Calculating Phi: A Sin Wave Explained

  • Thread starter Thread starter jkossis
  • Start date Start date
  • Tags Tags
    Phi
Click For Summary
To calculate phi from a sine wave graph, determine the duration of a complete cycle (T) and the time of the first peak (t0). The relationship can be expressed as φ = -(2π/T)t0, where ω = 2π/T. It's noted that the discussion initially confused sine and cosine waves, with the clarification that cosine peaks at the origin while sine starts at zero. For greater accuracy, measuring zero crossings rather than peaks is recommended for determining phase shifts. Understanding the elements of the equation A*cos(ωt + φ) is crucial for accurate interpretation.
jkossis
Messages
1
Reaction score
0
A question I have is how is one supposed to calculate phi when simply looking at a sin wave. The equation is of the form x(t) = A*cos(wt + phi). Once again, the only thing i am given is the graph. I appreciate any help you can give!
 
Engineering news on Phys.org
jkossis said:
A question I have is how is one supposed to calculate phi when simply looking at a sin wave. The equation is of the form x(t) = A*cos(wt + phi). Once again, the only thing i am given is the graph. I appreciate any help you can give!

Welcome to PF, jkossis! :smile:

Find the duration of a complete cycle, which is for instance the time between 2 peaks.
Let's call this T.

Now read off the time of the first peak and let's call this t0.

The first peak corresponds to an angle of zero, so you get: ωt0 + φ=0

Since ω=2pi/T, this yields:
(2pi/T)t0 + φ=0
φ = -(2pi/T)t0
 
jkossis said:
A question I have is how is one supposed to calculate phi when simply looking at a sin wave. The equation is of the form x(t) = A*cos(wt + phi). Once again, the only thing i am given is the graph.
Welcome jcossis !
Are you sure you need to calculate \phi? If you just wish to determine \phi, depending on what your graph looks like, there may be an easier method. You may be able to read \phi directly from your graph. Can you post your graph?
Have you figured out what each of the elements in your equation A cos(\omega + \phi) means? how it relates to the graphical representation? (hint: google is your friend).
 
Hello Bassalisk.

You should be discussing a cosine wave, not a sine wave.

go well
 
Studiot said:
Hello Bassalisk.

You should be discussing a cosine wave, not a sine wave.

go well

Oh darn :D Curse my fast reading ways
 
Cosine leads sine by 90 degrees... and sin begins at the origin, so cosine peaks at the origin.

where ever the peak on your cos signal is, is the measure of phi (phase shift)
 
if you're doing it graphically it's way easier and generally more accurate to measure zero crossings where slope is max instead of peaks where slope is zero...
 
Ouabache said:
Have you figured out what each of the elements in your equation A cos(\omega + \phi) means?
my typo, equation missing time variable t. It should read A cos(\omega t + \phi)
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
809
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 26 ·
Replies
26
Views
837
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K