RMS of complex waveform (acoustics)

[mattattack900]

Hey everyone, just got a quick question in acoustics. I am mainly looking for a mathematical understanding

Homework Statement

Consider two harmonic plane progressive waves of the form

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

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

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

1) Total acoustic pressure
2) Total mean squared pressurein the above expressions $\tilde{P(x)}$ represents the acoustic pressure and $\tilde{A}$ and $\tilde{B}$ are complex amplitudes

Homework Equations

The Attempt at a Solution

my solution for part 1 is due to linear superposition:

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

i know that the SOLUTION to the second part is:

|$\tilde{P(x)}$|² = |$\tilde{A}$e^-jkx|² + |$\tilde{B}$e^+jkx|² + 2Re{ $\tilde{A}$$\tilde{B}$*}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

 

[Simon Bridge]

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?

 

[mattattack900]

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

 

[Simon Bridge]

You may not need to though:

If: $z=a+jb$

Then: $|z|^2 = a^2+b^2 = z\cdot z^\star$