Hi all,
I'm confused about something which seems simple. And yet I can't wrap my head around it.
Say you have a sine wave with amplitude 2V, which is the input to a passive differentiator (or high pass filter, which is the same, correct?) with a 100k resistor and a 10 nF cap.
If you look at this from frequency perspective and work with complex impedances, the cutoff-frequency is 1/2piRC = 159 Hz. At this frequency, the output will be 45° out of phase with the input; the output voltage will be 1V. A 100 Hz sinewave will be more out of phase (some 57°), with an output voltage of 0.628V. Higher frequencies (10 kHz) will probably be completely in phase with the input and suffer no voltage loss, hence a high pass filter.
Now if you analyse this circuit in the time domain to look for its differential behaviours, (Vout = -RC*d(V)/dt), and you fill in, for example, a sine wave with 100 Hz frequency, you end up with: 0,628*cos2pi100t; the same amplitude as you'd analyse it in a frequency domain, 0.628V, but the differential of a sine wave is a cosine wave... which is 90 degrees out of phase with the original sine wave... which doesn't add up with the phase calculations I made above, using complex impedances.
If I look at this on an oscilloscope, the amplitude is as expected and the output wave is 57° out of phase with the input. So it's no pure cosine-wave. Then why does the differentiator-equation comes up with this (pure) cosine-wave as the solution? I'm filtering (the output voltage is reduced), but "how much" am I differentiating?
Thanks for clearing this up!
I'm confused about something which seems simple. And yet I can't wrap my head around it.
Say you have a sine wave with amplitude 2V, which is the input to a passive differentiator (or high pass filter, which is the same, correct?) with a 100k resistor and a 10 nF cap.
If you look at this from frequency perspective and work with complex impedances, the cutoff-frequency is 1/2piRC = 159 Hz. At this frequency, the output will be 45° out of phase with the input; the output voltage will be 1V. A 100 Hz sinewave will be more out of phase (some 57°), with an output voltage of 0.628V. Higher frequencies (10 kHz) will probably be completely in phase with the input and suffer no voltage loss, hence a high pass filter.
Now if you analyse this circuit in the time domain to look for its differential behaviours, (Vout = -RC*d(V)/dt), and you fill in, for example, a sine wave with 100 Hz frequency, you end up with: 0,628*cos2pi100t; the same amplitude as you'd analyse it in a frequency domain, 0.628V, but the differential of a sine wave is a cosine wave... which is 90 degrees out of phase with the original sine wave... which doesn't add up with the phase calculations I made above, using complex impedances.
If I look at this on an oscilloscope, the amplitude is as expected and the output wave is 57° out of phase with the input. So it's no pure cosine-wave. Then why does the differentiator-equation comes up with this (pure) cosine-wave as the solution? I'm filtering (the output voltage is reduced), but "how much" am I differentiating?
Thanks for clearing this up!