jan
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Most certainly not. While there has been some claims to break Shannon here and there, it usually turned out that the Shannon theorem was just applied in the wrong way. I've yet to see any such claim ...

A signal having a mean-value or DC component of zero is commonly referred to as mean-free or as having no DC component. It does not mean that it cannot be averaged, just that the average comes out as ...

Depends on what you are controlling. For DC-motors it is the inertia of the device that acts as a low-pass filter of the PWM modulated signal resulting in a continuous motion. For most LEDs it is the ...

Follows from the DFT definition. It's defined as $$X(k) = \sum_{n=0}^{N-1} x(n) e^{-j2\pi \frac{kn}{N}}$$ So $X(0)$ is X(0) = \sum_{n=0}^{N-1} x(n) e^{-...

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 ...

Unfortunately you did not mention what your channel is so I assume that merely AWGN is present. Furthermore I assume symbol-by-symbol detection is desired since a decoder for the Hammingcode is ...

The OFDM signal as a whole is affected by frequency selective filtering. It is usually designed such that the subcarrier bandwidth is smaller than the channel coherence bandwidth. This yields you ...

Yes, for example let $$x[k]=1$$ for all $k$ and $$y[k] = \begin{cases}1 & k=0\\-1 & k=1\\0 & otherwise \end{cases}$$ It is easy to see that in case of a convolution, the result will be ...

It comes down to latency vs. complexity. If your filter is 10 seconds long, you need to store the audio data of the last ten seconds and then you are able to calculate the current output audio sample ...

In addition to Peter's answer, if you have a nonlinear system that is well-behaved in a sense of being only mildly nonlinear or at least exhibiting no discontinuities, special variants of the Kalman ...

Hard to believe, that you actually looked. You probably don't even need a DSP for that, a normal microcontroller should be sufficient. For example, the STM32F4-Discovery board comes with an ARM ...

There is no general formula to calculate time constants for all signals. Time constants are defined for some signals. For your exponential signal e^{-a \cdot t} = e^{-t / c} \end{...

Since it will most likely not hurt to have the possibility for more streams, this might for one be just to future-proof the system. Nothing would be more stupid than to specify for 8 channels and ...

You are most likely not having a control vector $u$. Maybe you can somehow model it into the filter, but it is not required in your case and uncommon. Furthermore $\overset{.}{x}$ is not your gyro ...

Regarding the problem with high correlation values, this happens when both signals are not zero-mean. Assume two signals $x[k]$ and $y[k]$ of length $N$ each. Each signal can be decomposed into the ...

Well, a DC signal for sure. Other than that, there is probably more context from the book required (especially about the assumptions).

Well you could try to estimate a transformation matrix that maps between the coordinates in the two images. The idea is that \begin{bmatrix} x_a\\ y_a \end{bmatrix} = \begin{bmatrix} ...

In addition to Deve's very good answer above, it should be noted that the phase difference $\arg(r_n^* r_{n+N})=2\pi\beta$ is circular and you cannot tell the difference between $0, 2\pi, 4\pi, ...$. ...

Seems to me that the DSP position goes the same way as most jobs in large corporations. Join a big enough company and you will specialize yourself because for the other things there is a guy who ...

You could normalize the average amplitude, i.e. YData = YData / mean(abs(YData)); Or you could normalize the signal power to one, i.e. YData = YData / sqrt(mean(abs(YData).^2)); If just the peaks ...

What you describe is a direct-sampling architecture. Those are practically available for radio amateurs using frequencies up to 30MHz. They are also actively researched to open up their usage for much ...

Filtering can be done in the frequency domain which is actually a very efficient technique (and it can very well be, that Matlab does this internally). However, for very long signals it's not as ...

I can't tell about SciLab but if it is not implemented, you can always get the circular convolution by IDFT(DFT(x1) * DFT(x2)) where x1 and x2 are your signal vectors and the multiplication is ...

This can be split up in two questions for the sampling frequency and the window. 1) The required sampling frequency is given by the Nyquist theorem to be $f_s>2*f_{max}$ with $f_{max}$ being the ...

Since the signal is most certainly real-valued, you can ignore the FFT values after $f_s/2$ (or sample 300 in your plot) since they will mirror the first ones. With a sampling frequency of 2500Hz, the ...

Well, for a signal $x[k]$, the energy is: $$E=dt*\sum_{k=-\infty}^{\infty}x^2[k]$$ You want the energy to be one by dividing $x[k]$ through a value $a$, hence: \begin{...

Well, since you simulate you could calculate the residual frequency offset which is basically the remaining CFO after correction or the difference between the true CFO and the estimated CFO. Another ...

It's actually a band-pass characteristic, since very low frequencies will be inaudible on headphones. Firstly, searching for "headphone frequency response" brings up a lot of results, especially on an ...

No, that is not true. It is $$|a+b| \leq |a| + |b|$$ Edit: This is not true for the power which was asked for by the OP (see the comments). I over-read the re-definition ...
The inner product or dot product or scalar product will not output a vector, it will output a scalar which is (with exception of the $1/n$ in front of the sum) the same as the first equation that you ...