# Frequency in an oscilloscope simulation

1. Jul 24, 2011

### brekfast

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 [Broken]

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.

Last edited by a moderator: May 5, 2017
2. Jul 24, 2011

### MisterX

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.

Last edited: Jul 24, 2011
3. Jul 24, 2011

### brekfast

Derp! I knew that but I guess I had a brainfart.

Thanks! I got the correct answer (5e5 s^-1)

4. Jul 24, 2011

### ehild

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