AC signals #1

Background: This exercise will focus on measuring the phase difference between AC sinusoidal singals at the same frequency

ΔΦ=tx*ω*(180/π)

Part A.
;2.Vrms= 0.3536V
;3. DMM measured RMS Vrms=0.318
6. f= 1000Hz, C=100nF
theory(100nF)Zcap=  1/2*pi*1000Hz*100nF=1591.55Ω
real(102nF)Zcap= 1/2*pi*1000Hz*102nF=1560.34Ω

8.
a) Vcap,pak=pk=0.852V
b) Vcap,rms=0.256V
d) tx=105.41 μs
e) Φ=105.41μ*2π*1000*180/π=37.9476°
f) CH1 lead.

Part B
increase f from 1kHz to 10kHz
2. Zcap=156.034Ω

3
a)Vcap,pk-pk=0.154
b)Vcap.rms=0.033
d)tx=237.8μs
e)Φ=237.8μ*2π*1000*180/π=85.608°

Part C
change resistance to 10kΩ, and return the f to 1kHz
a)Vcap,pk-pk=0.166
b)Vcap.rms=0.049
d)tx=221.62μs
e)Φ=221.62μ*2π*1000*180/π=79.78°
f)Rbox=3700Ω
g)Vcap.rms=0.119
h)tx=189.19μs
i)Φ=189.19μ*2π*1000*180/π=68.11°

Part D
2 in low frequency range is the capacitor voltage amplitude greatest
3.in high frequency range is the capacitor voltage amplitude smallest
 Vcap=Vs*Zcap/(Zcap+Zres), so the Zcap increase, the Vcap will increase.
 Zcap=1/(ωC), so the ω decrease the Zcap will increase.
4. high pass filter, the CH2 amplitude is less than CH1.
5. ΔΦ=tx*ω*(180/π)=tx*2*f*180, so the frequence increase the Φ will increase.
6.