To see why the derivative of the angle function would be the instantaneous frequency, repeat the experiment of Section 3-7.2.
(a) Use the following parameters to define a chirp signal:
Determine α and β in (3.46) to define x(t) so that it sweeps the specified frequency range.
(b) The rest of this problem is devoted to a MATLAB experiment that demonstrates why the derivative of the angle function is the “correct” definition of instantaneous frequency. First, make a plot of the instantaneous frequency fi(t) (in Hz) versus time.
(c) Make a plot of the signal synthesized in (a). Pick a time-sampling interval that is small enough so that the plot is very smooth. Put this plot in the middle panel of a 3 × 1 by using subplot, subplot (3, 1, 2).
(d) Generate a 4 Hz sinusoid, and plot it in the upper panel of a 3 × 1 by using subplot, subplot (3, 1, 1).
(e) Generate an 8 Hz sinusoid, and then plot it in the lower panel of a 3 × 1 by using subplot, subplot (3, 1, 3).
(f) Compare the three signals and comment on the apparent frequency of the chirp in the time range 1.6 ≤ t ≤ 2 s. Which sinusoid matches the chirp in this time region? Compare the expected fi(t) in this region with 4 Hz and 8 Hz.
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