3015_Homework_9_Solutions

3015_Homework_9_Solutions - % Normalize coefficients %...

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Matlab Problem clear all close all load 3015_Discussion_7_Matlab % Fourier tansform f_oboe=fftshift(fft(oboe)); N=length(f_oboe); %Frequency axis (in Hertz) f=-fs/2:fs/(N-1):fs/2; plot(f,abs(f_oboe)); title ('FFT') xlabel('Frequency (Hz)') ylabel('Amplitude') k_min=round(f(1)/440); k_max=round(f(end)/440); %Find peaks for k=k_min:k_max index_f=find(f>=(k*440-220) & f<=(k*440+220)); %Find peak inside a window of 440Hz [value,pos]=max(abs(f_oboe(index_f))); %pos is the position of the peak coeff_oboe(k+k_max+1)=f_oboe(index_f(pos)); end %coeff_oboe=fftshift(coeff_oboe); %Negative frequencies % Find signal power t=0:1/fs:(length(oboe)-1)/fs; index=find(t>0.75 & t<=0.75+1/440); %Signal during sustain time power_oboe=sum(oboe(index).^2)./length(index);
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Unformatted text preview: % Normalize coefficients % Using Parseval, sum of squares of coefficients equals signal power norm=sum(abs(coeff_oboe).^2); coeff_oboe=coeff_oboe*sqrt(power_oboe/norm); %renormalize coefficients %Plot resulting coefficients figure stem(k_min:k_max,abs(coeff_oboe)); title ('Fourier Series Coefficients'); xlabel('k') ylabel('a_k') % Amplitude of coefficients may not coincide exacly with the one from % Discussion# 7. This is because here we used the whole signal to find the % coefficients, whereas in Discussion #7 only one period was used. -2.5-2-1.5-1-0.5 0.5 1 1.5 2 2.5 x 10 4 1000 2000 3000 4000 5000 6000 7000 8000 FFT Frequency (Hz) Amplitude-50-40-30-20-10 10 20 30 40 50 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Fourier Series Coefficients k a k...
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3015_Homework_9_Solutions - % Normalize coefficients %...

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