Computer Algebra Recipes

8.2. THE POWER SPECTRUM 333
trum will now reveal what they are. The sampling time interval is taken to be
Ts = T2=d and the sampling frequency Fs =1=Ts is calculated.
> T[s]:=T[2]/d; F[s]:=1/T[s];
Ts := 0:05000000000 Fs := 20:00000000
The continuous function x(t) is sampled at times t = nTs with n =0toN ¡ 1.
The sequence of sampled x values is then put into an array format.
> x:=Array([seq(x(n*T[s]),n=0..N-1)]):
The (discrete) Fourier transform of x is performed. Maple de¯nes this transform
such that the kth item in FT is Xk= p N in our notation.
> FT:=FourierTransform(x):
The following operator will allow us to calculate S(k) foragivenk value.
> S:=k->abs(FT[k])^2:
A plotting point operator pt is introduced that gives the power spectrum S(k)
at the frequency (k ¡ 1) Fs=N .
> pt:=k->[F[s]*(k-1)/N,S(k)]:
Taking the power spectrum to be zero at zero frequency, we produce the points
in the power spectrum for k ranging up to N=2.
> pts:=[[0,0],seq(pt(k),k=2..N/2)]:
The complete power spectrum is plotted, and shown in Figure 8.9. There are
clearly two frequencies present at 1 and 2 beats per second, as expected.
> plot(pts,labels=["f","S"],view=[0..5,0..0.3]);
300
250
S
150
100
50
0
1 2 3 4 5
f
Figure 8.9: Frequencies in Frank N. Stein's heartbeat.
PROBLEMS:
Problem 8-8: Third Harmonic
Suppose that Frank N. Stein's heartbeat also contains the term 0:1 sin(6¼t).
Plotting p S, show that the presence of the third harmonic is revealed.