Graph the magnitude of the CTFT of x(t) = 8 rect(3t). This signal is not bandlimited so it cannot be sampled adequately to exactly reconstruct the signal from the samples. As a practical compromise, assume that a bandwidth which contains 99% of the energy of x(t) is great enough to practically reconstruct x(t) from its samples. What is the minimum required sampling rate in this case?
X( f ) = (8/3) sinc( f /3)
The total signal energy can be found most simply in the time domain.
From MATLAB simulation and trapezoidal-rule integration the minimum possible frequency range that would contain 99% of the energy of the signal would be from -30.9 Hz to +30.9 Hz.
totalArea=64/3 ; %From analytical solution in time domain.
ptsPerLobe=40 ; df=3/ptsPerLobe ; %First zero at 3 Hz, 20 pts per lobe.
nLobes=4 ; nPts=ptsPerLobe*nLobes ;
f=[0:df:nPts*df] ; X=abs((8/3)*sinc(f/3)).^2 ;
p1=plot(f,S,’k’) ; grid ;
set(p1,’LineWidth’,2) ;
title(‘Problem 9.3.11’,’FontName’,’Times’,’FontSize’,18) ;
xlabel(‘Frequency, f (Hz)’,’FontName’,’Times’) ;
ylabel(‘|(8/3)*sinc(f/3)|^2’,’FontName’,’Times’) ;
set(gca,’Position’,[0.1,0.6,0.6,0.3],’FontName’,’Times’) ;
loop=’y’ ; area=0 ; f1=0 ; f2=df ;
while loop==’y’,
area=area+(abs(8/3)^2)*(sinc(f1/3)^2+sinc(f2/3)^2)*df/2 ;
disp([’f2 = ‘,num2str(f2),’, Area = ‘,num2str(area)]) ;
if area>.99*totalArea/2,
loop=’n’ ;
else
f1=f1+df ; f2=f2+df ;
end
end
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