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Homework Help: Envelope function for two sine curves

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data


    I am trying to get an envelope function for two sinusoidal curves (A, B) added up.
    a1,a2 are the amplitudes(metre), T1,T2 are the periods (hrs), B lags by dt from A at t=0.
    a1= 1, a2=0.5, p1=11,p2=10,dt=0
    A = a1*cos(w1*t), B=a2*cos(w2*t+dt)
    t varies from 0 to 650 hours

    2. Relevant equations

    using a envelope function sqrt(a1^2 + a2^2+2*a1*a2*(cos(w2-w1)*t+dt))
    i am getting a nice envelope as in the figure 1


    a1= 1, a2=0.5, T1=60, T2=10, dt=0
    using a envelope function

    a2 – a1 +2*a1*a2*(cos(w1)*t+dt))
    I can manage an envelope as in the figure 2

    3. The attempt at a solution

    Is there a way to combine these two conditions ( or any combination of T1 and T2)
    so that I can have one generalized envelope function that does not miss the peaks.

    Thanks in anticipation for any help
  2. jcsd
  3. Nov 8, 2015 #2


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    Just a reminder of a trigonometric identity: [itex]\cos(x)+\cos(y)= 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) [/itex].
  4. Nov 9, 2015 #3
    Thanks Svein,
    Could you please elaborate little more.
    From what ever I have tried out, it works only for a range and not for all combinations of T1 and T2.
    I want to identify the location of extreme amplitude analytically from added up sine curves.
    I believe the envelope could be represented by a set of sinusoids

    sorry if it is something too simple and i am getting lost.
  5. Nov 10, 2015 #4


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    OK. We start out with [itex]A_{1}\cos(\omega_{1} t)+A_{2}\cos(\omega_{2}t+\phi) [/itex]. Assume A1>A2. Then we can transform it into [itex](A_{1}-A_{2})\cos(\omega_{1} t)+A_{2}(\cos(\omega_{1} t)+\cos(\omega_{2}t+\phi))=(A_{1}-A_{2})\cos(\omega_{1} t)+2A_{2}\cos(\frac{\omega_{1}+\omega_{2}+\phi}{2})\cos(\frac{\omega_{1}-\omega_{2}-\phi}{2}) [/itex].
    In radio technology this shows us that adding an audio signal (ω2) to a carrier wave (ω1) gives us a carrier wave plus two side bands [itex] \frac{\omega_{1}+\omega_{2}+\phi}{2}[/itex] and [itex] \frac{\omega_{1}-\omega_{2}-\phi}{2}[/itex]. You can see this best when ω1>>ω2. In your case (when ω1≈ω2), the result is harder to see.
  6. Nov 10, 2015 #5


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    I don't know how you arrived at the envelope function you posted, so I may be just telling you what you already know here.
    First, lose the phase constant. If the two frequencies are different then the phase constant is equivalent to a shift along the time axis, so is not interesting.
    I would then apply the identity Svein mentioned, but in a more symmetric manner.

    Get F(t) into the form ##C\sin(\bar \omega t)\cos(\hat \omega t)+D\cos(\bar \omega t)\sin(\hat \omega t)## where ##\bar\omega = \frac{\omega_1+\omega_2}2## and ##\hat\omega = \frac{\omega_1-\omega_2}2##. Clearly the ##\bar \omega## factors are the higher frequency of the two, so we want to find the maximum of F over a cycle of that.

    If this frequency is much higher than ##\hat \omega##, we can treat the ##\hat \omega## as constant over such a cycle and find the max of F on that basis. This leads to the equation you posted.

    If the original frequencies are very different then ##\hat \omega## and ##\bar \omega## will be of similar magnitude, so this method does not work. Instead, you can treat the lower frequency function as roughly constant over a cycle of the higher frequency and arrive at your other equation.

    So each solution is just an approximation that works well in its own domain. The worst case, I guess, would be where the ratio of the original frequencies is the same as the ratio of the sum and difference frequencies: ##\omega_1:\omega_2=\bar\omega:\hat\omega##. Try plotting one of those to see if it gives any insight.
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