T-61.3010 Digitaalinen signaalinkäsittely ja suodatus T-61.3010 Digital Signal Processing and Filtering ============================================================ RATKAISUJA / SOLUTIONS PART 2 ============================================================ ------------------------------------------------------------ (16) 1. välikoe / 1st mid term exam, 7.3.2006 ------------------------------------------------------------ [A/B/C/D/-/Z] 1.1: [ABCD] --> [24/11/7/62/26/0] 1.2: [C] --> [3/14/87/3/23/0] 1.3: [C] --> [6/32/83/5/4/0] 1.4: [D] --> [3/2/10/109/6/0] 1.5: [C] --> [12/4/106/8/0/0] 1.6: [C] --> [5/4/117/1/2/1] 1.7: [A] --> [50/12/9/41/18/0] 1.8: [B] --> [1/61/13/22/33/0] 1.9: [B] --> [19/82/6/2/21/0] 1.10:[A] --> [54/2/12/15/46/1] 1.11:[A] --> [69/9/11/31/9/1] 1.12:[D] --> [17/7/8/41/57/0] 1.13:[A] --> [47/50/9/3/20/1] 1.14:[AD] --> [5/7/11/96/11/0] 2a) > >= 18 kHz (max 1p) 2b) 1600, 1601 (max 1p) 2c) [0..4] ok, [4..8] aliased to [4..0], [8..9] aliased to [0..1] (max 2p) 2d) LP [0..~2], Y_r(jW) [0..~2], muualla/else ~0 (max 2p) 3) h1[n] = K (-0.8)^(n-1) mu[n-1] h[n] = 0.09 (0.8^n mu[n] + (-0.8)^n mu[n]) + 0.09 (0.8^(n-2) mu[n-2] + (-0.8)^(n-2) mu[n-2]) h[0] = 0.18, h[1] = 0, h[2] ~ 0.30 OR simply with a FIR h1[n] so that the zero at the same position as h2-pole z=0.8 -> FIR h[n] ------------------------------------------------------------ (17) 1. välikoe / 1st mid term exam, 11.3.2006 ------------------------------------------------------------ [A/B/C/D/-/Z] 1.1: [AC] --> [70/4/23/3/20/0] 1.2: [BD] --> [8/14/9/67/22/0] 1.3: [CD] --> [16/9/77/6/12/0] 1.4: [C] --> [12/13/89/2/4/0] 1.5: [C] --> [10/2/99/2/7/0] 1.6: [D] --> [32/4/5/68/11/0] 1.7: [D] --> [6/2/21/66/25/0] 1.8: [B] --> [20/48/9/35/8/0] 1.9: [A] --> [36/28/10/1/45/0] 1.10:[D] --> [4/3/3/62/48/0] 1.11:[B] --> [9/29/6/7/69/0] 1.12:[C] --> [7/42/59/2/10/0] 1.13:[C] --> [7/29/63/9/12/0] 1.14:[B] --> [4/87/4/19/6/0] 2a) f0 = 200 Hz, f1=2f0, f2=3f0, f3=27f0, f4=46f0, f5=51f0 2b) Piikit/peaks 400Hz, 600Hz, 5400Hz, 9200Hz, 10200 Hz 2c) 400Hz-->400Hz, 600Hz-->600Hz, 5400Hz->4600Hz, 9200Hz-->800Hz, 10200Hz-->200Hz 2d) HP, esto/stop: [0..~2.5kHz], |Y_r(jOmega)|: ainut piikki / only peak 4600 Hz. S1(z) = [(1-1z^(-1))(1-1z^(-1)] / [1 - 0.18z^(-1)] S(z) = S1(z) + S2(z) S1(z) IIR!!! --> S1(z)=B1(z)/A1(z) S2(z) FIR!!! --> S2(z)=B2(z)/1 B1(z) + B2(z) B1(z)+B2(z)A1(z) S(z) = ----- ----- = ---------------- A1(z) + 1 A1(z) Now B1(z)+B2(z)A1(z) should have zeros at about omega=pi/2 - not necessary on a unit circle!!! B2(z) = 2z^(-1) S(z) = K [1 + 0.64 z^(-2)] / [1 - 0.18 z^(-1)] Max omega=0 <-> z=1 S(z=1) = K 1.64/0.82 = 2K --> K=0.5 OR B2(z) = -0.35 + 1.94 z^(-1) and K=0.63: S(z) = K [1 + z^(-2)] / [1 - 0.18 