# Tag Info

9

Because complex exponentials $e^{\jmath \omega t}$, which are results of Fourier transform, are the eigenfunctions for linear, time invariant (LTI) systems. See eigenfunction of LTI. Also see this answer on SP.SE. Thus, Fourier transform is useful for analyzing linear (not suitable for non-linear one) time invariant (can be intepreted as stationnary) system....

7

well i'm assuming you mean "conventional" DACs and not $\Sigma \Delta$ DACs. in a conventional DAC (like an R-2R ladder or something), there are the micro errors that occur between neighboring DAC codes. e.g. non-monotonicity. i think the DSP solution to that is adding a teeny amount dither noise to the value that is output to the DAC. there is a more ...

7

Yes the MATLAB code is correct. Be careful though, the bandwidth of the signal squared is twice that of the signal itself, which may lead to aliasing if the sampling frequency is too low compared to the signal bandwidth. This can be remedied by properly resampling the signal to a higher sampling frequency with a lowpass filter, then squaring, low-pass ...

6

I think you mean "images", not "aliases". They become aliases if there is foldover from resampling. It's because you are not adding two signals, $x(t)$ and $\operatorname{III}(t)$, you are multiplying them that these images appear. \begin{align} x_\text{s}(t) & \triangleq x(t) \cdot \operatorname{III}(t/T) \\ &= x(t) \cdot \sum\limits_{n=-\... 5 The system must be time invariant and smooth in the functional derivative sense. That doesn't guarantee that the Volterra series converges (like with Taylor series, there are pathological counter examples), but almost all systems that have these properties have a convergent Volterra series. The problem in practice is that the required expansion order for ... 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 If the system is nonlinear then if y_1(t) is the response to the signal x_1(t), and y_2(t) is the output given input signal x_2(t) then the response to the signalx(t)=a_1x_1(t)+a_2x_2(t)\tag{1}$$with arbitrary constants a_1 and a_2 will generally not be equal to$$y(t)=a_1y_1(t)+a_2y_2(t)\tag{2}$$However, for the given system an input ... 5 As Matt L. says you'll need to check for homogeneity and, possibly, additivity. Homogeneity That test says that if:$$ y[n] = f(x[n]) $$then$$ A \cdot y[n] = f(A \cdot x[n]) $$for all scalar A. Additivity This test says that if$$ y_1[n] = f(x_1[n]) $$and$$ y_2[n] = f(x_2[n]) $$then$$ y_{\rm tot}[n] = f(x_1[n] + x_2[n]) = y_1[n] + y_2[n] $$You ... 4 A transform being linear has very little to do with its ability to analyze linear or nonlinear systems. The wavelet transform W[s(t)] of a signal s(t) is linear because$$W[a s_1(t) + b s_2(t)]=a W[s_1(t)]+b W[s_2(t)]$$for real or complex a and b. The signal you're analyzing is just a signal, it has no concept of linearity. However, if you try to ... 4 For -1 <= x <= 1, let's compare Chebyshev polynomials of the first kind, T_n(x), and the basis functions of the Fourier cosine series, F_n(x): F_n(x)=\cos(n \pi x) T_n(x)=\cos(n\ \text{acos}\ x) Writing T_n(x_T) = F_n(x_F) and solving for x_F gives x_F = (\text{acos}\ x_T) / pi, revealing that the Chebyshev polynomial series is ... 4 A flavor of dynamic convolution (is that a trademark by the way?) has a different impulse response g_i associated with each range of instantaneous input. A number of ranges can be defined by fuzzy membership functions f_i(x) (Fig. 1). Figure 1. Amplitude ranges that each use a different convolution kernel. Omitting time indices, the input x and ... 4 Using those four basic elements will allow you to implement linear systems, which can change the magnitude and phase of the input signal, but which will not add the harmonics that are expected from a distortion effect. In order to create distortion in that sense (i.e., non-linear distortion) you will need some non-linear element. The most basic ... 4 Hints: Can an LTI system generate components in some frequency \omega_0 if the input signal x(n) was such that X(e^{j\omega_0})=0? Does aliasing do such thing? The answers to these questions are straightforward and, combined, they answer the original question. 3 For this problem you can't use the formula involving |H(f)|^2 because it only applies to linear time-invariant (LTI) systems, and a squarer is obviously a non-linear system. The only way to solve this problem that I can think of is to use the formula$$E\{x^2y^2\}=E\{x^2\}E\{y^2\}+2E^2\{xy\}\tag{1}$$which is valid for jointly Gaussian and zero mean ... 3 Another typical approach, that independently of my other answer works, is predistortion, for example with the look-up table mentioned by robert, or with a correction polynomial. If you can really pinpoint your nonlinearities to a simple digital-in/analog out curve, you can just find the inverse of that curve, and put it in a correcting mapping, and apply ... 3 After going through the literature regarding HHT and EMD, I found that the "Huang" part of HHT comes from the fact that he is the one who proposed EMD in the first place. That explains the name of the method... For more developments regarding EMD and HHT, I recommend the papers by Rilling et al. "On empirical mode decomposition and its algorithms". For the ... 3 Is this a well-known phenomenon? Yes, of course. You will see harmonics as soon as your clip point is lower than the maximum amplitude in the time domain. The latter is a function of the relative phases between the harmonic components. In your case the max amplitude is indeed 2.5 (plus whatever the noise adds). If you change the phases you will get a ... 3 I faced the same problem in the past. Perhaps there is a way without adding a delay but I haven't found it. You need to realize that your 3 first solutions (delay after vq, delay at the delta_freq and delay after the frequency) will yield the same result as omega_g is a constant and because your PI controller has fixed coefficients. Anyway, place the ... 3 It is very strange phenomena that one object is completely dropped out of attention of researchers. It is Urysohn operator. First of all Urysohn is equivalent to multiple parallel Hammersteins and Urysohn followed by static nonlinearity is a model of any deterministic dynamic object, it maps any given input to any provided output. I obtained Ph.D. in ... 3 You have$$ f\left(\mathbf x, u\right) = \begin{bmatrix}\frac{-1}{T}\tau+\frac{K}{T} u \\ \frac{\tau}{mr} \\ 0 \end{bmatrix} \tag a $$From which you (eventually) derive$$ \mathbf {A}_d=\begin{bmatrix} 1-\frac{\Delta T}{T} & 0 &0 \\ \frac{\Delta T}{m_{op} r} & 1 & \frac{-\tau_{op} \Delta T}{m_{op}^{2} r}\\ 0& 0 & 1 \end{bmatrix} \tag ...

