# Drawing a sine wave

Discussion in 'Homework Help' started by Gdrumm, Apr 5, 2009.

1. ### Gdrumm Thread Starter Distinguished Member

Aug 29, 2008
684
36
How do I determine Total Time, so I can correctly draw a sine wave?
I have the following:
Draw a 1.8 V AC sine wave, F= 300 K Hz

I am given V/Div= .1 V
I am given .5 micro S
I am given Coupling = 10:1

I don't know how to calculate total time.

Thanks,
Gary

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
The period (time from a peak to the next) of the sine wave is given by: T = 1/F

3. ### Gdrumm Thread Starter Distinguished Member

Aug 29, 2008
684
36
I think this will help me solve it, but I'm still lost on the time divisions.

Sinse 1.8V AC is actually 1.8 V RMS, to get P-P, I multiply that times 2.828; that = 5.09 V P-P. That I can show as 2.545 on the positive alternation, and 2.545 on the negative alternation.

1/300K Hz = Total time =3.3 micro S, and there are .5 micro S per div.

I'm not sure how to proceed from here.

Thanks,
Gary

4. ### inventorjack Member

Apr 4, 2009
14
0
If you're using .5uS per division, one full sine wave will be just under 6.5 divisions (horizontal). The vertical will depend on how many volts per division you're using. For instance, if you used 1 V/div, it'll be 2.828 divisions above and below the reference line.

I'm including an attachment of a sample sine wave drawing. In this example, I'm assuming a 25kHz frequency at 12.6V peak-to-peak. When I consider that Time=1/Frequency, I see that T=40uS.

Picture 1: Because I have my display set to 10uS/division, I know one iteration of my sine wave will take 4 horizontal divisions. Further, since my peak-to-peak voltage is 12.6, and my display is 3V/div, It'll take just over 2 divisions above my reference line, and just over 2 divisions below the reference line. I've drawn a box with these parameters. One sine wave must fit within this box.

Picture 2: Here is the sine wave. I started at 0V horizontal and also ended at 0 V. I could just have easily chosen any other point on the wave to start and end with, but it has to remain constant. If I started at the peak of the wave (+6.3V), I'd have to end at that same voltage level at the end of my sine wave.

Picture 3: Now just continue the wave, using the same parameters.

Picture 4: The end result. I'm now very good with these image editors, so hopefully your sine wave will look a little nicer than mine

Hope that helps. Just apply these fundamentals, and you should be set.

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5. ### Gdrumm Thread Starter Distinguished Member

Aug 29, 2008
684
36
Very cool, that is a big help.

Thanks,
Gary