This is what I did
I expected a nice square wave on my soundcard oscilloscope software but I'm getting this
What is happening ?
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\$\begingroup\$ Well written question. \$\endgroup\$– KalleMPCommented Jun 19, 2016 at 18:26
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1 Answer
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AC Coupling.
Sound cards have a capacitor in series with the input which acts as a high pass filter, removing frequencies lower than around 20Hz.
You appear to be feeding in a signal of only 2Hz which is basically going to get filtered into nothing more than an impulse response at each transition as you are seeing. The reason you see the impulse is that a step change in the signal voltage has a large bandwidth requirement - having many higher order harmonics which will not be filtered out.
answered Jun 19, 2016 at 18:06
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3\$\begingroup\$ Ah ! I see , Thanks , it works perfectly now that I have increased the frequency . \$\endgroup\$– VriskCommented Jun 19, 2016 at 18:09
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5\$\begingroup\$ @AnirudhGanesh AC Coupling. You signal that goes from 0-5V is essentially a +/-2.5V signal with a 2.5V DC offset. What frequency is DC? :) \$\endgroup\$ Commented Jun 19, 2016 at 18:13
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1\$\begingroup\$ .... 0 ? wouldn't that mean the DC part will be filtered out completely and nothing else would happen ?? \$\endgroup\$– VriskCommented Jun 19, 2016 at 18:24
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4\$\begingroup\$ @AnirudhGanesh correct, the DC part goes away. So you are left with just the +/-2.5V square wave, hence you see negative. \$\endgroup\$ Commented Jun 19, 2016 at 18:25
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2\$\begingroup\$ Not quite - the capacitor-resistor combo means that changes that are fast enough get passed through, but levels decay rapidly to 0V on the soundcard side of the capacitor. So the rising edge of the square wave turns into an upward spike. Then it decays. Then the falling edge turns into a downward spike. \$\endgroup\$– pjc50Commented Jun 19, 2016 at 19:09