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Find beats formula using imaginary parts

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the imaginary parts

    When using complex representation, it is customary to use the real parts. Instead use the imaginary part of [itex]e^{j\theta}[/itex] to calculate an expression for the sum:
    [tex]\sin(\omega t) + \sin((\omega + \Delta \omega)t)[/tex]

    Remember, it should come out to be the beats formula.

    2. Relevant equations
    e^{j\theta} = \cos(\theta) + j\sin(\theta)\\
    \omega_{beat} = |\omega_{1}-\omega_{2}|\\
    cos(\theta) = Re[e^{j\theta}]\\
    sin(\theta) = Im[e^{j\theta}]

    3. The attempt at a solution
    I'm having trouble understanding what the question is asking for. I tried rewriting the above equation using the complex exponential with the hope to take the imaginary part later but that hasn't seemed to bring me closer to the "beat formula". I'm also not sure what it means "it should come out to be the beats formula"
    \sin(\omega t) + \sin((\omega+\Delta \omega)t)\\
    \sin(\omega_{1}) + \sin(\omega_{2}t)\\
    e^{j\omega_{1}t} + e^{j\omega_{2}t}\\

    Using some trig I get an expression for the sum that does have look like it will exhibit beating behavior when the frequencies are close, but this wasn't using the complex exponential function.

    \sin(\alpha) + \sin(\beta) = 2 \sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\\
    \sin(\omega_{1}t)+\sin(\omega_{2}t)= 2\sin\left(\frac{\omega_{1}+\omega{2}}{2}t\right)\cos\left(\frac{\omega_{1}-\omega_{2}}{2}t\right)

    The biggest question I'd like to ask is: How do you use the imaginary part of [itex]e^{j\theta}[/itex] to calculate an expression for the sum?
  2. jcsd
  3. Oct 3, 2013 #2
    Consider ## e^{ia} e^{ib} ##.
  4. Oct 3, 2013 #3
    e^{j\alpha}e^{j\beta} = e^{j(\alpha+\beta)}

    but I don't see how this is helps me to calculate an expression. With trig I use a trig identity and then I calculate an expression. With the imaginary part of [itex]e^{j\theta}[/itex] I use __________ or do _________ and calculate an expression. But right now I don't see what goes in those blanks.
  5. Oct 4, 2013 #4
    $$ e^{ia} e^{ib} = (\cos a + i \sin b) \times ... $$
  6. Oct 4, 2013 #5
    Using Euler's equation I was able to go about it a way. I'm not sure if this is a reasonable answer, but it does seem to meet the requirements of using the imaginary part of [itex]e^{j\theta}[/itex] and calculating an expression.

    \sin(\omega_{1}t)+\sin(\omega_{2}t) = e^{j\omega_{1}t}+e^{j\omega_{2}t}\\
    = \cos(\omega_{1}t)+cos(\omega_{2}t) + j(\sin(\omega_{1}t)+sin(\omega_{2}t))\\
    Im[e^{j\omega_{1}t}+e^{j\omega_{2}t}] = \sin(\omega_{1}t)+sin(\omega_{2}t)\\
    = 2\sin\left(\frac{\omega_{1}+\omega_{2}}{2}t\right) \cos\left(\frac{\omega_{1}-\omega_{2}}{2}t\right)
  7. Oct 4, 2013 #6
    I do not see how that uses the complex representation to prove the formula. Consider the hint in #4.
  8. Oct 4, 2013 #7
    e^{ja}e^{jb} = (\cos(a) + j\sin(a))(\cos(b)+j\sin(b))\\
    = [\cos(a)\cos(b) - \sin(a)\sin(b)] + j[\cos(a)\sin(b)+\sin(a)\cos(b)]
    Thank you for the hint. Will you provide another hint? Because I don't see how this moves me closer to an expression for
    \sin(a) + \sin(b)
  9. Oct 4, 2013 #8
    Another hint is $$ e^{ia} e^{ib} = e^{i(a + b)} $$
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