AC Signals

PROBLEM
Consider two signals captured in the same scope display. We find the phase difference between the two signals by measuring the time difference(tx) between corresponding parts of the two waveforms - between the positive peaks. We convert this time measurement into an angle so we need to multiply it by the angular frequency(2*pi*f) and convert the result to degrees(180/pi):

     phase_angle = tx(2pi)(f)(180/pi) = tx(f)(360)

METHOD
Set Function Generator(FG) to CH1, set the FG to produce a 10V peak-to-peak sinusoid at 1kHz.
Calculate the expected Vrms - Expected Vrms = 7.04 V
Calculate the DMM value     - Measured Vrms = 6.7 V


Turn of FG. Set Rbox to 1kOhm. We calculate the complex impedance of the capacitor where Zcap=1/2*pi*f*C >>> Zcap = 1592 ohms

We construct a circuit similar to the figure.

Connect CH1 of the scope to the FG. Connect CH2 to the top of the capacitor (between Rbox and C). Connect Rbox and C as shown. Be sure to make connections as shown. Energize the FG; connect the DMM.
DATA

Measure the peak-to-peak capacitor voltage on CH2: Vcap = 18 V , Vrms(DMM) = 6.36 V
Measure Tx from the two waveforms (Tx is time difference) -- Tx= 0.06 s
Compute the phase angle: phase_angle = 34.3 degrees

CH1 leads CH2 by phase_angle


Increase the FG frequency to 10kHz. Calculate new complex impedance of capacitor.
Zcap = 159 Ohm

Measure the peak-peak capacitor voltage on CH2: Vcap = 3.75 V
Record Vrms from DMM: Vcap(rms) =  0.89 V
Measure Tx from the two waveforms (Tx is time difference) -- Tx= 8 us
Compute the phase angle: phase_angle = 28.8 degrees


RETURN THE FG frequency to 1kHz!!! Increase Rbox to 10kOhms.
Measure the peak-peak capacitor voltage on CH2: Vcap = 1.6 V
Record Vrms from DMM: Vcap(rms) =  1.12 V
Measure Tx from the two waveforms (Tx is time difference) -- Tx= 150 us
Compute the phase angle: phase_angle = 54 degrees

Adjust Rbox until capacitor voltage is 4V peak-to-peak. Record Rbox.  --- Rbox = 4kOhms
Record Vcap(rms) from DMM. --- Vcap(rms) = 2.56 V
Measure Tx from the two waveforms (Tx is time difference) -- Tx= 130 us
Compute the phase angle: phase_angle = 46.8 degrees


Finally, vary the FG from low to high frequency and observe the impact on the capacitor voltage amplitude.

The capacitor voltage amplitude is greatest when the frequency is in the low range, and voltage amplitude is smallest at higher frequencies. This circuit is a lowpass filter. The resistor will always lead even when we adjust the frequency. At high frequencies, the two signals tend to a Tx of 10.5 us.