Re: Simple Doppler effect

• To: mathgroup at smc.vnet.net
• Subject: [mg120796] Re: Simple Doppler effect
• From: Bert Aerts <bert.ram.aerts at gmail.com>
• Date: Wed, 10 Aug 2011 06:47:44 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j1ba5i\$ajk\$1@smc.vnet.net> <j1cl8u\$g5l\$1@smc.vnet.net>

```This is a question about the physics of the Doppler effect...

If you plot the frequency Plot[fDoppler[t], {t, 0, 12}], you get 440 Hz
at t=6 as expected, the frequency of the source at the shortest distance.

But the instanteneous frequency D[fDoppler[t] t,t] gives not 440 but 55 Hz.

And also Play[Sin[2 Pi fDoppler[t] t], {t, 0, 12}] sounds as a very low
frequency at t=6.

What is wrong?

Kind regards,

Bert

On 04/08/11 01:23, Kevin J. McCann wrote:
> The instantaneous frequency, which is what you hear, is the derivative
> of the argument of the sine along with the 2pi. For example:
>
> F(t) = sin(2pi f t)
> deriv(2pi f t)/2pi = f
>
> Simple enough. Now for your problem.
>
> G(t) = Sin[2 Pi fDoppler[t] t]
> deriv =  D[2 Pi fDoppler[t] t,t] = complicated stuff
>
> Now, plot it
>
> Plot[deriv,{t,0,12}]
>
> and you will see what you hear. The punchline is that if your signal is
>
> Sin[phi[t]]
>
> Then the instantaneous frequency is phi'[t]/(2Pi).
>
> Cheers,
>
> Kevin
>
> On 8/3/2011 7:07 AM, David Harrison wrote:
>> fSource = 1;
>> Plot[ Sin[2 Pi fDoppler[t] t], {t, 0, 12}, PlotPoints ->  100000]
>

```

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