Instantaneous Voltage--Math Question!

[wdcw]

Could someone help me out with some Instantaneous Voltage calculations please.

I need to calculate the value of the resultant waveform at 45deg 120deg & 240deg that includes a third harmonic content of 20v.

Vmax is 200V, third harmonic amplitude is 20v

I know that to find the instantaneous voltage the calculation is:

e = Vmax sin0

So

e = 200 x sin 45 = 141.4V
e = 200 x sin 120 = 173.2V
e = 200 x sin 240 = -173.2V

Do I just ADD the 20v to each result to include the third harmonic amplitude or is there more to it?

Cheers

 

[bob91343]

There is more to it. First, you need to know the phase relationship. The 20 V (peak?) third harmonic has to be evaluated at the instant in question. It too is sinusoidal and is defined by peak voltage times sin (3f) Plus the shift due to phase difference.

 

[wdcw]

bob91343,

Thanks for the reply, I think I get what your saying, I'll try to work it out & see how I go.

Cheers.

 

[zigzagflux]

Try this out.

Just an Excel 97 spreadsheet that plots out what the waveform would look like with a number of harmonics and phase shifts.

 

[wakibaki]

Its:

e=200*sin(45)+20*sin(45*3)
e=200*sin(120)+20*sin(120*3)
e=200*sin(240)+20*sin(240*3)

Here's a spreadsheet showing the graph of the actual values you're interested in:

View attachment 10percent_3rd_harm.zip

w

I've taken the phase relationship as 0 degrees.

 

[wdcw]

Thanks for the replies guy's & also the Excel spreadsheets,

I see how this is calculated, much appreciated.

If I could ask one more question.

How does the phase relationship factor into the equation, could you possibly give me a quick example, basically for future reference.

Thanks Guy's

Cheers

 

[bob91343]

The phase relationship causes the voltages to add in different ways. As the waveform changes you will get different results, although the rms will be unchanged.

Just pick a phase and plot it with graph paper, adding the voltages at each instant of time. The wave could be flattened or peaked but will be symmetrical with only odd harmonics.

 

[wakibaki]

OK, suppose the 3rd. harmonic was 60 degrees lagging the fundamental then the equations would become:

e=200*sin(45)+20*sin(-60+(45*3))
e=200*sin(120)+20*sin(-60+(120*3))
e=200*sin(240)+20*sin(-60+(240*3))

w

 

[wdcw]

Thanks very much, I understand.

The posts have been very useful.

Thanks again

Cheers