# Tag Info

## New answers tagged frequency-spectrum

0

You are on the right track. After upsampling the spectrum of X would look the same except the x axis would be divided by 2. So the limit of the spectrum would be at pi/4. Yes, that's correct. Since the filter has amplitude 2 would the amplitude of X be increased from 1 to 2 or will the amplitude of X remain unchanged? Recall that in the frequency ...

0

This seems to be a textbook case for a narrow median filter. Your $S$ is very broadband compared to your $N$, so a median filter will not affect your signal, as it can be designed quite narrow. And by using median instead of averaging (i.e. lowpass) you minimize the error introduced by the anomalies. I would give it a try.

0

To complement the other quite appropriate answers: Say you build a real signal $s(t)$ with bandwidth $B$ that transmits $R$ bits per second. Its spectrum $S(f)$ extends from $-B$ to $B$. The modulation property of the Fourier Transform tells you that the spectrum of $$s(t)\cos(2\pi f_c t)$$ is $$\frac{1}{2} \large[ S(f+f_c)-S(f-f_c) \large].$$ Note that ...

1

Because bit rate depends on the bandwidth and not on the carrier frequency. Of course at higher frequencies you have more bandwidth, and thus you can transmit more data. But 1 MHz in lower frequencies and 1 MHz at higher frequencies have no difference on the data rate. Other effects may need to be taken into consideration though at higher frequencies. For ...

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If you are a bass and sing A 220 Hz and B 247 Hz at 120 beats per minute, you can sing data at a certain rate. If you are a soprano and sing two quarter notes A 880 Hz and B 988 Hz at the same BPM, your data rate isn't any higher. If you sing more notes (higher bandwidth) you could communicate a more complex score (e.g. a higher data rate). So bandwidth (...

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Here is a link with a bunch of stuff relating to HVR and how to calculate various features using python discrete-heart-rate-signal-using-python-part-1. Hope it helps.

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Derivative spectroscopy was at its peak in the 70-80s. It is still very useful. The only problem with derivatives is that they enhance noise like crazy i.e., they suppress low frequency signals and enhance high frequency signals. Before you do anything with the derivatives, you have to apply a smoothing process on them, specially on the second derivative. ...

-1

$$R_{n}(\tau) = E(n(t)n(t-\tau))$$ $$R_{n_i}(\tau) = \cos(2\pi f_ct)\cos(2\pi f_c(t-\tau))E(n(t)n(t-\tau))=\cos(2\pi f_ct)\cos(2\pi f_c(t-\tau))R_{n}(\tau)$$ Using $\cos(a-b) = \cos a \cos b + \sin a \sin b$, we get $$R_{n_i}(\tau) = \cos(2\pi f_ct)[ \cos(2\pi f_ct ) \cos(2\pi f_c\tau)+ \sin(2\pi f_ct) \sin(2\pi f_c\tau)]R_{n}(\tau)$$ or R_n(\tau) = \...

1

I try to think of it this way: In the time domain, when downsampling occurs, the signal gets compressed; while on upsampling, the signal gets stretched. Then, from the Fourier transform we know that time stretching means frequency compression, and vice-versa. It may not be a rigorous answer, but hope this helps!

2

Much depends on what you want to deduce from your data. In general, if your sample rate is not uniform, your measurements should be accurately time stamped. The nearer to an average of 1/10 a second your sample intervals are the better. A typical hueristic used in Engineering is the 1/10 rule, so if your samples are within 1/100 of a second of 1/10 ...

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