If I have for example DC signal (voltage and current) and I select on my "scopecorde/datalogger" LPF (low pass filter, not digital! with 5 Hz) what happened on output signals? Is that the same as "mean value" of input signal?
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4\$\begingroup\$ I don't think "LPF" is either definitive or generic enough to answer this question without examining what your device actually does. My advice is read its data sheet. \$\endgroup\$– Andy akaCommented May 21, 2014 at 21:12
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\$\begingroup\$ Well "device" was mentioned just for example.... \$\endgroup\$– McJackCommented May 21, 2014 at 21:24
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3 Answers
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If it's a DC signal, then, yes, there should be no change in the measured signal. If, however, you have other signals, such as, for example, a 1 kHz signal, this would be significantly attenuated. If you want a bit more of the math here, try to think of your signal in the Fourier domain. The Fourier transform of a signal breaks a signal into a number of sinusoids at many different frequencies that, when added together, recreate your signal. The zero-th term corresponds to your DC signal. Higher terms represent AC components of the signal, starting at the fundamental frequency, a sinusoidal signal with period equal to your measurement duration. These higher terms will also be present in the output signal, until the frequency of the term is higher than the cutoff frequency of your low-pass filter. Higher frequencies are attenuated.
As for "mean value", you need to be careful of your word choice. The mean value of a signal is its average value. More commonly encountered is the root mean square (RMS) value. So, for a signal V(t): $$V_{RMS}= \sqrt{{1\over{T}}\int_0^TV(t)^2dt}$$ This is a bit different from your zero-th term DC signal.
answered May 21, 2014 at 22:27
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A low pass filter will take your input DC signal and attenuate any AC riding on it before outputting it.
In your case, if the "knee" of the filter is located at 5 Hz, then a 5 Hz signal on the input will be attenuated by 3 dB at the output with a rolloff of 6dB per octave - after that - as frequency increases.
answered May 21, 2014 at 22:30
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\$\begingroup\$ Actually, the rolloff slope is not specified, although a single pole (10 db/decade) is almost certainly what will happen. \$\endgroup\$ Commented May 21, 2014 at 22:43
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A low-pass filter is the indeed the same as some kind of weighted average of some window of the history of the signal.
An ordinary single pole RC filter, when regarded in the time domain, is basically an exponentially weighted average of the entire history of the input signal.
Exponentially weighted means that the recent values of the input signal matter a lot more than older values, according to an inverse exponential function.
Why the weight is exponential is because the charging or discharging of a capacitor is exponential. Think about the response of the system to a step function. If a step function is averaged with a exponential weight function, the result is exactly the charge/discharge curve of the capacitor. Since the system is linear, that step response "tells us everything".
