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]