T-61.3010 Digitaalinen signaalinksittely ja suodatus
T-61.3010 Digital Signal Processing and Filtering

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               RATKAISUJA / SOLUTIONS

                       PART 2
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(16) 1. vlikoe / 1st mid term exam, 7.3.2006
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                 [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]


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(17) 1. vlikoe / 1st mid term exam, 11.3.2006
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               [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)]


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(18) 2. vlikoe / 2nd mid term exam, 9.5.2006
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Tehtv 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]

Tehtv 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.

Tehtv 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}

Tehtv 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.), ...

Tehtv 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]


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(19) 2. vlikoe / 2nd mid term exam, 16.5.2006
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Tehtv 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]

Tehtv 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

Tehtv 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 hyvksikytten "alkuperinen" vli 0..3pi
     piirretn 0..pi :ksi. / 
     Using Figure 3 "original" range 0..3pi is drawn to 0..pi.

Tehtv 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}

Tehtv 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]

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(20) Kestentti / Exam, 5.6.2006
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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|
    <z == <p
3c) allpass

4) FIR-design
4a)^
   |________
   |        |
   |        |
   |        |  
   -------------->
   |        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)

