hwsolB17

# hwsolB17 - (iii) Estimate of R1/R2 equals ((a1^2 +...

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Sheet1 Page 1 HW 17 Solution ______________ (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)'), . ..

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Sheet1 Page 2 xlabel('t (microseconds)') % see hw17graph1.pdf
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Unformatted text preview: (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.) Sheet1 Page 3...
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## This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB17 - (iii) Estimate of R1/R2 equals ((a1^2 +...

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