hello
im using multisim to do fourier analysis I have this circuit and results


I put this into excel

I got a square wave from adding the odd harmonics and the fundamental
I have to find the thd
I used the square root of added harmonics (which are squared) over total voltage.

the problem,
I have a 10v square wave and im getting 6.36 for the fundamental. I don't know why, is this because the rest of the odd magnitudes add up. because when I did that I got more than 10 volts.
I got an answer of 54.45%
I used the values 2.12255, 1.27404, 0.910576 and 0.708795 all individually squared. under a square root which is over total voltage. which is the harmonic values added.

is this correct because the thd shown in excel and multisim is 42% though I thought this would be because of the even harmonics as well.

I also have this question,

• Research and determine the Fourier series formula used to construct a square wave

any help would be great
thanks

simon

 

I have this from a pdf



I am going to say that I have changed the cos to sin and using the formula Vsin(2PI()Ft) in excel created sine waves. im not sure if this answers the question,

• Research and determine the Fourier series formula used to construct a square wave

• 
  i don't know how else to do this.

 

bump

 

bump

Not much point in that. If nobody has the answer, it won't likely change overnight.

 

@ninjaman
I would assume the the harmonics go as normal for a square wave, where the sequence of relative amplitudes (starting with the fundamental) is:

1, 1/3, 1/5, 1/7, ..... etc

which is an infinite sequence of only the odd harmonics. Even harmonics all being zero.

I believe the preferred general THD form is given as a fractional value by

\(\small{\text{THD = \frac{\sqrt{V_2^2+V_3^2+V_4^2+\....}}{V_1}}}\)

while the form you appear to adopt

\(\small{\text{THD = \frac{\sqrt{V_2^2+V_3^2+V_4^2+\....}}{\sqrt{V_1^2+V_2^2+V_3^2+V_4^2+\....}}}}\)

is used less frequently.

I'm not sure which convention Multisim uses. The simulator I use adopts the former convention.

In any event, the result for THD will depend upon how many harmonics are included in the analysis.
The following hopefully elaborates the point...

Relative to the fundamental, I calculate that (odd only) harmonics 3 to 11 contribute a harmonic content of 43.833% - which I take as the preferred value of THD.
Increasing the number of harmonics to include 3 to 101, I obtain a harmonic content of 47.833% ....
Increasing the number of harmonics to include 3 to 1001, I obtain a harmonic content of 48.291% ....

Extending this to an infinite number of (odd only) harmonics would give

\(\small {\text{THD=\frac{\sqrt{\frac{\pi^2}{8}-1}}{1}\times{100} \ % \ = 48.3426 \%}}\)

For a useful collection of Pi related series see :

http://numbers.computation.free.fr/Constants/Pi/piSeries.html