Oscilloscope trigger homework problem

[Calcipher]

Oscilloscope trigger homework problem

Here is the problem.

Hopefully there's some OTFer out there who knows their oscilloscopes.

The first answer is 2.5V (not -2.5V in the picture I linked), but I'm having trouble getting ?t (for the second part)

I've just used an oscilloscope for the first time last week, and I don't know much about it. What does changing the trigger slope polarity affect in this case? Apparently the scope still triggers at 2.5V... please shed some light on me!

 

[P2blr]

Re: Oscilloscope trigger homework problem

well, the trigger is set on the unit itself so it will always trigger at 2.5V, and i've never seen one where you get to pick which part of the cycle it triggers at so that's what you've gotta find.
in the example, it shows exactly the same way to find it, except that when you bring V0 to 5V you further change V0 to -5V

     V = V0sin(wt)
  2.5V = -5Vsin(wt)
    wt = arcsin(-0.5)
    wt = -30
    wt = 150 degrees

edit: what i meant by "which part of the cycle it triggers at" i meant which exact angle, positive and negative edge triggering was implied

 

[Calcipher]

Re: Oscilloscope trigger homework problem

Thanks a million, P2.

 

[Corneo]

Re: Oscilloscope trigger homework problem

I use a scope pretty often at work (in fact I work for a company that manufactures scopes).

The slope of the trigger is important in some ways. But for a sine way or any periodic way it plays very little significance. Say you have a signal that starts from 0V and ramps to 10 V in a linear fashion (think power supply as you turn on your computer). If you set the scope to trigger at 2.5V with positive edge trigger, the scope will capture the waveform and display it for you. If you set the trigger to negative slope. The scope will never "trigger" because the signal never goes down in a negative manner. Since a periodic signal will go up and down, whether the trigger is positive or negative, the scope will capture the data and display it.