2

As I said in the comments, just follow the 1D case from Wikipedia and augment it with the extra $y$ and $z$ dimensions (and velocities): $$\mathbf{x}_k = \left [ \begin{array}{c} x\\ \dot{x}\\ y\\ \dot{y}\\ z\\ \dot{z} \end{array} \right]$$ You will also need to augment $\mathbf{F}$ and $\mathbf{G}$: $$\mathbf{F} = \left[ \begin{array}{cccccc} 1 & \... 2 Check out this paper. I would have made a comment but not high enough rep. http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=1211087&url=http%3A%2F%2Fieeexplore.ieee.org%2Fiel5%2F81%2F27258%2F01211087 Looks like you need multiple in to get subharmonics in Volterra series The abstract states "Subharmonic generation is a complex nonlinear ... 2 So, the intuitive reaction to this situation is oversampling. Basically, if you use twice the sampling rate, you can always average to samples to get one "output sample value" (thanks, Nyquist!). That would give you one bit of additional per every oversampling factor of two, or$$\Delta b = \log_2\frac{f_\text{sample}}{f_\text{target}}$$Let's introduce ... 2 If your nonlinearity can be expressed as a polynomial (i.e., in terms of addition and multiplication), you can make use of: The linearity of the Fourier transform, i.e., if f and g are (benign) functions, a and b are numbers and ℱ denotes the Fourier transform, then:$$ℱ(a·f+b·g) = a·ℱ(f) + b·ℱ(g)$$The convolution theorem, which states that ... 2 Dynamic convolution is a model for systems that can be written as follows:$$ S[x] = \sum_k (L_k \circ N_k)[x]$$where L_k are linear time-invariant systems and N_k are non-linear memoryless time-invariant systems. The order of the factors is important! Such systems do not necessarily have a Volterra expansion, so you cannot directly compare them. ... 2 As already suggested by Robert and Olli, a system that maps x(t)=k\cos(2\pi f_0t) to y(t)=k\cos(4\pi f_0 t) can be formalized as$$y(t)=|x(t)|_{max}\left(2\left(\frac{x(t)}{|x(t)|_{max}}\right)^2-1\right)\tag{1}$$which is a time-invariant non-linear system. However, I doubt that this system works well (i.e., sounds good as a distortion effect) when ... 2 Well, any input-output representation obviously admits a state-sapce form. for your equation in y[k] you can easily construct one as follows. Create a "shift" system (an integrator chain) as$$ \begin{aligned} x_1[k+1] &= x_2[k],\\ x_2[k+1] &= x_3[k],\\ &\vdots\\ x_n[k+1] &= y[k] \end{aligned}  In this way indeed you have $x_n[k] = y[k-1]$...

2

After skimming through the paper, I can see more clearly now. The measure $D_3$ quantifies the relative strength of the 3rd order intermodulation product. If two sinusoidal signals with frequencies $f_1$ and $f_2$ are input to a non-linear device (such as a microwave amplifier, as referred to in the paper), there will be intermodulation products at the ...

2

In general there is no systematic way and you simply have to analyze the given system. In the case of the system in your question, it's easy to see that it can't be invertible, because the output is just a constant, namely the integral over the input function (assuming this integral exists). There are infinitely many functions which will result in the same ...

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