowing the amplitude (and phase) for several frequencies allows you to fit a model with as many parameters, hoping the system is linear. With little information you cannot observe the whole system behavior accurately, but just a simplified model. This might be enough for your purpose. Knowing the internal system structure should help you select an ... 3 If the sensor is linear and invariant in time (an LTI system for linear and time-invariant), the output to a sine should be a sine with the same frequency, and a different phase and amplitude. Assuming that you will only probe the sensorin its linearity range (e.g. outside saturation), and that you only have access to the magnitude of the output sine, you ... 3 A system is memoryless if its output ($y(t)$) for each value of the independent variable ($t$in this case) at a given time is dependent only on the input at that same time ($x(t)$). Every system that consists of a delay (like the one in your example) or an accumulator, for example, are systems with memory. This can be seen just by replacing$t$by some ... 3 There's a very simple way to check controllability, indeed if you define the reachability matrix $$R = \begin{pmatrix}B & AB & \dots & A^{n-1}B\end{pmatrix}$$ then the reachable subspace is the image of R. Hence to check complete controllability you just have to check that$R$is full rank. First, I think there's an error in the question,$B$... 3 First, please read this answer of mine for a detailed description of matched filters for real-valued signals. In particular, note that what I called the matched filter for a signal$x(t)$is a(n LTI) filter with impulse response$h(t) = x(-t)$which is better described as the time-reversed signal rather than the "inverse" of the impulse response as you call ... 3 They are system norms, a metric that you can compare two different systems in terms of their generalized gain and spread. You can look these up no need for attaching physical motivation. You don't. They are for assesing and using in the optimization programs. Yes, but only if you know what you are doing. Can be. But again, they are for assessment and ... 3 The system is nonlinear (bilinear in$x$), with a nonlinear law (square) on indices. Odds are the system is not invertible. One can try to prove it in its full generally, or try to find a counterexample. Let us try the lazy way, using properties at hand. The system is bilinear. Hence, the output for$k x[n]$,$k\in \mathbb{R} $, will be$k^2 y[n]$. ... 3 Output of a memoryless system depends only on the current input value and therefore every memoryless system is also causal; since a causal system's output cannot depend on the future input values. The converse in general is not true; causal systems can be memoryless as well as can exhibit memory (if their outputs depend on the past input values in addition ... 3 Long answer: Let's model the information flow from your "hidden" IIR$X$to your observable output$Y$as $$X \longrightarrow Y$$ Then, we call the amount of information you get per observation the *mutual information$I(X;Y)$; that information is the reduction of uncertainty about$X$to be achieved by observing$Y\$. We call the expected uncertainty ...

3

It is very strange phenomena that one object is completely dropped out of attention of researchers. It is Urysohn operator. First of all Urysohn is equivalent to multiple parallel Hammersteins and Urysohn followed by static nonlinearity is a model of any deterministic dynamic object, it maps any given input to any provided output. I obtained Ph.D. in ...

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