AC Circuits - peak to peak voltage

[pyroknife]

Let's say you have a sine wave V=2sin(2t) V. For this peak to peak voltage is 4V.
Then let's say that everything below V=0 is cut off.
Is the peak to peak value for this just from 0 to 2V? Making Vpp=2V?

 

[barryj]

yeppp

 

[rude man]

Let's say you have a sine wave V=2sin(2t) V. For this peak to peak voltage is 4V.
Then let's say that everything below V=0 is cut off.
Is the peak to peak value for this just from 0 to 2V? Making Vpp=2V?

Yes, but it's not equivalent to a sine wave with 2V pk-pk = sin(2t). Not even if it's biased upwards so that V = 1 + sin(2t).

 

[barryj]

You are correct. pk-pk is exactly what it says, the difference between the maximum and minimum voltage. The v = 2sin(2t) signal cut off below 0 would have a pk-pk of 2 volts, but the rms voltage of this signal would be different from the rms voltage of the v =- sin(2t) signal. The rms voltage I am speaking of is the "root mean square" of tghe signal and is the effective power of the signal.

 

[Nickie]

Yes, in this case, the peak to peak value would be 2V. This is because the peak to peak voltage is the difference between the maximum and minimum values of a wave. Since the wave is only defined from 0 to 2V, the maximum value is 2V and the minimum value is 0V, resulting in a peak to peak value of 2V. However, it is important to note that in AC circuits, the peak to peak voltage is typically calculated from the full wave, including both positive and negative values. Therefore, in this scenario, the peak to peak voltage would actually be 4V, as the full wave includes values from -2V to 2V.