In another lesson you should have learned about an electrical filter.  In that filter we did some mathematical analysis that started by assuming a sinusoidal output for the filter, and then worked backwards through the filter to find the input that caused the output.  In that process, however, we didn't give a very good feel for what happens physically in that filter.  We'll fix that omission in this note.

        Let's assume that we have the filter represented by the circuit diagram below.

In this filter, we can examine what happens.  We already should know how the input and output voltage are related.  Here is a graph of the input voltage (red) and the output voltage (blue).

• At the beginning of the cycle (t = 0 on the graph), we have:
• The input voltage is zero (0) volts - shown with an orange dot.
• The output voltage is negative - somewhere around -4 volts - shown with a purple dot.
• There is current flowing from the source, through the resistor, into the capacitor as shown in the circuit diagram below.  That means that the capacitor is charging, and we can see that the output voltage is increasing at that point.

        Now, consider a second point on the graph, as shown below.

• At this point (orange dot) on the graph:
• The input voltage and the output voltage are the same
• There is no current flowing from the source, through the resistor, into the capacitor.  That means that the capacitor is not charging, and it is not discharging, so that the output voltage is not increasing or decreasing at that point.
• That means that the output voltage is at a peak, and it also equals the input voltage exactly.
• We also should note that the current into the capacitor - through the resistor - is positive up until that point.

• At this time on the graph:
• The input voltage - shown with an orange dot, somwhere around -6v - is less than the output voltage - shown with a purple dot, somewhere around +2v.
• At this point, the capacitor is discharging, and we can see that the output voltage is decreasing at that point.  That means that there is really charge flowing out of the capacitor, back through the resistor into the source.
• In the circuit diagram below, the current, I, would be negative.

        Finally, we come to a point where the input voltage and the output voltage are both the same, but they are both negative.

• At this point (orange dot) on the graph:
• The input voltage and the output voltage are the same
• There is no current flowing from the source, through the resistor, into the capacitor and there is no current flowing out of the capacitor back to the source either.  That means that the capacitor is not charging, and it is not discharging, so that the output voltage is not increasing or decreasing at that point.
• That means that the output voltage is at a negative "peak", and it also equals the input voltage exactly.
• We also should note that the current, I, into the capacitor - through the resistor - is negative up until that point.