AC Signal #1

In this experiment, we will focus on how to measure the phase difference between AC sinusoidal signals at the same frequency. We will see the changes of the capacitor and resistors who are connect in series, while frequency changes.

concepts:

where frequency = 1/period
the phase angle =time difference*w*180/pi
where w = 2pi*f
 V_rms=V_pp/(2sqrt(2))


Next, we energize a function generator(FG) and connet to oscilloscope
1.set the FG to produce 10V peak to peak sin wave with 1kHZ
but since we did not use a function generator, we were not able to get 10V peak to peak value but 1V
 2.Center the waveform on the scope, adjust the horizontal time-base to display 1 to 2 period on the screen
3. Confirm its 10V peak-to-peak
4. The anticipated RMS value is V_rms=0.371V
when we connet to the DM(digital multimeter) we got a value of V_rms=0.318V

Next, we set the variable resistor box to 1kohm

we calculate the complex impedance of the 100-nF capacitor
Z_cap=(1/wc)=(1/(2pif*100*10^-9))=1591.55ohm

Next, we build the circuit below

Which will look like this

Next, we use the O-scope cursore to find the measure peak-to-peak capacitor voltage on Ch2
V_cap,pp=0.852V

Next, the rms value of the capacitor voltage provided by the DDM
V_cap,rms=0.3012

if we divide 0.852/(2sqrt(2))=0.3012, which is the same value from the reading of multimeter

Next, we measure the time difference between the two waveforms
we get t=105.41 us

where the phase angle will be 37.9 degree

by looking at the graph, CH1 leads CH2

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Next, we want to see the affect from changing the frequency
then, we increase the frequency from 1k to 10kHZ
we calculate the new complex impedance of the capacitor
Z_cap=1/wc=159.1ohm

then, we use the o-scope to measure the peak-to-peak capacitor voltage on CH2
V_cap,pp=0.154V

The V_rms value from the multimeter reading is 0.033V
when we divide 0.154/(2sqrt(2))=0.0544, which we can say the values are close

next, we find the time difference, which is 23.78us
the phase angle is 86.61 degree

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We return the FG frequency to 1kHZ
increase the resistance box to 10kohm
we do the same thing and find the same value
V_cap,pp=0.178V
V_cap.rms=0.049V
0.178/(2sqrt(2))=0.063V, which are consider close due to the lack of precision due to the equipment
next, the time difference is 221.62 us
where the phase angle is 79.78 degree
Next, we change the value of resistance box untill the capacitor voltage is 2 divisions on the o-scope
we obtain R_box = 3.7K

the V_rms from DDM = 0.119V
time difference=189.19 us
phase angle is 68.1 degree

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Finally  we set the FG frequency from low to high and observe the impact on the capacitor voltage value
when the frequency is low, the voltage is high
when the frequency is high, the voltage is low
by this observation, we can conclude that the circuit is a lowpass filter
when we adjust the frequency
the higher the frequency, the angle between the resistor and the capacitor is more


Last, we dissemble the circuit and put the parts back to where they belong

With the observations we found in this experiment, we were able to understand that change in frequency and the resistance is able to change the V_rms and the phase angle between the capacitor and resistor. As a result, we were able to control the circuit with desire value of capacitance, resistance and frequency to find a suitable values to do the demand of filter waves