Electronic – Phase Difference Between Two Signals (Problem)

phase

I am asked to calculate the phase difference between the two signals below.

$$i_1 = -4\sin(377t + 55^\circ) \hspace{0.2cm}\quad \mathrm{and}\quad \hspace{0.2cm} i_2 = 5\cos(377t – 65^\circ)$$

What I did was to convert $$i_1 = -4\sin(377t + 55^\circ)$$
to
$$i_1 = 4\cos(377t + 145^\circ)$$

Then find the difference by $$145 – (-65)=210$$
Then we conclude $$i_1\hspace{0.2cm} leads \hspace{0.2cm} i_2 \hspace{0.2cm}by \hspace{0.2cm}210^\circ $$
But, let's add 360 to (since the result won't change)
$$i_2 = 5\cos(377t – 65^\circ)$$
We get $$i_2 = 5\cos(377t + 295^\circ)$$

Now comparing the two we get difference to be 150 degrees,which means
$$i_2\hspace{0.2cm} leads \hspace{0.2cm} i_1 \hspace{0.2cm}by \hspace{0.2cm}150^\circ $$

My questions are that,

  • What is the difference between these two answers? Is the latter wrong?
  • How can both lead?
  • How could we rewrite the second answer in terms of i1 leading i2?

Best Answer

Both phase differences are correct and, in fact, indistinguishable. There is an infinite number of ways writing the same sine wave, just by adding any integer multiple of 2*pi (or 360 degrees) to the phase.

The problem is with trying to apply the concept of "leading" and "lagging" to sine waves. Strictly speaking a sine wave has infinite length, i.e. it has no start and no end. They have been going forever, so it's pointless to ask "which one comes first". Every peak of the first sine wave is preceded by a peak of the second one, which in turn is again preceded by another peak of the first one and so forth ad infinitum.

In practice, if the phase difference is reasonably small (say 90 degrees or less), you could call the one with the larger phase "leading" because the peaks are relatively close together and the one with the larger phase visually appears to come first. However with larger phase differences and especially 180 degrees that makes no sense any more.