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Homework Help: Lagging and leading

  1. Apr 27, 2006 #1
    Im having trouble comparing sinusoidal waves and their phases.
    As a sample problem I was given
    v_1=120cos(120*pi*t - 40deg)
    and i_1=2.5cos(120*pi*t +20deg)

    and I was asked to find the angle by which i_1 lags v_1.
    I have no clue on how to go through with this problem, I dont even know where to start..
  2. jcsd
  3. Apr 27, 2006 #2
    Given two signals S1 and S2 in phasor form:

    S1 = [itex]S_1 \angle \theta_1[/itex] and S2 = [itex]S_2 \angle \theta_2[/itex]

    where [itex]S_1, S_2[/itex] are the magnitudes of the signals and [itex]\theta_1, \theta_2 \in (-\pi, \pi][/itex] are the phases.

    Signal S1 is said to be leading signal S2 if [itex]\theta_1 > \theta_2[/itex]; it is said to be lagging signal S2 if [itex]\theta_1 < \theta_2[/itex]. Otherwise the two signals are said to be in phase.
    Last edited: Apr 27, 2006
  4. Apr 28, 2006 #3


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    Staff: Mentor

    Keeping lagging and leading straight can be a little confusing, but there is a trick that has helped me a lot. Draw the traditional sine amplitude versus theta graph with amplitude on the vertical axis and the angle theta on the horizontal. The sin(theta) graph of course goes through zero, rises to the right and oscillates along for a couple cycles, going through zero at Pi, 2Pi, etc. Now also draw cos(theta) on the same graph, and it starts at cos(0)=1 of course, and comes down and oscillates along, crossing the horizontal axis at Pi/2, 3Pi/2, etc.

    Now look at the two plots, and think of the horizontal axis as a time-related axis (like when theta is a function of time). Time is increasing to the right, so the waveform that is shifted to the right is shifted to later time, which is lagging. When you take the cos(theta) plot and shift it to the right by Pi/2, you get the sin(theta) plot, right? So the sin(theta) function *lags* the cos(theta) function by Pi/2. And since the sin and cos functions have a period of 2Pi, you can also say that the sin(theta) function *leads* the cos(theta) function by 3Pi/2. Makes sense?

    And finally, let's write sin(theta) as cos(theta-Pi/2). Look at the argument (theta-Pi/2) -- it is zero when theta is Pi/2. And cos(0)=1, so cos(theta-Pi/2) is a *right* shift of the cos(theta) function. Makes sense?
    Last edited: Apr 28, 2006
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