Oscilloscope trigger problem. The \'trigger\' settings determine when the trace

Question

Oscilloscope trigger problem. The 'trigger' settings determine when the trace begins. Suppose you have displayed a sine wave with V = Vosin(?t) on the screen, where Vo=1V and ?/2?=5kHz. Assume further that the signal triggers the scope when V=0.5Vand ?t=30 degrees.
EXAMPLE:
Suppose you double the amplitude of the wave without changing any oscilloscope settings. What are VV and ?t?t now when the scope is triggered? [Answer: the level hasn't changed so the scope still triggers when V=0.5V. V=0.5=2sin(?t) ?t=sin?1(0.5/2)=14.4775121851511 degrees
QUESTION:
Assume that you reduce VoVo back to 1 V and switch the trigger slope polarity from '+' to '-' without changing the trigger level dial. What are V and ?? t when the scope is triggered now?
WHAT ARE THE ___ Volts,
WHAT IS THE ___ degrees

Explanation / Answer

The scope will still trigger at 0.5 Volts.

Now to find wt , we need to look at down cycle.

Then we have

Sin wt = 1 / 2

wt = sin-1 1/2

wt = 30, However,  phase at the other side of the maximum where the slope is downward is given as

wt = 180 - 30

wt = 150 degrees

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