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Toets 1

Toets 1. Opgave 1. x=[2 1 3 1 2 1 3] xp=[2 1 3 1 ] % FFT periodic signal: y1=abs(fft(xp)) figure(1) stem(y1) % % define window w=window(@triang,7) % FFT windowed signal y2=abs(fft(w'.*x)) figure(2) stem(y2). a) Periodiek signaal:. b) arb. signaal:. Opgave 2. a).

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Toets 1

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  1. Toets 1

  2. Opgave 1 x=[2 1 3 1 2 1 3] xp=[2 1 3 1 ] % FFT periodic signal: y1=abs(fft(xp)) figure(1) stem(y1) % % define window w=window(@triang,7) % FFT windowed signal y2=abs(fft(w'.*x)) figure(2) stem(y2) a) Periodiek signaal: b) arb. signaal:

  3. Opgave 2 a) b) AC= FT-1 (PSDF)

  4. Opgave 3 a) b) a) Voldoende precies aangezien de breedte van de spreiding rond de 4 niveau;’s beduidend kleiner is als de afstand tussen de niveau’s b) 3 1/8 1 3/8 3 1

  5. Opgave 3 c) As a result of symmetry: odd moments are zero 3 Histogram 1 1 3 1 Histogram 2

  6. Opgave 4 a) b) b) a) autocorrelatie Ruis, geen correlatie S/N = 2/(6-2) = 1/2 Periodiciteit 40 S/N = 0.4/(1.6-0.4) = 1/3

  7. Opgave 4c kruiscorrelatie Ja, er is een relatie, verschil in fase van 10 (op periodiciteit van 40) Shift = 10

  8. Opgave 5 Butterworth laag-doorlaat filter orde n

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