Phase Difference Between Two Signals (Problem)

Question

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?

Answer 1

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.

Answer 2

Both of the results you got are right. You feel it confusing, because you imagine it as a running race, where one has to be first and the other the second, but in fact these are periodic signals. Which means that the leading is always relative.

Try to sketch the signals and select a local maximum of i1. The i1 leads i2 by the distance between your chosen i1 maximum and the closest i2 maximum to the left. But there is another i2 maximum to the right from the chosen i1 maximum, so i2 leads i1 by that distance. Its always true that if you add these two phase differences, you always get 360 degrees, what you can use to check your answer

Answer 3

Usually when you specify leading vs lagging phase difference you limit yourself to a "normalized" phase angle of ±180°.

More specifically, to be a mathematically nitpicker, any phase angle should be reduced to the normalized interval

$$ (-180°,180°] $$

so that leading vs. lagging would be uniquely defined.