Frequency in an oscilloscope simulation

AI Thread Summary
The discussion centers on calculating frequency from an oscilloscope simulation, where the user initially miscalculated the frequency as 4.5e6 s^(-1) based on observing 2 and 1/4 cycles over 0.5e-6 seconds. They realized their error stemmed from misunderstanding the time scale setting of the oscilloscope, which is measured in time per division. Upon reevaluation, they determined that one complete period corresponds to 4 units on the time scale, leading to the correct frequency of 5e5 s^(-1). The clarification highlights the importance of accurately interpreting oscilloscope measurements for frequency calculations. Ultimately, the user successfully corrected their mistake and confirmed the accurate frequency value.
brekfast
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Every time I try to find frequency on this problem I get the same answer - and according to the simulation it is incorrect.

Here's a picture of the simulation.

Please note: I am just trying to find frequency. I know the other two variables (.6A and 6V ; these are confirmed correct)

[PLAIN]http://img854.imageshack.us/img854/100/screenshot20110724at104.png



Looking at this image, I see only 2 and 1/4 cycles.

Frequency is, essentially, cycles per second (1/s) and period is the time it takes to complete a single cycle.

Based on the image I attached, I see 2 and 1/4 cycles (9/4) over a time of .5e-6 s. This is the same as dividing (9/4) by (.5e-6).

(9/4) / (.5e-6) = 4.5e6 s^(-1)

I do not understand where my mistake is.
 
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The time scale setting for an oscillisope is usually in terms of time/division. 0.5e-6 s would correspond to the distance between vertical lines.
 
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Derp! I knew that but I guess I had a brainfart.

Thanks! I got the correct answer (5e5 s^-1)
 
One period (minimum to minimum) is 4 units on the time scale. The oscilloscope shows the length of such units, it is 0.5 x 10-6 s.

ehild
 
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