Re: Simple Doppler effect

• To: mathgroup at smc.vnet.net
• Subject: [mg120683] Re: Simple Doppler effect
• From: "Kevin J. McCann" <Kevin.McCann at umbc.edu>
• Date: Wed, 3 Aug 2011 19:19:32 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j1ba5i\$ajk\$1@smc.vnet.net>

```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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