Sine Wave measurements vs equations

  • Thread starter Yarnzorrr
  • Start date
  • #1
Yarnzorrr
11
0
I have the equation E(t)=7*sin(11000t+∏/3) and I measured the following:

E(.22ms) = .422V
Frequency = 1.751kHz
Period = 572.0 μs
Peak = 7V
Peak-Peak = 14V
E(rms) = 4.8V
E(average) = 105mV

I've calculated the following to compare
E(.22ms) = .422V
Frequecy = 1.75 kHz
Period = 571.4μs
Peak = 7v
Peak - Peak = 14v
E(rms) = 4.95v
E(average)=4.459V--------------------This value is no where near the measured value.


I'm using the formula Peak*0.637. Is this wrong? I'm not really sure. please help!
 

Answers and Replies

  • #2
Windadct
1,433
400
Looks like you do not have a pure sine or not measuring TRUE RMS - and the measured "Average" includes the + and - side of the wave form ...~ 0v. Also "measured" RMS and Average totally depend on the instrument being used to measure with.
 
  • #3
yungman
5,624
226
I have the equation E(t)=7*sin(11000t+∏/3) and I measured the following:

E(.22ms) = .422V
Frequency = 1.751kHz
Period = 572.0 μs
Peak = 7V
Peak-Peak = 14V
E(rms) = 4.8V
E(average) = 105mV

I've calculated the following to compare
E(.22ms) = .422V
Frequecy = 1.75 kHz
Period = 571.4μs
Peak = 7v
Peak - Peak = 14v
E(rms) = 4.95v
E(average)=4.459V--------------------This value is no where near the measured value.


I'm using the formula Peak*0.637. Is this wrong? I'm not really sure. please help!

Average is average value over time. For a sine wave with no offset, the top half is equal and opposite to the bottom half. So if you average out over time, it should be zero. Your assumption is not correct of Peak*0.637.

Then fact you measure 105mV average might due to the sine wave is not pure, containing even harmonic that create DC offset when averaging out.
 
  • #4
uart
Science Advisor
2,797
21
As yungman said, the average should be zero (or close to it). The value of 0.637 times the peak is for the (ideal) full wave rectified sine wave.

BTW. 0.637 is a numerical approximation of [itex]2/\pi[/itex], the theoretically exact value.

[tex]\frac{1}{\pi} \int_0^\pi \sin(x) dx = \frac{2}{\pi}[/tex]
 

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