z^(-1)] ------------------------------------------------------------ (18) 2. välikoe / 2nd mid term exam, 9.5.2006 ------------------------------------------------------------ Tehtävä 1 yhteispisteet / Problem 1 total points: Kohdittain / Statement by statement: lkm/count N=105: [A/B/C/D/-/Z] 1.1: [C] [16/6/78/3/2/0] 1.2: [D] [7/4/11/43/40/0] 1.3: [B] [5/59/2/3/36/0] 1.4: [B] [2/58/15/4/26/0] 1.5: [B] [1/72/8/6/18/0] 1.6: [AD] [95/3/0/6/1/0] 1.7: [C] [31/7/40/3/24/0] 1.8: [C] [2/0/82/2/19/0] 1.9: [BC] [0/66/7/6/26/0] 1.10: [ABCD] [25/13/16/19/32/0] 1.11: [ABD] [30/28/0/20/27/0] 1.12: [C] [4/3/93/0/5/0] 1.13: [A] [92/1/5/0/7/0] 1.14: [B] [1/78/3/6/17/0] Tehtävä 2 / Problem 2 Vastauksia / Solutions: y1[n] = A cos(w0 n) mu[n] y2[n] = A sin(w0 n) mu[n] See lecture slides 37-44 / Chapter 7. Tehtävä 3 / Problem 3 Vastauksia / Solutions: h_ideal[n] = {0.1592, 0.2251, 0.25, 0.2251, 0.1592} w[n] = {0.08, 0.54, 1 , 0.54, 0.08 h**[n] = {0.0127, 0.1215, 0.25, 0.1215, 0.0127} h*[n] = h**[n-2] K = 1.9285 h[n] = K h*[n] h[n] = {_0.0246_, 0.2344, 0.4821, 0.2344, 0.0246} Tehtävä 4 / Problem 4 Vastauksia / Solutions: H(z) = [1 - z^(-1)] / [1 + 2a z^(-1) + 2a^2 z^(-2)] Navat / poles: --> p = -a +- aj a = 0, ..., a =~ 1/sqrt(2) (stab.), ... Tehtävä 5 yhteispisteet / Problem 5 total points: Kohdittain / Statement by statement: lkm/count N=17: [A/B/C/D/-/Z] 5.1: [CD] [2/1/10/3/1/0] 5.2: [C] [0/3/14/0/0/0] 5.3: [C] [12/3/1/1/0/0] 5.4: [BD] [0/6/2/5/4/0] 5.5: [D] [1/0/0/16/0/0] 5.6: [C] [0/1/13/0/3/0] 5.7: [C] [0/1/5/1/10/0] 5.8: [BC] [0/8/3/3/3/0] 5.9: [A] [15/0/2/0/0/0] 5.10: [B] [0/3/0/0/14/0] 5.11: [AD] [3/1/1/10/2/0] 5.12: [C] [0/0/15/0/2/0] 5.13: [B] [4/8/2/0/3/0] 5.14: [B] [1/10/0/2/4/0] ------------------------------------------------------------ (19) 2. välikoe / 2nd mid term exam, 16.5.2006 ------------------------------------------------------------ Tehtävä 1 yhteispisteet / Problem 1 total points (max 12p) lkm/count N = 65: ka/mean = 7.20 histogrammi/histogram: p 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 n 0 0 0 1 2 0 2 3 4 3 2 4 5 p 6.5 7.0 7.5 8.0 8.5 9.0 9.510.010.511.011.512.0 n 3 2 4 8 1 4 4 3 2 2 1 5 Kohdittain / Statement by statement: [A/B/C/D/-/Z] 1.1: [A] [32/3/16/8/6/0] 1.2: [AD] [5/2/1/34/23/0] 1.3: [A] [24/3/0/5/33/0] 1.4: [A] [49/8/3/0/5/0] 1.5: [B] [10/15/3/3/34/0] 1.6: [C] [0/2/62/1/0/0] 1.7: [B] [2/28/2/3/30/0] 1.8: [C] [7/1/39/6/12/0] 1.9: [A] [42/10/1/5/7/0] 1.10: [BC] [0/6/3/8/48/0] 1.11: [ABCD] [14/8/7/22/14/0] 1.12: [ABC] [18/12/4/2/29/0] 1.13: [C] [2/0/59/0/4/0] 1.14: [A] [50/1/2/3/9/0] Tehtävä 2 / Problem 2 Vastauksia / Solutions: omega_c = 0.75pi [Katso "LTI filter design" taulukosta / see "LTI filter design" in formula paper] Omega_pc = k