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Toets 2

Toets 2. Opgave 1 a. Opgave 1 b. B1=firwd(7,1,pi*1/7,0,1). B2=firwd(7,1,pi*1/7,0,5). Used unwrap for continuous phase. Opgave 1 c d. d) A notch filter located at 0.14 (and its multiples). Opgave 2a. Convolutie in tijdsdomein is product in z-domein:. Opgave 2b. b1=[1 -4 3] a1=[1 2 4]

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Toets 2

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  1. Toets 2

  2. Opgave 1 a

  3. Opgave 1 b B1=firwd(7,1,pi*1/7,0,1) B2=firwd(7,1,pi*1/7,0,5) Used unwrap for continuous phase

  4. Opgave 1 c d d) A notch filter located at 0.14 (and its multiples)

  5. Opgave 2a Convolutie in tijdsdomein is product in z-domein:

  6. Opgave 2b b1=[1 -4 3] a1=[1 2 4] [r1,p1,k1]=residue(b1,a1) r1 = -3.0000 - 1.4434i -3.0000 + 1.4434i p1 = -1.0000 + 1.7321i -1.0000 - 1.7321i k1 = 1 Table 5.1- nr 15:

  7. Opgave 2c Impulsrespons: Step respons: b2a=[1 -1] a2a=[1 4 1 0] [r2a,p2a,k2a]=residue(b2a,a2a) b2b=[1] a2b=[1 4 1 ] [r2b,p2b,k2b]=residue(b2b,a2b) r2a = -0.3660 1.3660 -1.0000 p2a = -3.7321 -0.2679 0 k2a = [] r2b = -0.2887 0.2887 p2b = -3.7321 -0.2679 k2b = []

  8. Opgave 3 wa=1.200*pi/sqrt(2) num=[1 2*wa 2*wa^2] den=[1 4*wa 2*wa^2] figure(1) freqs(num,den) fs=2.000 [numd,dend]=bilinear(num,den,fs/2) figure(2) freqz(numd,dend) num = 1.0000 5.3315 14.2122 den = 1.0000 10.6629 14.2122 fs = 2 numd = 0.7303 0.5166 0.1909 dend = 1.0000 0.5166 -0.0787

  9. Opgave 3 alternatief fs=2000; fc=600; wc=2*pi*fc; awc=2*fs*tan(wc/fs/2) [b,a]=bilinear([1/awc^2 2/awc 2],[1/awc^2 4/awc 2],fs) figure(1) freqz(b,a,512,fs) sys = tf([1/wc^2 2/wc 2],[1/wc^2 4/wc 2]); figure(2) bode(sys)

  10. Opgave 4 a) b) Maak twee filters: een laagdoorlaat 2e orde (6dB/oct) bij fc een hoogdoorlaat 2e orde ook bij fc Pole moet in de buurt liggen van 0.7, dan geen resonantie.

  11. Opgave 4c

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