timer switch

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i see.. so here the frequency would be so high so the signal transfers with a phase difference ~= 0? and the transfer function is 1?
 
A pulse edge has a very high (fourier) frequency associated with it, approximately 0.35/Tr, where Tr is the edge rise-time. Thus the edge goes through the capacitor with no significant phase delay or attenuation. But the flat part of the pulse looks like a low frequency which causes the resistor-capacitor combination to generate an exponential decay of the voltage toward the DC level. Therefore the positive going pulse edge (0 to V+) will generate a positive output spike decaying to the DC level, and the negative going edge (V+ to 0) will generate a negative output spike, also decaying to the DC level.
 
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