0
\$\begingroup\$

I was reading a book which says:

Phase describes the position of the waveform relative to time 0. It indicates the status of the first cycle. and below is a picture

enter image description here

I am a little bit confused. According to the picture, at time 0, the waveform is just a point. Later it starts to form a waveform shape, so how come a waveform is relative to a point at time 0?

\$\endgroup\$

3 Answers 3

1
\$\begingroup\$

Definition:

waveform /ˈweɪvfɔːm/
noun PHYSICS
a curve showing the shape of a wave at a given time.
Oxford Dictionaries.

I think that Oxford have it slightly wrong and that it should be "a curve showing the shape of its graph as a function of time". Wikipedia is better.

The waveform represents the variation in a value with respect to time. This could be a voltage (electrical), a height (wave on the sea), pressure (sound), etc. and simply shows the instantaneous value at any point on the time axis.

According to the picture, at time 0, the waveform is just a point.

Not quite. We can deduce two things from the graph:

  • At t = 0 the amplitude is zero.
  • At t = 0 the amplitude is rising at a rate given by the slope of the curve at that point.

Later it starts to form a waveform shape, so how come a waveform is relative to a point at time 0?

The diagram is comparing the relative phase of three different waves of the same frequency with a periodic time of T.

  • The first has been taken as reference and t = 0 has, for convenience, been chosen as the point where the wave crosses the time axis. (It's easy to visually see and compare the intersection of the curve with the axis.)
  • The second is is at maximum at t = 0. By sketching in the dotted section we can see that it crossed the time axis earlier than the first trace and is leading the first by 90° or T/4.
  • The third trace is leading by 180° or T/4.

... so how come a waveform is relative to a point at time 0?

Typically t = 0 is just chosen for convenience to make the illustration clear.

\$\endgroup\$
0
\$\begingroup\$

"Waveform" describes the whole thing, and not a "shot picture" at some point in time t.

So, a system overall has a waveform, and at any point in time that means that the signal has a value (that's what you meant when you said "it's just a point").

\$\endgroup\$
0
\$\begingroup\$

One can only define a phase of a waveworm relative to another (base) waveform (not directly relative to t=0).
The base waveform is considered at a convinient point (where it cuts the time axis) or a random point in time which is then called t=0.
The other waveforms are seen with respect to this waveform and the phase is determined by the shift seen from t=0.

In your question, the base waveform is a sine, so, \$A \sin(2\pi f t) \$. So, the first shown waveform has no phase delay.
Were the base waveform a cosine, then the second waveform would have no phase delay.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.