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Suppose you have displayed a sine wave with V = Vosin(wt) on the screen, where Vo=5V and 2=5kHz. Assume further that the signal triggers the scope when V=2.5V and wt=30.

EXAMPLE:
Suppose you double the amplitude of the wave without changing any oscilloscope settings. What are V and wt now when the scope is triggered? [Answer: the level hasn't changed so the scope still triggers when V=2.5V. V=2.5=10sin(wt); wt=sin^–1(2.5/10)=14.4775121851511.]

QUESTION:
Assume that you reduce Vo back to 5 V and switch the trigger slope polarity from '+' to '-' without changing the trigger level dial. What are V and wt when the scope is triggered now?

2007-01-16 19:14:14 · 3 answers · asked by Anonymous in Science & Mathematics Physics

3 answers

My O-scope would still trigger at 2.5V, but the phase would change, because changing the slope polarity makes it trigger on a downward, instead of upward, slope.
wt=sin^-1(5/10)=30, but you have to look for the phase at the other side of the maximum (where the slope is downward)-- so it's 150 degrees.

2007-01-17 03:05:21 · answer #1 · answered by Anonymous · 1 0

Basically, it allows a person to stabilize the display of a signal on an oscilloscope screen. Sometimes you need to measure signals at the channel input connector that are hard to display because their frequency is varying or are composed of multiple frequencies. The oscilloscope usually has a variable level control that might be able to synchronize it's own sweep circuitry to stabilize the channel input signal for viewing. But if that doesn't help, then you can input a fixed frequency signal (like a sawtooth or square wave waveform) into the trigger input of the scope that will force the oscilloscope to synchronize to this signal for a stable display. However, the trigger input must be of the same frequency, or some multiple frequency, of the channel input signal that you are trying to measure. Click on the source link I posted below for an explanation along with illustrations.

2016-05-23 23:23:57 · answer #2 · answered by Anonymous · 0 0

The scope will trigger when the wave is at -2.5v, which will occur exactly one half-wave later. So the trigger is wt0 + π, where wt0 is the trigger point in the example.

2007-01-16 19:18:57 · answer #3 · answered by gp4rts 7 · 0 0

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