1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

RMS of complex waveform (acoustics)

  1. Jun 4, 2013 #1
    Hey everyone, just got a quick question in acoustics. Im mainly looking for a mathematical understanding


    1. The problem statement, all variables and given/known data

    Consider two harmonic plane progressive waves of the form

    [itex]\tilde{P(x)}[/itex] = [itex]\tilde{A}[/itex]e-jkx

    and


    [itex]\tilde{P(x)}[/itex] = [itex]\tilde{B}[/itex]e+jkx

    traveling in opposite directions. Showing all workings, derive expressions for:

    1) Total acoustic pressure
    2) Total mean squared pressure


    in the above expressions [itex]\tilde{P(x)}[/itex] represents the acoustic pressure and [itex]\tilde{A}[/itex] and [itex]\tilde{B}[/itex] are complex amplitudes

    2. Relevant equations





    3. The attempt at a solution

    my solution for part 1 is due to linear superposition:

    [itex]\tilde{P(x)}[/itex] = [itex]\tilde{A}[/itex]e-jkx + [itex]\tilde{B}[/itex]e+jkx

    i know that the SOLUTION to the second part is:

    |[itex]\tilde{P(x)}[/itex]|2 = |[itex]\tilde{A}[/itex]e-jkx|2 + |[itex]\tilde{B}[/itex]e+jkx|2 + 2Re{ [itex]\tilde{A}[/itex][itex]\tilde{B}[/itex]*}cos(kx)

    where Re{} denotes the real part ( I couldn't find the actual symbol ), * denotes the complex conjugate and k is the wave number ( k= ω/c )

    like i said above i am trying to get a mathematical understanding of the second part. I do not understand how this solution is derived. Thanks
     
  2. jcsd
  3. Jun 4, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Note: \Re gives ##\Re## ... it's easier to just type out the LaTeX than use the equation editor.

    What do the tildas indicate here?

    $$\tilde{P}\!_A(x) = \tilde{A}e^{-jkx}\\

    \tilde{P}\!_B(x)=\tilde{B}e^{jkx}\\

    \tilde{P}(x)=\tilde{A}e^{-jkx}+\tilde{B}e^{jkx}\\$$

    You want to understand this:
    $$\left | \tilde{P}(x) \right |^2=\left | \tilde{A}e^{-jkx}\right |^2+\left | \tilde{B}e^{jkx}\right |^2 = \left |\tilde{A}\tilde{B}^\star\right | \cos(kx)$$

    ABcosθ would normally be a scalar product right - so how does that work if A and B are complex valued?

    What happens if you expand the complex amplitudes into ##\tilde{A}=a+jb\; , \; \tilde{B}=c+jd## and expand the exponentials into trig functions?
     
    Last edited: Jun 4, 2013
  4. Jun 4, 2013 #3
    Hey, thanks for the reply.

    i believe the tilda is the nomenclature used to represent a Complex number.
    I will try what you have suggested
     
  5. Jun 5, 2013 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You may not need to though:

    If: ##z=a+jb##

    Then: ##|z|^2 = a^2+b^2 = z\cdot z^\star ##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: RMS of complex waveform (acoustics)
Loading...