# Tag Info

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I will introduce some terminology and intuition that will be helpful when reading other references. It will be neither complete nor completely rigorous. The measures that we first encounter in real analysis assign sizes (non-negative real numbers) to measurable subsets of $\mathbb{R}$; Lebesgue measure is the measure that agrees with the intuition we build ...

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Ultimately if your noise is white then the matched filter would be the best approach (multiply by one frequency and sum-- if it is complex you would multiply by the complex conjugate- both a form of correlation). This is equivalent to computing a single bin in the DFT (assuming your frequency is on bin center, otherwise equivalently the operations in the DFT ...

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There is no problem in your plot if by fft you mean DTFT (or Discrete Time Fourier transform) and it is true to say "the points on this dash line are the DTFT result of corresponding frequency": If your function is a cosine: $$x[n] = \cos(\omega_0n) \Longrightarrow X(e^{j\omega}) = \pi\delta(\omega-\omega_0)+\pi\delta(\omega+\omega_0)\ ; \qquad|\... 1 Everything is possible with advanced technology, due diligence and perseverance. To discuss the means to reach your goals requires even more detailed description of target scenarios. If you have a pure tone -- I mean, a musically pure single note performed by a good musician with a quality instrument or an opera singer -- you can analyze this sound by ... 1 After reading the linked article by Julius Smith, flipping the spectrum entirely as the OP is requesting is NOT the intention of that author. That article is shifting the spectrum to the left by only half a DFT bin as an alternate approach to the complexity of creating an analytic signal with the Hilbert Transform for purposes of making an octave band filter ... 1$$ y[n] = (-1)^n \cdot x[n] $$This will flip the spectrum in linear frequency. Easy to do, easy to recover. 1 You're right about the relation between the imaginary part of the frequency response and the odd part of the impulse response:$$\textrm{DTFT}\big\{h_o[n]\big\}=jH_I(e^{j\omega})\tag{1}$$So from the given H_I(e^{j\omega}) you can obtain h_o[n]. Now note that the odd part of h[n] is defined by$$h_o[n]=\frac12\big(h[n]-h[-n]\big)\tag{2} The important ...

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The answer depends on the statistical characterization of the signal. If the random EEG signal is stationary over when seen over 5 second cycles, then you are better off averaging over 5 seconds cycles. If not, then averaging over these 5 seconds cycles might not yield the best estimate. In that case better off averaging over the entire duration.

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I think that with a 250 MHz sampling clock, you should be able to generate a signal with a 125 MHz max frequency. Also, what type of DDS do you use? Is it based on a table of sine/cosine or is it a CORDIC algorithm? It has an impact on precision. Is the signal path from DDS to DAC ok? Do they work at the same sampling frequency? If not, a resampling module ...

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Since sines and cosines can be expressed as sums of complex exponentials, you can always think of any signal as complex. For real valued signals it just works out that the imaginary parts cancel. If the sines and cosines in a real-valued signal use frequencies from 0 to B, the corresponding sum of complex exponentials will include frequencies from -B to B. ...

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