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Math Help - sine wave amplitude calculation

  1. #1
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    sine wave amplitude calculation

    Hi,


    I'm having some troubles determining the amplitude/magnitude of the following equation.


    A\cos(2\omega t+\beta_1)+B\cos(3\omega t+\beta_2)+C\cos(5\omega t+\beta_3)


    Since each part is at a different frequency, i cannot sum the magnitudes of each part.


    I have also thought about using variations of the double/triple angle formulae and some basic trigonometric identities, so that I can write the equation under a single frequency, but by doing so, I am introducing some higher order terms, which seem to negate the ability to sum the amplitudes together.


    For example, the term with 5\omega would look like
    16\cos^5(\omega t)-20\cos^3(\omega t)+5\cos(\omega t)


    I suppose another method would be to plot the first equation and then record the amplitude, but i would like a more generalised approach to solve this problem. All input is welcome and appreciated.
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  2. #2
    Junior Member ignite's Avatar
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    Re: sine wave amplitude calculation

    I will show how to do it using an example and then you can generalize it.
    f(t)=cos(2t)+2sin(3t)
    f(t)=\sqrt{1^2+2^2}(\frac{1}{\sqrt{1^2+2^2}}cos(2t  )+\frac{2}{\sqrt{1^2+2^2}}sin(3t))
    sin(u)=\frac{1}{\sqrt{1^2+2^2}};cos(u)=\frac{2}{ \sqrt{1^2+2^2}}
    \Rightarrow f(t)=\sqrt{5}(sin(u)cos(2t)+cos(u)sin(3t))=\sqrt{5  }sin(u+2t)

    So amplitude in this case would be \sqrt{5}.
    Thanks from suzuki
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  3. #3
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    Re: sine wave amplitude calculation

    Hi,

    Thanks for your answer. I actually have a question about this. When you found that it is \sqrt(5), is this a peak value or the average value? I plotted your example in matlab, and found that the peak amplitude value is 3. What is this method that you are using? Could you please further explain your method?

    Thanks for your help.
    Last edited by suzuki; April 29th 2012 at 09:41 PM.
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  4. #4
    Junior Member ignite's Avatar
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    Re: sine wave amplitude calculation

    \sqrt{5} is the maximum value.Please check the graph at Wolfram|Alpha
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  5. #5
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    Re: sine wave amplitude calculation

    Hi,
    Sorry I must have made a mistake there.

    I'm having some trouble visualizing how to do it for three terms. im assuming the magnitude can be found by \sqrt(A^2+B^2+C^2). But when you have to break it down into cosine and sine terms, it seems to be quite difficult. If A = 1, B = 2 and C = 3, i can see that maybe you can break the C term into A+B, but for the general case, this is not so simple. Please advise.

    Thanks again.
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  6. #6
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    Re: sine wave amplitude calculation

    Seems like another problem i run into is that the simplification does not necessarily work if all my terms are cosines. Is there any further help that can provided for this? I am really stuck on how to solve this...
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  7. #7
    Junior Member ignite's Avatar
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    Re: sine wave amplitude calculation

    I am extremely sorry.The method I described above works only when the frequencies are same,but amplitudes are different.
    In general case,it is not possible to write it as a simple sine or cosine formula.
    Last edited by ignite; April 30th 2012 at 02:24 PM.
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  8. #8
    Junior Member ignite's Avatar
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    Re: sine wave amplitude calculation

    Quote Originally Posted by ignite View Post
    I will show how to do it using an example and then you can generalize it.
    f(t)=cos(2t)+2sin(3t)
    f(t)=\sqrt{1^2+2^2}(\frac{1}{\sqrt{1^2+2^2}}cos(2t  )+\frac{2}{\sqrt{1^2+2^2}}sin(3t))
    sin(u)=\frac{1}{\sqrt{1^2+2^2}};cos(u)=\frac{2}{ \sqrt{1^2+2^2}}
    \Rightarrow f(t)=\sqrt{5}(sin(u)cos(2t)+cos(u)sin(3t))=\sqrt{5  }sin(u+2t)

    So amplitude in this case would be \sqrt{5}.
    Read sin(2t) instead of sin(3t).
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