tan(0.75pi/2) H(z) = H(s)|_{s= k (1-z^(-1))/(1+z^(-1))} H(z) = 0.293 (1 - z^(-1))/(1 + 0.414 z^(-1)) zeros = +1, poles = -0.414 Tehtävä 3 / Problem 3 Vastauksia / Solutions: [Katso "LTI filter analysis" taulukosta / see "LTI filter analysis" in formula paper] a) H(z) = G (PROD{(1 - d_m z^(-1))/(1 - p_n z^(-1))} = G (1-1.8^(-1)+0.9^(-2)) / (1+1.4z^(-1)+0.53z^(-2)) b) http://lyhytlinkki.net/?gkyriaxb c) Kuva 3 hyväksikäyttäen "alkuperäinen" väli 0..3pi piirretään 0..pi :ksi. / Using Figure 3 "original" range 0..3pi is drawn to 0..pi. Tehtävä 4 / Problem 4 Vastauksia / Solutions: kausaalinen ja stabiili: a = 0 & e = 0 / causal and stable: a = 0 & e = 0 x[0] != 0, y[0] = 0 => b = 0 => h[n] = c d[n-1] + d d[n-2] => h[n] = 3 d[n-1] - d[n-2] => y[n] = {_0_, 3, -1, -6, 5, -1} Tehtävä 5 yhteispisteet / Problem 5 total points (max 12p) lkm/count N = 42: ka/mean = 5.38 histogrammi/histogram: p 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 n 6 0 0 2 4 1 0 0 0 2 2 2 4 p 6.5 7.0 7.5 8.0 8.5 9.0 9.510.010.511.011.512.0 n 1 3 5 4 1 0 1 1 1 1 0 1 Kohdittain / Statement by statement: [A/B/C/D/-/Z] 5.1: [C] [1/10/14/2/15/0] 5.2: [B] [5/30/0/3/4/0] 5.3: [A] [14/4/1/18/5/0] 5.4: [AD] [19/2/4/4/13/0] 5.5: [AB] [11/13/6/0/12/0] 5.6: [C] [1/12/27/0/2/0] 5.7: [C] [2/1/26/3/10/0] 5.8: [D] [6/2/8/24/2/0] 5.9: [B] [9/22/2/4/5/0] 5.10: [D] [1/7/2/11/21/0] 5.11: [C] [7/2/21/4/7/1] 5.12: [B] [0/27/2/1/12/0] 5.13: [C] [1/2/21/7/11/0] 5.14: [A] [29/1/1/2/9/0] ------------------------------------------------------------ (20) Kesätentti / Exam, 5.6.2006 ------------------------------------------------------------ N=16 1.1: [B] 1.2: [CD] 1.3: [B] 1.4: [B] 1.5: [C] 1.6: [D] 1.7: [CD] 1.7 also C accepted: there are some components in that range, but not in all range. 2) Deconvolution 2a) causal, 2nd order FIR with no delay: h[n] = a d[n] + b d[n-1] + c d[n-2] = 0.5 d[n] + 0.5 d[n-2] 2b) H(z) = 0.5 + 0.5 z^-2 2c) Zeros +-j, bandstop filter at w=pi/2 2d) n>=18: ... 3) All-pass system, see Problem 43 in "exercise material Spring 2006" 3a) H(z) = K [0.57 - 0.28 z^-1 + z^-2] / [1 - 0.28 z^-1 + 0.57 z^-2] 3b) z = .246 +- j 1.302, |z| = 1.325 p = .14 +- j 0.742, |p| = 0.755, |z|=1/|p| | 4 5 f ^ ________|________ | | | | | | | | | -----------------------> -5 -4 4 5 f ^ |________ ________ | | | | | | | | | ------------------------> | 4 5 6 10 f 4b) h_ideal[n] = sin(0.8pin)/(pi n) = 0.8 sinc(0.8n) = {..., -.15, .19, .80, .19, -.15, ...} 4c) w_Hamming[n] = ... = {.08, .54, 1, .54, .08} h_FIR[n] = h_ideal[n] . w_Hamming[n] 4d) Delta_w = 3.32pi / 2 ~ 1.6pi (?!?) 5) Error shaping / quantization. See Mitra or Problem 56 in "exercise material Spring 2006". Y(z) = [1-z^-2]/[1+.81z^-2] X(z) + [1+k^-2]/[1+.81z^-2] E(z)