# Resultant sine wave from two sine waves

Discussion in 'General Electronics Chat' started by Philbilly, Aug 19, 2009.

1. ### Philbilly Thread Starter New Member

Aug 18, 2009
7
0
I am looking for the equation that will give the resultant voltage of two sine waves. I have tried using E= A*Sin(θ)*B*Cos(θ)+A*Cos(θ)*B*Sin(θ). This does not appear correct when plotted. Should I be converting to rectangular coordinates and then adding?

2. ### Ratch New Member

Mar 20, 2007
1,068
4
Philbilly,

How about disclosing the two sine waves so we can see if they have the same period or phase. We cannot very well help you if we don't know what the parameters of your problem are. You have posted the attempted solution, but not the the particular problem.

Ratch

3. ### Wendy Moderator

Mar 24, 2008
21,770
3,023
I've done this for software, it shows how and why AM modulation looks like it does.

Basically figure out what the instantaneous voltage for each sine wave for a specific period of time (count up in microseconds), the do a straight algebraic addition of the two calculated points. Time in this case is directly equivalent to phase, so it is much simplier than you are thinking.

If the period of the first wave is 30ms, and the period of the second sine wave is 40ms, then 1 ms would be 1/30 of the 360°, while 1 ms would be 1/40 of the 360° of the second sine wave.

4. ### russ_hensel AAC Fanatic!

Jan 11, 2009
825
57
best way is to express a a complex function and take the real part of the result.

5. ### davebee Well-Known Member

Oct 22, 2008
539
47
What your words say isn't quite the same as what your formula says.

The formula you've listed shows adding angles, then taking the sine of their sum, sin (a + b). The formula is perfectly correct, if that is what you want to do.

What is it you're working on?

6. ### count_volta Active Member

Feb 4, 2009
434
24
You can convert to rectangular coordinates and add but This will only work when the frequency ω is the same for both.

For example (if you forgot how) I will use cosines here. sine is just 90 degrees off from cosine.

56cos(5t+45°) + 18cos(5t+90°)

Convert to phasors (polar)

56 /45° + 18/90° Convert them to complex and real part.

(39.5+39.5j) + (18j) = 39.5+ 57.5j

Now convert the result to polar and you get. sqrt( 39.5^2+57.5^2) = 69.7 (magnitude) and inv tan (57.5/39.5)= 55.4° (angle)

Now the result is 69.7cos(5t+55.4°) But notice that the frequency is still 5 radians/sec.

Last edited: Aug 20, 2009
7. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,206
423

AM is actually the MULTIPLICATION of the two waves. Simple addition just gives you two superimposed waves, with no interactiion between them.

8. ### Wendy Moderator

Mar 24, 2008
21,770
3,023
I know, but if you graph two waveforms, say 990Khz and 1Mhz, the 1Khz envelope pops out. A perfect demonstration of a single sided AM signal modulated with 1Khz.

You might think about using it for a class sometime. I found it very enlightening, even though I knew it intellectually. I did this on a old TRS80 computer using BASIC, the graphics is much better now.

9. ### Philbilly Thread Starter New Member

Aug 18, 2009
7
0
Hey everyone,

Thanks for all of your replies, this was my first post and I did not expect so many responses. Correctly, someone stated that I did not provide enough info. I am troubleshooting a synchro circuit which uses only the X and Y leg, Z leg grounded. The excitation is 400Hz to the rotor. The measured wave forms appear normal and steady, but intermittently the S2D will set an invalidity bit. Because of it random nature, I believe the S2D is setting this fault due to a difference between the digital angle and the analog angle (>6° in this application).

That said, I just wanted to mathematically calculate the analog input to the S2D before recommending the addition of a software persistence filter. (>6° for two seconds)

Thanks again