17 HW Solution
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(i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond)) :
f1 = 3 ; f2 = 2.5; gamma = 1.45
t = (0 : 0.1 : 10).' ;
u1 = cos(2*pi*f1*t) ;
w1 = sin(2*pi*f1*t) ;
u2 = cos(2*pi*f2*t) ;
w2 = sin(2*pi*f2*t) ;
V = [u1 w1 u2 w2] ;
c = (V'*V)\(V'*s)% s as in data17.txt
Obtain:
c =
2.2657 (=a1)
-9.8499 (=b1)
4.8427 (=a2)
-8.1701 (=b2)
(ii)
plot(t,s,t,V*c,'-.'), ...
title('Plot of s (solid) and s^{hat} (dotted)'), ...
xlabel('t (microseconds)') see ;
% hw17graph1.pdf
(iii) Estimate of R1/R2 equals
((a1^2 + b1^2)/(a2^2 + b2^2))^(-gamma) = 0.8349
(iv)
S = fft(s) ;
plot(abs(S)), title('Plot of abs(fft(s))')
% see hw17graph2.pdf
The ratio of the two peaks is given by
450.7/411.6 = 1.0951
Note that the quantity
(a1^2 + b1^2)/(a2^2 + b2^2))^(1/2)
which estimates the ratio of the two amplitudes in the expression for s(t),
equals 1.06. (The taller peak corresponds to f1 = 3 MHz, the shorter
peak to f2 = 2.5 MHz.)