Re: Simple Doppler effect

• To: mathgroup at smc.vnet.net
• Subject: [mg120981] Re: Simple Doppler effect
• From: Bert Aerts <bert.ram.aerts at gmail.com>
• Date: Sat, 20 Aug 2011 06:16:52 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```This is an answer to the message of Dana DeLouis.

Thanks for the simple examples like:

f = 440;
z = Integrate[2*Pi*f, t];
Play[Sin[z], {t, 0, 10}]

And

f = 400 - 40*t;
z = Integrate[2*Pi*f, t];
Play[Sin[z], {t, 0, 10}]

I learned that the phase is the indefinite integral of the instantaneous
frequency.

So now the Doppler one:

vSound = 343;(*speed of sound*)
vSource = 50;(*speed of source*)
xObserver = 300;(*x coord of observer*)
yObserver = 100;(*y coord of observer*)
fSource = 440;(*frequency of source*)
x[t_] := vSource*t
fDoppler[t_] := Module[{xPosn, r}, xPosn = x[t];
r = Sqrt[(xObserver - xPosn)^2 + yObserver^2];
fSource*vSound/(vSound - vSource*(xObserver - xPosn)/r)]
Plot[fDoppler[t], {t, 0, 12}]
phase = Integrate[2 Pi fDoppler[t], t,
Assumptions -> {t \[Element] Reals, t >= 0, t <= 12}]

This delivers a complex result, but an indefinite integral has infinite
solutions, all different on another constant, as the derivative of a
constant is zero. So lets state that the phase at t=0 must be zero.

phase0 = phase /. t -> 0
Plot[Re[phase - phase0], {t, 0, 12}]
Plot[Im[phase - phase0], {t, 0, 12}]

The imaginary part is smaller than 1E-12.
So subtracting the complex constant makes our indefinite integral
solution real!

Play[Sin[ Re[phase - phase0] ], {t, 0, 12}]

This gives the expected Doppler sound.

As a check, we calculate the instantaneous frequency again:

deriv = D[phase/(2 Pi) , t]
N[deriv /. t -> 6]
Plot[deriv, {t, 0, 12}]

Another way to calculate the Doppler frequency is via Cos and ArcTan:

fDopplerBert[t_] := fSource*vSound/(vSound -
vSource*Cos[ArcTan[(xObserver - x[t]), yObserver]])
Plot[fDopplerBert[t], {t, 0, 12}]
phaseBert[t_] = Integrate[2 Pi fDopplerBert[t], t]
phaseBert0 = phaseBert[0]

But we get in trouble at t=6

phaseBert[6] - phaseBert0

gives Indeterminate because of division by zero. We can avoid this by

Play[Sin[If[t == 6, 0, Re[phaseBert[t] - phaseBert0]]], {t, 0, 12}]

The solution without ArcTan is more elegant.

Kind regards,

Bert

\$Version
8.0 for Linux x86 (64-bit) (February 23, 2011)

```

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