# Tag Info

5

The reason is Euler's formula, from which you get $$\cos(\omega)=\frac12\big(e^{j\omega}+e^{-j\omega}\big)\tag{1}$$ and $$j\sin(x)=\frac12\big(e^{j\omega}-e^{-j\omega}\big)\tag{2}$$ If you have symmetric or anti-symmetric coefficients, the corresponding frequency response can always be decomposed in purely real-valued cosine terms $(1)$ or purely ...

4

Here's Octave code for a (digital) RRC filter: pkg load signal; % Octave needs this; MatLab doesn't Fs = 16000; % sample rate Rs = 400; % symbol rate sps = Fs/Rs; % samples per symbol % % Root raised cosine pulse filter % https://www.michael-joost.de/rrcfilter.pdf % r = 0.22; % bandwidth factor ntaps = 8 * sps + 1; % Pulse duration is 8 symbols st = [-...

2

When you upsample a signal by a factor of 4, then, you have to churn out 4 times the number of samples in the same time in order to maintain the same Symbol Rate. Otherwise your symbol rate will be $\frac{1}{4}$ times the original symbol rate. This means your DAC needs to operate at least at 4 times the sampling frequency now. This can be costly. Also faster ...

2

For each sample, we note $1$ if error and $0$ otherwise. Then, for bit error rate $q$, the tests are independent Bernoulli random variables $\{X_i\}$ with probability $q$. The estimate is $\hat{p} = \frac{S}{n}$ where $$S = \sum^n_i X_i \tag{1}$$ and $\mu = \mathbb{E}[X] = np$. Chernoff bound (see Corollary 5) tells us that for every $0 < t < \mu$, \...

2

Based in my comment, here are quite accurate coefficients for A-Weighting filter to be used with 33kHz sampling only: b = [0.759941332414235 -1.931718891800229 0.5144488202077094 2.040523943539552 -1.283344336756084 -0.5073918947145036 0.3880093863750892 0.03272320276137633 -0.01319156202714533] a = [1 -2.840785425332851 1.688214038430988 1.889562806030876 -...

2

Analyzing signals per segments, with proper windowing, is a way to cope with non-stationary in audio samples. With full-size analysis, features can get mixed. Segment-splitting is thus at play in many algorithms (mp3, shazam). The length of window is often a matter of trade-offs, between data information and computing advantages: signal sampling (window ...

1

20Vrms is the maximum voltage you can apply across the transmitting transducer without the risk of immediately damaging it. The amplitude of sound it produces is determined by the driving voltage. The transmitter is characterized at 10Vrms, so about 28Vp-p assuming a sine wave, probably where you would prefer to use it for reliability and long life.

1

If you have a pure tone signal that has a frequency higher than the Nyquist frequency, it doesn't "disappear", it looks like it has a different frequency (mirror image around the Nyquist). This is called aliasing. If it happens to be exactly at Nyquist, it could disappear if your sampling happens to hit the zero crossings. The same is true for a ...

1

Nyquist didn't say that. The Nyquist-Shannon sampling theorem says that the sampling rate asymptotically approaches twice the bandwidth of the signal, more or less no matter where the signal's center frequency is. So you care about the width of the signal, not its maximum frequency. But where you got the expression wrong in one direction pertaining to ...

1

Probably no, but... maybe. The classical sampling theorem is often misstated: if a continuous signal is band-limited, it can be recovered exactly (in theory) if the sampling frequency is twice its maximum frequency. However, there are cases where signals can be recovered above the Nyquist limit, provided that we have more information, because ambiguities may ...

1

First, learn about digital differentiators and how to implement them with software. Then apply your accelerometer output signal to a digital differentiator. Next, use software to detect a large positive-valued sample followed by a large negative-valued sample in the differentiator's output sequence. That sample will correspond to your "first selected ...

1

If the signal is oversampled and the PSF variation corresponds (approximately) to a smooth local compression/expansion, perhaps you can resample y so as to make the PSF approximately LTI, then apply conventional methods (somewhat akin to homomorphic processing) If the input signal is convolved with a small discrete set of PSFs, perhaps you can devonvolve ...

1

The power spectral density (PSD) is a natural measure of the signal's power content with respect to frequency. A central part of non-parametric signal processing is to provide a "best" estimate of the "true" PSD from knowing only one or some "realizations" with finite length. By taking into account the influence of stationary ...

1

The typical method is to use the cross correlation function to recognize the start of each pulse sequence. Typically pulse patterns are chosen for good autocorrelation properties for this very purpose (likely the start of the packet as a header if there is a random data payload also involved). "Good autocorrelation properties" refer to an ...

1

if the implementation preserves bits until quantization must occur at the final output, the Direct Form is far simpler than the Transposed Form. using the transposed form requires that your states have double-wide word widths unless you quantize each of those states back to single width. but that is more quantization error than if you just add up a bunch ...

1

I don't see how memory or computational burden can possibly be reduced between the two implementations (note this is not implying simplicity, just number of computations required, please read on...) Usually the decision to use transposed-form is for high speed FPGA implementations with a large number of taps as you can eliminate a long adder tree which would ...

1

Have you considered reading the documentation ? https://www.mathworks.com/help/curvefit/fit.html Goodness-of-fit statistics, returned as the gof structure including the fields in this table. sse: Sum of squares due to error rsquare: R-squared (coefficient of determination) dfe: Degrees of freedom in the error adjrsquare: Degree-of-...

1

Hi Nouali Ibrahim Yassine, Without any additional information about the two signals, what you are trying to do is a blind source separation. This means that you will have to use a set of advanced statistical tools ( principal components analysis or independent component analysis for example, I would go more for the Independant component analysis) to try to ...

1

Hint: (1) $sinc(\frac{t}{T})$ is the inverse fourier transform of a $rect$ function in frequency domain with $f=[-\frac{1}{2T}, \ \frac{1}{2T}]$. Time-shift will only result in a phase term in Fourier transform which does not impact the bandwidth. (2) $z(t)$ is product of two $sinc(\frac{t}{T})$ in time-domain. Hence, its frequency domain representation will ...

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