R3 -- Matlab Round #2

Starting...
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[We start XX.15 by overview, creating subdirectories and
opening Matlab.]

Problem 1. 
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[Problem 1, XX.25 ~ XX.50]

16400 Hz aliases down to 400 Hz, because of too low sampling
frequency.  Consider sampling as taking samples of continuous t by
interval T = 1/fT. In other words, substitute t by n/fT.

x(t)   = cos(2 pi 16400 t) 
x[n]   = cos(2 pi 16400/16000 n), where t <- n/fT
       = cos(2 pi 16400/16000 n - 2 pi 16000/16000 n)
       = cos(2 pi 400/16000 n)
x_r(t) = cos(2 pi 400 t), where n/fT <- t


Here normalized angular frequency, cos(omega n):

omega1 = 2 pi (400/16000) = 0.05 pi, and 
omega2 = 2 pi (16400/16000) = 2.05 pi 
       = 0.05 pi + 2 pi

Missing code: 

hold on;
y2 = 2.5 * cos(2*pi*16400*t + pi/6);
plot(t, y2)


Problem 2. 
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[Problem 2, ~XX.50 ~ XY.25]

Find /i/ and /s/: see difference in Figure 1 and 3.

fT = 22050 Hz
M  = 29001
--> length in seconds 29001/22050 ~ 1.3 s

Figure 2: frequency content of x is changing during 1.3 seconds. DFT
of the whole x is "sum" of /i/ and /s/ - not good.

Spectrogram = "Short-time Fourier Transform" in Mitra's book. Take a
small time window in x[n] (20-50 ms), compute spectrum, plot the
spectrum with colors as a narrow column in a spectrogram, and advance
to the next time window...


Problem 3. 
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[Problem 3, ~XX.25 ~ XY.45]

See also [P21]. Here we apply filter h for input signal x and
get the filtered signal y as shown in the basic scheme:

                           x -> [h] -> y

Missing line in ma2.m:

y(k) = ( x(k) + x(k-1) ) / 2;

This is now 1st order FIR filter. Even if there is some difference
in the figure, it is hard to _hear_ any difference. In practice,
the filter order can be, e.g., 100 instead of 1, and then you really
can notice the effect of filtering.

Extra problem
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Return the spectra, code and the fundamental frequency/period
as attachment with other mid term exam exercises.